mm-477 === Subject: : Re: volume of ball in R^4>Help!!! I'm really confused since I cannot visualize this:>a) Find the volume in R^4 of a ball of radius a>b) Find the volume in R^5 of a ball of radius aWhen you don't know how to do a problem, one useful question is Can I solve a related but simpler problem?Can you find the volume of a ball of radius a in R^3? Then use the same technique in R^4 and R^5.=== === Subject: : How to solve an exponential ciphers? How to solve an exponential ciphers?How to solve an exponential ciphers?p=2633 , C=423 , e=29plain text: EX (0423)how to solve 423^29 mod 2633 without using calculator?=== === Subject: : Vector subspace test on polynomialsHere's an interesting question. Can someone tell me what I did wrong?Determine if the following is a subspace of P3 (polynomials of degree 3or less):Set of polynomials a0 + a1x + a2x^2 + a3x^3 for which a0 + a1 + a2 + a3 = 0I got this problem wrong. The answer sheet says this is a subspace. Iworked it out such that it is not closed under addition. Perhaps Imisunderstood this but I did the following:Let p1(x) = 6 + 4x - 2x^2 - 8x^3 (so (6 + 4 - 2 - 8 = 0))and I let p2(x) = 1 - x^3 (so (1 + 0 + 0 - 1 = 0))Adding p1(x) + p2(x) gives me: 7 + 4x - 2x^2 - 9x^3so my new a0,a1,a2,a3 for the result is 7,4,-2,-9Well 7 + 4 - 2 - 9 is not = 0. So I concluded that it was not asubspace. Yet the answer sheet says it's closed under both addition andscalar multiplication. I'm sure I'm doing something idiotic. Pleasetell me what I'm doing wrong.Here's what the answer sheet says:Suppose p(x);q(x) 2 S and a 2 R.There exist a0;a1;a2;a3;b0;b1;b2;b3 such that p(x) = a0 +a1x +a2x2 +a3x3where a0+a1+a2+a3 = 0 and q(x) = b0+b1x+b2x2+b3x3 where b0+b1+b2+b3 = 0:Then p(x)+q(x) = (a0+a1x+a2x2+a3x3)+(b0+b1x+b2x2+b3x3) = (a0+b0)+(a1+b1)x+(a2+b2)x2+(a3+b3)x3.Now, (a0+b0)+(a1+b1)+(a2+b2)+(a3+b3) = (a0+a1+a2+a3)+(b0+b1+b2+b3) = 0+0= 0 and so p+q 2 S.Did I get this problem wrong because I was NOT treating the polynomialsas a function of x (and thus perform the function space tests ratherthan the vector space tests)?? Seems to me like that's what I didwrong.=== === Subject: : Re: Vector subspace test on polynomials> 7 + 4 - 2 - 9 is not = 0Actually, it is.meerohIf this message helped you, consider buying an itemfrom my wish list: Here's an interesting question. Can someone tell me what I did wrong?> Determine if the following is a subspace of P3 (polynomials of degree 3> or less):> Set of polynomials a0 + a1x + a2x^2 + a3x^3 for which > a0 + a1 + a2 + a3 = 0> I got this problem wrong. The answer sheet says this is a subspace. I> worked it out such that it is not closed under addition. Perhaps I> misunderstood this but I did the following:> Let p1(x) = 6 + 4x - 2x^2 - 8x^3 (so (6 + 4 - 2 - 8 = 0))> and I let p2(x) = 1 - x^3 (so (1 + 0 + 0 - 1 = 0))> Adding p1(x) + p2(x) gives me: 7 + 4x - 2x^2 - 9x^3> so my new a0,a1,a2,a3 for the result is 7,4,-2,-9> Well 7 + 4 - 2 - 9 is not = 0. It's not?=== === Subject: : Re: Vector subspace test on polynomialsalt.math.undergrad:> Here's an interesting question. Can someone tell me what I did wrong?> Determine if the following is a subspace of P3 (polynomials of degree 3> or less):> Set of polynomials a0 + a1x + a2x^2 + a3x^3 for which > a0 + a1 + a2 + a3 = 0> I got this problem wrong. The answer sheet says this is a subspace. I> worked it out such that it is not closed under addition. Perhaps I> misunderstood this but I did the following:> Let p1(x) = 6 + 4x - 2x^2 - 8x^3 (so (6 + 4 - 2 - 8 = 0))> and I let p2(x) = 1 - x^3 (so (1 + 0 + 0 - 1 = 0))> Adding p1(x) + p2(x) gives me: 7 + 4x - 2x^2 - 9x^3> so my new a0,a1,a2,a3 for the result is 7,4,-2,-9> Well 7 + 4 - 2 - 9 is not = 0.Sure it is: 7 + 4 = 11; 11 - 2 = 9; 9 - 9 = 0. (In otherwords, your reasoning was entirely correct, but yourarithmetic let you down badly.)[...]=== === Subject: : Re: How to solve an exponential ciphers?> how to solve 423^29 mod 2633 without using calculator?29 = 16 + 8 + 4 + 1 (that's 29 in binary)423^29 = 423^16 * 423^8 * 423^4 * 423Now you need to figure out:423 mod 2633 = 423423^2 mod 2633 = 423 * 423 = -115423^4 mod 2633 = -115 * -115 = 60423^8 mod 2633 = 60 * 60 = 967423^16 mod 2633 = 967 * 967 = 374So 423^29 mod 2633 = 423 * 60 * 967 * 374 = -950 * 967 * 374 = 267 * 374 = -196Each of the steps is at worst the multiplication of two tree-digit numbers, or division of a five-digit number by 2633, which you can do by hand without too much trouble.=== === Subject: : Re: I'm TERRIBLY confused, linear algebra help requestedHi Virgil:wondering if you could clear something up for me.I posted this same reply to Miro's reply in this thread and will repostit here in-case you did not have a chance to view his reply. Waswondering if you could help identify the error in my understanding or ifthere was an honest mistake made (I suspect it's an error in myunderstanding)Here goes:> Question 3:> u1 = (1,3,4), u2 = (4,0,1) and u3=(3,1,2).> b) determine if (1, 2, 3) lies in the span {u1, u2, u3}I received 2 different answers for this question:One answer I received somewhat contradicts another answer. Eitherthey're both equivelant or one may be an honest mistake...One answer was:In matrix terms you are trying to solve matrix equation[[ 1 3 4 ] [[x] [[1] [[x] [ 4 0 1 ] times [y] equals [2] for column vector [y] [ 3 1 2 ]] [z]] [3]] [z]]--------------Second answer was:For the span, look at 1*X + 4*Y + 3*Z = 13*x + 0*Y + 1*Z = 24*X + 1*Y + 2*Z = 3===============1 4 3 13 0 1 24 1 2 3The matrix is written differently in both cases. I could be VERY wrongon this (and in-fact I probobly am) but I believe the first answer maybe an honest mistake? To check to see if a vector B is in the span,isn't basically saying determine if 'B' can be expressed as a linearcombination of the vectors in A which goes back to the:C1X1 + C2X2 + C3X3 = X check for linear combination? Thus: (1) (4) (3) (1)C1(3) + C2(0) + C3(1) = (2) (4) (1) (2) (3)which would NOT be the same as saying:[[ 1 3 4 ] [[x] [[1] [[x] [ 4 0 1 ] times [y] equals [2] for column vector [y] [ 3 1 2 ]] [z]] [3]] [z]]Please let me know if I'm incorrect on this.