$\displaystyle \int_T \ t^n \phi(t) \, dt = \left( \int_T \ t \phi(t) \, dx \right)^n $

n is an integer .

what could $\displaystyle \phi(t) \$ be ?

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- Apr 26th 2008, 02:34 PMmmzajis there such a thing ??
$\displaystyle \int_T \ t^n \phi(t) \, dt = \left( \int_T \ t \phi(t) \, dx \right)^n $

n is an integer .

what could $\displaystyle \phi(t) \$ be ?

- Apr 26th 2008, 02:40 PMPlato
$\displaystyle \phi(t) = 0 ? \$

- Apr 26th 2008, 02:42 PMtopsquark
- Apr 26th 2008, 02:46 PMmmzaj
i'm talking about all positive integers .

my bad (Wink)

$\displaystyle

\phi(t) = 0 \

$

is rather a trivial solution that i'm not interested in

i know that there will be a family of solutions , i'm interested in this family . - Apr 26th 2008, 03:12 PMmr fantastic
Consider:

$\displaystyle \int_0^t x^n \phi(x) \, dx = \left( \int_0^t x\, \phi(x) \, dx \right)^n$ and differentiate both sides with respect to t:

$\displaystyle t^n \phi(t) = n \left( \int_0^t x\, \phi(x) \, dx \right)^{n-1} (t \, \phi(t) \, )$

$\displaystyle \Rightarrow t^{n-1} = n \left( \int_0^t x\, \phi(x) \, dx \right)^{n-1}, ~ \phi(t) \neq 0$

$\displaystyle \Rightarrow \frac{t\, \phi(t) }{n^{n-1}} = \int_0^t x\, \phi(x) \, dx$ ..... - Apr 26th 2008, 03:32 PMmmzaj
nice !! but i was hoping to get $\displaystyle \phi(t)$ more explicitly ... such as in terms of special functions , or even a series expansion or so ...

- Apr 26th 2008, 07:51 PMmr fantastic
Well, since this contains two such careless mistakes, I though I'd better set the record straight.

The result in the last line should of course be $\displaystyle \frac{t}{n^{1/(n-1)}} = \int_0^t x\, \phi(x) \, dx$.

Then:

Let $\displaystyle k = \frac{1}{n^{1/(n-1)}} = n^{1/(1-n)}$ and differentiate both sides wrt t:

$\displaystyle k = t \phi(t) \Rightarrow \phi(t) = \frac{k}{t}$. - Apr 27th 2008, 08:38 AMmmzaj
thanks , but there is something wrong , 'cause $\displaystyle \phi(t) $ should be independent of n . the equality is required to be true for any positive integer ... so - obviously - $\displaystyle \phi(t) $ should be independent of n .

here is what i did :

assume $\displaystyle \phi(t) $ is smooth and analytic at $\displaystyle t=t_0 $ , then it can be expanded in terms of a power series . now setting the correct relations on both RH and LH parts , and integrating over T , we end up with something like this :

$\displaystyle \sum^{\infty}_{r=0}\frac{b_r}{r+n+1}= \left(\sum^{\infty}_{r=0}\frac{b_r}{r+2}\right)^{n }$

$\displaystyle n=0,1,2,3 ...... $

now the program is to solve for $\displaystyle b_r $ in general .. so , is this doable ? - Apr 27th 2008, 07:20 PMmr fantastic
- Apr 28th 2008, 10:09 AMmmzaj
formally - if i'm not mistaken (Wondering) - the problem transforms to finding a set of measure spaces whose measure ($\displaystyle \ ds =\phi(t)dt $) and nth norm $\displaystyle

\left\|t\right\|_n = \left( \int \ t^n\ ds \right)^\frac{1}{n}

$ satisfy :

1- $\displaystyle \int ds =1$

2- $\displaystyle \left\|t\right\|_n$=$\displaystyle \left\|t\right\|_1$ ,$\displaystyle n=2,3,4 .... $

i think the problem went harder , but more formal . - Apr 28th 2008, 10:17 AMmmzaj
- Apr 28th 2008, 03:28 PMmmzaj
here is another equivalent formulation .

$\displaystyle \int \ e^t \phi(t) \, dt = \exp\int \ t \phi(t) \, dt $ - Apr 29th 2008, 11:42 AMmmzaj
so , i have been discussing the problem with the guys in physics forums , here is the link

a problem in Lp spaces .

i think it helps to look at it .