mm-301 === Subject: : Re: Calculus StuffContent-transfer-encoding: 8bit> maybe i didn't put it clearly, here's the original if i'm missing> anything...> http://roasted.homelinux.net/~helpinghand/1.jpg> http://roasted.homelinux.net/~helpinghand/2.jpgAh. That's _much_ better.I presume the question is how2A(a^2 + ab + b^2) + 3B(b + a) + 6C =(Aa^2 + Ba + C) + 4(A((b+a)/2)^2 + B((b+a)/2) + C) + (Ab^2 + Bb^2 + C) ?If so,2A(a^2 + ab + b^2) + 3B(b + a) + 6C =(Aa^^2 + Ba + C) + (Ab^2 + Bb + C) + Aa^2 + 2Aab + Ab^2 + 2Bb + 2Ba + 4CAnd...Aa^2 + 2Aab + Ab^2 + 2Bb + 2Ba + 4C = A(a^2 + 2ab + b^2) + 2B(b +a) + 4C =A((b + a)^2) +2B(b + a) + 4C =4[(1/4)A((b + a)^2) + (1/2)B(b + a) + C] =4(A(((b +a)/2)^2 + B((b + a)/2) + C)-- === Subject: : Re: Calculus Stuffthx much, but what I was looking to find was how the first line on 2.jpgequal to the integral [a,b] of p(x)dx = (...the point of 1.jpg was to show from where did that equation came from. SorryI guess I said the wrong thing. In the second line of equation on 2.jpg, Idon't know where from or how is p(x) related to the original, especiallywhen p(x) is not anywhere from the original. === Subject: : Re: Calculus StuffI think I found a possible solution, after reading it more thoroughly, I didit backward, here's my solution, see what you guys think...http://roasted.homelinux.net/~helpinghand/ solution.jpgto look into it when I'm feeling better.P.S.S. My scanner will crop the page if I don't open the lit, cuts of mostof the work. === : Re: Calculus StuffContent-transfer-encoding: 8bit> I think I found a possible solution, after reading it more thoroughly, I did> it backward, here's my solution, see what you guys think...http://roasted.homelinux.net/~helpinghand/ solution.jpgto look into it when I'm feeling better.P.S.S. My scanner will crop the page if I don't open the lit, cuts of most> of the work.Starting from where you say you are working backwards, the firstequality is wrong - among other things ((a + b)/2)^2 will have an abterm.Your last equality is also wrong (for the same reason). It seems thatin both places you used the incorrect (a + b)^2 = a^2 + b^2.It is amusing that the two errors canceled.My previous post was intended to get you from the bottom of page 306free.-- === Subject: : Re: Calculus Stuffmy algebra is horrible! : How can this be correct?Consider the following 2nd order non-homogenous diff'l equation:mu'' + cu' + ku = FoCos(wt).if the damping constanct c = 0, then, mu'' + ku = FoCos(wt).Now, for the case where wo is not equal to w ( w = frequency), then y = C1Cos(wot) + C2 Sin(wot) + (FoCos(wt))/m(wo^2-w^2), [1] is a solution.Now, if we treat wo equal to w, then the above solution is the same with theexeption of the last term. The last term is FotSin(wt)/2mw, [2]. This isknown as Resonance.Anyway, what is interesting is that you can apply L'Hopital's rule to thelast term of [1] and get [2]. Try it! Then take the Limit as wo approachesw. Any ideas as to why this works when in fact the last term in [1] is NOTindeterminate?Gord Bramfield. === Subject: : Re: How can this be correct?> Consider the following 2nd order non-homogenous diff'l equation:> mu'' + cu' + ku = FoCos(wt).> if the damping constanct c = 0, then, mu'' + ku = FoCos(wt).> Now, for the case where wo is not equal to w ( w = frequency), then y = C1> Cos(wot) + C2 Sin(wot) + (FoCos(wt))/m(wo^2-w^2), [1] is a solution.> Now, if we treat wo equal to w, then the above solution is the same with the> exeption of the last term. The last term is FotSin(wt)/2mw, [2]. This is> known as Resonance.> Anyway, what is interesting is that you can apply L'Hopital's rule to the> last term of [1] and get [2]. Try it! Then take the Limit as wo approaches> w. Any ideas as to why this works when in fact the last term in [1] is NOT> indeterminate?Mere concidence?