mm-49 === Dear all,Can I understand the limit in complex analysis in the following way?In complex analysis, all infinities are one point. Hence limit=-inf isthe same as limit=+inf, and if we consider only numbers in complexplane, limit=inf means limit doesnot exist; if we consider numbers inextended complex plane, limit=inf means limit does exist.Contrastingly, in real analysis, lim=inf always means that limit doesnot exist...Is my understanding correct?Please give me a hand!-Walala === > Can I understand the limit in complex analysis in the following way? In complex analysis, all infinities are one point.The one-point compactificiation is often used, yes. (But it's not the onlypossibility.)> Hence limit=-inf is the same as limit=+inf,Right, because -(inf) = inf.> and if we consider only numbers in complex> plane, limit=inf means limit doesnot exist; if we consider numbers in> extended complex plane, limit=inf means limit does exist.Correct.> Contrastingly, in real analysis, lim=inf always means that limit does> not exist...There is not such a stark contrast. In real analysis, we could use theone-point compactification of the reals, for example, in which case wewould again have limit=inf means limit does exist. However,the two-point compactification, [-oo, +oo], is more often used in realanalysis. In [-oo, +oo], limit=-oo means limit does exist and limit=+oomeans limit does exist. But if the limit were -oo from one side and +oofrom the other side, then the limit would not exist.David Cantrell === Can you take a look at lim exp(z) when z->inf?Does the limit exist or not?-WalalaDavid W. Cantrell .91.b9> Can I understand the limit in complex analysis in the following way? In complex analysis, all infinities are one point. The one-point compactificiation is often used, yes. (But it's not the only> possibility.) Hence limit=-inf is the same as limit=+inf, Right, because -(inf) = inf. and if we consider only numbers in complex> plane, limit=inf means limit doesnot exist; if we consider numbers in> extended complex plane, limit=inf means limit does exist. Correct. Contrastingly, in real analysis, lim=inf always means that limit does> not exist... There is not such a stark contrast. In real analysis, we could use the> one-point compactification of the reals, for example, in which case we> would again have limit=inf means limit does exist. However,> the two-point compactification, [-oo, +oo], is more often used in real> analysis. In [-oo, +oo], limit=-oo means limit does exist and limit=+oo> means limit does exist. But if the limit were -oo from one side and +oo> from the other side, then the limit would not exist. David Cantrell === Can you take a look at lim exp(z) when z->inf? Does the limit exist or not?Assuming that your inf denotes the infinity of the one-pointcompactification of C, the limit does not exist. [To see why it doesn'texist, consider what exp(z) does as z->inf via different paths.]David Cantrell === Can you take a look at lim exp(z) when z->inf?Does the limit exist or not?Certainly not.Lee Rudolph === Can you take a look at lim exp(z) when z->inf?Does the limit exist or not?questions you've asked in this thread.)**David C. Ullrich === Dear all,This is a book problem but not a homework problem... I am doing it tocram for exam...I need to find a p(z), harmonic within annulus(ring domain) abs(z)=1and abs(z)=2, such that p(z)=0 on the inner circle andp(2*exp(i*theta))=5*cos(3*theta) on the outer circle...(hint: think ofz^n and z^(-n))My question is what is the procedure finding harmonic function forthis question and for any harmonic function in general?-Walala === > Dear all, This is a book problem but not a homework problem... I am doing it to> cram for exam... I need to find a p(z), harmonic within annulus(ring domain) abs(z)=1> and abs(z)=2, such that p(z)=0 on the inner circle and> p(2*exp(i*theta))=5*cos(3*theta) on the outer circle...(hint: think of> z^n and z^(-n)) My question is what is the procedure finding harmonic function for> this question and for any harmonic function in general?> -WalalaThe harmonic function has to have the formp(r e^(it)) = sum [from -infinity to infinity] ( a_n r^n e^(int) )for some doubly-infinite sequence {a_n}. === > I need to find a p(z), harmonic within annulus(ring domain) abs(z)=1> and abs(z)=2, such that p(z)=0 on the inner circle and> p(2*exp(i*theta))=5*cos(3*theta) on the outer circle...(hint: think of> z^n and z^(-n))Both r^3*cos(3*theta) and (1/r^3)*cos(3*theta) are harmonic on C {0}. A linear combination of these two should work. === Dear all,This problem haunted me for long. I am quite headache with it:Suppose a polynomial P(z) has its all roots in a convex closed polygonD; show that its derivative P'(z) also has all roots belong to thesame polygon D...Please give me a hand, thanks a lot,-Lalila === Dear all,This complex analysis problem huanted me for long:Suppose a polynomial P(z) has all roots in a convex polygon D... howto prove all the roots of its derivative P'(z) also located in thesame D?-Lalila === > Dear all, This complex analysis problem huanted me for long: Suppose a polynomial P(z) has all roots in a convex polygon D... how> to prove all the roots of its derivative P'(z) also located in the> same D?> -LalilaThere was a thread on this not long ago. You can find it in Google bysearching for roots derivative polynomial, search only in sci.math.It was in, or related to that thread that Ignacio mentioned the delightfulfact that if you start with a triangle in the complex plane and form thepolynomial which has its vertices as roots, then the roots of thederivative of the polynomial are the foci of the Steiner ellipse, thelargest ellipse that can be inscribed in the triangle. === >Dear all,This complex analysis problem huanted me for long:Suppose a [nonconstant] polynomial P(z) has all roots in a convex polygon D... how>to prove all the roots of its derivative P'(z) also located in the>same D?Hmm, this one took me a few minutes (actually I came close tosaying it was clearly false...)Since a convex polygon is the intersection of a family of closed half-planes it's sufficient to prove the result for D = a closedhalf-plane. And now that result follows from this:Thm. If P is a non-constant polynomial and all the zeroes of Phave strictly positive real part then P'(0) <> 0.Pf: Say the zeroes are a_1, ... a_n. We can assume thatP is monic: P(z) = (z - a_1)...(z - a_n). Let A be the productof all the roots.If you think about it for a second you see that the coefficientof x in P is plus or minus A (1/a_1 + ... + 1/a_n).The fact that a_j has strictly positive real part shows thatthe real part of 1/a_j is also strictly positive. Hence thesum of the 1/a_j is non-zero. So the coefficient of x isnon-zero, hence P'(0) <> 0. QED.-Lalila**David C. Ullrich === Dear all, This complex analysis problem huanted me for long: Suppose a polynomial P(z) has all roots in a convex polygon D... how> to prove all the roots of its derivative P'(z) also located in the> same D?> -LalilaIt is a consequence of the Luca's theorem:If all zeroes of a polynomial P(x) lie in a half plane, then all zeroes ofthe derivative P'(x) lie in the same half planeWe have that is a1, a2, ..., an are the roots of P(x), thenP(z) = c(z - a1)...(z - an)P'(z)/P(z) = 1/(z - a1) + ... + 1/(z - an).Suppose that the half Plane H is defined as he part of the plane whereIm((z - a)/b) < 0. If ak is in H and z not, we have thenIm((z - ak)/b) = Im((z - a)/b) - Im((ak - a)/b) > 0But the imaginary part of reciprocal numbers have opposite sign. Therefore,under the same assumption, Im(b/(z - ak)) < 0If this is true for all k we conclude thatIm(bP'(z)/P(z)) = Sum(Im(b/(z - ak)), k, 1, n) < 0and consequently P'(z) =/= 0 === > explain why the function Log|z| is harmonic in every domain excluding {0}?Dear all,Please help me [snip]... harmonic on domain C{0}, but we cannot find its harmonic conjugate?Dear all,Please help me on this problem as I am practicing some [snip]...>Dear all,This is a book problem [snip]...>[snip][snip][snip]DEAR ALL... whoops I mean, Dear walala,Why, don't you learn how to do your own damn work. The number of messagesyou have posted is approaching infinity faster than Exp[ Gamma[z] ]-oh, but you have no idea what that means-and I, for one, find it rather annoying because it appears as if youare trying to get sci.math to complete a whole course for you. That's rather unethical, wouldn't you say.You, walala, are the embodiment of an essential singulatity... veryugly, and to be avoided at all costs.That's all I have to say... oh, one more thing. The professor ofyour course HAS been notified as to your unethical online activities.Cheerio, === Dear all,Please help me on this problem as I am practicing some text bookproblems cramming for exam. The problem statement is:prove that although u=ln(abs(z)) is harmonic on domain C{0}, but wecannot find its harmonic conjugate v such that ln(abs(z))+i*v(z) isanalytic on domain C{0}...-Walala === > Dear all, Please help me on this problem as I am practicing some text book> problems cramming for exam. The problem statement is: prove that although u=ln(abs(z)) is harmonic on domain C{0}, but we> cannot find its harmonic conjugate v such that ln(abs(z))+i*v(z) is> analytic on domain C{0}...> -WalalaIf z = r e^(it) then u = ln(r). Show that if you remove the negative realnumbers from the domain, then you can easily find a harmonic conjugate v foru, that is, you can easily find an analytic function of which u is the realpart. If there were a function on C{0}, it would have to agree with the vyou found on the common domain, except for an additive constant. But lookhow v behaves as you approach the negative real axis from above versus frombelow. === >Dear all,Please help me on this problem as I am practicing some text book>problems cramming for exam. The problem statement is:prove that although u=ln(abs(z)) is harmonic on domain C{0}, but we>cannot find its harmonic conjugate v such that ln(abs(z))+i*v(z) is>analytic on domain C{0}...Probably the best way is more or less what Will said: There is a verywell-known, very important analytic function L in C{0} such thatu = Re(L). But this L is not really a function - it can't be defined so as to be continuous in all of C{0} (it's what used to be calleda multi-valued function.) I say this is probably the best way todo the problem because it's very important to understand howthat function L works - look in the book to find what L I'm talkingabout.There are more interesting ways to do the problem. For example,suppose that such a v existed, and let f = u + iv. Then f hasan isolated singularity at the origin, while Re(f(z)) -> -infinityas z -> 0. Consider the various kinds of isolated singularity(removable, pole, essential) and show that it's impossibleto have Re(f) -> -infinity at an isolated singularity.Or: again assume there is such a v and let f = u + iv. LetC_r be the circle with center at the origin and radius r.Then Cauchy's Theorem implies that the integral off(z)/z dz over C_r should be independent of r, but ifyou parametrize that curve and look at what the integralis you see that it's _not_ independent of r.Or (this one is actually a version of the first solution):Suppose that such a v exists and let f = u + iv. Youcan use the Cauchy-Riemann equations to find whatf'(z) is, even though you don't know v. Now recallthat the integral of f' over a closed path is _always_equal to 0 (the integral of f over a closed path canbe non-zero, depending on whether various hypothesesin Cauchy's Theorem are satisfied, but the FundamentalTheorem of Calculus shows that the integral of f' is 0regardless). But here it turns out that there is a closedpath over which the integral of f' is non-zero, which isa contradiction.equations, the calculations will be simpler if you rewriteu as ln(abs(z)^2)/2.)-Walala**David C. Ullrich === Dear all,Please help me on this problem as I am practicing some text book>problems cramming for exam. The problem statement is:prove that although u=ln(abs(z)) is harmonic on domain C{0}, but we>cannot find its harmonic conjugate v such that ln(abs(z))+i*v(z) is>analytic on domain C{0}...[..]> There are more interesting ways to do the problem. For example,[...]> Or (this one is actually a version of the first solution):> Suppose that such a v exists and let f = u + iv. You> can use the Cauchy-Riemann equations to find what> f'(z) is, even though you don't know v. Now recall> that the integral of f' over a closed path is _always_> equal to 0 (the integral of f over a closed path can> be non-zero, depending on whether various hypotheses> in Cauchy's Theorem are satisfied, but the Fundamental> Theorem of Calculus shows that the integral of f' is 0> regardless). But here it turns out that there is a closed> path over which the integral of f' is non-zero, which is> a contradiction.Yes, this way is especially appealing. === How to prove that? -Walala===>How to prove that? By using the definition (which will require you to understandlim in this context; you have previously expressed some confusion about limits as z -> infinity, but here you don'tneed any limits but finite limits). What is the differencequotient of (abs(z))^2 at a given point (with a given increment Delta z), to start with? It will help to write (abs(z))^2 in another way--do you know any?Lee Rudolph === Dear all,Please help me before the exam begins... The puzzle here is how toshow Log(|z|) is harmonic without verifying Laplace equation... (TheLog means principal branch of logirithms...)Can you give me a hand on this?-Walala === >Dear all,Please help me before the exam begins... The puzzle here is how to>show Log(|z|) is harmonic without verifying Laplace equation... (The>Log means principal branch of logirithms...)It's harmonic because (locally) it is the real part of the analyticfunction ___.>Can you give me a hand on this?>You're posting really a lot of questions. I suspect you willdo better on the exam if you work out some of the solutionsyourself...>-Walala**David C. Ullrich === Let a_n be a sequence such that :For all n=>1, 0<= a_n <= cFor all i,j, such that i<>j, |a_i - a_j | => 1/(i+j)Prove that c=>1. === >Let a_n be a sequence such that :>For all n=>1, 0<= a_n <= c>For all i,j, such that i<>j, |a_i - a_j | => 1/(i+j)Prove that c=>1.Nice problem! Here's my solution:For any n >= 2, we define the weighted graph G_n with vertices 1,...,nwhere the edge (i,j) has weight exactly 1/(i+j). We will find a lowerbound for the least-weight Hamiltonian path in G_n (open TSP tour).Let p = (p_1 p_2 ... p_n) be any permutation of {1,...,n} representing aHamiltonian path, and define r_i = p_i + p_{i+1}, 1 <= i <= n-1, so thatthe weight of the path p, denoted W(p), is exactly sum(1/r_i). We have: sum(r_i) = 2*sum(p_i) - p_1 - p_n = n(n+1) - p_1 - p_n >= n^2 + n - 3.Since all r_i's are positive, a convexity argument shows that for a givenvalue of sum(r_i), sum(1/r_i) is minimised when all r_i's are equal, thus sum(1/r_i) >= (n-1)^2/(n^2+n-3) = 1-(3n-4)/(n^2+n-3) = 1 - 3/n + O(1/n^2).[ Note: this bound is pretty close to tight, since the permutation (1 n 2 n-1 3 n-2 ... 1+[n/2]) has a weight of 1 - 5/2n + O(1/n^2). ]Now let {a_n} be a sequence satisfying the above. Fix m > 1, and sortthe elements {a_1,a_2,...,a_m} in ascending order: a_{i_1} < a_{i_2} < ... < a_{i_m}.Now | a_{i_m} - a_{i_1} | = | a_{i_2} - a_{i_1} | + ... + | a_{i_m-1} - a_{i_m} | >= 1/(i_1+i_2) + 1/(i_2+i_3)... + 1/(i_m-1 + i_m) = W(i_1 i_2 ... i_m) >= 1 - O(1/m).Thus a_{i_m} = max(a_1,...,a_m) >= 1 - O(1/m). Taking limits, we getc >= sup{a_n} >= 1. Finis.Anyone have thoughts on what the real lower bound on c is? It seems to methat the minimal TSP is given by a different path for each m, which leavesenough room for a significant gap. I found a simple construction that Ithink works for c = 2: { 0, 1, 1/2, 3/2, 1/4, 3/4, 5/4, 7/4, ... }. -- Erick === > sum(r_i) = 2*sum(p_i) - p_1 - p_n = n(n+1) - p_1 - p_n >= n^2 + n - 3.Minor correction: this should be <=. Typo, doesn't affect the argument. -- Erick === >Let a_n be a sequence such that :>For all n=>1, 0<= a_n <= c>For all i,j, such that i<>j, |a_i - a_j | => 1/(i+j)Prove that c=>1.> Nice problem! Here's my solution: For any n >= 2, we define the weighted graph G_n with vertices 1,...,n> where the edge (i,j) has weight exactly 1/(i+j). We will find a lower> bound for the least-weight Hamiltonian path in G_n (open TSP tour).> Let p = (p_1 p_2 ... p_n) be any permutation of {1,...,n} representing a> Hamiltonian path, and define r_i = p_i + p_{i+1}, 1 <= i <= n-1, so that> the weight of the path p, denoted W(p), is exactly sum(1/r_i). We have:> sum(r_i) = 2*sum(p_i) - p_1 - p_n = n(n+1) - p_1 - p_n >= n^2 + n - 3.> Since all r_i's are positive, a convexity argument shows that for a given> value of sum(r_i), sum(1/r_i) is minimised when all r_i's are equal, thus> sum(1/r_i) >= (n-1)^2/(n^2+n-3) = 1-(3n-4)/(n^2+n-3) = 1 - 3/n + O(1/n^2).> [ Note: this bound is pretty close to tight, since the permutation> (1 n 2 n-1 3 n-2 ... 1+[n/2]) has a weight of 1 - 5/2n + O(1/n^2). ]> Now let {a_n} be a sequence satisfying the above. Fix m > 1, and sort> the elements {a_1,a_2,...,a_m} in ascending order:> a_{i_1} < a_{i_2} < ... < a_{i_m}.> Now | a_{i_m} - a_{i_1} |> = | a_{i_2} - a_{i_1} | + ... + | a_{i_m-1} - a_{i_m} |>= 1/(i_1+i_2) + 1/(i_2+i_3)... + 1/(i_m-1 + i_m)> = W(i_1 i_2 ... i_m) >= 1 - O(1/m).> Thus a_{i_m} = max(a_1,...,a_m) >= 1 - O(1/m). Taking limits, we get> c >= sup{a_n} >= 1. Finis.> Anyone have thoughts on what the real lower bound on c is? It seems to me> that the minimal TSP is given by a different path for each m, which leaves> enough room for a significant gap. I found a simple construction that I> think works for c = 2: { 0, 1, 1/2, 3/2, 1/4, 3/4, 5/4, 7/4, ... }.> -- ErickI think you can get c=1 by letting the a_i's be the dyadic numbers on[0,1].a_1=0, a_2=1, a_3=1/2, a_4=3/4, a_5=1/4 and in generala_(2n+1)=1/(2^n), a_(2n)=1-(1/(2^n)), a_(2^n-1)=1/2 - 1/(2^n),a_(2^n-2)=1/2 - 1/(2^n) etc.Karin === > sum(r_i) = 2*sum(p_i) - p_1 - p_n = n(n+1) - p_1 - p_n >= n^2 + n - 3. Minor correction: this should be <=. Typo, doesn't affect the argument. -- ErickErick, another WELL DONE! If you keep this up you will scare everyone elseaway. === I got interested in multidimensional geometry lately, and I wondered,what would happen to a 3D object if it would be rotated around a forthaxis, which is perpendicular to the x, y and z axes and not x, y , orz?Tal P === > I got interested in multidimensional geometry lately, and I wondered,> what would happen to a 3D object if it would be rotated around a forth> axis, which is perpendicular to the x, y and z axes and not x, y , or> z?> Tal PIn 4D space, you would fix a 2D axis and rotate about that.Just as in 3D space you use a line as axis, and in 2D you use a point. === > I got interested in multidimensional geometry lately, and I wondered,> what would happen to a 3D object if it would be rotated around a forth> axis, which is perpendicular to the x, y and z axes and not x, y , or> z? Tal PA=(a1,a2,a3,a4) objectA'=rotated AD=(0,0,0,1) axis of rotationC= angle of rotationFind some point P on the plane(P-(A*D)D)*D=0 such that|(P-(A*D)D)*(A-(A*D)D)/[|P-(A*D)D||A-(A*D)D|]!