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Q: Challenging calculus problem ( No Answer,   4 Comments )
Subject: Challenging calculus problem
Category: Science > Math
Asked by: eloi-ga
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 non-infinite.
* Non-zero, 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
There is no answer at this time.

Subject: Re: Challenging calculus problem
From: berkeleychocolate-ga 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: eloi-ga on 04 Jul 2006 16:41 PDT
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,
Subject: Re: Challenging calculus problem
From: berkeleychocolate-ga 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

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

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 = a1-b1)

0    	-a3/(a1^2*D)			0		a3/(b1^2*D)

1	(b1*a2-3*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

x2 = c1^2/(c2*(c1-1))* x1

x3=d2/c2 * x1,

where cj=aj/bj and d2=b2/b1.

So g? = c1^2/(c2*(c1-1))*g. This means that g = k*exp[c1^2/(c2*(c1-1)) * 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

Hope this is readable.
Subject: Re: Challenging calculus problem
From: berkeleychocolate-ga 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
non-zero 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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