=== === Subject: : Re: I'm TERRIBLY confused, linear algebra help requested> Question 3:> u1 = (1,3,4), u2 = (4,0,1) and u3=(3,1,2).> b) determine if (1, 2, 3) lies in the span {u1, u2, u3}> I received 2 different answers for this question:> One answer I received somewhat contradicts another answer. Either> they're both equivelant or one may be an honest mistake...> One answer was:> In matrix terms you are trying to solve matrix equation> [[ 1 3 4 ] [[x] [[1] [[x]> [ 4 0 1 ] times [y] equals [2] for column vector [y]> [ 3 1 2 ]] [z]] [3]] [z]]> Second answer was:> For the span, look at > 1*X + 4*Y + 3*Z = 1> 3*x + 0*Y + 1*Z = 2> 4*X + 1*Y + 2*Z = 3That is definitely correct, since it is merelythe equation-centric way of writing (x)U1 + (y)U2 + z(U3) = [1 2 3]' (' = transpose)which is of course the definition of saying that[1 2 3]' lies in the space spanned by U1, U2, U3 (if youcan find an (x, y, z) that makes the equation true).> be an honest mistake? To check to see if a vector B is in the span,> isn't basically saying determine if 'B' can be expressed as a linear> combination of the vectors in A which goes back to the:> C1X1 + C2X2 + C3X3 = X check for linear combination? Thus:> (1) (4) (3) (1)> C1(3) + C2(0) + C3(1) = (2)> (4) (1) (2) (3)That is correct.So we know the second answer is correct.But is the first answer correct?The second answer wants you to solve Ax = b (x = [x y z]' and b = [1 2 3]')The first answer wants you to solve A'x = bThis can only be true when A=A', i.e. A is a symmetric matrix.Since A=A' is not true in general and the second answer iscorrect, the first answer must be mistaken.=== === Subject: : Re: I'm TERRIBLY confused, linear algebra help requested> Find a basis of the subspace W of R4 spanned by > (1, -4, -2, 1), (1, -3, -1, 2), and (3, -8, -2, 7)> Then the solutions would have looked like:> Take some vectors ...> .... make a matrix out of those vectors as columns...> .... column-reduce the matrix...> .... look at the result -- look for non-zero columns, zero columns, > whatnot.> With this approach, I col reduce> (1 1 3)> (-4 -3 -8)> (-2 -1 -2)> (1 2 7)> to get:> (1 1 3)> (0 1 4)You shouldn't be shedding elements when you row-reduce or column-reduce, as far as I know; you may end up with some zero entries, but that's ok. Column-reducing the matrix (1 1 3)(-4 -3 -8)(-2 -1 -2)(1 2 7)should look exactly the same as row-reducing the matrix(1 -4 -2 1)(1 -3 -1 2)(3 -8 -2 7)but, of course, transposed. Assuming I remember how to do this, that should yield:(1 -4 -2 1)(0 1 1 1)(0 4 4 4)(1 -4 -2 1)(0 1 1 1)(0 0 0 0)If we are looking at row-reduction or( 1 0 0)(-4 1 4)(-2 1 4)( 1 1 4)( 1 0 0)(-4 1 0)(-2 1 0)( 1 1 0)if w are looking at column-reduction. So, the first part is wrong (or it's right and you are doing something other than what I expected).> Being that the pivots are the first 2 columns, I can say that the first> 2 vectors (1, -4, -2, 1), (1, -3, -1, 2) are a basis for the subspace> and the dimension of the subspace W is equal to 2. > Ok, so please correct me if I'm wrong on the following steps:> However, the solution they give you really goes like this:> Take some vectors...> .... make a matrix out of those vectors as columns> I take (1, -4, -2, 1), (1, -3, -1, 2), and (3, -8, -2, 7)> and convert to:> (1 1 3)> (-4 -3 -8)> (-2 -1 -2)> (1 2 7)> TRANSPOSE EVERYTHING> Now becomes > (1, -4, -2, 1)> (1, -3, -1, 2)> (3, -8, -2, 7)> .... row-reduce the matrix> Now becomes (I omit my row operations and just give the ref form)> (1 -4 -2 1)> (0 1 1 1)> (0 0 0 0)> .... look at the result -- look for non-zero rows, zero rows, whatnot> Non-zero rows = first 2:> TRANSPOSE EVERYTHING> CONFUSED HERE --V> I'm confused by this transpose, if I transpose, I get:> 1 0 0> -4 1 0> -2 1 0> 1 1 0> I'm not sure what that's telling me (is this column reduced form then)? Yes> If I rref the above form I get (steps omitted):> 1 0> 0 1> So where I'm confused is that I get 2 different answers for the basis> depending on how I perform the steps. When doing them as columns, I> get {(1, -4, -2, 1), (1, -3, -1, 2)} as the basis. When doing it as> rows, I get {(1 -4 -2 1), (0 1 1 1)}.This is because you did two slightly different things. When you were doing the column reduction, you took the rows in the _original matrix_ corresponding to the pivots in the reduced matrix. When you were doing the row reduction, you took the rows in the _reduced_ matrix corresponding to the pivots. If you do the same in both cases, you get the same answers (but first you have to correctly column-reduce the original matrix, as I did above).> I am sure that they're both equivelant but I wanted to verify that one will > (pretty much always but I suspect there's a few exceptions) get 2 different > answers but both will still end up being a basis. So if the Linear Algebra > book would have done that problem the first way, they would have come up > with a solution that looked like {(1, -4, -2, 1), (1, -3, -1, 2)} rather than > what they came up with {(1 -4 -2 1), (0 1 1 1)}.> If you look at the steps they are doing between the two transpositions, > they > could have just as easily been done in column form, rather than in row > form, but > the book insists on doing them in row form. There is no profound reason why > this > is so, but row-reduction is a bit easier to write out, and just happens to > be > the way people are usually taught matrix operations.