-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice. === Subject: Subject: Re: Recommendation for book on Integration methods?> Can anyone recommend a book that presents different techniques for > calculating integrals? I'm not really looking for theory-- although a > little is fine-- just something that might help me to depend less on my > TI-89 or integral tables :)>I didn't mean basic techniques like change of variable, trig. > substitution, integration by parts, etc. Those only get one so far. For > instance, what is the integral of cos(x)ln(x), if it exists in terms of > elementary functions? And how do I know when there is no integral? I've > heard of the Risch algorithm-- if there is a book that teaches that with a > minimum of theory, then that's what I'm looking for.> AlexWhat about Joel Moses book Symbolic integration: the stormy decade?Or: http://www.cl.cam.ac.uk/users/mywyb2/publications/ becker01part2project.pdfMads === Subject: : vector field - div - but in cylindrical polarsThe three unit vectors in cylindrical polars are Er, Et, Ez(where Et stands for E-theta)Am I right in saying that:div Er = 1/rdiv Et = 0div Ez = 0If these are wrong, then why?.Kev === Subject: : L^1([0,1]) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i27D2Id04526;Why isn't L^1([0,1]) the dual of any normed linear vector space? Does this have something to do with L^infinity([0,1]) has extreme points of the unit sphere are those x that satisfy |x(t)|=1 a.e. whereas the unit sphere in L^1([0,1]) has no extreme pts? === Subject: : Question about linear algebra theorem (ie, determinant for determining independence)X-No-Archive: ?yesMy book has a theorem which reads:3 vectors in R^3 are independent if their determinant is not = 0.Then they turn around and give a problem of a 4x4 and ask if that isindependent in R^41 3 0 5-2 0 4 01 2 -1 31 -2 1 -1Well doing it the long way (of elimination and solving for a,b,c,d) Ihave a solution (and determined it's independent)Now for my question:The book seemed to take great strides in specifically mentioning thatThe determine of 3 vectors in R^3 determine if they're independent. They did not mention any other spaces where a square matrix is involved.For the above 4x4, the determinant is not = 0. Am I wrong in thinkingthe book meant to say If the determinant of any square matrix in R^M(where matrix is an mxm) is not = 0, then the vectors are linearlyindependant?. Somewhat unrelated, I know there's a special case of finding thedeterminant of a 3x3 that only works on a 3x3 (ie, the diagonalproduct method after repeating the first 2 columns at the end). I wasnot sure if the above theorem only applied to 3x3 in R^3 and if I wasjust getting lucky that the 4x4 was independant and also itsdeterminant was not = 0.!!! === Subject: : Re: Question about linear algebra theorem (ie, determinant for determining independence)Dear Mrs. Madness,> My book has a theorem which reads:> 3 vectors in R^3 are independent if their determinant is not = 0.> (...)> They did not mention any other spaces where a square matrix is> involved.n vectors from R^n are linearly independent iff the corresponding matrixhas a non-zero determinant.To be more general: some vectors are linearly independent iff the onlylinear combination of theses vectors that yield the 0-vector is thetrivial one (with all coefficients equal to zero).> Somewhat unrelated, I know there's a special case of finding the> determinant of a 3x3 that only works on a 3x3 (ie, the diagonal> product method after repeating the first 2 columns at the end). I was> not sure if the above theorem only applied to 3x3 in R^3 and if I was> just getting lucky that the 4x4 was independant and also its> determinant was not = 0.I'm not sure what you ask here, but you cannot compute the determinantof a 4x4-matrix with the you would use for the 3x3 case. For n>3 Iwould use Gauss-elemination or Laplace.Greetings,Oswald-- _/_/ _/ | _/ _/ _/ | Oswald Jaskolla _/ _/ _/ _/ | http://www.jaskolla.net/oswald/ _/_/ _/ _/_/ _/ | === Subject: : Re: Infinite cyclic groupsHi!