

Subject:
Challenging calculus problem
Category: Science > Math Asked by: eloiga List Price: $100.00 
Posted:
03 Jul 2006 18:51 PDT
Expires: 02 Aug 2006 18:51 PDT Question ID: 743121 
Find the functions f, g, u, and v such that: c(p) = f(x)*g(h+p) + u(x)*v(w+p) where h and w are defined by the definite integrals (D): h(x) = D[0..x](f(x)dx) w(x) = D[0..x](u(x)dx) and c is a function of p, but a constant in x, i.e: dc/dp != 0 dc/dx = 0 The functions f, g, u, and v each have the following properties: * Continuous, real, and noninfinite. * Nonzero, i.e. always positive or always negative. * All functions are harmonic with the same period, a, i.e.: f(x) = f(x+n*a), g(x) = g(x+n*a) u(x) = u(x+n*a), v(x) = v(x+n*a) where n is an integer. The solution may be symbolic or numeric. However if a numerical solution is found, a solution must be provided along with the numerical method that produces it. Note that this problem is for my personal research as in not part of a homework problem or anything like that. For a formatted version, see http://www.pisymbol.com/eloi/functionsconstant.pdf 

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Subject:
Re: Challenging calculus problem
From: berkeleychocolatega on 04 Jul 2006 13:02 PDT 
Here is a sketch of a proof that for nice functions (representable by power series), the only solutions are when f, g, u, v are constants. The periodicity was not used. Let G(t) be the integral from 0 to t of g(s) ds, V(t) = the integral from 0 to t of v(s) ds. So G'=g, V'=v,h'=f,w'=u. So d/dx [ h'G'(h+p) + w'V'(w+p)] = 0, This can be rewritten as d^2/dx^2 [ G(h+p) + V(w+p) ] = 0. So G(h+p) + V(w+p) = c(p)*x + d(p) for some functions c(p) and d(p). Let x= 0 to get d(p) = G(p) + V(p) since h(0) = w(0) = 0. Also differentiating with respect to x and then replacing x with 0, we get: c(p) = h'(0)*G'(p) + w'(0)*V'(p). Now substitute for c(p) and d(p) to get: G(h+p) + V(w+p) = [h'(0)G'(p) + w'(0)V'(p)]*x + [ G(p) + V(p) ] Now use the Taylor expansion of G(h+p) and V(w+p) about p to get: G(p) + G'(p)h(x) + G"(p)h^2(x)/2! + ... + V(p) + V'(p)w(x) + V"(p)w^2(x)/2! + .. = G(p) +V(p) + [h'(0)G'(p) + w'(0)V'(p)]x . Simplifying, we get: G'(p) [ h(x)  h'(0)x ] + G"(p)h^2(x)/2! + ... + V'(p) [ w(x)  w'(0)x ] + V"(p)w^2(x)/2! + ... = 0 . If G and V are not constants then h(x) = h'(0)*x, w(x) = w'(0)*x, G"(p) = 0, and V"(p) = 0. The result follows. 
Subject:
Re: Challenging calculus problem
From: eloiga on 04 Jul 2006 16:41 PDT 
berkeleychocolate, Your manipulation of this equation is masterful. Thank you for your time. It is beyond my limited calculus toolbox to have come up with this on my own, but I worked through and mostly understand the steps of the proof you provided. I have a couple questions/clarifications: your last step says that if G & V are not constants (i.e. G'(p) and V'(p) are not zero) then [h(x)h'(0)*x]=0 and [w(x)w'(0)*x]=0. How did you eliminate the possibilty of G'(p)*[h(x)h'(0)*x] = V'(p)*[w(x)w'(0)*x] ? I suspect that the equation would degrade because g=v or something of the sort, but I can't see it. If periodicity is not considered, then wouldn't the following functions be valid? * f(x) = u(x) = 1/x * g(x) = v(x) = exp(x) Is it because 1/x is not representable by the power series? Thanks again, eloi 
Subject:
Re: Challenging calculus problem
From: berkeleychocolatega on 06 Jul 2006 15:41 PDT 