> 4. Put them in a matrix in column form (taught in step 2)> 5. Realize you have to column-reduce the matrix> 6. Realize you never taught anyone how to column-reduce> So basically, nothing happens in step 4,5,6In step 4 you create a matrix out of the vectors. Nothing but bafflement happens in steps 5 and 6> 7. Transpose everything> 8. Row-reduce (taught in step 1)> Correct me if I'm wrong but 4-5-6 involve row reduction techniques on> a matrix that's in column form (ie, what I have been doing since day one> in class)Yes.> 9. Transpose back> When transposing back (see my CONFUSED HERE --V above), I'm not sure> what that particular transposed result is telling me You started with a matrix in column form, which you then transposed into row form, and row reduced. You now have a reduced matrix, still in row form. You need to read a basis from that matrix. Alas, you only need to know how to read a basis out of a matrix in column form! All is lost! Except... if you transpose it, like you do in step 9, your matrix is in the column form and you can read the basis from the columns! Transposition saves the day!> A little more but it will make more sense once I understand what the> very last transpose-back is doing.All it's doing is putting the matrix back in the column from so that you can read the basis from its columns (because that's where you were taught to find the basis)meerohIf this message helped you, consider buying an itemfrom my wish list: You shouldn't be shedding elements when you row-reduce or column-reduce, as far > as I know; you may end up with some zero entries, but that's ok. Column-reducing > the matrix > (1 1 3)> (-4 -3 -8)> (-2 -1 -2)> (1 2 7)> should look exactly the same as row-reducing the matrix> (1 -4 -2 1)> (1 -3 -1 2)> (3 -8 -2 7)Oh shoot. Something that I did not realize was that column-reduction istotally different from row reduction. Now I kind of understand why thetransposition needed to occur. I'm basically taking an elementarylinear algebra course and I don't recall being taught anything likecolumn reduction. Everything we have done thus far has involved rowreduction. Reason I got the wrong answer above was that I wasapparently row-reducing the matrix that I had written in column form. I did not realize they meant 2 different things. My bad...One last (yet unrelated) thing, please let me know if this isequivelant...In my previous post to the following:> Question 3:> u1 = (1,3,4), u2 = (4,0,1) and u3=(3,1,2).> b) determine if (1, 2, 3) lies in the span {u1, u2, u3}One answer I received somewhat contradicts another answer. Eitherthey're both equivelant or one may be an honest mistake...One answer was:In matrix terms you are trying to solve matrix equation[[ 1 3 4 ] [[x] [[1] [[x] [ 4 0 1 ] times [y] equals [2] for column vector [y] [ 3 1 2 ]] [z]] [3]] [z]]--------------Second answer was:For the span, look at 1*X + 4*Y + 3*Z = 13*x + 0*Y + 1*Z = 24*X + 1*Y + 2*Z = 3===============1 4 3 13 0 1 24 1 2 3The matrix is written differently in both cases. I could be VERY wrongon this (and in-fact I probobly am) but I believe the first answer maybe an honest mistake? To check to see if a vector B is in the span,isn't basically saying determine if 'B' can be expressed as a linearcombination of the vectors in A which goes back to the:C1X1 + C2X2 + C3X3 = X check for linear combination? Thus: (1) (4) (3) (1)C1(3) + C2(0) + C3(1) = (2) (4) (1) (2) (3)which would NOT be the same as saying:[[ 1 3 4 ] [[x] [[1] [[x] [ 4 0 1 ] times [y] equals [2] for column vector [y] [ 3 1 2 ]] [z]] [3]] [z]]Please let me know if I'm incorrect on this.=== === Subject: : Re: I'm TERRIBLY confused, linear algebra help requested> I could be VERY wrong on this (and in-fact I probobly am) but I believe the > first answer may be an honest mistake?I believe you are right.meerohIf this message helped you, consider buying an itemfrom my wish list: my teacher is difficult when it comes to proofs. I'm studying and I>can't for anything figure these ones out can any one help me?>1. Prove that the det of the following is equal to zero> w1 x1 y1 z1> w2 x2 y2 z2> 2w1-3w2 2x1-3x2 2y1-3y2 2z1-3z2> w3 x3 y3 z3>2. Slove the following for x and y>det of the following is equal to zero> x+y x x> x x+y x> x x x+y>3. Prove that the det of the following is equal to>(a-b)(b-c)(c-a)(a+b+c)> 1 1 1> a b c> a3 b3 c3 (the 3's mean cubed)=== === Subject: : Re: Linear Algebra Proofs> my teacher is difficult when it comes to proofs. I'm studying and I> can't for anything figure these ones out can any one help me?Well, to prove these you need to start from some basic facts you know about matrices and then show why those facts imply the statements below. We can't give you proofs because we don't know what basic facts you know (but we can make good guesses).> 1. Prove that the det of the following is equal to zero> w1 x1 y1 z1> w2 x2 y2 z2> 2w1-3w2 2x1-3x2 2y1-3y2 2z1-3z2> w3 x3 y3 z3You can use either:The determinant of a matrix is unchanged by adding one row (column) to another row (column).Multiplying all elements or a row (column) by N multiplies the determinant by N.or:If any row of a matrix can be written as a linear combination of other rows, then the determinant of the matrix is zero. (A linear combination of rows R1, R2, ... is a1 * R1 + a2 * R2 + ..., where a1, a2, ... are non-zero real numbers)> 2. Slove the following for x and y> det of the following is equal to zero> x+y x x> x x+y x> x x x+yWhat is the determinant of that matrix in terms of x and y?> 3. Prove that the det of the following is equal to> (a-b)(b-c)(c-a)(a+b+c)> 1 1 1> a b c> a3 b3 c3 (the 3's mean cubed)Figure out the determinant in whatever way you have been taught to do that, and then we can help you prove that's equal to what the problem asks...meerohIf this message helped you, consider buying an itemfrom my wish list: Write the simplest polynomial function with integral coefficientsthat>has the given zeros>19.) 