# is there such a thing ??

• Apr 26th 2008, 02:34 PM
mmzaj
is 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 PM
Plato
$\displaystyle \phi(t) = 0 ? \$
• Apr 26th 2008, 02:42 PM
topsquark
Quote:

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

n is an integer .

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

Are you talking about for all values of n? Or specific values of n? Obviously if it's the last one, every function fits n = 1....

-Dan
• Apr 26th 2008, 02:46 PM
mmzaj
i'm talking about all positive integers .
$\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 PM
mr fantastic
Quote:

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

n is an integer .

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

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 PM
mmzaj
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 PM
mr fantastic
Quote:

Originally Posted by mr 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:

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

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

$\displaystyle \Rightarrow \frac{t\, \phi(t) }{n^{n-1}} = \int_0^t x\, \phi(x) \, dx$ .....

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 AM
mmzaj
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 PM
mr fantastic
Quote:

Originally Posted by mmzaj
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 .
[snip]

I disagree. A solution containing n is perfectly fine ...... I think the only solution you'll find that's independent of n will be the trivial solution $\displaystyle \phi(t) = 0$ ....
• Apr 28th 2008, 10:09 AM
mmzaj
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 AM
mmzaj
Quote:

Originally Posted by mr fantastic
I disagree. A solution containing n is perfectly fine ...... I think the only solution you'll find that's independent of n will be the trivial solution $\displaystyle \phi(t) = 0$ ....

i agree (Rofl) , but - and it's a big fat but - the problem in hand requires the equality to hold for every +ive integer . if 0 is the only solution , i need a proof .
• Apr 28th 2008, 03:28 PM
mmzaj
here is another equivalent formulation .

$\displaystyle \int \ e^t \phi(t) \, dt = \exp\int \ t \phi(t) \, dt$
• Apr 29th 2008, 11:42 AM
mmzaj
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 .