=1(A,P and (A*D)D cannot be colinear)call this point RB=R-[(A*D)D+(R-(A*D)D)*(A-(A*D)D)(A-(A*D)D)/|A-(A*D)D|^ 2result of any angle of rotation can be found,A'=(A*D)D+(A-(A*D)D)cosC + (B/|B|)|A-(A*D)D|sinCThe 3D slice of this 4D situation can be found by the graphsof each triple === I have the following question when writing Fortran program codes:How to efficiently solve the equation,([A11] [A12]){x}=lambda([B11] [0]){x}([A21] [A22]) ( [0] [0])Both matrix [A] and [B] are symmetric and sparse.Jian === Dear all,Just a little note to inform you that you can now order your copy online andmake the payment by credit card.For further information please go to:http://www.gene-expression-programming.com/gep/Books/ index.aspCandida Ferreira------------------------------------------------------ ---------Candida Ferreira, Ph.D.Chief Scientist, Gepsoft73 Elmtree DriveBristol BS13 8NA, UKph: +44 (0) 117 330 9272www.gepsoft.com/gepsoftwww.gene-expression-programming.com /author.asp--------------------------------------------------- ------------ === Suppose A, B, C are continuous functions from R->R such that, for all x and y, A(B(x)+y)=C(y)+xHow could one prove that no non-trivial solution exists? Eugene Shuberthttp://www.everythingimportant.org === Example: g(x) = ax + ab, f(x) = x/a - b f(g(x) + y) = y/a + x = f(g0) + y) + x-- f(g(x)) = f(g0) + xf(g(x) + g(x)) = f(g0 + g(x)) + x = f(g0 + g0) + 2xf(3g(x)) = f(g0 + 2g(x)) + x = f(g0 + g0 + g(x)) + 2x =f(ng0 + mg(x)) = f((n+1)g0 + (m-1)g(x)) + xf(ng(x)) = f(ng0) + nxf(g(x) - g(x) - g0) = f(g0 - g(x) - g0) + xf(-g(x)) = f(-g0) - x---- === I need to draw a graphical representation of a bunch of data points. All of the data points, in theory, correspond to some y=f(x)+C function. However, the data points are in sets of around 4, and each set has a different C value. This data is experimental, and as such, not necessarily completely accurate.I devised a weird home brewed method of doing this (described below) and I tested it with randomly generated data (for which I know the function).My questions are: is my method sound? Is there a better established method of doign this?The notation below is some odd fusion of mathematical and programming notation... I have to apologize, I am not a mathematician.I generated a super set S of 319 sets S[j] with 2 to 5 x-values each.Each x-value = (random fraction between 0 and 1) * 10^(random integer -6 to 6)Each set of x values = S[j][x] with values S[j][x][i] where S[j][x][i] = I[x][n+1]Where this is true for S[j][x][i], {the length correction factor at thepoint S[j][i], I[n]} = lcf(S[j][i],I[n]) equals one of two things (depending on the graph being generated - I tried both methods, and they seem to be variably accurate depending on what f(x) is):linear proportion: (I[x][n+1] - I[x][n])/(S[j][x][i+1] - S[j][x][i]) orlogarithmic proportion [log( I[x][n+1] ) - log( I[x][n] )]/[log(S[j][x][i+1] ) - log( S[j][x][i] )]Then the weighted slope for S[j][i] at I[n] is wslope(S[j][i],I[n]) =S[j][slope][i] * lcf(S[j][i],I[n])Then, for each x interval in I, the average slopeI[slope][n]=SUM(wslope(S[*j][*i],I[n]) )/SUM( lcf(S[*j][*i],I[n]) ) Theasterisks indicate all of the S[j][i] combinations for which there is awslope(S[j][i], I[n]) and an lcf([S[j][i], I[n])Finally the graph is reconstructed as follows:I[y][0] = 0then I[y][n] = {( I[x][n] - I[x][n-1] )*( I[slope][n-1] )} + I[y][n-1]If you would like to see the data points I randomly generated and the two graphs I created out of those data points using the above method, EdPSBelow is a textual key, in case my notation is not clear.S: super set of setsS[j]: a set of sets of x, y, and slope valuesS[j][x]: a set of x valuesS[j][x][i]: an x value that corresponds with S[j][y][i] y value andS[j][slope][i] slope valueI: set of x y and slope values for the final graphI[x]: set of x values for the final graph, also referred to as intervals.This set is all of the unique x values from all of S.I[x][n]: A particular x intervalI[slope][n]: the averaged weighted slopes that fall into interval betweenI[slope][n] and I[slope][n+1]lcf(,): a cross sectional space where one axis is the point of S[j][i] andthe point of I[n] and where they intersect is the length correction factorfor the particular S[j][slope][i] slope in the interval I[n]wslope(,): a space like lcf(,) but the intersecting points are the weightedslopes for slope S[j][slope][i] in interval I[n] === What I've done is find myself needing to consider other thanpolynomial factors of a polynomial P(x).The situation is analogous to when earlier people found reason toconsider other than integer factors of integers.So today it's taken for granted that you have factors of 2, likesqrt(2) that aren't integers.But I've been facing a lot of resistance when I've talked about nonpolynomial factors of polynomials.It seems to me that something happened and mathematicians are behind.By now mathematicians should have considered non polynomial factors atleast to the extent that they couldn't sustain the flawed thinkingI've seen.Let me get more into specifics.Consider x^2 + 3x + 2 = (x+1)(x+2), where you see a simple quadraticpolynomial, and its factorization into linear terms.It shouldn't faze any of you for me to talk about factors g_1 g_2 g_3g_4 g_5 of that *same* polynomial.If it bothers you then something is wrong.Consider that I can talk of 2^{1/5} as a factor of 2, and many of youshouldn't care if I say x^5 + 12x^4 - 17x^2 + 3x - 2, which will giveroots that are factors of 2 in the ring of algebraic integers.So then with polynomials, what's the equivalent of rings?I know the answer and you should as well, but if you knew the answer Iwouldn't see the arguments I'm seeing.Mathematicians need to develop further. Something happened andapparently mathematics has fallen behind. Mathematicians are herebyinstructed to fix this problem as soon as possible.James Harris === [snip]> Mathematicians are hereby> instructed to fix this problem as soon as possible. James HarrisYou are hereby instructed to seek psychiatric counseling from a competenttherapist.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > [snip]> Mathematicians are hereby>instructed to fix this problem as soon as possible.>James Harris> You are hereby instructed to seek psychiatric counseling from a competent> therapist.Forget counseling. JSH needs strong drugs.Gib === > What I've done is find myself needing to consider other than> polynomial factors of a polynomial P(x).> The situation is analogous to when earlier people found reason to> consider other than integer factors of integers.> So today it's taken for granted that you have factors of 2, like> sqrt(2) that aren't integers.> But I've been facing a lot of resistance when I've talked about non> polynomial factors of polynomials.The resistance is only when you use terminology whichyou won't define, or use it inconsistently or wrong.Nobody is resisting the concept of factoring, you loon.> Let me get more into specifics.> Consider x^2 + 3x + 2 = (x+1)(x+2), where you see a simple quadratic> polynomial, and its factorization into linear terms.> It shouldn't faze any of you for me to talk about factors g_1 g_2 g_3> g_4 g_5 of that *same* polynomial.> If it bothers you then something is wrong.> Consider that I can talk of 2^{1/5} as a factor of 2, and many of you> shouldn't care if I say x^5 + 12x^4 - 17x^2 + 3x - 2, which will give> roots that are factors of 2 in the ring of algebraic integers. So then with polynomials, what's the equivalent of rings?The equivalent of rings for polynomials would be rings. Inparticular, the ring of polynomials of [define what specifictype of polynomials you're talking about].For instance, the polynomials with integer coefficients forma ring. A factor in that ring is another polynomial withinteger coefficients. The polynomials with algebraic integer coefficients forma larger ring.The polynomials with [something] coefficients may also forma ring, depending on what you choose for [something], butthat ring won't include functions like sqrt(x^5+3) whichyou seem to want. - RandyMime-version: 1.0 === > [snip]> Mathematicians are hereby> instructed to fix this problem as soon as possible.> James Harris> You are hereby instructed to seek psychiatric counseling from a competent> therapist.> Forget counseling. JSH needs strong drugs.> GibActually, many of us wouldn't be surprized to find that the problem wascaused by strong drugs. === Could anyone suggest how I might model affine transformations usingClifford/Geometric algebraMy background is more of a programmer than a mathematician and what I amtrying to do is to model solid objects with simple Newtonian mechanics. WhatI would like to do is model all the quantities using multivectors so thatthe linear and rotational properties can all be handled in one equation, Idon't want to do anything advanced like space-time or quantum theory, justsimple mechanics of solid objects. As I say I'm not a mathematician but Iwould like to understand the principles behind this.equivalent to using a quaternion made up of the scalar and the bivector andthe second vector is equivalent to using a quaternion made up of thetrivector and the vector.My working for this is here:http://www.martinb.com/maths/algebra/clifford/transforms/ My question is that I want to be able to represent the position of solid (6degree of freedom) objects, rotation + translation, but instead of the sumof a rotational and a linear component I have ended up with the sum of tworotational components?So how do I model affine transformations? What I have done so far does notseem to have any advantage over quaternions? What use are two quaternionsadded together?I gather that the same trick can be used as with matrices, i.e. add a dummy4th dimension to represent the translation component (I have also seen itsuggested that as well as the 4th dimension to represent the origin point, a5th dimension can be added to represent a point at infinity?).However this is a problem, firstly I don't like the idea of using 16 (or 32)scalar values to represent a 6 degree of freedom object. This seems like abig overhead in memory usage and in normalising all these dummy variables.Also I can't find any information on the arithmetic of 4D multivectors andhow to represent rotation, translation and scaling using them.Can anyone suggest how I could take this forward?Martin === given 5 vectors (each of size 1x5) i would like to test whether the fifth is a linear combination of the first four OR almost linear combination of them. what can be a good and efficient way to do this ? determinant is usefull here, but i have problems with itsince in case of almost linear dependency, determinant is notnecessarily close to 0.Anyone has a suggestion ? -- Direct access to this group with http://web2news.comhttp://web2news.com/?sci.mathTo contact in private, remove nn0oosp+a8m === > given 5 vectors (each of size 1x5) i would like to test> whether the fifth is a linear combination of the first four> OR almost linear combination of them.> what can be a good and efficient way to do this ?> determinant is usefull here, but i have problems with it> since in case of almost linear dependency, determinant is not> necessarily close to 0.> Anyone has a suggestion ?I dare not guess what exactly you mean by almost linear dependency,but if it means that adding a short vector to the fifth vector willmake the system dependent, then the following should do the trick:Let X_1,X_2,...,X_5 be your vectors (in that order). Carry out theusual Gram-Schmidt orthogonalization to arrive at the orthogonal setY_1,Y_2,...,Y_5. Now Y_5 will measure the almost linear dependency in the followingprecise sense. Namely Y_5 is the shortest possible difference vecetor between X_5 and a vector that can be expressed as a linear combinationof X_1,...,X_4. In particular, Y_5 is the zero vector, iff X_5 is a linear combination of the other 4.Would this fit the bill?Jyrki Lahtonen, Turku, FinlandMime-version: 1.0 === > given 5 vectors (each of size 1x5) i would like to test> whether the fifth is a linear combination of the first four> OR almost linear combination of them.> what can be a good and efficient way to do this ?> determinant is usefull here, but i have problems with it> since in case of almost linear dependency, determinant is not> necessarily close to 0.> Anyone has a suggestion ?Does this come close to what you need:Divide the determinant, if non-zero, by the product of the lengths, toelimiate the effects of vectors of different lengths.Or, perhaps, if the above is no help, you could be more precise about whatalmost linear dependence means to you? === First of all, thank you for your response.I'll try to formulate my problem better:I have two sets:S_1 = { {x11,x12,x13,x14}, {y21,y22,y23,y24}, }S_2 = { v1, v2 , } each element in set S_1 is a set of 4 linearly independent vectors(x_{ij} is 1x5 vector), each element in S_2 is 1x5 vector.I would like to find a measure that will give me some special value in case v_j approximately equal to sum_{i=1}^{4} a_i * x_{ji}or some similar values ...and some different value in case v_j is very far from being sum_{i=1}^{4} a_i * x_{ji}Determinant is not a good measure since it gives 0 in case the vectors are lin. dependent however in all other cases it goes crazy.I think this is due to the fact that determinant checks if all vectorslin. indep. with all vectors ... too general measureWhat do you think ? therefore, normalizing won't help either since its only a scale factor.Concerning Gram-Schmidt, I'm not sure how it will act, but it might be a good idea.Do you have any additional suggestions, now when the problem explained a bit better ? -- Direct access to this group with http://web2news.comhttp://web2news.com/?sci.mathTo contact in private, remove nnoo+sp4amm === By almost linear dependent I mean:find coefficients 'a' (1x4) such that:min_{a} || v_j - a*X ||where vj is 1x5 and X is 4x5, let a' be the min coefficientsif ||v_j - a'*X|| / || v_j || is about 5%then v_j is almost lin. dependent on X-- Direct access to this group with http://web2news.comhttp://web2news.com/?sci.mathTo contact in private, remove n5osp6p+a-9mm === > given 5 vectors (each of size 1x5) i would like to test> whether the fifth is a linear combination of the first four> OR almost linear combination of them. what can be a good and efficient way to do this ? determinant is usefull here, but i have problems with it> since in case of almost linear dependency, determinant is not> necessarily close to 0. Anyone has a suggestion ?Try singular values: if V={v_1, ..., v_k} is an ordered list ofvectors in an inner product space, set up the Gram matrix G: G_(i,j) = and if its smallest eigenvalue of G is s^2, with thecorresponding k-by-1 normalized singular vector w, thenthe smallest singular value of V is s (>=0), and u = V*wis the shortest possible linear combination of vectors form V,subject to norm(w)=1; it has norm s.To discover the suspected dependence, isolate the vector form Vcorresponding to the absolutely largest entry og w.If you want the search for approximate dependence to beinsensitive to scaling of the vectors, normalize them beforecalculating G.Determinant can be misleading for many reasons: for example itcan be far from 0 while the vectors can be near-dependent.For example, the determinant of[ 1, 0, -5, 0, 1][ 1, 2, 0, 1, 0][ 0, -2, -3, 1, 0][ 0, 4, 2, 1, -2][ 1, 3, 5, 1, 0]is 1, but the first column is close to a linear combination ofthe other four columns, relatively within 0.62% .Hope it helps,ZVK(Slavek). === try a search on condition number along with matrix?cdj> First of all, thank you for your response.> I'll try to formulate my problem better:> I have two sets:> S_1 = { {x11,x12,x13,x14}, {y21,y22,y23,y24}, ? }> S_2 = { v1, v2 , ?}> each element in set S_1 is a set of 4 linearly independent vectors> (x_{ij} is 1x5 vector), each element in S_2 is 1x5 vector.> I would like to find a measure that will give me some special value > in case v_j approximately equal to sum_{i=1}^{4} a_i * x_{ji}> or some similar values ...> and some different value in case v_j is very far from being > sum_{i=1}^{4} a_i * x_{ji}> Determinant is not a good measure since it gives 0 in case the vectors > are lin. dependent however in all other cases it goes crazy.> I think this is due to the fact that determinant checks if all vectors> lin. indep. with all vectors ... too general measure> What do you think ? > therefore, normalizing won't help either since its only a scale factor.> Concerning Gram-Schmidt, > I'm not sure how it will act, but it might be a good idea.> Do you have any additional suggestions, now when the problem explained > a bit better ? === You can use a feature of a matrix known as the CONDITION NUMBER. Thecondition number may be interpreted as being inversely proportional tohow close a matrix is to a noninvertible matrix.Given a matrix A consisting of liearly independent column vectors, youfind the condition number by first finding A^(-1). Then for somedefinition of matrix norm you find the norm of A and the norm of A^(-1),and multiply them together. The result is always >= 1, and the higherit is the closer te matrix is to being singular.There are two subtleties that must be considered. First, if vectorshappen to be of different SIZE that will affec honditinnum er eventhough thevector are ot inherenty nearly dependent. If we have two2-vectors(0)(1)and(10^10)(0)then the resulting matrix(0 10^10)(1 0 ) (sorry if it doesn't line up)has a condition number of 10^10 (using typical norm definitions) andthat's misleading since the vectors are literally far from collinear(they're orthogonal). Normalizing vectors to have unit norm preventsthis problem.The second problem is inchoosing the norm. The condition number of amatrix depends on the definition of the norm in all interesting cases.The math works out most neatly with the Euclidean norm but this involveseigenvalues of a matrix times its adjoint and is hard to get at. Foryour purposes it's easier to use the L1 or taxicab norm. Conditionnumberbased on the L1 norm is not exactly aperfect measure of how nearlydependent the vectors are, but it will do for your application.Begin by normalizing your vectors so that the sum of absolute values ofall components -- which is what the L1 or taxicab norm is -- equals 1for each vector. Make these normalized vectors column vectors in yourmatrix A. The L1 norm of your matrix is now the maximum L1 norm ol allthe column vectors, hence 1 for the A matrix you constructed. Now findA^(-1) and compute its taxicab norm. Since the norm of A is 1, the normyou compute for A^(-1) is your condition number. If it's over 100 for a5x5 matrix, you ought to judge the vectors to be close to dependent.