3, 3i> Either there's a misprint in your book, or you copied the problem > wrong.> A polynomial function with integral coefficients can't have an odd > number of non-real roots. For that matter, it also can't have anodd > number of irrational roots.>She didn't say those are the only roots.>meeroh No, but that would be the most reasonable interpretation of having the GIVEN zeros. I didn't notice that word myself until after I had answered.=== === Subject: : Re: Star of david by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3A1OC809180;>Draw a star of david using two triangles. There are 6 different>points at which lines intersect and 6 outer points of the the star.>Using the numbers 1-12 and using each number once on those 12points,>is there a way to make all straight lines (and there are 6 of them)>equal to 26?>I'm asking this for a student whose in sixth or seventh grade. I'm>stumped, any ideas?>Well, in a complete fluke I found an answer; it looks like this:> 8> / > 12-11 - 2 - 1> / /> 3 9> / / > 4 - 5 - 10- 7> /> 6>I'm not sure how you would solve it mathematically (i.e., without>trial and error). I attempted to do this and got this far:>First, label each point in the star.> c> / > a - b - d - e> / /> f g> / / > h - i - j - k> /> l>Since all straight lines add up to 26, the following system of linear>equations is derived from the picture:>a + b + d + e = 26>a + f + i + l = 26>h + f + b + c = 26>h + i + j + k = 26>l + j + g + e = 26>c + d + g + k = 26>This system can be solved down to this point:>a = 26 - C_1 - C_3 - C_6>b = -26 + C_2 + C_3 + C_4 + C_5 + C_6>c = 26 - C_1 - C_2 - C_6>d = C_1 - C_5 + C_6>e = 26 - C_2 - C_4 - C_6>f = C_1>g = C_2>h = 26 - C_3 - C_4 - C_5>i = C_3>j = C_4>k = C_5>l = C_6>with C_1, C_2, C_3, C_4, C_5, C_6 arbitrary.>Then if you applied the restrictions that a, b, c, ..., l must be the>numbers 1 to 12, conceivably you could solve it for specific numbers,>but I'm not sure at the moment how you would do that.>At any rate, since this is for an student in elementary school, I'd>imagine that the answer at the top is all you really need, so there>you go.for hours over several days. My wife on the other hand just drew thediagram and by trial and error, woman's intuition or whatever, solvedthis in a matter of minutes.=== === Subject: : order of inf by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3A1OBu09154;Hi everybody!!!I'll be grateful for you if you could help me such problem:i have to prove that order of inf (u^(2n+1)+1-u^2) is ln/n . u from (0;1) I'll be grateful for someone who could help me.Bye!!!=== === Subject: : Re: Graph theory by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3A1ODD09199;I just have a question about the solution that you have posted:in this part where you have said If v = 2, e > 0, then G is e-1 edge connected and e > (e-1)v/2For induction step add vertex, thus is added at least k edges.e + k > kv/2 + k > k(v+1)/2I don't understand how you have replaced (e-1)v/2 by kv/2 in theinequality.------------------------------------------------ --------------Since you have been so kind I was wondering if you can help on thesetwo.....I really appreciate your help, thanx again If a k-chromatic graph G has a coloring in which each color isassigned to at least two vertices, show that G has a k-coloring ofthis type.Show that if G is simple and 3-regular, then k=k'?=== === Subject: : Re: Graph theory> I just have a question about the solution that you have posted:> in this part where you have said> If v = 2, e > 0, then G is e-1 edge connected and e > (e-1)v/2> For induction step add vertex, thus is added at least k edges.> e + k > kv/2 + k > k(v+1)/2> I don't understand how you have replaced (e-1)v/2 by kv/2 in the> inequality.That's in the wrong place. Put it after> If v = 2, e > 0, then G is e-1 edge connected and e > (e-1)v/2As G is e-1 edge connected, then by definition k = e-1.> Since you have been so kind I was wondering if you can help on these> two.....I really appreciate your help, thanx againYou're taking the class, I don't know anything about graph theory exceptwhat I learn solving problems posted by students.> If a k-chromatic graph G has a coloring in which each color is> assigned to at least two vertices, show that G has a k-coloring of> this type.I'll skip on that unless the night is slow.> Show that if G is simple and 3-regular, then k=k'?What's k' ?=== === Subject: : Re: Implications of square-summabilityI don't think this is true.Sum (n = 1 to infinity) (1/n)^2 converges, butSum (n = 1 to infinity) (1/n) does not converge.Here a_n = 1/n.Is there something that I am missing?>Can someone prove, or at least point me in a direction, of how to>prove the following?> infinity infinity>If SIGMA((a_n)^2) converges, then SIGMA(|a_n|) converges (i.e. it> 1 1> converges absolutely).>Michael === === Subject: : Re: Subsequences in Metric Spaces> 2) Let M1 and M2 be metric spaces. Let {(x_n),(y_n)} be a sequence> of points in the product metric M1 X M2. Let (x,y) be an element on> M1 X M2. Prove that lim (x_n, y_n) = (x,y) iff lim (x_n) = x and lim> (y_n) = y. (All limits are taken as n -> infinity).Has nothing to do with metrics, holds in general.If (xj,yj) -> (x,y)The projections p1:M1xM2, (x,y) -> x p2:M1xM2, (x,y) -> yare continuous. As continuous functions preserve limits:lim_j (xj) = lim_j p1((xj,yj) = p1(lim_j (xj,yj)) = p1((x,y)) = xConversely when (xj) -> x, (yj) -> y:If UxV is a open base set of M1xM2 containing (x,y)then x in U, y in V, xj eventually in U, yj eventually in V(xj,yj) is eventually in UxVAs that occures for all open nhoods of (x,y)(xj,yj) -> (x,y)> There are several different ways to define the metric on> M1 x M2; the details of the proof depend heavily on which> definition you're using. If you use the Pythagorean> formula, the proof here is almost identical to the special> case in which M1 and M2 are both R. As a hint, if m_1, m_2,> and m are the metrics on M1, M2, and M1 x M2, respectively,> and if m is defined from m_1 and m_2 using the Pythagorean> formula, the following observation is easy to prove and may> be useful:> If m_1(x, x') < e and m_2(y, y') < e, then> m((x, y), (x', y')) < e*sqrt(2).