> Are mappings from infinite cyclic groups into infinite cyclic groups> necessarily injective?Of course not. Take f:Z->Z with f(0)=f(1)=0 and f(x)=x for all other x.> If not, is there some general criteria (relating to the> infinite cyclic groups in question) that can tell you if such a> mapping is injective?How could that be? You don't know nothing about the mapping.Greeting,Oswald-- _/_/ _/ | _/ _/ _/ | Oswald Jaskolla _/ _/ _/ _/ | http://www.jaskolla.net/oswald/ _/_/ _/ _/_/ _/ | === Subject: : Re: Infinite cyclic groupsContent-transfer-encoding: 8bit> Hello. Could someone please help me with the following: Are mappings from> infinite cyclic groups into infinite cyclic groups necessarily injective?> If not, is there some general criteria (relating to the infinite cyclic> groups in question) that can tell you if such a mapping is injective?> in advance. We may as well let both groups be the additive group of integers Z.Let f : Z -> Z be a homomorphism and suppose f(n) = f(m) with n > m.Then, 0 = f(n) - f(m) = f(n - m) = (n - m)*f(1) so f(1) = 0 and thus f(x) = x*f(1) = 0 for all x.We conclude that either f = 0 or f is injective.-- === Subject: : Cubic polynomialI have been set a problem to prove that a function f(x) = x^3+ax^2+bx+c hasat least one root, with a,b and c being real coefficients.A hint is given, and it says For x > max{|a|+|b|+|c|,1}, show that x^3 >|ax^2+bx+c|.I have shown this, but I can't see how this helps me solve the problem. Anyhelp to get me going again in the right direction would be most appreciated.!Rob === Subject: : Re: Cubic polynomialI have been set a problem to prove that a function f(x) = x^3+ax^2+bx+c has> at least one root, with a,b and c being real coefficients.A hint is given, and it says For x > max{|a|+|b|+|c|,1}, show that x^3 |ax^2+bx+c|.I have shown this, but I can't see how this helps me solve the problem. Any> help to get me going again in the right direction would be most appreciated.!RobCan you show that there is some x for which f(x) > 0 and another for which f(x) < 0? === Subject: : Re: Cubic polynomialI have been set a problem to prove that a function f(x) = x^3+ax^2+bx+c has> at least one root, with a,b and c being real coefficients.A hint is given, and it says For x > max{|a|+|b|+|c|,1}, show that x^3 |ax^2+bx+c|.I have shown this, but I can't see how this helps me solve the problem. Any> help to get me going again in the right direction would be most appreciated.The usual way to show this is to consider what happens as the magnitude of x gets very large in the positive direction, and very large in the negative direction. === Subject: : axiom of choice by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i27MXFi27637;could someone give me a clue on how to prove that the axiom of choiceis equivalent to Zerlelo's axiom????Zermelo's axiom says:For every set C consisting of nonempty, pairwise descrete sets, existsa set B whose intersection with every element of C is a singleton.thank you. === Subject: : Re: axiom of choice Adjunct Assistant Professor at the University of Montana.>could someone give me a clue on how to prove that the axiom of choice>is equivalent to Zermelo's axiom????>Zermelo's axiom says:>For every set C consisting of nonempty, pairwise descrete sets,I assume that is supposed to be pairwise distinct...> exists>a set B whose intersection with every element of C is a singleton.But you didn't tell us what your version of the Axiom of Choice is...I shall assume either of:The product of a nonempty family of nonempty sets is nonempty.orGiven a family {A_i} i in I of sets such that I is nonempty and foreach i, A_i is nonempty, there exists a function f:I -> Union(A_i)such that for each i, f(i) is in A_i.