--OL === Planck's ConstantPreviously in the thread Angular Momentum in Rotating Bodies, Ipresented an analytical framework for the interpretation of dr/dt incircular rotation of a point mass m at velocity v and radius r. No oneI know of agrees with my interpretation of dr/dt. However, in theinterests of further establishing this general framework, I would liketo pursue general developement of the idea which culminates in theanalytical definition of Planck's constant.We begin by noting that in cases of circular rotation at constantangular velocity we have a centripetally directed dr/dt acting onpoint mass m of a magnitude equal to tangential velocity v. This iswhat causes the rotation of v and produces r as a consequence ofrotation.We then integrate dr/dt along r which produces 1/2 mvr/2pi with unitsof measure equal to rr/t. Now, I have been cautioned on severaloccasions not to suggest that this quantity represents angularmomentum in conventional terms and I agree. Perhaps we should simplycall it rotational momentum to prevent confusion.What we notice immediately however is that it bears the same form asthe quantity mvr corresponding to Planck's constant. However, we haveto straighten certain things out in this connection.In conventional macro angular rotation such as flywheels we have acentripetal dr/dt and tangential v which are equal to each other. Theyare effectively bound up through tensile forces internal to the bodyundergoing rotation. In celestial angular mechanics on the other handwe have a wide variety of potential dr/dt's and tangential orbitalvelocities operating in various combinations.different situation. The tangential velocity of rotation v is constantunder all circumstances. In other words, v = c. Thus dr/dt operatesmass.second) times an analytical masslet, m0 (kg-sec) and interpret thequantity mvr as a multiple of nm0vr. Further we can interpret r as afunction of c/n such that Planck's constant = m0cc. In other words, m0is roughly on the order of 10^-50 kg-sec in magnitude and Planck'sconstant corresponds to the multiple of m0 and the square of thevelocity of light.We notice several things about rotational momentum. In linear motionat constant velocity rotational momentum is zero because dr/dt and mvrare both zero. And in circular rotation at a constant angular velocityrotational momentum is constant because mvr is constant. Thisrepresents the analytical distinction between circular and linearmotion.Further we notice that dr/dt can be of any magnitude. It is not boundby the constancy of the velocity of light as an upper limit because itdoesn't go anywhere. It only produces rotation in relation to actualtangential motion v = c.mass and radius of rotation are inversely proportional, that is that === > Planck's Constant Previously in the thread Angular Momentum in Rotating Bodies, I> presented an analytical framework for the interpretation of dr/dt in> circular rotation of a point mass m at velocity v and radius r. No one> I know of agrees with my interpretation of dr/dt. Lester--You should take courses in differential and vector calculus. === > Planck's Constant> Previously in the thread Angular Momentum in Rotating Bodies, I> presented an analytical framework for the interpretation of dr/dt in> circular rotation of a point mass m at velocity v and radius r. No one> I know of agrees with my interpretation of dr/dt.People you don't know of are also disgusted. You are a crank, aloud ass, and a boring ineducable lout.http://w0rli.home.att.net/youare.swf[snip]> mass and radius of rotation are inversely proportional, that is thatRes ipsa loquiter, moron. Hey Zick - electrons are point masses andprotons are not. Why don't you tell us how a 1.67262158 x 10^(-27) kgproton is smaller than a 9.10938188 x 10^(-31) kg electron.If empirical reality says your are an ass, Zick, you are an empiricalass.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === >Your first paragraph is just nonsense. You knew that r was of constant>length and that it is directed outward radially from the initial>assumptions of uniform circular motion. . . .> So far I'm with you.> . . . Getting to an inward>pointing vector of constant length is a simple matter of taking the >second derivative, i.e. finding the acceleration. There is no need to >fudge a differential, hoping to a achieve a what amounts to a variable >constant. Pure silliness.> But here I've lost you. I just don't see what you're getting at.>Your second paragraph is just plain wrong. If a, b and c are arbitrary>vector functions, it is not true that b' always lies in the direction>of b and the c' always lies in the direction of c. For example, as>functions of time if b = (1,1,t) then b' = (0,0,1). They are NEVER in>the same direction for finite t.> Think it through, Joe. A cross product such as a = b x c is always> proportional to area. So, the derivative of a cross product is> necessarily defined in terms of changes in area. Which means you can> only evaluate the derivative of a cross product in terms of changes to> the area subtended by b x c. Which means b' has to point normal to c.> That's by the definition of a cross product and the derivative of a> cross product.> This is why b' has to be radially directed. The product b' x c has to> define a change in area to b x c even if the change is zero. But a b'> in the direction of c does not correspond to any area zero or not> because they don't intersect and don't subtend or define an area of> any kind. Consequently, b' x c with b' in the direction of c does not> correspond to the derivative of a cross product. It's as simple as> that.> More nonsense. You're just playing games.I'll spell it out for you and then I'm giving up.You claim that the angular momentum of a mass undergoing circular motion varies with time. This is false.Proof:Vectors are indicated with underscores (e.g. r_ ) or as ordered expressions: (x,y)m = constant mass of body in KgR = constant radius of circle in metersw = constant angular velocity in radians/secondt = time in secondsr_ = position vectorv_ = vector (NOT TANGENTIAL VELOCITY)p_ = momentum vectorL = angular momentumX = vector cross product* = scalar multiplication|x_| = magnitude (i.e. the length) of vector x_Note that any vector of the form (cos(wt),sin(wt)) has magnitude = 1.For circular motion:r_ = R*(cos(wt),sin(wt))v_ = dr_/dt = Rw*(-sin(wt),cos(wt))Note that r_ and v_ are perpendicular for all t (Proof: if you compute the dot product between them you get zero).p_ = m*v_ = mRw*(-sin(wt),cos(wt))Note that p_ is parallel to v_ and they are both perpendicular to r_L = r_ X p_ = |r_| * |p_|* sin(90 degrees) = |r_| * |p_| = R*mRw = mwR^2Each of m, w, and R is a constant scalar, therefore L is a constant scalar.-- Joe Legris === > Your first paragraph is just nonsense. You knew that r was of constant> length and that it is directed outward radially from the initial> assumptions of uniform circular motion. . . .> So far I'm with you.> . . . Getting to an inward> pointing vector of constant length is a simple matter of taking the > second derivative, i.e. finding the acceleration. There is no need to > fudge a differential, hoping to a achieve a what amounts to a > variable constant. Pure silliness.> But here I've lost you. I just don't see what you're getting at.> Your second paragraph is just plain wrong. If a, b and c are arbitrary> vector functions, it is not true that b' always lies in the direction> of b and the c' always lies in the direction of c. For example, as> functions of time if b = (1,1,t) then b' = (0,0,1). They are NEVER in> the same direction for finite t.> Think it through, Joe. A cross product such as a = b x c is always> proportional to area. So, the derivative of a cross product is> necessarily defined in terms of changes in area. Which means you can> only evaluate the derivative of a cross product in terms of changes to> the area subtended by b x c. Which means b' has to point normal to c.> That's by the definition of a cross product and the derivative of a> cross product.> This is why b' has to be radially directed. The product b' x c has to> define a change in area to b x c even if the change is zero. But a b'> in the direction of c does not correspond to any area zero or not> because they don't intersect and don't subtend or define an area of> any kind. Consequently, b' x c with b' in the direction of c does not> correspond to the derivative of a cross product. It's as simple as> that.> More nonsense. You're just playing games.> I'll spell it out for you and then I'm giving up.> You claim that the angular momentum of a mass undergoing circular motion > varies with time. This is false.> Proof:> Vectors are indicated with underscores (e.g. r_ )> or as ordered expressions: (x,y)> m = constant mass of body in Kg> R = constant radius of circle in meters> w = constant angular velocity in radians/second> t = time in seconds> r_ = position vector> v_ = vector (NOT TANGENTIAL VELOCITY)> p_ = momentum vector> L = angular momentum> X = vector cross product> * = scalar multiplication> |x_| = magnitude (i.e. the length) of vector x_> Note that any vector of the form (cos(wt),sin(wt)) has magnitude = 1.> For circular motion:> r_ = R*(cos(wt),sin(wt))> v_ = dr_/dt = Rw*(-sin(wt),cos(wt))> Note that r_ and v_ are perpendicular for all t (Proof: if you compute > the dot product between them you get zero).> p_ = m*v_ = mRw*(-sin(wt),cos(wt))> Note that p_ is parallel to v_ and they are both perpendicular to r_> L = r_ X p_> = |r_| * |p_|* sin(90 degrees)> = |r_| * |p_|> = R*mRw> = mwR^2 Each of m, w, and R is a constant scalar, therefore L is a constant scalar.> Oops, L is a constant vector perpendicular to r_ and p_. The final expression is its magnitude. It's direction is constant because r_ and p_ are restricted to the assumed plane of circular motion.-- Joe Legris === [snip]>I can't make sense of v dt = dr/dt dt = dr. Do you mean v * dt = (dr/dt) *>dt = dr?> I mean that the integral of a over the period of time dt as dt --> 0> produces v and that the integral of v over the period of time dt as dt> --> 0 produces dr and that all are centripetal in direction.Hmm. When I was taught about integration and stuff, we were warned>Very Sternly that if we thought we could treat dr and dt as though>they were ordinary quantities, we would soon produce nonsense. I'm>glad I got that warning.I think the terms I was looking for to describe what I had in mindwere antiderivative and indefinite integral.If you have a centripetally directed acceleration, by definition youalso have a centripetally directed antiderivative dv + kv. And youalso have a centripetally directed antiderivative dr + kr bydefinition. These are matters of definition because that's how thederivatives are taken to begin with.So, it would seem that dv, dr, dt, etc. are a lot more ordinary thanis commonly believed.> I would like to stay away from the question as to whether the results> are vectors or scalars or scalar multiples of vectors because I'm justCan you point to a vector that is not a scalar multiple of a>vector?As opposed to vector multiples of vectors perhaps?> Unless we are talking about infinitesimal values of r and t in which> case we are not talking about finite magnitudes of v, r, or t. The> integral is a function. It's just that taking the integral over the> same dt as is used in forming the derivative produces an instanteous> value for the result rather than the more conventional finite> integration over a range.It does? Hmm, an interesting idea. OK, so consider this body moving>round this circle, a snail gliding around the periphery of a clock, if>you like.[nf] stands for normal finite - this is the one the rest of us talk>about.At 12 o'clock, r, the [nf] position vector, points up, v the [nf]>velocity vector points left, a, the [nf] acceleration vector is>centripetal, and points down.>However, you say, integrating dv/dt [down] over dt gives the>instantaneous velocity, which must of course be in the same direction>- i.e. down. By exactly the same argument, integrating dr/dt [left]>over dt gives the instantaneous position, which must of course be in>the same direction - i.e. left. So although the [nf] snail is munching>a [nf] lettuce at the top of the clock, its instantaneous position>is somewhere along the hour hand. (I omitted to mention that it>happens to be nine o'clock.)I think your argument could also easily show that a braking car is>actually going backwards. Somewhere I fancy there should be a useful>legal application of all this.Brian Chandler>---------------->geo://Sano.Japan.Planet_3>Jigsaw puzzles from Japan at:>http://imaginatorium.org/shop/ === Your first paragraph is just nonsense. You knew that r was of constant>length and that it is directed outward radially from the initial>assumptions of uniform circular motion. . . .> So far I'm with you.> . . . Getting to an inward>pointing vector of constant length is a simple matter of taking the >second derivative, i.e. finding the acceleration. There is no need to >fudge a differential, hoping to a achieve a what amounts to a variable >constant. Pure silliness.> But here I've lost you. I just don't see what you're getting at.>Your second paragraph is just plain wrong. If a, b and c are arbitrary>vector functions, it is not true that b' always lies in the direction>of b and the c' always lies in the direction of c. For example, as>functions of time if b = (1,1,t) then b' = (0,0,1). They are NEVER in>the same direction for finite t.> Think it through, Joe. A cross product such as a = b x c is always> proportional to area. So, the derivative of a cross product is> necessarily defined in terms of changes in area. Which means you can> only evaluate the derivative of a cross product in terms of changes to> the area subtended by b x c. Which means b' has to point normal to c.> That's by the definition of a cross product and the derivative of a> cross product.> This is why b' has to be radially directed. The product b' x c has to> define a change in area to b x c even if the change is zero. But a b'> in the direction of c does not correspond to any area zero or not> because they don't intersect and don't subtend or define an area of> any kind. Consequently, b' x c with b' in the direction of c does not> correspond to the derivative of a cross product. It's as simple as> that.>More nonsense. You're just playing games.Well, if you really think I'm just playing games then I'll give youthe last word.I'll spell it out for you and then I'm giving up.You claim that the angular momentum of a mass undergoing circular motion >varies with time. This is false.Proof:Vectors are indicated with underscores (e.g. r_ )> or as ordered expressions: (x,y)>m = constant mass of body in Kg>R = constant radius of circle in meters>w = constant angular velocity in radians/second>t = time in seconds>r_ = position vector>v_ = vector (NOT TANGENTIAL VELOCITY)>p_ = momentum vector>L = angular momentum>X = vector cross product>* = scalar multiplication>|x_| = magnitude (i.e. the length) of vector x_>Note that any vector of the form (cos(wt),sin(wt)) has magnitude = 1.>For circular motion:r_ = R*(cos(wt),sin(wt))>v_ = dr_/dt = Rw*(-sin(wt),cos(wt))Note that r_ and v_ are perpendicular for all t (Proof: if you compute >the dot product between them you get zero).p_ = m*v_ = mRw*(-sin(wt),cos(wt))Note that p_ is parallel to v_ and they are both perpendicular to r_L = r_ X p_ = |r_| * |p_|* sin(90 degrees) = |r_| * |p_| = R*mRw = mwR^2Each of m, w, and R is a constant scalar, therefore L is a constant scalar.-- >Joe Legris> === >I think your argument could also easily show that a braking car is>actually going backwards. Somewhere I fancy there should be a useful>legal application of all this.Brian ChandlerZINNNNGGGG!-- Best Wishes,Wolf Kirchmeir, Blind River ONNot that brains are everything --you'll also need a skull to put them in. (Nancy Franklin, 1997) === > On Nov 15, 2002 in a discussion with Mati Meron the following points> were raised in connection with angular momentum and the circular> rotation of a point mass at distance r and constant tangential> velocity v.> My contention was and is that this situation represents constantly> changing angular momentum and his response was that in this situation> angular momentum is constant because dL/dt = 0.> [1] L = r x p by definition and that> [2] dL/dt = dr/dt x p + r x dp/dt.> He further states that dr/dt = v so that the cross product> [3] dr/dt x p = 0> because v and p are co linear.> Now, my question is whether this represents the correct interpretation> of the direction of dr/dt. The magnitude of dr/dt is certainly equal> to v. But I would suggest that the direction of dr/dt should be taken> radially in the direction of r.If that was true, then the point would fly away from the>center. It does not: it keeps the same distance from the>center of the circle.Fix a point O. Mark the place of the point on its circle>with center O with a vector r. Look a little bit later and>draw a vector r' from O to the point. Now draw the>vector Delta(r) = r'-r. Divide this vector by the time>difference Delta(t) and what you get is the average>velocity between the two points. You see that this>vector is not perpendicular to r.>But look less later than before, and draw a new r' and>ditto Delta(r) and divide again by a smaller Delta(t).>Again you get the average velocity between the two>points, but now this vectore is somewhat more>perpendicular to the vector r. Continue drawing vectors>r' closer and closer to r, and you notice that Delta(r)/Delta(t)>gets more and more perpendicular. Finally, in the limit>where Delta(t) goes to zero (which can be defined in>a perfectly strict mathematical way), you get something>that is abbreviated as dr/dt and that is perfectly>perpendicular to r.> Consequently, the cross product> dr/dt x p would not be zero nor would dL/dt.hthDirk Vdm You know, I'm afraid I was rather abrupt in the previously cited> discussion, which involved not only Meron, but Heymann, Dogfrey, and> Poe in somewhat heated terms. So, rather than just reply off the top> of my head I would rather read your reply in detail before answering. I do have one other question in the same regard. Do you happen to know> the origin of these angular mechanical concepts in terms of who> actually devised them? I'm aware that Newton did the conceptual work> for linear mechanics but never heard that he did the same for angular> mechanics.>I believe it was Euler.> === >Ah, c'mon, Wolf. A time derivative is just the ratio dr/dt as t --> 0.No, it's as dt --> 0, which is not quite the same thing.> I think you're correct here. But I don't see that it changes the case.>So, I think that integral of v dt = dr/dt dt = dr over the same>interval. That's what I mean by infinitessimal or instanteous>integral.I can't make sense of v dt = dr/dt dt = dr. Do you mean v * dt = (dr/dt) *>dt = dr?Back from vacation. This seems as good a place to jumpback in as any.> I mean that the integral of a over the period of time dt as dt --> 0> produces vNo, it doesn't. The integral of anything non-singular overa period of time dt as dt->0 goes to zero.The relationship between v(t), a and v(0) is thatv(t) - v(0) is the integral of a(t) from 0 to time t,not the integral over a tiny time interval. Theinterval from time 0 to time t could be decades.