> Brian=== === Subject: : Re: Subsequences in Metric Spacesalt.math.undergrad:> 2) Let M1 and M2 be metric spaces. Let {(x_n),(y_n)} be a sequence> of points in the product metric M1 X M2. Let (x,y) be an element on> M1 X M2. Prove that lim (x_n, y_n) = (x,y) iff lim (x_n) = x and lim> (y_n) = y. (All limits are taken as n -> infinity).> Has nothing to do with metrics, holds in general.The *theorem* has nothing to do with metrics. The *proof*may depend on metrics, however; it all depends on whatMichelle already knows. It's a waste of your time and hersto assume that she knows about general product topologies;she may be taking a course that deals (so far, at least)exclusively with metric spaces and defines a product metricrather than a product topology.[...]=== === Subject: : Vector Subspace test on polynomialsHere's an interesting question. Can someone tell me what I did wrong?Determine if the following is a subspace of P3 (polynomials of degree 3or less):Set of polynomials a0 + a1x + a2x^2 + a3x^3 for which a0 + a1 + a2 + a3 = 0I got this problem wrong. The answer sheet says this is a subspace. Iworked it out such that it is not closed under addition. Perhaps Imisunderstood this but I did the following:Let p1(x) = 6 + 4x - 2x^2 - 8x^3 (so (6 + 4 - 2 - 8 = 0))and I let p2(x) = 1 - x^3 (so (1 + 0 + 0 - 1 = 0))Adding p1(x) + p2(x) gives me: 7 + 4x - 2x^2 - 9x^3so my new a0,a1,a2,a3 for the result is 7,4,-2,-9Well 7 + 4 - 2 - 9 is not = 0. So I concluded that it was not asubspace. Yet the answer sheet says it's closed under both addition andscalar multiplication. I'm sure I'm doing something idiotic. Pleasetell me what I'm doing wrong.Here's what the answer sheet says:Suppose p(x);q(x) 2 S and a 2 R.There exist a0;a1;a2;a3;b0;b1;b2;b3 such that p(x) = a0 +a1x +a2x2 +a3x3where a0+a1+a2+a3 = 0 and q(x) = b0+b1x+b2x2+b3x3 where b0+b1+b2+b3 = 0:Then p(x)+q(x) = (a0+a1x+a2x2+a3x3)+(b0+b1x+b2x2+b3x3) = (a0+b0)+(a1+b1)x+(a2+b2)x2+(a3+b3)x3.Now, (a0+b0)+(a1+b1)+(a2+b2)+(a3+b3) = (a0+a1+a2+a3)+(b0+b1+b2+b3) = 0+0= 0 and so p+q 2 S.Did I get this problem wrong because I was NOT treating the polynomialsas a function of x (and thus perform the function space tests ratherthan the vector space tests)?? Seems to me like that's what I didwrong.=== === Subject: : Re: Vector Subspace test on polynomials> Well 7 + 4 - 2 - 9 is not = 0.7 + 4 - 2 - 9 does equal 0. The book's answer is correct.=== === Subject: : Re: Vector Subspace test on polynomials> Let p1(x) = 6 + 4x - 2x^2 - 8x^3 (so (6 + 4 - 2 - 8 = 0))> and I let p2(x) = 1 - x^3 (so (1 + 0 + 0 - 1 = 0))> Adding p1(x) + p2(x) gives me: 7 + 4x - 2x^2 - 9x^3> so my new a0,a1,a2,a3 for the result is 7,4,-2,-9Expanding my understanding of polynomial addition (unless I've beenwrong all these years):p1(x) + p2(x) = p3(x) = 7 + 4x - 2x^2 - 9x^3 where p3(x)'s a=7, p3(x)'s b = 4, p3(x)'s c = -2 and p3(x)'s d = -9and my conclusion was based on the notion that p3(x) is not in thesubspace that p1(x) and p2(x) is due to the sum of the coefficiants ofp3(x) not equaling 0.Yet if the answer sheet is saying that p3(x) *IS* in the subspace (wherethe sum of the coefficiants of p3(x) = 0), then I apparently need acrash course in polynomial addition. Why the answer sheet is treating this as a function space rather than apolynomial space is beyond my comprehension.=== === Subject: : Re: HOW TO INTEGRATE X*e^(x^4) ??> Hi guyz, just wondering if anyone can help me integrate the function> x*e^(X^4), it x times e to the power of x to the 4.> ThanxIt's been a while, but I don't think e^(x^4) has an elementaryantiderivative. Try looking at a power series expansion for e^(x^4).Pat=== === Subject: : Re: HELP!!! Y-intercept form!!!!> I need a creative and challenging math question that has to do with> y-intercept form for a project. Does anyone have one? If you've just recently met this at school, then I think that by far the best thing to work on is applications. Here are a couple of examples. Taxi fares are usually made up of a fixed flag-fall amount plus so much per Kilometre travelled. Electricity bills usually contain a fixed basic charge plus so much per unit consumed. Do you see how those can be shown by straight-line graphs? You should be able to find plenty of other such examples to show how the theory makes contact with practical things. .=== === Subject: : Row Space/Col Space revisited (this time theory)First of all, I want to thank everyone who helped me the other day whenI posted about my terrible confusion as to when to use column vectors asopposed to row vectors.myself to think first about what it is I'm trying to do and thenfigure if it makes more sense to put the vectors in row or column form.Now for another question(s), I'm looking for non-formulas here butrather a description in layman's terms for someone like me who's takingtheir first Linear Algebra course.What I'm looking for is for someone to tell me how they understand thecolumn space and how they understand the row space.Also, how does one geometrically view a vector space in terms of itsrow space as opposed to its column space? Again, I can't seem to findthis in my books.What follows is my limited/rudimentary understanding of both ofthese subspaces.