To eloi: Thanks for your nice comment. You are correct that the last step was not justified. I?ve looked at it further and have some information for you. Firstly f(x) = ln x is not defined at 0, and this violates the above conditions, but this is not a serious problem. One could just as easily define h(x) = the integral from some constant k to x of f(t) dt. Then h(k) = 0 instead of h(0). Then carry through with this change. The last step is trickier than I thought when I first looked at it. Here?s a complicated way to proceed. There may be a better way. Take the basic equation G(h+p) + V(w+p) = c(p)*x + d(p) and differentiate it twice with respect to x to obtain: (*) h?g(h+p)+(h?)^2 g?(h+p) + w?v(w+p)+(w?)^2 v?(w+p)=0. Let aj=the jth derivative of h at 0 and bj=the jth derivative of w at 0. Let x1=g, x2=g?, x3=v, x4=v?. (1)Substitute x=0 into (*).(2) Differentiate (*) with respect to p. Then substitute x=0.(3) Differentiate (*) with respect to x. Then substitute x=0. We get the following three equations: (1) a2*x1 + a1^2*x2 + b2*x3 + b1^2 *x4 = 0. (2) a2*x2 + a1^2*x2? + b2*x4 + b1^2*x4?=0 (3) a3*x1 + 3 a1*a2*x2 + a1^3*x2? + b3*x3 + 3*b1*b2*x4 + b1^3*x4? = 0 Also obviously, (4) x1?=x2 (5) x3?=x4. Using equations (2) and (3) we can solve for x2? and x4?. These plus equations (4) and (5) create a system of 4 linear differential equations with constant coefficients which can be solved by taking the eigenvalues and eigenvectors of the matrix: (Note:D = a1b1) 0 a3/(a1^2*D) 0 a3/(b1^2*D) 1 (b1*a23*a1*a2)/(a1^2*D) 0 2*a1*a2/(b1^2*D) 0 b3/(a1^2*D) 0 b3/(b1^2*D) 0 2*b1*b2/(a1^2*D) 1 (a1*b2+3*b1*b2)/(b1^2*D) This is standard practice if you look up systems of linear first order differential equations. Once one gets the 4 eigenvalues e1, e2, e3, e4 and four eigenvectors v1, v2, v3, v4 one can write down the solution in the vector form exp(e1*p)*v1 + exp(e2*p)*v2+... (4 terms ? assuming no repeated eigenvalues). I looked at it a little and it was mostly a mess. Except that the above matrix has determinant=0 since the first and third rows are multiples. So the characteristic equation is a cubic * p. So 0 is an eigenvalue. I tried to find the eigenvector and got x2 = c1^2/(c2*(c11))* x1 x3=d2/c2 * x1, where cj=aj/bj and d2=b2/b1. So g? = c1^2/(c2*(c11))*g. This means that g = k*exp[c1^2/(c2*(c11)) * p] and v=d2/c2 * g. I could have easily made a mistake here. Also equation (1) must check out. I looked at it quickly and it didn?t, but I could have made an algebra error, or it could be a side condition on all the a?s and b?s. I?m not sure. Also there are other solutions by solving for the other eigenvalues and eigenvectors if you want to go through the trouble. Every cubic can be solved, so it all can be done theoretically. Hope this is readable. 
Subject:
Re: Challenging calculus problem
From: berkeleychocolatega on 08 Jul 2006 13:22 PDT 
To eloi: I was going to put this problem behind me, but I realized that it only takes a little more work to finish it off without having to solve the cubic referred to in the last comment. Without actually going through the work of calculating eigenvalues and eigenvectors one can see that x2 will be a multiple of x1, say x2=k*x1, and x3 will be a multiple of x1, say x3=m*x1. (In fact for a nonzero eigenvalue one can see that m will be b3/a3 by comparing the 2nd and 3rd rows of the matrix mentioned earlier.) Just this fact is enough to finish off the problem since this says that g? = k*g, so g(p) = A*exp(k*p) . Also v(p) = A*m*exp(k*p). We can now find the conditions on h and w (and so f and u) by substituting the above g and v into the original equation: h?*A*exp(k*h)*exp(k*p) + w?*A*m*exp(k*w)*exp(k(p) is independent of x. So h?*exp(k*h)+ w?*m*exp(k*w) is a constant. Since h(0)=0 and w(0)=0, the constant is a1+m*b1 (Recall a1=h?(0) and b1=w?(0). ). This can be rewritten as: [exp(k*h)+exp(k*w)]? = a1+m*b1. So exp(k*h)+m*exp(k*w) = (a1+m*b1)*x + a constant of integration, which we see is 1+m by letting x=0. So exp(k*h)+exp(k*w) = (a1+m*b1)*x + (1+m). The final answer is: Let w(x) be an arbitrary function, k and m arbitrary constants. Let h(x) be defined to be 1/k * ln[(a1+m*b1)*x + (1+m) ? m*w(x) ]. Let f(x) = h?(x) and u(x) = w?(x). Let the functions g and v be defined as above. This is the general solution to your question. (Note: I didn?t really consider the question of periodicity.) 
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