> I would like to stay away from the question as to whether the results> are vectors or scalars or scalar multiples of vectors because I'm just> trying at this point to establish the values involved. I think the> results are scalar multiples of vectorsThe term for scalar multiples of vectors is vectors.> but am just trying to> concentrate on the general idea of instantaneous integral values,> which are just the integrals over the same period of time dt used to> form derivatives as dt --> 0.There is no period of time. The derivative is a limitprocess. There's no period of time dt used toform derivatives. There's a decreasing sequence ofsuch intervals. And the inverse is most certainlynot an integral over a tiny interval. - Randy === >Ah, c'mon, Wolf. A time derivative is just the ratio dr/dt as t --> 0.>No, it's as dt --> 0, which is not quite the same thing.> I think you're correct here. But I don't see that it changes the case.>So, I think that integral of v dt = dr/dt dt = dr over the same>interval. That's what I mean by infinitessimal or instanteous>integral.>I can't make sense of v dt = dr/dt dt = dr. Do you mean v * dt = (dr/dt) *>dt = dr?Back from vacation. This seems as good a place to jump>back in as any.> I mean that the integral of a over the period of time dt as dt --> 0> produces vNo, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero.If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?>The relationship between v(t), a and v(0) is that>v(t) - v(0) is the integral of a(t) from 0 to time t,>not the integral over a tiny time interval. The>interval from time 0 to time t could be decades.> I would like to stay away from the question as to whether the results> are vectors or scalars or scalar multiples of vectors because I'm just> trying at this point to establish the values involved. I think the> results are scalar multiples of vectorsThe term for scalar multiples of vectors is vectors.> but am just trying to> concentrate on the general idea of instantaneous integral values,> which are just the integrals over the same period of time dt used to> form derivatives as dt --> 0.There is no period of time. The derivative is a limit>process. There's no period of time dt used to>form derivatives. There's a decreasing sequence of>such intervals. And the inverse is most certainly>not an integral over a tiny interval. - Randy === >[. . .]> I do have one other question in the same regard. Do you happen to know> the origin of these angular mechanical concepts in terms of who> actually devised them? I'm aware that Newton did the conceptual work> for linear mechanics but never heard that he did the same for angular> mechanics.>I believe it was Euler.Yes. At least it was mentioned by Edward Green that Euler had a handin it.> <3f043ec1.38463188@netnews.att.net> <3F04948D.2000709@xympatico.ca> <3f04c637.43017153@netnews.att.net> <3F04FB21.8070404@xympatico.ca> <3f0590c5.46325859@netnews.att.net> <3f074717.1885256@netnews.att.net> <3f077ccb.8655246@netnews.att.net> <585ab5d8.0307071130.3fd690b7@posting.google.com> <3f09f6d3.19319908@netnews.att.net> === > I mean that the integral of a over the period of time dt as dt --> 0> produces vNo, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero. If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?No, that's not how calculus works. You're taking dv/dt to be dv divided by dt, this isn't true at all.dv/dt is a derivative, while dv and dt are differential forms, 1-forms to be more exact. -- William C. Hogg === >Ah, c'mon, Wolf. A time derivative is just the ratio dr/dt as t --> 0.>No, it's as dt --> 0, which is not quite the same thing.> I think you're correct here. But I don't see that it changes the case.>So, I think that integral of v dt = dr/dt dt = dr over the same>interval. That's what I mean by infinitessimal or instanteous>integral.>I can't make sense of v dt = dr/dt dt = dr. Do you mean v * dt = (dr/dt) *>dt = dr?>Back from vacation. This seems as good a place to jump>back in as any. I mean that the integral of a over the period of time dt as dt --> 0> produces v>No, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero.If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?Wow, there sure are a lot of misstatements there.1. No, because the antiderivative process is not the process oftaking the area under an infinitesimally thin slice of a curve.2. The antiderivative of a dt doesn't mean anything, since a dt isnot a function of time. a is.3. dv is not the antiderivative of anything.dt and dv have certain meanings, but they can't be treated asfunctions themselves. - Randy === > I mean that the integral of a over the period of time dt as dt --> 0> produces v>No, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero.> If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?>No, that's not how calculus works. You're taking dv/dt to be dv divided >by dt, this isn't true at all.>dv/dt is a derivative, while dv and dt are differential forms, 1-forms to >be more exact. As I noted long back in this thread, LZ simply does not understand what aderivative is. And he doesn't want to; he wants people to keep trying toexplain things to him, explanations that he will not accept. That's hispurpose in this NG, not finding the truth. He will, of course, deny this.There is a term for this type of poster. Why people spend so much timetrying to straighten him out is a mystery. To him it's simply a game inwhich he gets a lot of attention. Rather a sad case. Denial to follow.But where does he show that he understands what a derivative is?Check out his performance in other NGs.>-- >William C. Hogg--John E. PrussingUniversity of Illinois at Urbana-ChampaignDepartment of Aerospace Engineeringhttp://www.uiuc.edu/~prussing === > I mean that the integral of a over the period of time dt as dt --> 0> produces v>No, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero. If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?>No, that's not how calculus works. You're taking dv/dt to be dv divided >by dt, this isn't true at all.>dv/dt is a derivative, while dv and dt are differential forms, 1-forms to >be more exact. As I noted long back in this thread, LZ simply does not understand what a>derivative is. And he doesn't want to; he wants people to keep trying to>explain things to him, explanations that he will not accept. That's his>purpose in this NG, not finding the truth. He will, of course, deny this.Do I crave your attention? It's really hard to know what to say.Should I say I absolutely concur with your flat assertions? I seem torecall that you recently took umbrage at my attacks on academics andneo scholastics. You apparently know what you know. But I doubt verymuch you know what you don't know.There is a term for this type of poster. Why people spend so much time>trying to straighten him out is a mystery. To him it's simply a game in>which he gets a lot of attention. Rather a sad case. Denial to follow.>But where does he show that he understands what a derivative is?Check out his performance in other NGs.I'm not sure what performances in which other newsgroups you'rereferring to. There are a number.>-- >William C. Hogg-->John E. Prussing>University of Illinois at Urbana-Champaign>Department of Aerospace Engineering>http://www.uiuc.edu/~prussing === > I mean that the integral of a over the period of time dt as dt --> 0> produces v>No, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero.> If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?No, that's not how calculus works. You're taking dv/dt to be dv divided >by dt, this isn't true at all.dv/dt is a derivative, while dv and dt are differential forms, 1-forms to >be more exact.I think I am taking dv/dt to be dv divided by dt. You say this isn'thow calculus works. But I'm really unclear as to what the differencewould be between dv/dt as a derivative and dv divided by dt as dt -->zero.And further I would like to understand what the converse of aderivative would be over dt as dt --> zero? We get the derivativesomehow, so it seems there should be some converse antiderivativeprocess. If antiderivative is not the appropriate term, is there someother way to describe the reverse of the derivative process withrespect to infinitely short dt's? === >[. . .] > I mean that the integral of a over the period of time dt as dt --> 0> produces v>No, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero.>If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?Wow, there sure are a lot of misstatements there.Well, it's good to have you back from vacation. I'm sure you'll behappy to see that nothing much has changed in your absence.1. No, because the antiderivative process is not the process of>taking the area under an infinitesimally thin slice of a curve.I'm not trying to describe the area under a curve. You take aderivative dv/dt as dt --> 0. I would just like to understand how totake the reverse of that derivative as dt --> 0.>2. The antiderivative of a dt doesn't mean anything, since a dt is>not a function of time. a is.So, what is the converse of dv/dt as t --> 0?>3. dv is not the antiderivative of anything.dt and dv have certain meanings, but they can't be treated as>functions themselves.> === >And further I would like to understand what the converse of a>derivative would be over dt as dt --> zero? There is no derivative over dt. That's not what a derivative is.dv/dt is a derivative of a function that relates t and v; label it F(v.t).Unless you specify what that function is, you can't ell what dv/dt is.The antiderivative of dv/dt is F(v,t)The reason you're confused is that you're using Newtonian notation. InLeibnizian notation, we write The derivative of F(v,t) is F'(v,t).See?-- Best Wishes,Wolf Kirchmeir, Blind River ONNot that brains are everything --you'll also need a skull to put them in. (Nancy Franklin, 1997) === >If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?>Wow, there sure are a lot of misstatements there.Well, it's good to have you back from vacation. I'm sure you'll be>happy to see that nothing much has changed in your absence.Yeah, I see that.>1. No, because the antiderivative process is not the process of>taking the area under an infinitesimally thin slice of a curve.I'm not trying to describe the area under a curve. You take a>derivative dv/dt as dt --> 0. I would just like to understand how to>take the reverse of that derivative as dt --> 0.By the fundamental theorem of calculus, the reverse of the derivativeis the area under the curve.The derivative as dt->0 is some weird misstatement or misconception.The derivative is the slope of a tangent line. Its value is found by alimit process of taking secant lines over smaller and smallerintervals. Your language makes it sound like there's a differencebetween derivative and derivative as dt->0. There's only onederivative, it involves a limit process. Its inverse involvescalculation of an area over a finite region.You want to know how to reverse the process? That's how to reverse theprocess.>2. The antiderivative of a dt doesn't mean anything, since a dt is>not a function of time. a is.So, what is the converse of dv/dt as t --> 0?The converse of dv/dt is to take the area under a curve.The value of dv/dt at t=0 is the slope of a curve at t=0. It's nodifferent from taking the slope at t=10. Again, you seem to want towillfully ignore the difference between t and dt. - Randy === >In wchogg derivative is. And he doesn't want to; he wants people to keep trying to>explain things to him, explanations that he will not accept. That's his>purpose in this NG, not finding the truth. He will, of course, deny this.There is a term for this type of poster. Why people spend so much time>trying to straighten him out is a mystery. They're talking through him. There are others who don't know./BAH === >And further I would like to understand what the converse of a>derivative would be over dt as dt --> zero? There is no derivative over dt. That's not what a derivative is.dv/dt is a derivative of a function that relates t and v; label it F(v.t).>Unless you specify what that function is, you can't ell what dv/dt is.The antiderivative of dv/dt is F(v,t)Okay, I'll buy this. The question I have then is whether in thepresence of dv/dt, do you necessarily have F(v,t)? The reason you're confused is that you're using Newtonian notation. In>Leibnizian notation, we write The derivative of F(v,t) is F'(v,t).See?> === > The derivative as dt->0 is some weird misstatement or misconception.Or a truncation combined with abuse of notation: something like [theexpression dx/dt goes to] the derivative as dt->0Notational abuse. The hidden shame, hidden no longer! Call theInternational Taskforce Succoring Battered Symbols (ITsBS) today! <3f043ec1.38463188@netnews.att.net> <3F04948D.2000709@xympatico.ca> <3f04c637.43017153@netnews.att.net> <3F04FB21.8070404@xympatico.ca> <3f0590c5.46325859@netnews.att.net> <3f074717.1885256@netnews.att.net> <3f077ccb.8655246@netnews.att.net> <585ab5d8.03070 === > I mean that the integral of a over the period of time dt as dt --> 0> produces v>No, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero.> If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?>No, that's not how calculus works. You're taking dv/dt to be dv divided >by dt, this isn't true at all.>dv/dt is a derivative, while dv and dt are differential forms, 1-forms to >be more exact. > As I noted long back in this thread, LZ simply does not understand what a> derivative is. And he doesn't want to; he wants people to keep trying to> explain things to him, explanations that he will not accept. That's his> purpose in this NG, not finding the truth. He will, of course, deny this.> There is a term for this type of poster. Why people spend so much time> trying to straighten him out is a mystery. To him it's simply a game in> which he gets a lot of attention. Rather a sad case. Denial to follow.> But where does he show that he understands what a derivative is?> Check out his performance in other NGs.Well, it's my policy that if a person isn't clearly insane or doesn't respond with only insults, then I try to set them straight at least once.I recently had a try with Laurent, which I've given up on, and seeing LZ's response to my post where he asserts the algebraic division of dv and dt *is* a derivative, I've given up on him now too.There was even a guy awhile back who refused to understand the concept of a non-Riemmanian metric. I gave up on him after one post as well.-- William C. Hogg === >There is no derivative over dt. That's not what a derivative is.>dv/dt is a derivative of a function that relates t and v; label it F(v.t).>Unless you specify what that function is, you can't ell what dv/dt is.>The antiderivative of dv/dt is F(v,t)Okay, I'll buy this. The question I have then is whether in the>presence of dv/dt, do you necessarily have F(v,t)? Yes, but it may be very difficult to determine what it is.The integral/derivative concept is hierarchical.EG, if you have function F(x), then there is an integral of F(x), call itf(x). There is also a derivative, call it F'(x). Now, F(x) is the derivativeof f(x), and F'(x) is the derivative of F(x). See?Problem is, given f(x), it may not be possible to determine f(x) or F'(x), orboth. There's a whole branch of calculus that deals with whether or not agiven class of functions can be differentiated or integrated. IIRC (it's beena very long time), the differentiation question is generally more difficultto answer than the integration question. Then there's partialdifferentiation.....HTH-- Best Wishes,Wolf Kirchmeir, Blind River ONNot that brains are everything --you'll also need a skull to put them in. (Nancy Franklin, 1997) === >Notational abuse. The hidden shame, hidden no longer! Call the>International Taskforce Succoring Battered Symbols (ITsBS) today!LOLHave to remember that one.-- Best Wishes,Wolf Kirchmeir, Blind River ONNot that brains are everything --you'll also need a skull to put them in. (Nancy Franklin, 1997) === [. . .]By the fundamental theorem of calculus, the reverse of the derivative>is the area under the curve.The derivative as dt->0 is some weird misstatement or misconception.>The derivative is the slope of a tangent line. Its value is found by a>limit process of taking secant lines over smaller and smaller>intervals. Your language makes it sound like there's a difference>between derivative and derivative as dt->0. There's only one>derivative, it involves a limit process. Its inverse involves>calculation of an area over a finite region.So, what is the difference between the slopes of successive tangentlines over smaller and smaller differences for a circle? Thats's allI'm trying to put a name to.I've seen the terminology used dr/dt as dt-> 0 used frequently todescribe differential approximations. So, I don't see anythingespecially weird about it.You want to know how to reverse the process? That's how to reverse the>process.[. . .] === >2. The antiderivative of a dt doesn't mean anything, since a dt is>not a function of time. a is.> So, what is the converse of dv/dt as t --> 0?Let's do a simple example.Suppose I have a uniform acceleration of 2 m/sec^2,so a = 2. Suppose my velocity starts out at 0 at timet=0. Thus, as a function of time my velocity is v = 2t.You can plot this. It is a straight line with slope2, that goes through the origin. The derivative is2 everywhere for all t>=0. What is dv/dt at t=0? 2.What is dv/dt at t=10? 2. What is dv/dt at t=100? 2.What is v at t=0? 2*t = 0What is v at t=10? 2*t = 20Waht is v at t=100? 2*t = 200.Now perhaps you can say what you mean by dv/dt ast-->0 or dv/dt as dt-->0 in this case and whetherwhat you get by it is that acceleration = 2 for allpositive times.If I plot a vs. t, it's just a horizontal line ata=2. You ask, how do I invert this process to obtainv from this curve?I'm trying to tell you, by calculating the area. Inthis case it's particularly easy, since the curveis horizontal. All slices from t=0 to any positivetime t form rectangles of height 2 and width t. Thearea of such a rectangle is 2t which, I hope you'llnotice, is the correct expression for velocity thatI started with.You are asking for some infinitesimal process andclaiming that somehow doing something on thishorizontal line with infinitesimal slices is goingto give me v = 2t. You're welcome to tell me whatthat process is. - Randy === >The derivative as dt->0 is some weird misstatement or misconception.>The derivative is the slope of a tangent line. Its value is found by a>limit process of taking secant lines over smaller and smaller>intervals. Your language makes it sound like there's a difference>between derivative and derivative as dt->0. There's only one>derivative, it involves a limit process. Its inverse involves>calculation of an area over a finite region.> So, what is the difference between the slopes of successive tangent> lines over smaller and smaller differences for a circle? Thats's all> I'm trying to put a name to.I don't know what that means. Here's how I draw atangent line to a circle: 1. Pick a point on a circle.2. Draw a line lying outside the circle which touchesthe circle only at that point.Where do smaller and smaller differences come in? Howdo I do successive tangent lines over smaller andsmaller distances? Give me a point, there's one tangent,with no differences.I can't guess what name to give your concept when I don'teven know what your concept is.