==================================================== ==================I know that row vectors span the row space and column vectors span thecolumn space but a question now comes:- What really is the difference between the row space and the columnspace? Here's what I know right now (perhaps someone can fill inaddition):Vectors in row space:- Span row space (ie, denote the range of the row span). But what really is in the row space other than all possible linearcombinations of vectors in R^n (where 'n' is the 'n' portion of an mxn matrix)Vectors in column space:- Allow you to get the range of the matrix (Ax=b) via the basis inthe column span. So this spans the column space which is all linearcombinations of columns in R^m (where 'm' is the 'm' portion of an mxnmatrix)- Allow you to determine a subset of the original vector set thatforms a basis for the column space of a matrixThe books (and the internet including the MIT lecture video site)basically tell you formulas about how to find the basis of a row space,column space and a null space but I have yet to find a really gooddescription of what the main difference (from an applicationstandpoint) is between a row space/column space (ie, when does it makesense to view the matrix of vectors in terms of its row space as opposedto its column space). Maybe that's a topic for an advanced math theorycourse? Because I sure don't see it in my textbook. They just sayhere's how to find the basis for the rowspace (and blah, blah) andhere's how to find a basis for the column space (and blah, blah) when Iwish they'd go off on a tangent and say ok here's where the columnspace is different in terms of the row space and here's the cases wherethe column space handles this differently than the row space.=== === Subject: : Re: Row Space/Col Space revisited (this time theory)[...]> Also, how does one geometrically view a vector space in terms of its> row space as opposed to its column space? Again, I can't seem to find> this in my books.Well, vector spaces don't _have_ row or column spaces. Also, IMO,geometric intuition is of little or no help in dimensions larger thanthree. [...]> I know that row vectors span the row space and column vectors span the> column space but a question now comes:> - What really is the difference between the row space and the column> space? Here's what I know right now (perhaps someone can fill in> addition):> Vectors in row space:> - Span row space (ie, denote the range of the row span). > But what really is in the row space other than all possible linear> combinations of vectors in R^n (where 'n' is the 'n' portion of an mxn > matrix)Nothing is in the span other than the linear combinations.> Vectors in column space:> - Allow you to get the range of the matrix (Ax=b) via the basis in> the column span. So this spans the column space which is all linear> combinations of columns in R^m (where 'm' is the 'm' portion of an mxn> matrix)> - Allow you to determine a subset of the original vector set that> forms a basis for the column space of a matrixLinear algebra is, well, _algebra_ in the same sense as pre-calcalgebra. At the beginning level, you are learning some algebraictechniques/facts which will, it is hoped, be put to use later. Sinceeveryone, it is again hoped, who takes linear algebra is familiar withsystems of linear equations, those are used as one example of theapplication of linear algebra.The situation is somewhat analogous to the quadratic formula: it is ofno particular interest in itself - the trick is knowing when to use it.Let's look at the system1X + 2Y + 3Z = a4X + 5Y + 6Z = b7X + 8Y + 9Z = cFor which a, b and c does the system have at least one solution? You think for a second and say Aha! I'm being asked for the span of(1, 4, 7), (2, 5, 8) and (3, 6, 9). To say what that span isconveniently, I'll get a basis for that span. So I'll look at thecolumn space of 1 2 34 5 67 8 9because I know how to find a basis for that.But maybe you are feeling perverse and decide to find a basis for therow space of1 4 72 5 83 6 9.That's OK too.If the above system has a solution what are they and how many arethere? You want the null space of the first matrix and one of itsbases.Now suppose, for my own nefarious reasons, I ask you for a basis of thesubspace of the space of polynomials spanned by f(x) = 1 + 2x + 3x^2,g(x) = 4 + 5x + 6x^2 and h(x) = 7 + 8x + 9x^2. You think:af(x) + bg(x) + ch(x) = (a + 4B + 7c) + (2a + 5b + 8c)x + (3a + 6b + 9c)x^2 is a typicalelement of the span. In other words, A + Bx + Cx^2 is in the span iff(A, B, C) is a linear combination of (1, 4, 7), (2, 5, 8) and (2, 5, 8). Look for the column space of the first matrix (or the rowspace of the second one). You can even find a subset of {f, g, h} whichis a basis.The point of all this is that you use the same _tools_ for two_different_ questions.=== === Subject: : number of roots of unityHello.I'm stuck on this...Is there an easy proof I'm not seeing?Let K be a finite field extension of Q (the rationals). Show that Kcontains only a finite number of roots of unity.=== === Subject: : Re: number of roots of unity> Let K be a finite field extension of Q (the rationals). Show that K> contains only a finite number of roots of unity.The roots of unity from x^n = 1for different values of n are linear independent over Q.Thus infinitely many roots of unity, can't be in a finite vector spaceover Q, hence the extension containing them wouldn't be finite.=== === Subject: : Re: Zeros of higher degree polynomials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkZZ11658; 1) degree 5, with zeros -2,-1,0,1,22) Q has degree 3 and zeros -3, and 1+i=== === Subject: : Re: Zeros of higher degree polynomials>1) degree 5, with zeros -2,-1,0,1,2P=[x-(-2)][x-(-1)][x-0][x-1][x-2]Expand if you want.