/sci/ - Science & Math » Thread #11733428

8KiB, 346x98, minimizethis.png

View Same Google iqdb SauceNAO

Anonymous Sat 30 May 08:56:52 2020 No.11733428 View Reply Original Report

Quoted By: >>11733469 >>11733861 >>11733983 >>11742587 >>11743266

Find the function f from R to R that minimizes this integral. You must show that the integral exists in the first place.

This is not my homework.

Anonymous

Anonymous Sat 30 May 2020 09:17:57 No.11733469 Report

Quoted By: >>11733492

>>11733428
>not a Kurisu thread
Yeah I don't care.

Anonymous

Anonymous Sat 30 May 2020 09:25:14 No.11733492 Report

Quoted By:

>>11733469
"You should be able to solve this"
Are you happy now?

Anonymous

Anonymous Sat 30 May 2020 10:38:38 No.11733621 Report

Quoted By: >>11733627 >>11742770

It's a troll question. There is no function for which the integral exists.

Anonymous

Anonymous Sat 30 May 2020 10:42:47 No.11733627 Report

Quoted By: >>11733983

>>11733621
>You must show that the integral exists in the first place.
hence, prove your statement.

Anonymous

Anonymous Sat 30 May 2020 10:57:23 No.11733658 Report

Quoted By: >>11734143

$\displaystyle \int_{a-2\infty}^{b+2\infty} \frac{1}{f} d^{3}ba$

Anonymous

Anonymous Sat 30 May 2020 12:21:32 No.11733861 Report

Quoted By: >>11734148

>>11733428
For any f you pick, 2f will give a smaller result.

Anonymous

Anonymous Sat 30 May 2020 13:11:47 No.11733983 Report

Quoted By: >>11734114

>>11733428
>>11733428
>>11733627
if f exists, then $| \int_a^b 1/f | = | \int_0^b 1/f - \int_0^a 1/f |$
if there exist two values x and y such as f(x) is not equal to f(y), then the integral doesn't exist.
and if f is constant, then the integral also doesn't exist.

so f doesn't even exist

Anonymous

Anonymous Sat 30 May 2020 13:51:08 No.11734114 Report

Quoted By:

>>11733983
>if there exist two values x and y such as f(x) is not equal to f(y), then the integral doesn't exist.
how ?

Anonymous

Anonymous Sat 30 May 2020 14:00:11 No.11734143 Report

Quoted By:

>>11733658
It's a joke yes?

Anonymous

Anonymous Sat 30 May 2020 14:01:15 No.11734148 Report

Quoted By: >>11734418

>>11733861
Maybe you can achieve 0 though.

Anonymous

Anonymous Sat 30 May 2020 15:06:15 No.11734418 Report

Quoted By: >>11735743

>>11734148
how can you achieve 0 by summing a strictly positive number?

Anonymous

Anonymous Sat 30 May 2020 17:11:37 No.11735628 Report

Quoted By:

Can it be a distribution? Then 1/delta'(x), where delta'(x) is the derivative of the Dirac delta distribution.

Anonymous

Anonymous Sat 30 May 2020 17:18:49 No.11735743 Report

Quoted By:

>>11734418
Ok right, and since I implicitely disallowed complex functions I guess the challenge is kind of boring.

Anonymous

Anonymous Sat 30 May 2020 22:57:11 No.11742587 Report

Quoted By: >>11742685

>>11733428
>suppose you found a function f(x) that minimizes the integral
>multiply that function by C, where C is a real number greater than 1 (g(x) = C*f(x))
>g(x) either produces a smaller minimum value of the integral, or the minimizing function is not uniquely defined

Anonymous

Anonymous Sat 30 May 2020 23:12:29 No.11742685 Report

Quoted By: >>11742730 >>11742869

>>11742587
Let's accept that it's not unique - find at least one function that minimizes the integral.

Anonymous

Anonymous Sat 30 May 2020 23:23:31 No.11742730 Report

Quoted By:

>>11742685
it's not a minimum if there's a function that integrates to a smaller number

Anonymous

Anonymous Sat 30 May 2020 23:33:24 No.11742770 Report

Quoted By:

>>11733621
I'm learning integrals. Is there anything in the picture that gives it away as a troll?

Anonymous

Anonymous Sat 30 May 2020 23:53:44 No.11742869 Report

Quoted By:

>>11742685
If the minimizing function is not uniquely defined, then the minimum must be 0.
But, because there is an absolute value sign in the integrand, the only way for the whole integral to be 0 is for 1/f(x) to be 0 almost everywhere. There is no function f(x) that satisfies this requirement, so there are no minimizing functions f(x).

Anonymous

View Same Google iqdb SauceNAO tripleintegral.png, 477KiB, 932x917

Anonymous Sun 31 May 2020 01:43:53 No.11743266 Report

Quoted By:

>>11733428

Anonymous

Anonymous Sun 31 May 2020 02:13:30 No.11743395 Report

Quoted By:

It's a bit trickier than what people are saying. It reduces to finding a function $f$ that is not 0 almost everywhere whose inverse integrates 0 at every interval. Well maybe it is not trickier, but for a precise proof I really don't see a way without using measure theory. The thing is becuase the integral is 0 on all the intervals it must be 0 on all measurable subsets and so it must be equal to the 0 function almost everywhere.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches