is there such a thing ??

**mmzaj** · April 26th 2008, 02:34 PM

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

n is an integer .

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

**Plato** · April 26th 2008, 02:40 PM

$\phi(t) = 0 ? \$

**topsquark** · April 26th 2008, 02:42 PM

Originally Posted by mmzaj

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

n is an integer .

what could $\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

**mmzaj** · April 26th 2008, 02:46 PM

i'm talking about all positive integers .
my bad

$<br /> \phi(t) = 0 \<br />$
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 .

**mr fantastic** · April 26th 2008, 03:12 PM

Originally Posted by mmzaj

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

n is an integer .

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

Consider:

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

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

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

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

**mmzaj** · April 26th 2008, 03:32 PM

nice !! but i was hoping to get $\phi(t)$ more explicitly ... such as in terms of special functions , or even a series expansion or so ...

**mr fantastic** · April 26th 2008, 07:51 PM

Originally Posted by mr fantastic

Consider:

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

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

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

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

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

Then:

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

$k = t \phi(t) \Rightarrow \phi(t) = \frac{k}{t}$ .

**mmzaj** · April 27th 2008, 08:38 AM

thanks , but there is something wrong , 'cause $\phi(t)$ should be independent of n . the equality is required to be true for any positive integer ... so - obviously - $\phi(t)$ should be independent of n .

here is what i did :

assume $\phi(t)$ is smooth and analytic at $t=t_0$ , 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 :

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

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

now the program is to solve for $b_r$ in general .. so , is this doable ?

**mr fantastic** · April 27th 2008, 07:20 PM

Originally Posted by mmzaj

thanks , but there is something wrong , 'cause $\phi(t)$ should be independent of n . the equality is required to be true for any positive integer ... so - obviously - $\phi(t)$ 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 $\phi(t) = 0$ ....

**mmzaj** · April 28th 2008, 10:09 AM

formally - if i'm not mistaken

- the problem transforms to finding a set of measure spaces whose measure ( $\ ds =\phi(t)dt$ ) and nth norm $<br /> \left\|t\right\|_n = \left( \int \ t^n\ ds \right)^\frac{1}{n}<br />$ satisfy :

1- $\int ds =1$

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

i think the problem went harder , but more formal .

**mmzaj** · April 28th 2008, 10:17 AM

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 $\phi(t) = 0$ ....

i agree

, 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 .

**mmzaj** · April 28th 2008, 03:28 PM

here is another equivalent formulation .

$\int \ e^t \phi(t) \, dt = \exp\int \ t \phi(t) \, dt$

**mmzaj** · April 29th 2008, 11:42 AM

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 .

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