>2) Q has degree 3 and zeros -3, and 1+iThe fundamental theorem of algebra says the complex roots come asconjugate pairs, so 1+i is accompanied by 1-i:Q=[x-(-3)][x-(1+i)][x-(1-i)]=== === Subject: : Re: Help with Function and log systems by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkWL11560;>There are a few problems I am having trouble finding an answer for. >Im sure I am missing something simple but any help on these would be>great.>1. a) Function f(x) has property that f(2x+3) = 2f(x)+3 for all x. >If f(0)=6, what is the value of f(9)? 9= 2(6)+ 3 so f(9)= f(2(6)+ 3= 2f(6)+ 3.> b) Suppose fuctions f(x) and g(x) satisfy the system of equations:>f(x) + 3g(x) = x^2 + x + 6>2f(x) + 4g(x) = 2x^2 + 4 Okay, I suppose that and? Are you asking us to solve for f and g? If so, just treat it like any pair of linear equations:Multiply the first equation by 2 (to get 2f(x) as in the secondequation) and subtract the second equation from it:2f(x)+ 6g(x)- 2f(x)- 4g(x)= 2x^2+ 2x+ 12- 2x^2- 42g(x)= 2x+ 8 so g(x)= x+ 4. Putting that back into the first equation,f(x)+ 3(x+ 4)= f(x)+ 3x+ 12= x^2+ x+ 6 sof(x)= x^2+ x+ 6- 3x- 12= x^2- 2x- 6.>And finally,>2. Solve the system of equations>log10(x^3) + log10(y^2) = 11>log10(x^2) - log10(y^3) = 3 The hard part: log(x^3)= 3log(x), log(y^2)= 2log(y), etc.These equations are just 3log(x)+ 2log(y)= 11 2log(x)- 3log(y)= 3 Multiply the first equation by 3 and the second by 2 to get 9log(x)+ 6log(y)= 33 4log(x)- 6log(y)= 6 and add: 13 log(x)= 39 so log(x)= 3. x= 10^3= 1000. Since log(x)= 3, the first equation is 3(3)+ 2log(y)= 11 or 2log(y)= 8 so log(y)= 4 and y= 10^4= 10000.=== === Subject: : Re: Help with Function and log systems by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkZd11665;>There are a few problems I am having trouble finding an answer for. >Im sure I am missing something simple but any help on these would be>great.>1. a) Function f(x) has property that f(2x+3) = 2f(x)+3 for all x. >If f(0)=6, what is the value of f(9)?> b) Suppose fuctions f(x) and g(x) satisfy the system of equations:>f(x) + 3g(x) = x^2 + x + 6>2f(x) + 4g(x) = 2x^2 + 4>And finally,>2. Solve the system of equations>log10(x^3) + log10(y^2) = 11>log10(x^2) - log10(y^3) = 3Alright for now I will only reply to the 2nd question as I do not havea lot of time and it does not look to hard.for example log10(x^3) = 3log10x that is a general log rulelets say log10x = z , log10y = mDoing the same for all the 2 equations, I like to see it as a matrix:|3 2|*|z| = |11||2 3| |m| | 3|nao just reduce the matrix to the Identity matrix and you get|0 1|*|z| = |-2.6||1 0| |m| |-2.4|logx = -2.4logy = -2.6x = 10^-2.4 y = 10^-2.6that will give you a small number=== === Subject: : Re: differential equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkWE11549;>rewrite the following equation without dy/dx>2^y = 7 dy/dxSeparate the variables: 7*2^(-y) dy = dxIntegrate: -7*2^(-y)/ln(2) = x + CIs that what you wanted?(We can solve for y, but it gets very messy.)=== === Subject: : Re: differential equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkWX11577;>rewrite the following equation without dy/dx>2^y = 7dy/dx>thanx>paul What??? Rewrite it how? Would 2^y= 7 y' do? :) I suspect what you mean is to separate x and y, putting theequation in differential form: dx= 7(2^(-y))dy which can then be integrated. x= (7/ln(2))2^(-y)+ C or 2^y= (ln(2)/7)(x-C) so that y= ln((ln(2)/7)(x-C)/ln(2).=== === Subject: : Re: differential equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkZ911699;>rewrite the following equation without dy/dx>2^y = 7dy/dx To interpret the derivative: if x varies 7 how much does y variejust divide 2^y by 7 and that should give you answer which is (2^y)/7=== === Subject: : Re: Invertible Matrices by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CFkas11735;>Explain why columns of A^2 span Rn whenever the columns of A are>linearly independent.>OK this is my idea>A^2 = AA>det(A^2) = det(A)* det(A)= [det(A)]^2>So if columns of A are linearly independent then det(A) not zero.>so det(A^2) not zero. So columns of A^2 are L.I. then columns of A^2>span Rn.It does not matter if A is square or not A^2 will always be square.And if the rows are lineary independent in A they will also beindependent in A^2 (you actually multiply A by A, where A is thetranspose), it is logical if you have 2 vectors that are not paralelthen if you mupliply both by their base they still will never beparalel.So you have all the necessary conditions, a square matrix and itsdeterminants is not zero, you have proved it. Not very correctly, butlogically it makes cense=== === Subject: : Elementary Linear Algebra by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CJUxo12383;Recently there have been many notes about problems involving linearindependence, spanning, and row-reducing a matrix. Perhaps somereaders will find the following way of thinking about such problemsuseful.Suppose that we have matrices where: A is nXn, B is nXp, C is pX1, andA is invertible. For example, A might be the product of elementarymatrics that you could accumulate while row reducing B, i.e. U=(A*B)is row-reduced. And C might represent a linear dependence relationamong the columns of B, i.e. B*C=0.By the associative property, (A*B)*C = A*(B*C) for any compatibleA,B,C. So, when A is invertible, C represents a linear relation on thecolumns of B iff C also represents a linear relation on the columns ofU=(A*B). So, for example, columns 1,2,4,7 of B are linearly indep. iff columns1,2,4,7 of U are lin. indep., which would be obvious if 1,2,4,7happened to be pivot columns. === === Subject: : Elementary Linear Algebra by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3CJVHG12461;Recently there have been many notes about problems involving