> I've seen the terminology used dr/dt as dt-> 0 used frequently to> describe differential approximations.You've abused some notation which might have made senseoriginally, distorted it to the point where it doesn'tany more.> So, I don't see anything especially weird about it.The symbols dr and dt when talking about a derivativedon't stand for finite things. The symbol dr/dt can alsobe read this way: d/dt (r), that is the operator d/dtapplied to the function r(t). Then it should be muchclearer that the d is not a number, and neither isdt. d/dt is an operation.You keep jumping back and forth between things, treatedthem as interchangeable when they're not. Sometimes yousay as t->0, sometimes as dt->0. Those are entirelydifferent things. You just said differential approximation.That's different from derivative, and the symbolsused in such approximations are different from thesymbols used in derivatives.If you want to talk about finite differences, fine. Maybethat will remove some of the clutter from this conversation.Usually, to avoid this kind of confusion, the symbolsused aren't dr and dt which have reserved meanings,but delta-r and delta-t with upper or lowercasegreek deltas. - Randy === >The derivative as dt->0 is some weird misstatement or misconception.>The derivative is the slope of a tangent line. Its value is found by a>limit process of taking secant lines over smaller and smaller>intervals. Your language makes it sound like there's a difference>between derivative and derivative as dt->0. There's only one>derivative, it involves a limit process. Its inverse involves>calculation of an area over a finite region.> So, what is the difference between the slopes of successive tangent> lines over smaller and smaller differences for a circle? Thats's all> I'm trying to put a name to.I don't know what that means. Here's how I draw a>tangent line to a circle: >1. Pick a point on a circle.>2. Draw a line lying outside the circle which touches>the circle only at that point.Where do smaller and smaller differences come in? How>do I do successive tangent lines over smaller and>smaller distances? Give me a point, there's one tangent,>with no differences.Well, I guess what's confusing is that above you use the terminologysmaller and smaller intervals. This is what I'm referring to withthe phrase smaller and smaller distances. I assume that tangentstaken at successive smaller and smaller distances would bedifferent. And all I'm asking is what the difference betweensuccessive tangents would amount to for circles as those smaller andsmaller differences approach zero? I can't guess what name to give your concept when I don't>even know what your concept is.> I've seen the terminology used dr/dt as dt-> 0 used frequently to> describe differential approximations.You've abused some notation which might have made sense>originally, distorted it to the point where it doesn't>any more.> So, I don't see anything especially weird about it.The symbols dr and dt when talking about a derivative>don't stand for finite things. The symbol dr/dt can also>be read this way: d/dt (r), that is the operator d/dt>applied to the function r(t). Then it should be much>clearer that the d is not a number, and neither is>dt. d/dt is an operation.You keep jumping back and forth between things, treated>them as interchangeable when they're not. Sometimes you>say as t->0, sometimes as dt->0. Those are entirely>different things. You just said differential approximation.>That's different from derivative, and the symbols>used in such approximations are different from the>symbols used in derivatives.If you want to talk about finite differences, fine. Maybe>that will remove some of the clutter from this conversation.>Usually, to avoid this kind of confusion, the symbols>used aren't dr and dt which have reserved meanings,>but delta-r and delta-t with upper or lowercase>greek deltas.You know, I understand what you are saying here. I've made notationalmistakes and am trying to correct those. I also agree that deltanotation would be more appropriate. But what about infinitesimallysmall differences? - Randy === > I mean that the integral of a over the period of time dt as dt --> 0> produces v>No, it doesn't. The integral of anything non-singular over>a period of time dt as dt->0 goes to zero.> If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?>No, that's not how calculus works. You're taking dv/dt to be dv divided >by dt, this isn't true at all.>dv/dt is a derivative, while dv and dt are differential forms, 1-forms to >be more exact. As I noted long back in this thread, LZ simply does not understand what a> derivative is. And he doesn't want to; he wants people to keep trying to> explain things to him, explanations that he will not accept. That's his> purpose in this NG, not finding the truth. He will, of course, deny this.> There is a term for this type of poster. Why people spend so much time> trying to straighten him out is a mystery. To him it's simply a game in> which he gets a lot of attention. Rather a sad case. Denial to follow.> But where does he show that he understands what a derivative is?> Check out his performance in other NGs.Well, it's my policy that if a person isn't clearly insane or >doesn't respond with only insults, then I try to set them >straight at least once.I recently had a try with Laurent, which I've given up on, and seeing LZ's >response to my post where he asserts the algebraic division of dv and dt >*is* a derivative, I've given up on him now too.Well, I certainly appreciate your consideration in giving up on meafter only one inept try of apparently not reading my reply post. Asshould be intuitively obvious to the casual observer I was trying toagree with your observation regarding my analytical approach to thetaking of derivatives and asking what the alternative might be.And I find myself still asking the same question. I'm rather surprisedthat you gave me even one try. I would have supposed that you wouldhave tried me in absentia and pronounced me insane merely for making asuggestion with which you disagreed. Do you teach students the sameway? I'm frankly somewhat concerned that you condescend to get offyour fat academic laurels long enough to read posts that are soclearly beneath contempt.There was even a guy awhile back who refused to understand the concept of >a non-Riemmanian metric. I gave up on him after one post as well.-- >William C. Hogg === >In wchogg > As I noted long back in this thread, LZ simply does not understand what a>derivative is. And he doesn't want to; he wants people to keep trying to>explain things to him, explanations that he will not accept. That's his>purpose in this NG, not finding the truth. He will, of course, deny this.>There is a term for this type of poster. Why people spend so much time>trying to straighten him out is a mystery. > They're talking through him. There are others who don't know.> (Although, I knew this stuff pretty well already. I hope Lester can beconfused about some slightly more advanced calculus soon.)-Laurel === >Well, I guess what's confusing is that above you use the terminology>smaller and smaller intervals. This is what I'm referring to with>the phrase smaller and smaller distances. I assume that tangents>taken at successive smaller and smaller distances would be>different. And all I'm asking is what the difference between>successive tangents would amount to for circles as those smaller and>smaller differences approach zero?GAACK!!!!Lester, a tangent is a line that touches a curve at a point. There ain't nosuccessive tangents taken at smaller and smaller distances. You seem to bethinking of secants or chords, which are a different animal entirely.Where the hell did you learn geometry, anyhow?-- Best Wishes,Wolf Kirchmeir, Blind River ONNot that brains are everything --you'll also need a skull to put them in. (Nancy Franklin, 1997) === >There is no derivative over dt. That's not what a derivative is.>dv/dt is a derivative of a function that relates t and v; label it F(v.t).>Unless you specify what that function is, you can't ell what dv/dt is.>The antiderivative of dv/dt is F(v,t)>Okay, I'll buy this. The question I have then is whether in the>presence of dv/dt, do you necessarily have F(v,t)? Yes, but it may be very difficult to determine what it is.The integral/derivative concept is hierarchical.EG, if you have function F(x), then there is an integral of F(x), call it>f(x). There is also a derivative, call it F'(x). Now, F(x) is the derivative>of f(x), and F'(x) is the derivative of F(x). See?Problem is, given f(x), it may not be possible to determine f(x) or F'(x), or>both. There's a whole branch of calculus that deals with whether or not a>given class of functions can be differentiated or integrated. IIRC (it's been>a very long time), the differentiation question is generally more difficult>to answer than the integration question. Then there's partial>differentiation.....>I appreciate your candor. All I'm trying to suggest is that givenacceleration, there has to be a velocity, and that given a velocitythere has to be an r - because velocity is the time rate of change inr and acceleration the time rate of change in velocity.Of course, this is just plain language. === >In wchogg > As I noted long back in this thread, LZ simply does not understand what a>derivative is. And he doesn't want to; he wants people to keep trying to>explain things to him, explanations that he will not accept. That's his>purpose in this NG, not finding the truth. He will, of course, deny this.>There is a term for this type of poster. Why people spend so much time>trying to straighten him out is a mystery. > They're talking through him. There are others who don't know.> (Although, I knew this stuff pretty well already. I hope Lester can be>confused about some slightly more advanced calculus soon.)>disabused about SR and angular momentum, I'm sure I'll be thoroughlyconfused about Planck's constant very shortly. === >Well, I guess what's confusing is that above you use the terminology>smaller and smaller intervals. This is what I'm referring to with>the phrase smaller and smaller distances. I assume that tangents>taken at successive smaller and smaller distances would be>different. And all I'm asking is what the difference between>successive tangents would amount to for circles as those smaller and>smaller differences approach zero?GAACK!!!!Lester, a tangent is a line that touches a curve at a point. There ain't no>successive tangents taken at smaller and smaller distances. You seem to be>thinking of secants or chords, which are a different animal entirely.Where the hell did you learn geometry, anyhow?>different posters on the same subject is rather like taking on a wholesquad of tag team wrestlers by oneself. Oh, well.But I think you might be wrong on this one. There are different pointson the circumference of a circle. And there are tangents withdifferent slopes at each point. And as the points get closer andcloser the slopes of those tangents become closer as well. So as theintervals between points approaches zero, the difference betweentangents should approach some value. Nothing ridiculous here that Ican tell.So? Should we make the best two out of three falls? Or should I justtake a standing eight count? === >In wchogg > As I noted long back in this thread, LZ simply does not understand what a>derivative is. And he doesn't want to; he wants people to keep trying to>explain things to him, explanations that he will not accept. That's his>purpose in this NG, not finding the truth. He will, of course, deny this.>There is a term for this type of poster. Why people spend so much time>trying to straighten him out is a mystery. > They're talking through him. There are others who don't know.>(Although, I knew this stuff pretty well already. I hope Lester can be>confused about some slightly more advanced calculus soon.)> disabused about SR and angular momentum, I'm sure I'll be thoroughly> confused about Planck's constant very shortly. Looking forward to it.-Laurel === : But I think you might be wrong on this one. There are different points: on the circumference of a circle. And there are tangents with: different slopes at each point. And as the points get closer and: closer the slopes of those tangents become closer as well. So as the: intervals between points approaches zero, the difference between: tangents should approach some value. Nothing ridiculous here that I: can tell.The difference should approach 0. That should be pretty obvious. If theta is the position of a point on the circle in radians,then the slope of the tangent to that point is theta+PI/2.For a nearby point at position theta+d, the tangent is theta+d+PI/2.You seem to be asking what is lim (theta+PI/2) - (theta+d+PI/2) d->0which clearly is 0.Stephen === >The derivative as dt->0 is some weird misstatement or misconception.>The derivative is the slope of a tangent line. Its value is found by a>limit process of taking secant lines over smaller and smaller>intervals. Your language makes it sound like there's a difference>between derivative and derivative as dt->0. There's only one>derivative, it involves a limit process. Its inverse involves>calculation of an area over a finite region.> You said this:> So, what is the difference between the slopes of successive tangent> lines over smaller and smaller differences for a circle?Notice whose phrase is smaller and smaller differences.I said it doesn't make sense here. I don't know what successivetangent lines means. You give me a point on a circle, there's onetangent through it. No successive.>Where do smaller and smaller differences come in? How>do I do successive tangent lines over smaller and>smaller distances? Give me a point, there's one tangent,>with no differences.Well, I guess what's confusing is that above you use the terminology>smaller and smaller intervals. This is what I'm referring to with>the phrase smaller and smaller distances.That's your phrase. I don't know what you mean by it. I don't have asuccession of tangents, and I don't have any intervals or distancesassociated with any tangent. I have a point, and one tangent.> I assume that tangents>taken at successive smaller and smaller distances would be>different.I don't know how to take tangents at ANY distance, let alone differentones at different distances. I don't have any clue what you're talkingabout.> And all I'm asking is what the difference between>successive tangents would amount to for circles as those smaller and>smaller differences approach zero? I don't know. I've never talked about a succession of tangents.There's only one tangent. I've never talked about intervals inquestion about a terminology which is unique to you.>You know, I understand what you are saying here. I've made notational>mistakes and am trying to correct those. I also agree that delta>notation would be more appropriate. But what about infinitesimally>small differences?It's an abstraction. There's no such thing. There's a limit process,but it's a logical process that does not involve ever looking atinfinitesimally small differences.However, we can get to common ground if you stay away from infinitiesand talk about FINITE differences. If they are small (notinfinitesimal) then I think we can talk about what you want to talkabout with the understanding that it is a first-order approximation.When distances are very small, first-order is a very goodapproximation indeed. - Randy === >Well, I guess what's confusing is that above you use the terminology>smaller and smaller intervals. This is what I'm referring to with>the phrase smaller and smaller distances. I assume that tangents>taken at successive smaller and smaller distances would be>different. And all I'm asking is what the difference between>successive tangents would amount to for circles as those smaller and>smaller differences approach zero?>GAACK!!!!>Lester, a tangent is a line that touches a curve at a point. There ain't no>successive tangents taken at smaller and smaller distances. You seem to be>thinking of secants or chords, which are a different animal entirely.>Where the hell did you learn geometry, anyhow?>different posters on the same subject is rather like taking on a whole>squad of tag team wrestlers by oneself. Oh, well.But I think you might be wrong on this one. He's not wrong, he just said what I said.>There are different points>on the circumference of a circle. And there are tangents with>different slopes at each point.Yes. And each single point defines one tangent, with one slope.> And as the points get closer and>closer the slopes of those tangents become closer as well.That's true. So what?> So as the>intervals between points approaches zero, the difference between>tangents should approach some value.It approaches zero. So?> Nothing ridiculous here that I can tell.Except when you try to claim that somehow this has something to dowith estimating the slope of a tangent. You now appear to be doingsomething related to a second derivative (the rate of change of aslope) when you haven't sorted out the first derivative (the slope)yet. - Randy === >In wchogg > As I noted long back in this thread, LZ simply does not understand what a>derivative is. And he doesn't want to; he wants people to keep trying to>explain things to him, explanations that he will not accept. That's his>purpose in this NG, not finding the truth. He will, of course, deny this.>There is a term for this type of poster. Why people spend so much time>trying to straighten him out is a mystery. > They're talking through him. There are others who don't know.>'ey, Laurel.>(Although, I knew this stuff pretty well already. I hope Lester can be>confused about some slightly more advanced calculus soon.)I can't imagine people who don't want to learn math. I loveddoing algebra and then discovered that calculus made difficultalgebraic problems easy. It was a PITA to calculate areasthat involved curvy lines using algebra./BAH === > The derivative as dt->0 is some weird misstatement or misconception.Or a truncation combined with abuse of notation: something like [the>expression dx/dt goes to] the derivative as dt->0Notational abuse. The hidden shame, hidden no longer! Call the>International Taskforce Succoring Battered Symbols (ITsBS) today!Are you going to trademark that one? The guys trying to getsomething accomplished in IT would lap it up./BAH === > disabused about SR and angular momentum, I'm sure I'll be thoroughly> confused about Planck's constant very shortly. > Now you're starting to make some sense. Keep it up!-- Joe Legris === OK, I see what's going on in this exchange.