(The function is called a choice function; the equivalence betweenthe two statements is that the chice function gives an element of thedirect product, and an element of the direct product defines a choicefunction).So, First, assume Zermelo's axiom is true. Let {A_i} i in I be afamily of sets with I nonempty and A_i nonempty for each i. We definea new family, {B_i} i in I, where for each i, B_i = A_i x {i}. Thenthe B_i are nomepty, pairwise distinct. Let C = {B_i : i in I}. ByZermelo's Axiom, there is a set B such that the intersection of B witheach B_i is a singleton. We define f:I -> Union(A_i) as follows: giveni, let {(a_i,i)} be the element of C intersect B_i; then a_i is inA_i, and we let f(i) = a_i. Conversely, assuming the Axiom of Choice, let C = {f(i): i in I} wherethe family is given I = C, and A_c = c, for each c in C.-- === Subject: : Banach spaces by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i27MXFC27645;I need some help with the following:If X is a Banach space and Y a closed subspace of X, then prove thatthe quotient space X/Y is a Banach space.What i need is not the usual proof that can be found in almost everyfunctional analysis handbook, but a proof based exclusevely on thefollowing criterion:X is a Banach space if and only if SUM_{n=1}^{infinity}||x_{n}||converges, implies that SUM_{n=1}^{infinity} x_{n} converges too,where || || is the norm on X. === Subject: : Time-Dependent Inhomogeneous Diffusion Equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i281qOk15945;I have a good problem... And I'm looking to see if someone can checkmy solution.. Please forward along... ... dpath2oProblem:PDE: du/dt = d^2u/dx^2 + (e^(-2t))Sin(pi x) - x Sin(t); 0 I have a good problem... And I'm looking to see if someone can check> my solution.. Please forward along... ... dpath2o> Problem:> PDE: du/dt = d^2u/dx^2 + (e^(-2t))Sin(pi x) - x Sin(t); 0 0 BC's: u(0,t) = 1 t ?0> u(1,t) = Cos(t) t ?0> IC: u(x,0) = Sin(3 pi x) + 1 + (Sin(pi x) / (pi^2 - 2))I attempted to solve this by separation of variables; I concluded that theboundary conditions given are inconsistent:If u(x,t) = sum(n=0, +oo) [a_n exp(-(n*pi)^2 t) cos(n*pi*x)] + sum(n=1, +oo) [b_n exp(-(n*pi)^2 t) sin(n*pi*x)] + exp(-2t)*sin(pi*x)/(pi^2 - 2) + x*cos(t)then the equation is satisfied and the boundary & initial conditionsreduce tou(0,t) = 1 => a_0 = 1, a_n = 0 else.u(x,0) = sin(3*pi*x) + 1 + sin(pi*x)/(pi^2 - 2) => b_n = { - int_0^1 x*sin(n*pi*x) dx, n =/= 3 [1] { 1 - int_0^1 x*sin(3*pi*x_ dx, n = 3u(1,t) = cos(t) => 1 + exp(-2t)*sin(pi)/(pi^2 - 2) + cos(t) = 1 + cos(t) => 1 + cos(t) = cos(t)There is no way to satisfy this condition for all t.Conversely, demanding a_0 = 0 to satisfy the last condition means that theconditions on u(0,t) and u(x,0) cannot be satisfied.[1]: these may be out by a factor of 2; this is not important for thepresent purpose.-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice. === Subject: : application of definite intregral by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i28BgF306922;The problem:Find the volume of the solid rotated around the y-axis. The solid is adonut with the center of the donut at 0,0. The center of one side ofthe donut is at 4,0 (x=4, y=o) with a radius of 1 (center of otherside at -4,0). It will be bound by x=3, and x=5. Does this make sense?Can you picture the solid? - any help with setting up theformula would be much appreciated! === Subject: : Re: application of definite intregralContent-transfer-encoding: 8bit> The problem:> Find the volume of the solid rotated around the y-axis. The solid is a> donut with the center of the donut at 0,0. The center of one side of> the donut is at 4,0 (x=4, y=o) with a radius of 1 (center of other> side at -4,0). It will be bound by x=3, and x=5. Does this make sense?