My Math Forum A fun result on integral

 Real Analysis Real Analysis Math Forum

 June 4th, 2017, 02:31 PM #1 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 A fun result on integral Hi, $\displaystyle n \in \mathbb N, n>1, \{f\in C^0([0.1],\mathbb R) | \int_0^1 f^{2n}\text{ d}x=(\int_0^1 f\text{ d}x)^{2n}\}=\{f|\exists c\in \mathbb R, f=c\}$ ? Cordially. Last edited by Dattier; June 4th, 2017 at 02:46 PM.
 June 4th, 2017, 04:36 PM #2 Global Moderator   Joined: May 2007 Posts: 6,540 Thanks: 591 I am not sure what you are asking, but the equation will hold for any c. Both integrals = $c^{2n}$.
 June 4th, 2017, 05:06 PM #3 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 Ok, for the first inclusion, but it would miss the other.
 June 4th, 2017, 05:15 PM #4 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 Another form : Let $\displaystyle n\in\mathbb N, n>1$ $\displaystyle \{f\in C^0([0,1],\mathbb R^*)|\int_0^1f^n\text{ d}x=(\int_0^1 f\text{ d}x)^n\}=\{f|\exists c\in \mathbb R, f=c\}$ ?
June 5th, 2017, 06:01 PM   #5
Global Moderator

Joined: May 2007

Posts: 6,540
Thanks: 591

Quote:
 Originally Posted by Dattier Ok, for the first inclusion, but it would miss the other.
What is first inclusion? What is other?

Your last statement still looks like equality ($c^n$).

 June 6th, 2017, 06:36 AM #6 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 $\displaystyle \subset$ It's another form.
June 6th, 2017, 05:32 PM   #7
Global Moderator

Joined: May 2007

Posts: 6,540
Thanks: 591

Quote:
 Originally Posted by Dattier $\displaystyle \subset$ It's another form.

 June 7th, 2017, 12:12 AM #8 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 $\displaystyle \{f|\exists c\in \mathbb R^*, f=c\}\subset \{f\in C^0([0.1],\mathbb R^*) | \int_0^1 f^{2n}\text{ d}x=(\int_0^1 f\text{ d}x)^{2n}\}$ and $\displaystyle \{f\in C^0([0.1],\mathbb R^*) | \int_0^1 f^{2n}\text{ d}x=(\int_0^1 f\text{ d}x)^{2n}\} \subset \{f|\exists c\in \mathbb R^*, f=c\}$
 June 7th, 2017, 05:30 PM #9 Global Moderator   Joined: May 2007 Posts: 6,540 Thanks: 591 I finally get your point. The first statement is true. The second is not (for 2n exponent) since only |f|=c is required. f may be a step function oscillating between 2 values of opposite sign, but same magnitude. For odd n, the first is true. The second could have a function (like $\displaystyle sin2\pi x$) where negative part is mirror image of positive part. Last edited by mathman; June 8th, 2017 at 12:48 PM.
 June 9th, 2017, 04:02 AM #10 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 No, $\displaystyle f\neq 0$, it can't change of sign. I think this equality is true.

 Tags fun, integral, result

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jzh Number Theory 4 June 15th, 2012 06:09 AM Solarmew Applied Math 7 May 30th, 2012 07:58 PM ZardoZ Real Analysis 7 June 2nd, 2011 05:39 AM bigsnail Advanced Statistics 0 January 6th, 2010 07:04 AM bgbgbg Calculus 4 October 28th, 2009 04:07 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top