> I said this:>The derivative is the slope of a tangent line. Its value is found by a>limit process of taking secant lines over smaller and smaller>intervals.Lester apparently thought he was paraphrasing it backto me when he said this:> So, what is the difference between the slopes of successive tangent> lines over smaller and smaller differences for a circle? Thats's all> I'm trying to put a name to.I'm talking about secant lines, or chords, that hita curve at (at least) two points. I'm talking about asequence of such chords where the two points drawever closer together as the secant moves ever closerto a tangent line. But except in the case where thecurve is straight over a finite region, none of thesesecants/chords IS a tangent.Lester thought he could replace this careful phrasing,this sequence of lines none of which is a tangent, with sequence of tangent lines.Lester, you can't. Sequence of secant lines orsecant of chords makes sense, particularly withsmaller and smaller intervals. Sequence of tangentlines makes no sense and is not at all the same thing.Words are important. You can't willy-nilly replace someof my words with different ones and then think we're talkingabout the same thing. - Randy === > Yes. And each single point defines one tangent, with one slope.No it certainatly does not.A point on a curve merely describes a coordinate. It is thedifferentiable triangle AT THAT POINT that describes the tangent andslope.f(x) = y = some expression with xdy/dx = the derivative for yConsider the differentiable triangle. ds^2 = dx^2 + dy^2 ds/dx = sqrt(1 + dy^2/dx^2)ds = int sqrt(1 + f(x)^2) . dxIt is ds that describes the slope and the tangent. === I might as well get this on-record too.Let y, m, and n be positive reals such that n >= m.Then:/n | y |! |---| f(x) dx =! |_x_|/m! y |!---|!_m_|--- /min(y/k,n) |/ | f(x) dx--- /mk=1In linear-mode:integral{m to n} floor(y/x) f(x) dx =sum{k=1 to floor(y/m)} integral{m to min(y/k,n)} f(x) dx(Yay!)Leroy Quet === I have been kindly reminded that my posts tend to favor opinion andnot so much fact. That I have more imagination than training may bethe reason. I invite you to interact with me so I may learn and grow. Ernst Berg === Let p be a polynomial with p(0)=0. Prove that if p'(x)<=p(x) for all x, then>p must be identically 0.>I can only prove it for degree n=1,2,3.>Anyone?>There may be an easy argument, but I don't see it.> SOLUTION.> e^x(p(x)e^{-x})' =< 0 for all real x . Therefore> === ===========> g_1(x):=p(x)e^{-x} , (Note that g_1(0)=0),> === ===========> is non-increasing on (-infty,infty). Writing > g_1(infty) =< g_1(x) =< g_1(0) , x in [0,infty),> So what's important is not that P is a polynomial,> but that it is sub-exponential (p(x)/exp(x) -> 0 as x -> inf).> The result then can become:> If p is subexponential, p(0) = 0, and p'(x) <= p(x) for x >= 0,> then p(x) = 0 for x >= 0. === ======Nice remark ! Alex. === ======Alex.Lupas scribe:(...)> If p is subexponential, p(0) = 0, and p'(x) <= p(x) for x >= 0,> then p(x) = 0 for x >= 0.> === ======> Nice remark ! Alex.> === ===Yes, of course, it is a very nice way. And I like it, toknow/to learn fresh ideas. But there seems to be another(not so elegant) ODE-way by using the lemma of Gronwall. === haven't been able to work out a proof:Let f be a continuous real valued function on [0,1] with f(0)=0, andlet d>0 be given. ThenP(|B_t-f(t)| 0where B_t is a Brownian motion. That is to say, with positiveprobability Brownian motion will approximate any function as close asyou like on the unit interval(and hence on any interval, by scaling).I even am having trouble with the case f(t)=0 for all t. I havereduced it, using the reflection principle, to showing that theintegral from d to 3d plus the integral from 5d to 7d plus from 9d to11d plus... of the density of BM at t=1 is less than 1/4 for all d>0,and I'm sure that this is true, because the probabilisticinterpretation above shows that this sum increases as d decreases,which means if it were ever 1/4 for some d then it would always be 1/4for any smaller d, and I'm sure that doesn't happen, but can't quitework out the details. Any help or references will be appreciated. === > haven't been able to work out a proof: Let f be a continuous real valued function on [0,1] with f(0)=0, and> let d>0 be given. Then P(|B_t-f(t)| 0 where B_t is a Brownian motion. That is to say, with positive> probability Brownian motion will approximate any function as close as> you like on the unit interval(and hence on any interval, by scaling).>I am a little rusty and away from my books, but isn't it true that, forany e > 0, P{B(t)=0 for infinitely many t in (0,e)} = 1? If so,doesn't this contradict your conjecture for nonzero f?-- Direct access to this group with http://web2news.comhttp://web2news.com/?sci.mathTo contact in private, remove nn-oosp0am === > haven't been able to work out a proof:> Let f be a continuous real valued function on [0,1] with f(0)=0, and> let d>0 be given. Then> P(|B_t-f(t)| 0> where B_t is a Brownian motion. That is to say, with positive> probability Brownian motion will approximate any function> as close as> you like on the unit interval(and hence on any interval,> by scaling).> I am a little rusty and away from my books, but isn't it> true that, for> any e > 0, P{B(t)=0 for infinitely many t in (0,e)} = 1? If so,> doesn't this contradict your conjecture for nonzero f?I take it back - it is not clear that property of the zeroes contradictsyour conjecture.-- Direct access to this group with http://web2news.comhttp://web2news.com/?sci.math === >haven't been able to work out a proof:Let f be a continuous real valued function on [0,1] with f(0)=0, and>let d>0 be given. ThenP(|B_t-f(t)| 0where B_t is a Brownian motion. That is to say, with positive>probability Brownian motion will approximate any function as close as>you like on the unit interval(and hence on any interval, by scaling).I don't know, although I sort of doubt it (this proves _nothing_because I don't know anything about these things).What is true, and easy to prove, is that you can approximate fwith positive probability by Brownian motion composed witha change of variable in t: The probability that there exists anincreasing continuous function T(t) such that(*) |B_T(t) - f(t)| < d for all t in [0,1]is positive, and this is easy to show:The idea is this: Divide R into countably many disjointintervals I_n, each of length d, and consider the order inwhich B_t visits the I_n. That doesn't quite work becauseB_t is likely to jump back and forth infinitely many timesbetween two neighboring I_n's in finite time, so you don'tget a sequence of n's. But you can fix that:Consider a tiling of R into countably many disjointintervals, each of length d, and let I_n be the middlethird of the n-th one of these intervals. (Say the originis a point of I_0.) Now the continuity of B_t shows that it can't jump back and forth between I_n's infinitely many times in finite time, so you get a sequence n_1, n_2, ... such that each n is the previous one plus or minus 1, and such that B_t visits I_n_1, then I_n_2 (before any other I_n), then I_n_3, etc.Since going left is exactly as likely as going right,if n_1, n_2, ... n_N is any finite sequence (withn_1 = 0 and each n equal to the previous oneplus or minus 1), the probability that this sequenceis the one you get for the first N steps is 2^(N-1).And given f continuous on [0,1] and d > 0 thereexists N and a choice of n_1, ... n_N such thatif that's the sequence generated by B_t then(*) holds (for some time change T(t).)>I even am having trouble with the case f(t)=0 for all t. I have>reduced it, using the reflection principle, to showing that the>integral from d to 3d plus the integral from 5d to 7d plus from 9d to>11d plus... of the density of BM at t=1 is less than 1/4 for all d>0,>and I'm sure that this is true, because the probabilistic>interpretation above shows that this sum increases as d decreases,>which means if it were ever 1/4 for some d then it would always be 1/4>for any smaller d, and I'm sure that doesn't happen, but can't quite>work out the details. Any help or references will be appreciated.**David C. Ullrich === > haven't been able to work out a proof:> Let f be a continuous real valued function on [0,1] with f(0)=0, and> let d>0 be given. Then> P(|B_t-f(t)| 0> where B_t is a Brownian motion. That is to say, with positive> probability Brownian motion will approximate any function as close as> you like on the unit interval(and hence on any interval, by scaling).> I even am having trouble with the case f(t)=0 for all t. I have> reduced it, using the reflection principle, to showing that the> integral from d to 3d plus the integral from 5d to 7d plus from 9d to> 11d plus... of the density of BM at t=1 is less than 1/4 for all d>0,> and I'm sure that this is true, because the probabilistic> interpretation above shows that this sum increases as d decreases,> which means if it were ever 1/4 for some d then it would always be 1/4> for any smaller d, and I'm sure that doesn't happen, but can't quite> work out the details. Any help or references will be appreciated.The case f(t)=0 is easy. Let $T$ be the first exit time of the Brownian motion from the interval (-d,d). ThenP(|B_t-f(t)|= P(T>1)The distribution of the random variable T is known; this distribution is absolutely continuous with a continuous strictly positive density function q. (See, for example, the book of Karatzas & Shreve; a formula for q appears at the foot of page 99 in the first edition of the book.) Therefore P(T>1) = int_1^infty q(t) dt > 0.The case of general f amounts to the statement that the Wiener measure has full support in the space of continuous paths that vanish at time 0. This is a well-known fact. Here is a proof based on the Cameron-Martin theorem. First observe that it suffice to consider continuously differentiable f. Indeed, given continuous f and d>0, pick a continuously differentiable h with h(0)=0 such that |f(t)-h(t)|= P(|B_t-h(t)|0.So we may as well suppose that f is continuously differentiable. The theorem of Cameron and Martin now tells us that the distribution ofB-f = (B_t-f(t))_{t in [0,1]} is absolutely continuous with respect to Wiener measure: If A is a set of paths then P(B in A) > 0 if and only if P(B-f in A) > 0. Taking A to be those paths that stay within d units of 0 throughout the time interval [0,1] we haveP(|B_t-h(t)|0 because P(B in A) > 0 by the earlier discussion of the case f=0.-- A. === Ignacio Larrosa Ca.96estro escribi.97 en el mensaje|nbdvpff$119lro$1@ID-137122.news.dfncis.de:> Julien Santini escribi.97 en el> mensaje|nbdvfb1$3k4$1@news-reader2.wanadoo.fr:> Would there be an easy way to find all the positive integers a,b,c> (we may suppose they're relatively prime, ie> gcd(a,b)=gcd(a,c)=gcd(b,c)=1) such that:> a divides (b+c)^2, b divides (a+c)^2 and c divides (a+b)^2 ?> It must be infinitely many. With 1 <= a < b < c <= 1000 there are a b c> = = => 1 2 5> 1 5 13> 1 13 34> 1 34 89> 1 89 233> 1 233 610> 2 5 29> 2 29 169> 2 169 985> 5 13 194> 5 29 433> All of that values of a, b and c are sum of two squares. For a = 1, b = Fib(2n+1) and c = Fib(2n+3) in the nth solution. For a = 2, b and c are n as 2n^2 - 1 is a square. For a = 5, b and c satisfying the requrrence d(n) = 15*d(n-2)-d(n-4)Disregard that message, I obviously missunderstood the problem ...-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === > Disregard that message, I obviously missunderstood the problem ...thanks anyways! === Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:a^2+b^2=c^2x^2+y^2=c^2u^2+v^2=c^2ab+xy=uv?I.R. === > Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:> a^2+b^2=c^2> x^2+y^2=c^2> u^2+v^2=c^2> ab+xy=uv> ?> I.R.That's not arithmetic! That's a diophantine equation you've got there! === Starblade Darksquall a .8ecrit dans le message de> Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:> a^2+b^2=c^2> x^2+y^2=c^2> u^2+v^2=c^2> ab+xy=uv> ?> I.R. That's not arithmetic! That's a diophantine equation you've got there!I.J.R. === These are all from, if all limits converge absolutely,limit{m-> oo} (1/m)*sum{k=1 to m} (sum{j|k} a(j))= sum{k=1 to oo} a(k)/k.(Inner sum on top is over positive divisors, j, of k.)They are obvious using the above theorem. But they seem to, on theirown, be impressive to me (considering the number-theory aspects, aswell as the mysterious nature that some of the general terms have tosome mathematical people).I will only mention a few.Let L = m 1 --- ---limit --- m->oo m / ( / a(j) ) , --- --- k=1 j|kthe limit of the averages of the sum of a's over divisor-indexes.In linear-mode:L = limit{m->oo} (1/m) sum{k=1 to m} (sum{j|k} a(j) )So:1) If a(j) = binomial(n,j)(-1)^(j+1),then L = H(n), the n_th harmonic number = 1+1/2+1/3+...+1/n.2) If a(j) = 1/(n+j),then L = H(n)/n.3) If a(j) = s(j,n)/j! , where s(j,n) is the unsigned Stirling numberof the 1st kind,then L = zeta(n+1) = Riemann zeta function.4) If a(j) = B(j-1) y^j /(j-1)!, where B(j) is a Bernoulli number (sum{j=0 to oo} B(j)/j! x^j = x/(e^x-1)),then L = sum{k=1 to oo} (1-e^(-y))^k /k^2.5) If a(j) = B(j) y^j /(j-1)!,then L = y/(e^y -1) - 1.6) If a(j) = 1 if j = composite power of prime, and = 0 otherwise,then we can get:limit{m->oo} (1/m) *sum{k=1 to m} (M(k) - N(k))= sum{p=primes} 1/(p(p-1)) (which converges),where M(k) is sum of all exponents in prime factorization of k, andN(k) is the number of distinct primes dividing k.Leroy Quet === I'd be very glad if someone can cure my misundestanding of the followingsimple tensor query.If {e1,e2} and {f1,f2} are bases of vector spaces U,V then a basis of thetensor product space UoV is {e1 o f1, e1 o f2, e2 o f1, e2 o f2}. And so is{e1 o f1, e1 o f1 + e1 o f2, e2 o f1, e2 o f2}.By a change of bases: e1->ae1+be2, e2->ce1+de2 and f1->a'f1+b'f2,f2->c'f1+d'f2I feel I should (by appropriate values of a,b etc) be able to obtain thesecond basis in UoV from the first - but this doesn't seem possible?!What's wrong or what am I not understanding here please?Ron Jones === > I'd be very glad if someone can cure my misundestanding of the following> simple tensor query.> If {e1,e2} and {f1,f2} are bases of vector spaces U,V then a basis of the> tensor product space UoV is {e1 o f1, e1 o f2, e2 o f1, e2 o f2}. And so> is> {e1 o f1, e1 o f1 + e1 o f2, e2 o f1, e2 o f2}.> By a change of bases: e1->ae1+be2, e2->ce1+de2 and f1->a'f1+b'f2,> f2->c'f1+d'f2> I feel I should (by appropriate values of a,b etc) be able to obtain the> second basis in UoV from the firstWhy do you feel that?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === >I think the original poster meant the following (I'm omitting the description>of G and H this time, but they're in previous posts):Why not include the previous posts? >What is an automorphism f: G ----> G such that, given a coset C, f(C) = C, and>such that given c' in C, there exists c in C such that f(c) = c'.Any automorphism f of G has the property that given c' in C, there existsc in C such that f(c) = c' , so the identity automorphism always satisfiesyour condition.Derek Holt. === Try integrating from j to T (instead of 0 to T) and then take the limit ofthis answer as j goes to 0+.> Howdy everybody, I've been reading Quantum Mechanics and Path Integrals> well... except there is this one integral I have no idea> how to do. Here's the integral (in mathematica-speak): Integrate[((1/x)^(3/2)) Exp[(I (a/x)) + (I b x)], {x,0,T}] Where a,b,T are real positive numbers But even though I shout at mathematica and say, Do this> damn integral so I don't have to think! Mathematica refuses. So, here's what I think.> I know that the nastiness at x=0 is not a big problem> because as x goes to zero the integrand starts wiggling> faster and faster and tends to cancel. And I was trying> to use the method of stationary phase, but it's not> always true that the stationary point is within [o,T]...> and besides, I would much rather have a unapproximated> answer. Anyways, I was wondering if anyone here knows how to> integrate this, or if someone can suggest a good method> to try?> adam> === >I wonder to what extent the use of graphics has been inhibited by>mathematicians themselves. We are all taught that a proof must not rely on aFWIW, my personal opinion is that in fact stressing on the graphicaland specifically on computer-based teaching systems, while appealingon the one hand, is extremely *risky*, even at the very first stagesof mathematics teaching.The actual risk I see is a boost back to doing counts with matchesor little stones, event though in a digital environment. Of coursekids *have* to learn elementary arithmetic before they delve intoanything more serious, but even at their level their attention canbe drawn on to the conceptual sides of what they're learning.Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === You're not ADD as you show below. You've great concentration abilities.> What you've got is early training of entertain me, excite me to motivate> me. This comes of the media and thence to the schools who have grade> schoolers who are raised upon I've got to be entertained. William, aren't you by any chance trying to force your own perception> of the OP upon him? If he says he's got ADD then I guess he does.You may think so. Twice I've had experience that by denying theirdisorder, it wasn't manifested at least in my presence. One lateradmitted she indulged in hysteria just one and that politely by herself.In that period of time, once was major achievement for her.> Denigrating the media (whether deservedly or not doesn't matter> here) and the OP doesn't seem to me to be the best possible answer to> his questions.>I think the media is guilty of misguiding society, especially youngsociety. To know it's malfeasance is to know how we've been mistreated,misdirected and, whoops, less I wax off topic, bring this to newsgroupor.politics where such umbrage is decorous, where I lurk forth asapprentice Ginsberg attempting to emulate Mark Twain and even myself.> As for your remarks about philosophy, I found them really depressing> to read, especially considering that they come from a person of your> stature.>Indeed, you may consider wise cracks about wisdom unwise. === > You're not ADD as you show below. You've great concentration abilities.> What you've got is early training of entertain me, excite me to motivate> me. This comes of the media and thence to the schools who have grade> schoolers who are raised upon I've got to be entertained. William, aren't you by any chance trying to force your own perception> of the OP upon him? If he says he's got ADD then I guess he does.> You may think so. Twice I've had experience that by denying their> disorder, it wasn't manifested at least in my presence. One later> admitted she indulged in hysteria just one and that politely by herself.> In that period of time, once was major achievement for her.> Denigrating the media (whether deservedly or not doesn't matter> here) and the OP doesn't seem to me to be the best possible answer to> his questions. I think the media is guilty of misguiding society, especially young> society. To know it's malfeasance is to know how we've been mistreated,> misdirected and, whoops, less I wax off topic, bring this to newsgroup> or.politics where such umbrage is decorous, where I lurk forth as> apprentice Ginsberg attempting to emulate Mark Twain and even myself.