> Can you picture the solid? - any help with setting up the> formula would be much appreciated!> You are revolving the circle (x - 4)^2 + y^2 = 1 around the y-axis.Any Calculus text should tell you about volumes of revolution. Usingthe shell method (although I think the washer method works just aswell), I got the volume to be 4*Pi*int(x*sqrt(1 - (x - 4)^2), x = 3..5) = 8*(Pi)^2.-- === Subject: : Re: application of definite intregralContent-transfer-encoding: 8bitThe problem:> Find the volume of the solid rotated around the y-axis. The solid is a> donut with the center of the donut at 0,0. The center of one side of> the donut is at 4,0 (x=4, y=o) with a radius of 1 (center of other> side at -4,0). It will be bound by x=3, and x=5. Does this make sense?> Can you picture the solid? - any help with setting up the> formula would be much appreciated!> You are revolving the circle (x - 4)^2 + y^2 = 1 around the y-axis.Any Calculus text should tell you about volumes of revolution. Using> the shell method (although I think the washer method works just as> well), I got the volume to be > 4*Pi*int(x*sqrt(1 - (x - 4)^2), x = 3..5) = 8*(Pi)^2.PS: You might want to see:-- === Subject: : Algebra questionIf N is a normal subgroup of a group G,Then Can we say G/N is a subgroup of G ?I read definition of Quotient group from book, When N is a normal subgroup of a group G,The quotient group G/N is defined as G/N={gN | g in G}.... etcNow, I wonder why we call this group as Quotient groupor Factor group in spite of G/N does not meanordinary division of G by N.I even heard N cut G but, I don't know how can we say that.Could someone please explain me that reason in detail.Z/0*Z = {x+0*Z | x in Z} = {x+0 | x in Z} = {x | x in Z} = ZThus, Since Z/0*Z = Z, Z/0*Z is isomorphic to Z.Is this process correct ?I would like to construct isomorphism between Z/Z and 0*Z.If f is defined as f:Z/Z -> 0*Z , f(aZ)=0*Z,Can we say this is correct isomorphism between Z/Z and 0*Z? for reading. === Subject: : Re: Algebra question Adjunct Assistant Professor at the University of Montana.>If N is a normal subgroup of a group G,>Then Can we say G/N is a subgroup of G ?Certainly not; the elements of G/N are subsets of G. (Sometimes, it may be isomorphic to a subgroup of G, but that is infact rare)>I read definition of Quotient group from book,> When N is a normal subgroup of a group G,>The quotient group G/N is defined as G/N={gN | g in G}.... etc>Now, I wonder why we call this group as Quotient group>or Factor group in spite of G/N does not mean>ordinary division of G by N.Well, the orders do divide: the order of G/N is the order of G dividedby the order of N. But really it is because you are taking the quotient set modulo anequivalence relation; and those names date back to the equivalencerelation a=b (mod n) of Gauss, which ->was<- related to division.>Z/0*Z = {x+0*Z | x in Z}> = {x+0 | x in Z}Not quite; remember that x+N means x+N = {x + n : n in N}So x+0*Z = {x+a : a in 0*Z} = {x + 0} = {x}.So you should haveZ/0*Z = {x+0*Z : x in Z} = { {x} : x in Z}.> = {x | x in Z} = Z>Thus, Since Z/0*Z = Z, Z/0*Z is isomorphic to Z.They are not equal formally speaking, since the elements of Z/0Z aresets of the form {0}, {-1}, {1}, {2}, {-2},... etc., while theelements of Z are just the numbers, but yes, they are isomorphic.>Is this process correct ?>I would like to construct isomorphism between Z/Z and 0*Z.Since 0*Z = {0}, there is only one possible map from Z/Z to 0*Z.>If f is defined as f:Z/Z -> 0*Z , f(aZ)=0*Z,>Can we say this is correct isomorphism between Z/Z and 0*Z?It's the only isomorphism, so yes.--