linearindependence, spanning, and row-reducing a matrix. Perhaps somereaders will find the following way of thinking about such problemsuseful.Suppose that we have matrices where: A is nXn, B is nXp, C is pX1, andA is invertible. For example, A might be the product of elementarymatrics that you could accumulate while row reducing B, i.e. U=(A*B)is row-reduced. And C might represent a linear dependence relationamong the columns of B, i.e. B*C=0.By the associative property, (A*B)*C = A*(B*C) for any compatibleA,B,C. So, when A is invertible, C represents a linear relation on thecolumns of B iff C also represents a linear relation on the columns ofU=(A*B). So, for example, columns 1,2,4,7 of B are linearly indep. iff columns1,2,4,7 of U are lin. indep., which would be obvious if 1,2,4,7happened to be pivot columns. === === Subject: : Re: Medication> I know it is off topic, so apologies for that, but I would appreciate>anyone's further information.Get on-topic ....in another newsgroup...., or you'll have the rest ofus on anti-depressants.=== === Subject: : Need help with partial differential equationMucho appreciated for any help you might care to provide for solving thisPDE:dc/dt = D*(d2c/dx2)+s*(dc/dx)where c(x,t) and D, s are constants. As you may have guessed, this is aform of the Nernst-Planck equation, but rewritten to express the second(electric) term directly in terms of the drift speed s. D is of course thediffusion coefficients.Gregory=== === Subject: : Re: Need help with partial differential equation>Mucho appreciated for any help you might care to provide for solving this>PDE:>dc/dt = D*(d2c/dx2)+s*(dc/dx)>where c(x,t) and D, s are constants. As you may have guessed, this is a>form of the Nernst-Planck equation, but rewritten to express the second>(electric) term directly in terms of the drift speed s. D is of course the>diffusion coefficients.>GregoryDepending on the boundary data, you could(a) use a Laplace transform on the variable t to get an ordinarydifferential equation in x;or(b) use separation of variables, and use an eigenfunction expansionin functions of xor(c) use a Fourier transform in the variable x.:Alan Jones It is a bright cold day, and the clocks are striking thirteen.=== === Subject: : does this look okay?Hello everyone I was wondering if someone could look at this proof andtell me if I need to change anything or add something to make itQuestion: Let G be a group, let H be a subgroup of G. If each leftcoset of H in G is some right coset of H in G. (ie for each a E G itis the case that aH=Hb for some b E G), then prove that H is a normalsubgroup of G.Proof: Since H is given as a subgroup of G we need only show that H isnormal to G. To show that aH=Hb for all a E G and some b E G. We picksome element of aH, a*h1, and show that it's in Hb. Since 1 is in H,there is an element h E H such that a*1=h*b, or h^-1*a=b. So b is inHb. Now, let a*h1, be an element of aH, then there is an h2 E H, sothat a*h1=h2*b because aH=Hb. But we knowb=h^-1*a so a*h1=h2*b=h2(h^-1*a) or a*h1=(h2*h^-1)*a. Given any a*h1 EaH, we have that a*h1 E Hb. So aH is a subset of Hb.Conversely we pick some element of Hb, h1*b, and show that it's in aH.Since 1 is in H, there is an element h E H such that a*h=1*b, ora=b*h^-1. So a is in aH. Now, let h1*b, be an element of Hb, thenthere is an h2 E H, so that a*h2=h1*b because aH=Hb. But we knowa=b*h^-1 so a*h2=h1*b => (b*h^-1)h2=h1*b or b*(h^-1*h2)=h1*b. Givenany h1*b E Hb, we have that h2*b E aH. So Hb is a subset of aH.=== === Subject: : Re: Teaching multiplication tablesYou are right Dan, kids do love to rhyme and singalong but it helps if therhymes and singalongs are correct!! What does 3x2 equal children?> Kids love to rhyme and singalong, so have them recite ....> 1 x 1 = 1> 1 x 2 = 2> 1 x 3 = 3> 1 x 2 = 2> 2 x 2 = 4> 3 x 2 = 3> etc.> No real difference to us looking back, but a real difference to those> who are just beginning to count.> Dan> [email screwed to stop abuse]=== === Subject: : Re: Teaching multiplication tables>You are right Dan, kids do love to rhyme and singalong but it helps if the>rhymes and singalongs are correct!! What does 3x2 equal children?> 1 x 2 = 2> 2 x 2 = 4> 3 x 2 = 3Dan.=== === Subject: : Chinese Annals of Mathematics - Vol 25 No 2Chinese Annals of MathematicsView table-of-contents and abstracts athttp://www.worldscinet.com/cam.htmlContents:Homogenization Of Semilinear Parabolic Equations In Perforated DomainsP. Donato and A. NabilRegular Relations And Monotone Normal Ordered SpacesXiaoquan Xu And Yingming LiuPolynomial Recurrence For Levy ProcessesMinzhi Zhao And Jiangang YingProperties Of The Boundary Flux Of A Singular Diffusion ProcessJingxue Yin And Chunpeng WangEstimate Of The Upper Critical Field And Concentration ForSuperconductorK. I. Kim and Zuhan LiuThe Law Of The Iterated Logarithm Of The Kaplan-Meier Integral And ItsApplicationShuyuan He And Yanhua WangOn The Topology, Volume, Diameter And Gauss Map Image Of SubmanifoldsIn A SphereBingye WuUnconditional Stable Difference Methods With Intrinsic Parallelism ForSemilinear Parabolic Systems Of Divergence TypeYulin Zhou, Longjun Shen And Guangwei YuanIdeal Structure Of Uniform Roe Algebras Over Simple CoresXiaoman Chen And Qin WangSaddleMaoan Han And Ping BiNon-Constant Positive Steady-States Of A Predator-Prey-Mutualist ModelWenyan Chen And Mingxin WangGlobal Exponential Stability In Hopfield And Bidirectional AssociativeMemory Neural Networks With Time DelaysLibin Rong, Wenlian Lu And Tianping ChenA Central Limit Theorem For Strong Near-Epoch Dependent RandomVariablesZhengyan Lin And Jin QiuExistence, Multiplicity And Stability Results For Positive SolutionsOf Nonlinear P-Laplacian EquationsLi Ma And Ning SuFor more information, go to http://www.worldscinet.com/cam.html