> As for your remarks about philosophy, I found them really depressing> to read, especially considering that they come from a person of your> stature. Indeed, you may consider wise cracks about wisdom unwise.William, the point I was trying to make is that your reply hadlevelled accusations (and in quite caustic terms) at the OP instead ofanswering even one of his questions. Why not give the man a chancewith a more constructive approach (as exhibited in some other postsiin this thread)?Apropos, please be more careful with attributions - in your reply yourphrase Indeed...unwise is attributed (wrongly) to me. Or at leastthat's what I see in Google.As for the crack itself, I didn't find it epsilon funny, for I thinkthat some jokes are inappropriate in some contexts. Apparently we hadbetter agree to differ on these issues since I guess we both wouldrather be discussing some mathematics instead.Felix. === > I have a B.S. in Computer science and started my P.hd. program in> Mathematics last January. Before starting gradaute school, I worked as a> software developer for 4 years. > Maybe there is a more enjoyable way to study Mathematics, than just reading.> Any sort of feedback would be helpful.> Bill Well Bill you have one up on me. At 40 I started to see maththings. I have always liked the idea of being a computer programmer and theidea of returning to school is now the project I am preparing for. Soyou have one up on me if you are there now. When I went back to jr. College in 1991 for a few classes I used torecord the sound of the class and the talks of the teacher. I wouldlisten in the head phones at home while doing homework but not reallylisten as I worked the math.It put my mind in a mode by filling the hearing stimuli with sounds ofthe class. Nice thing is we can go digital now so the motor hum is nolonger an issue. So if I can save enough this year I will start back at Jr. College inthe spring of 2004 and carry a recorder. Best of luck to you.Ernst === > How in general should one study Mathematics? What makes reading Mathematics> books exciting? What goes on in your mind as you are reading? When was it> that you truly found Mathematics very exciting?I can't really say how should one study math, as everyone has his own way.Personally, I found that pondering on the links between different notions,whether obvious or not makes the notions more interesting as it gives you amore general view, which you then try to expand.It gets a bit like the feature craze that one can experience when hackingsoftware.As others have pointed out, you may like to look out for problems that combinemath and computer science. Artifical intelligence, but also cryptography andcompression are examples (I have been told there are interesting applicationsof fractal theory in video compression).I hope this helps.Sam-- One cannot be betrayed if one has no people - Kobayashi in The Usual Suspects, 1995 <49fff586.0307060936.24c96efc@posting.google.com> === > Indeed, you may consider wise cracks about wisdom unwise.> Apropos, please be more careful with attributions - in your reply your> phrase Indeed...unwise is attributed (wrongly) to me. Or at least> that's what I see in Google.>'Unwise' is attributed to the subject 'cracks' of the objective clause(that) wise cracks about wisdom (are) unwise.> As for the crack itself, I didn't find it epsilon funny, for I think> that some jokes are inappropriate in some contexts. Apparently we had> better agree to differ on these issues since I guess we both wouldNarry a delta more.> rather be discussing some mathematics instead.>Ok, isn't delta * epsilon < delta + epsilon ?X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === >The boredom of the endless definition-theorem-proof-homework system>is grad school's way of weeding out the geniuses from the rest of>us. What boredom? Grad school, for the most part, was fun.-- Shmuel (Seymour J.) Metz, SysProg and JOATto spamtrap@library.lspace.orgX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === at 04:18 PM, Bill said:>How in general should one study Mathematics?There are as many answers as there are students. What works for me isto find an area that grabs me and to plow in. Others might do betterwith a more structured approach.>What makes reading Mathematics books exciting? What makes reading anything exciting? I found Bourbaki exciting, butA Tale of Two Cities was a far, far duller book than I had everread. I find Topology exciting but Combinatorics leaves me cold. Thattells you more about me than about those; the excitement resides inthe reader, and every reader is different.>What goes on in your mind as you are reading?Pieces falling into place. An aha! reaction. A feeling ofenlightenment and awe. It varies widely.>When was it that you truly found Mathematics very exciting?Around 12.>Maybe there is a more enjoyable way to study Mathematics, than just>reading.Find a problem that seems interesting and solve it. Find a theorem andsee whether you can make it more general. In general, look forsomething Mathematical that you want to do, and do it.Mind you, there will be areas that you want to enter that requirebackground that you may not enjoy; you'll have to either slog throughthe preliminaries or give up those areas. But don't be surprised ifyou find things in the required background that are also engaging.Also, the Math Deparment will sponsor various colloquia and socialfunctions. Go to them. Talk to the faculty and other grad students.Even if a lot goes over your head, you will learn from it and it willbe time well spent. Ask for advice when appropriate.>Any sort of feedback would be helpful.Go for it. There will be some rough spots, but the field isfascinating.-- Shmuel (Seymour J.) Metz, SysProg and JOATto spamtrap@library.lspace.org === > I have a B.S. in Computer science and started my P.hd. program in> Maybe there is a more enjoyable way to study Mathematics, than just reading.> Any sort of feedback would be helpful.Do the excercises.Bob Kolker> === I am seeking recommendations for books on probability theory using theStieltjes integral in order to get a unified treatment of discrete andcontinuous spaces. Any suggestions?TiaGC === >I am seeking recommendations for books on probability theory using the>Stieltjes integral in order to get a unified treatment of discrete and>continuous spaces. Any suggestions?Instead learn a little measure theory - using measures alsogives a unified treatment of discrete and continuousprobability distributions on the line, and much more.I know that's not the answer you wanted, but measure theoryreally is the right way to do probability, and you don't needto know all that much of it.>Tia>GC**David C. Ullrich === >I am seeking recommendations for books on probability theory using the>Stieltjes integral in order to get a unified treatment of discrete and>continuous spaces. Any suggestions?> Instead learn a little measure theory - using measures also> gives a unified treatment of discrete and continuous> probability distributions on the line, and much more.> I know that's not the answer you wanted, but measure theory> really is the right way to do probability, and you don't need> to know all that much of it.> You can start with Volume 1 ofhttp://www.essex.ac.uk/maths/staff/fremlin/mt.htm === >I am seeking recommendations for books on probability theory using the>Stieltjes integral in order to get a unified treatment of discrete and>continuous spaces. Any suggestions? Instead learn a little measure theory - using measures also> gives a unified treatment of discrete and continuous> probability distributions on the line, and much more.I am, one of the books I'm reading is Jacod & Protter ProbabilityEssentials which uses measure theory. First they deal withprobabilities on countable (including finite) spaces, then they dealwith probability measures generally and on R specifically (specific ->general -> different specific, hope that makes sense :-). Pedagogically,I'm sure that's the right thing to do, but I thought it would be fun tosee finite sums and integrals dealt with together.(I'm also looking at the better-known Feller & Doob books. Vol I ofFeller seems to be discrete and vol II continuous but I've not looked atit closely yet.)GC> I know that's not the answer you wanted, but measure theory> really is the right way to do probability, and you don't need> to know all that much of it.>Tia>GC> **> David C. Ullrich === >I am seeking recommendations for books on probability theory using the>Stieltjes integral in order to get a unified treatment of discrete and>continuous spaces. Any suggestions?> Instead learn a little measure theory - using measures also> gives a unified treatment of discrete and continuous> probability distributions on the line, and much more. I know that's not the answer you wanted, but measure theory> really is the right way to do probability, and you don't need> to know all that much of it.> You can start with Volume 1 of> http://www.essex.ac.uk/maths/staff/fremlin/mt.htmvolumes yet!GC === >I am seeking recommendations for books on probability> theory using the>Stieltjes integral in order to get a unified treatment of> discrete and>continuous spaces. Any suggestions?> Instead learn a little measure theory - using measures also> gives a unified treatment of discrete and continuous> probability distributions on the line, and much more. I am, one of the books I'm reading is Jacod & Protter Probability> Essentials which uses measure theory. First they deal with> probabilities on countable (including finite) spaces, then they deal> with probability measures generally and on R specifically> (specific - general -> different specific, hope that makes sense :-).> Pedagogically,> I'm sure that's the right thing to do, but I thought it> would be fun to> see finite sums and integrals dealt with together. (I'm also looking at the better-known Feller & Doob books. Vol I of> Feller seems to be discrete and vol II continuous but I've> not looked at> it closely yet.)If you are completely new to probability, I recommend very strongly thatyou go through a less mathematically rigorous text first to developintuition about the subject. Ross's first course is a good book forthis. You could supplement this with study of the Stieltjes integralfrom baby Rudin, say. Afterwards, a measure-theoretic coverage is thebest way to go.Sorry, I have no strong recommendations in this regard. Billingsely isencyclopaedic and not very good to learn from. Whittle, Probability viaExpectation is a very interesting book.-- Direct access to this group with http://web2news.comhttp://web2news.com/?sci.mathTo contact in private, remove no2s6ppa+mm === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions?> For the Stieltjes integral approach there is Feller volume II. Butthat would be viewed as quite ald fashioned.Modern probability texts use measures for their unified treatment.That way they unify not only discrete and continuous random variables;but they also unify real, complex, vector, and other abstract randomobjects.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === >I am seeking recommendations for books on probability theory using the>Stieltjes integral in order to get a unified treatment of discrete and>continuous spaces. Any suggestions?> Instead learn a little measure theory - using measures also> gives a unified treatment of discrete and continuous> probability distributions on the line, and much more.I am, one of the books I'm reading is Jacod & Protter Probability>Essentials which uses measure theory. First they deal with>probabilities on countable (including finite) spaces, then they deal>with probability measures generally and on R specifically (specific -general -> different specific, hope that makes sense :-). Pedagogically,>I'm sure that's the right thing to do, but I thought it would be fun to>see finite sums and integrals dealt with together.??? Yes, it's great fun to see finite sums and integrals dealt withtogether. That's exactly why (or an example of why) you want tobase your probability on measure theory: Finite sums _are_integrals (with respect to counting measure).(Of course finite sums can often also be regarded as Stieltjesintegrals...)>(I'm also looking at the better-known Feller & Doob books. Vol I of>Feller seems to be discrete and vol II continuous but I've not looked at>it closely yet.)GC I know that's not the answer you wanted, but measure theory> really is the right way to do probability, and you don't need> to know all that much of it.>Tia>GC> **> David C. Ullrich**David C. Ullrich === >I am seeking recommendations for books on probability theory using the>Stieltjes integral in order to get a unified treatment of discrete and>continuous spaces. Any suggestions?> Instead learn a little measure theory - using measures also> gives a unified treatment of discrete and continuous> probability distributions on the line, and much more.I am, one of the books I'm reading is Jacod & Protter Probability>Essentials which uses measure theory. First they deal with>probabilities on countable (including finite) spaces, then they deal>with probability measures generally and on R specifically (specific -general -> different specific, hope that makes sense :-). Pedagogically,>I'm sure that's the right thing to do, but I thought it would be fun to>see finite sums and integrals dealt with together.> ??? Yes, it's great fun to see finite sums and integrals dealt with> together. That's exactly why (or an example of why) you want to> base your probability on measure theory: Finite sums _are_> integrals (with respect to counting measure).> (Of course finite sums can often also be regarded as Stieltjes> integrals...)Ok, I'll take your advice: first measure theory, then probability; then(just for fun) the Stieltjes integrals.GC === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions? > For the Stieltjes integral approach there is Feller volume II. But> that would be viewed as quite ald fashioned.> Modern probability texts use measures for their unified treatment.> That way they unify not only discrete and continuous random variables;> but they also unify real, complex, vector, and other abstract random> objects.text?GC> --> G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === >I am seeking recommendations for books on probability> theory using the>Stieltjes integral in order to get a unified treatment of> discrete and>continuous spaces. Any suggestions?> Instead learn a little measure theory - using measures also> gives a unified treatment of discrete and continuous> probability distributions on the line, and much more. I am, one of the books I'm reading is Jacod & Protter Probability> Essentials which uses measure theory. First they deal with> probabilities on countable (including finite) spaces, then they deal> with probability measures generally and on R specifically> (specific - general -> different specific, hope that makes sense :-).> Pedagogically,> I'm sure that's the right thing to do, but I thought it> would be fun to> see finite sums and integrals dealt with together. (I'm also looking at the better-known Feller & Doob books. Vol I of> Feller seems to be discrete and vol II continuous but I've> not looked at> it closely yet.)> If you are completely new to probability, I recommend very strongly thatI did an undergraduate course on probability but never took to it. Theappeal to intuition didn't work for me, hence I'm seeking a moremathematical treatment.GC> you go through a less mathematically rigorous text first to develop> intuition about the subject. Ross's first course is a good book for> this. You could supplement this with study of the Stieltjes integral> from baby Rudin, say. Afterwards, a measure-theoretic coverage is the> best way to go.> Sorry, I have no strong recommendations in this regard. Billingsely is> encyclopaedic and not very good to learn from. Whittle, Probability via> Expectation is a very interesting book.> --> Direct access to this group with http://web2news.com> http://web2news.com/?sci.math> To contact in private, remove no2s6ppa+mm === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions?>For the Stieltjes integral approach there is Feller volume II. But>that would be viewed as quite ald fashioned.>Modern probability texts use measures for their unified treatment.>That way they unify not only discrete and continuous random variables;>but they also unify real, complex, vector, and other abstract random>objects.Stieltjes invented his integral so that one could study themoment problem in full generality. While various types ofmeasures had been used, the general idea had not yet beeninvented. It is too simple.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions?> For the Stieltjes integral approach there is Feller volume II. But> that would be viewed as quite ald fashioned.> Modern probability texts use measures for their unified treatment.> That way they unify not only discrete and continuous random variables;> but they also unify real, complex, vector, and other abstract random> objects.text?Kai Lai Chung has at least two books on probability, one of which iswhat you say elsewhere you don't want, and one of which may bethe one you want. I think the title is A Course in ProbabilityTheory.>GC> --> G. A. Edgar http://www.math.ohio-state.edu/~edgar/** David C. Ullrich === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions?> For the Stieltjes integral approach there is Feller volume II. But> that would be viewed as quite ald fashioned.> Modern probability texts use measures for their unified treatment.> That way they unify not only discrete and continuous random variables;> but they also unify real, complex, vector, and other abstract random> objects.text?> Kai Lai Chung has at least two books on probability, one of which is> what you say elsewhere you don't want, and one of which may be> the one you want. I think the title is A Course in Probability> Theory.>GC === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions?> For the Stieltjes integral approach there is Feller volume II. But> that would be viewed as quite ald fashioned.> Modern probability texts use measures for their unified treatment.> That way they unify not only discrete and continuous random variables;> but they also unify real, complex, vector, and other abstract random> objects.text?> Kai Lai Chung has at least two books on probability, one of which is> what you say elsewhere you don't want, and one of which may be> the one you want. I think the title is A Course in Probability> Theory.> There's this: A Course in Probability Theory Revised by Kai Lai Chung,Academic Press; 2nd edition (January 15, 2000) GC === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions?>For the Stieltjes integral approach there is Feller volume II. But>that would be viewed as quite ald fashioned.>Modern probability texts use measures for their unified treatment.>That way they unify not only discrete and continuous random variables;>but they also unify real, complex, vector, and other abstract random>objects.