Need ideas with a problem

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Gunnar G - 13 Jun 2007 22:54 GMT

I'm trying to find one(or more) positive function f(x) and a constant h > 0
such that

0.45 * f(h+19)^2 = f(h)^2 and
\int( f(x)^2, x=0..h+19) = 5 \int( f(x)^2, x=0..h)

(I hope you understand my integrals)

By assuming f(x)=x

By solution I mean any kind of solution, even if a simple explicit formula
for the function f(x) would be best :-)
I guess there isn't much litterature on this?
So far I've only tried with f(x)=kx^n and found that it gives only
contradictions. Is there any such function and constant?
Can it be proven that there are/aren't any solutions?

Best wishes
Gunnar.

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Ken Pledger - 14 Jun 2007 03:51 GMT

> I'm trying to find one(or more) positive function f(x) and a constant h > 0
> such that
[quoted text clipped - 11 lines]
> So far I've only tried with f(x)=kx^n and found that it gives only
> contradictions....

If I understand you aright, there are lots of solutions. For
example, applying the two conditions to your suggestion f(x) = k(x^n)
leads to

(A) ((h + 19)/h)^(2n) = 1/(0.45)

(B) ((h + 19)/h)^(2n+1) = 5.

Dividing (B) by (A) gives (h + 19)/h = 2.25

whence h = 15.2. Substitute that into (A) and solve for n using
logarithms.

You could start from various other formulae for f(x) and solve for
parameters in a similar way. But perhaps I may be missing some other
condition which you forgot to mention.

Ken Pledger.

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Gunnar G - 14 Jun 2007 06:16 GMT

> (A) ((h + 19)/h)^(2n) = 1/(0.45)
>
> (B) ((h + 19)/h)^(2n+1) = 5.
>
> Dividing (B) by (A) gives (h + 19)/h = 2.25

> whence h = 15.2. Substitute that into (A) and solve for n using
> logarithms.

I got that too, but I must have made some error when controling the answers.

Thanks.

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Math Forum / Mathematics / Undergraduate Math / June 2007

Need ideas with a problem