> Stieltjes invented his integral so that one could study the> moment problem in full generality. While various types of> measures had been used, the general idea had not yet been> invented. It is too simple.>It is my understanding that every distribution function on R^n givesrise to a Lebesgue-Stieltjes that = 1 on the whole space and iscomplete, and hence is a probability measure. Also, all interestingprobability measures on R^n can be defined in this way.GC === > I am seeking recommendations for books on probability theory using the> Stieltjes integral in order to get a unified treatment of discrete and> continuous spaces. Any suggestions? >For the Stieltjes integral approach there is Feller volume II. Butthat would be viewed as quite ald fashioned. >Modern probability texts use measures for their unified treatment.That way they unify not only discrete and continuous random variables;>but they also unify real, complex, vector, and other abstract random>objects. Stieltjes invented his integral so that one could study the> moment problem in full generality. While various types of> measures had been used, the general idea had not yet been> invented. It is too simple. > It is my understanding that every distribution function on R^n gives> rise to a Lebesgue-Stieltjes that = 1 on the whole space and is ^^^^^^^^^^^^^^^^^^Meaning Lebesgue-Stieltjes measure of course.GC> complete, and hence is a probability measure. Also, all interesting> probability measures on R^n can be defined in this way.> GC === I'm having trouble solving part of the below problem. Any comments, hints oradvice would be appreciated. R denotes the reals.Problem: Let f be defined on RxR to R byf(x,y) = (x^2)*y/(x^3-y^2), if x^3 /= y^2 = 0, if x^3 = y^2Then f has a directional derivative at (0,0) in every direction, but f isnot continuous at (0,0). However, f is bounded on a neighborhood of (0,0).I believe I've shown the existence of all directional derivatives and that fisn't continuous at (0,0). However, I can't get the part about f beingbounded on a neighborhood of (0,0). In fact, it seems to me that f isn'tbounded on any nbhd of (0,0).The intuitive reason that I don't think f is bounded on any nbhd of (0,0) isas follows: Any such nbhd contains a portion of the curve x^3 = y^2. We canfind points close to, but not on, the curve such that the denominator of fis arbitrarily small and the numerator isn't 0. That makes f as large as welike.Here's a formal argument: Any nbhd N of (0,0) contains an open disk D(s)centered at (0,0) of radius s, for some s > 0. Then there exists an r > 0such that the points (r, r^(3/2)) and (r, r^((3/2) + t)) are all containedin D(s) for all t sufficiently close to 0. (For example, take 0 < r < Min{s/2, 1}.) The point (r, r^(3/2)) is on the curve x^3 = y^2 and the point(r, r^((3/2) + t) is close to the curve if t is close to 0. We compute thevalue of f at this point:f(r, r^((3/2) + t) = (r^((1/2) + t))/(1 - r^(2*t))As t --> 0, the numerator --> r^(1/2) and the denominator --> 0. Therefore,given any M > 0, we can make f(r, r^((3/2) + t) > M by choosing tsufficiently close to 0. This shows that f isn't bounded on the nbhd N.Pete Klimek === >Do you even have any idea what you or anybody else is saying? So centripetal acceleration produces this centripetally directed> change vector that is a velocity and is combined vectorially with v to> produce rotation is directed along r perchance? Kinda what I've been> saying all along isn't it? :-))))))))Kinda like Gilgamesh is really the same as Beowulf, once we translatefrom Akkadian to Old English?Oh really ... this is too much! Nobody else was able to understandmuch in your confused word play, but of course you weren't wrong orconfused about anything, just misunderstood! And who's to refute you,since you can say you meant whatever you like?Could be ... I said early on you had tyro disease: that's the diseasecontracted from poring word for word over elementary sources inisolation.> Are you sure you're not agreeing avec moi?> Certainly looks like it. But then maybe you have some clintonesque> definition of agree? Too much!!! Too much!!! Stop it, you're killing me! You senseimpending defeat, and turn it around into victory by claiming it waswhat you were saying all along, then have the unlaundered gall topre-accuse _Randy_ of sophistry if perchance he doesn't agree he wasreally agreeing with you!Yep ... you definitely have what it takes to be a politician. Go forit, Lester. But don't go for science: if you think you are foolinganybody here with your word-play, you are only fooling yourself. Ordon't go for elementary science, anyway, where understanding is widelydistributed and gobbledegook is much more quickly outed. You want toattack at the level of the Bogdanov brothers. === (11-8(cosx)^4 - 4sinx * cosx)^0.5=3sinx + cosxHelp. === .be.83.83 [Times][EGrave].83 escribi.97 en elmensaje|nbegend$188i$1@av7254.comex.ru:> (11-8(cosx)^4 - 4sinx * cosx)^0.5=3sinx + cosx> Help.Squaring,- 8cos^4(x) - 4sin(x)cos(x) + 11 = 6sin(x)cos(x) + 8sin^2(x) + 1 ==>8cos^4(x) + 8sin^2(x) + 10sin(x)cos(x) -10 = 08cos^4(x) + 8(1 -cos^2(x)) + 10sin(x)cos(x) -10 = 08cos^4(x) - 8cos^2(x) + 10sin(x)cos(x) -2 = 08cos^2(x)(cos^2(x) - 1) + 10sin(x)cos(x) - 2 = 0-8cos^2(x)sin^2(x) + 10 sin(x)cos(x) - 2 = 02sin^2(2x) - 5sin(2x) + 2 = 0sin(2x) = (5 +/- sqrt(25 - 16))/4 = (5 +/- 3)/4 === > 2 or 1/2.If we want real values,only must to take sin(2x) = 1/2 ==>2x = 30.bc + k*360 ==> x = 15.bc + k*180 ==> x = 15.bc, x = 195.bc2x = 150.bc + k*360 ==> x = 75.bc + k*180.bc ==> x = 75.bc, x = 255.bcin [0, 360.bc)Because the squaring, you must check that solutions. Then x = 15.bc and x =75.bc are solutions, while for x = 195.bc and x = 175.bc de RHS is negative.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === I know that the Cauchy- Schwartz inequality for vectors in a finitespace becames an equality if the vectors are parallel.Is this true also if I consider a space of finite dimensionalmatrices?Many thanks!Fabio. === > I know that the Cauchy- Schwartz inequality for vectors in a finite> space becames an equality if the vectors are parallel. Is this true also if I consider a space of finite dimensional> matrices?>Who said matrices can't be vectors too?--Julien Santini,CMI Technop.99le de Ch.89teau-Gombert, France.Home page: http://www.analgebra.com === I know that the Cauchy- Schwartz inequality for vectors in a finite> space becames an equality if the vectors are parallel. Is this true also if I consider a space of finite dimensional> matrices?Of course, since matrices are also vectors.BTW, it's Cauchy-Schwarz not Cauchy-Schwartz.Jose Carlos Santos-- Direct access to this group with http://web2news.comhttp://web2news.com/?sci.mathTo contact in private, remove noo-s8p5a+mm === > I know that the Cauchy- Schwartz inequality for vectors in a finite> space becames an equality if the vectors are parallel.If you are using the usual Euclidean norm for vectors, and the usualinner product.> Is this true also if I consider a space of finite dimensional> matrices?> Many thanks!> Fabio.Depends what norm you use for vectors. And what you use for the innerTrace(A* B), as I recall.I thought I posted this to sci.math.research a while back, but I can't seem tofind it there. Anyway, the following was quoted on the thread Hamiltoniancycles in Z_a x Z_b; all theorems and lemmas invoked are from the paper byBarone, Mauntle, and Miller, located athttp://www.nd.edu/~ndreu/research/barone_mauntel_ miller.pdf.Suppose that Cay(G;x,y) is a 2-generated Cayley graph on an abelian group G,that His a subgroup of G, and that Cay(G;x,y) - H has a hamiltonian cycle C.The proof of Lemma 3.2 shows that + H contains both x and y, so + H = G. (In other words, the quotient group G/H is generatedby x-y.) As in Corollary 3.3, we conclude that C is unique, and thatall vertices of g + H travel in the same direction, for each x in G - H.In other words, C is the lift of a hypohamiltonian cycle in G/H.The conclusion is that Cay(G;x,y) - H has a hamiltonian cycle if andonly if there is a hamiltonian cycleg, g+a_1, g+a_1+a_2, ..., g+a_1+a_2+...+a_n (with each a_i = x or y)in Cay(G/H;x,y) - {0}, such that the sum a_1 + ... + a_n generates H.This can easily be translated into an explicit number theoreticcondition, if we are given a presentation of G,and a generator ux+vy of H. (Both H and G/H must be cyclic ifCay(G;x,y) - H has a hamiltonian cycle.)Of course, although the letters x and y are used when discussing G/H, theyreally mean the _coset_ of x and the _coset_ of y. But, if the quoted sequence is indeed a hamiltonian cycle, then g and g + a_1 +... + a_n are the same point, thus a_1 + ... + a_n = 0, which would mean that H= < 0 >, which is clearly not true in general. I believe that H must be cyclic, but I don't see how what has been shown in thequoted passage proves that. Hopefully someone here can shed some light on that(or show, using some other method, that H must be cyclic). === Can some please give me an algebraic, that is non combinatatorical argument,why n choose r is an integer.Mime-version: 1.0 === > Can some please give me an algebraic, that is non combinatatorical argument,> why n choose r is an integer.Even more restrictive, n choose r is a natural number (inclucive of zero).Because you are counting things. === > Can some please give me an algebraic, that is non combinatatorical argument,> why n choose r is an integer.>Even more restrictive, n choose r is a natural number (inclucive of zero).Because you are counting things.I am guessing that because you are counting things is precisely whatthe OP meant by combinatorial argument...Perhaps one could use an argument involving the highest power of aprime p which divides n!, based on n and p; that should suffice toprove that n!/((n-r)!r!) is an integer. What is the highest power of p that divides n!?Well, you get one power of p for every multiple of p less than orequal to n; that is floor(n/p). You get another one for each multipleof p^2, floor(n/p^2); another one for the multiples of p^3, etc.So write n in base p,n = a_0 + a_1*p + ... + a_m*p^mwith m a nonnegative integer, and 0<= a_0 < p.Then the highest power of p that divides n! should befloor(n/p) + floor(n/p^2) + ... + floor (n/p^m).We can keep going, but then you start getting summands equal to 0.So, the highest order of p that divides n! is given above.The heighest order of p that divides (n-r)! isfloor((n-r)/p) + ... + floor( (n-r)/p^m);the heighest order of p that divides r! isfloor(r/p) + ... + floor(r/p^m).So the heighest order of p that divides (n-r)!*r! isfloor((n-r)/p) + floor(r/p) + ... + + floor((n-r)/p^m) + floor(r/p^m).We want this to be less than or equal to the heighest order of p thatdivides n!; but this follows from floor(a) + floor(b) <= floor(a+b)since then floor((n-r)/p^i) + floor(r/p^i) <= floor (n/p^i), givingthe desired inequality.Clearly, any prime dividing (n-r)!r! divides n!, so this proves that,after cancellation is done, n!/((n-r)!r!) must be an (non-negative)integer. === =========================== === ====================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu === mensaje|nUS_Na.22129 $1i1.37@news02.roc.ny:> Can some please give me an algebraic, that is non combinatatorical> argument, why n choose r is an integer.I suppose that you want to prove that Comb(n, r) = n!/(r!(n-r)!) is aninteger for 0 <= r <= nUse induction on n, taking in account thatComb(n+1, r+1) = Comb(n, r) + Comb(n, r+1)andComb(n, 0) = Comb(n, n) = 1, in particular Comb(1, 0) = Comb(1, 1) = 1Alternativally, analize the exponent of each prime p < r inComb(n, r) = n!/(r!(n-r)!) = n(n-1)...(n-(r-1))/(1*2*...*r)where we can suppose r <= n/2-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === > Can some please give me an algebraic, that is non combinatatoricalargument,> why n choose r is an integer.>First prove that (n+1 choose r) = (n choose r-1) + (n choose r)(This is the rule for generating Pascal's triangle.)Then you can prove it by induction on n. === >Can some please give me an algebraic, that is non combinatatorical argument,>why n choose r is an integer.-- Spammers: I don't want a small digital camera to post photos of a large, lowweight, penis on a re-financed Nigerian domain site. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === >Can some please give me an algebraic, that is non combinatatorical argument,>why n choose r is an integer.> That's combinatorial: each coefficient counts the number of wayssomething happens. === > Can some please give me an algebraic, that is non combinatatorical argument,> why n choose r is an integer.>Even more restrictive, n choose r is a natural number (inclucive of zero).>Because you are counting things.This is combinatorial, not algebraic.compute the prime factorization of the numerator and thedenominator. The highest power of a prime p dividing j!is sum_1 [j/p^k], where [q] is the largest integer lessthan or equal to q. It is not difficult to see that, termby term, the numerator is divisible by at least as higha power of p as the denominator for each p.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue UniversityCome On Let's Get More Partners Signed Up - S. Enteprize Co.'s- freeindependent business or science websites I need about 100 or 1000, or 10000 new people to sign up for their own freebusiness or science website and free website support...http://www.s-enterprize.com/100 megs free space for your business or science research or theories, withfree website editor software development online on internet.... . And if youpay about $15.00/month, you can get up to 1 gig Internet space 100 Megs free to start, what you complaining about? There is room for 10,000's +, new people.... .S. Enterprize Co. (Membership)http://www.s-enterprize.com/S. Enterprize (Science Journal)http://smart1234.s-enterprize.com/ === Some of you may now see comments from mathematicians trying to hidethe logical flaw in Wiles's work where they try to claim that you*can* indeed compare infinite sets and thereby prove a condition.However, consider the following link which shows by the work of Cantorthat the set of rationals and the set of integers are in a sense thesame size.Then using the reasoning of Wiles, you could prove that everyrational is an integer based on the equal size of the sets. Rememberwhen elliptic curves and modular forms are discussed that there areINFINITE number of them:http://mathforum.org/isaac/problems/cantor2.htmlSo why would mathematicians still argue?Well, ask yourself, do you find yourself still believing them?Or if you're finally hesitant, are you *still* hesitant despite beingan experienced logician?Why?Come on, admit it, as long as mathematicians keep *telling* youthere's more to it, you want to give them the benefit of the doubt,against logic in which you may have thought you had complete belief.But I've shown you how you have a weakness inside your own head, interms of your own rationality, as human beings are *wired* to believebased on social pressures. You're possibly a victim of that veryversatile but sometimes intriguingly limited human brain of yours.James Harris === [snip] > Then using the reasoning of Wiles, you could prove that every> rational is an integer based on the equal size of the sets.I believe it would be more accurate to say that for each rationalthere is a corresponding integer, not that the rational *is* aninteger.> Remember> when elliptic curves and modular forms are discussed that there are> INFINITE number of them:Does it really matter how many total there are as long as there isa method to correlate the sets? Consider the linear equation in 'x'ax+b and the sinusoidal waves a*sin(x/b). There are an infinite numberof each, but there is also a fairly clear one-to-one correspondancebetween the two sets. === You have recently posted a link to http://mathworld.wolfram.com/ Taniyama-ShimuraConjecture.htmlwhich talks about his proof. I'm using this as my information.Note: I found http://cgd.best.vwh.net/home/flt/flt01.htm which at first glance appears to be a far more thorough discussion of the topic. Perhaps James would care to poke at some of the details there? Hopefully with specific references?> Some of you may now see comments from mathematicians trying to hide> the logical flaw in Wiles's work where they try to claim that you> *can* indeed compare infinite sets and thereby prove a condition.That would depend entirely on the condition you wish to prove. If you want to prove they have the same cardinality, it would be quite sufficient.> However, consider the following link which shows by the work of Cantor> that the set of rationals and the set of integers are in a sense the> same size.Why say in a sense? They are the same size.> Then using the reasoning of Wiles, you could prove that every> rational is an integer based on the equal size of the sets. Remember> when elliptic curves and modular forms are discussed that there are> INFINITE number of them:Would you care to point out _which_ reasoning this would be?> http://mathforum.org/isaac/problems/cantor2.htmlThis is a nicely written paper.> So why would mathematicians still argue?Because you don't write as clearly as the poster of the link you referenced.> Well, ask yourself, do you find yourself still believing them?Pretty much.> Come on, admit it, as long as mathematicians keep *telling* you> there's more to it, you want to give them the benefit of the doubt,> against logic in which you may have thought you had complete belief.Ok, perhaps a little less vagueness would help someone coming in on the middle of this conversation. More to _what_? Wiles' proof?-- Will Twentyman === Is there any program of computing Fibonacci numbers on a Turing Machine?One of them is Computing a Fibonacci number.http://sourceforge.net/projects/turing-machinehttp:// alexvn.freeservers.com/s1/turing.html------------------------- -------------------------The program computes a Fibonacci number.A number 'n' is represented by n 1-s.Sample : 5 is represented as 1 1 1 1 1 3 is represented as 1 1 1Input : number 'n'Sample : 1 1 1 1 1 1 1Output : Fibonacci#nSample : 1 1 1 1 1 1 1 1 1 1 1 1 1---------------------------------------------------- === ============================= Alex Vinokur http://www.simtel.net/pub/oth/19088.html http://sourceforge.net/users/alexvn === ================================> Is there any program of computing Fibonacci numbers on a Turing Machine? One of them is Computing a Fibonacci number.> http://sourceforge.net/projects/turing-machine> http://alexvn.freeservers.com/s1/turing.html --------------------------------------------------> The program computes a Fibonacci number.> A number 'n' is represented by n 1-s.> Sample :> 5 is represented as 1 1 1 1 1> 3 is represented as 1 1 1 Input : number 'n'> Sample :> 1 1 1 1 1 1 1 Output : Fibonacci#n> Sample :> 1 1 1 1 1 1 1 1 1 1 1 1 1> -------------------------------------------------->Raw Logs : http://groups.google.com/groups?th=1e653c4ef60faa44 === ========================= Alex Vinokur http://mathforum.org/library/view/10978.html === =========================