/sci/ - Science & Math » Thread #12989790

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Anonymous Sat 17 Apr 14:03:12 2021 No.12989790 View Reply Original Report

Quoted By: >>12989837 >>12989861 >>12989952 >>12989979 >>12990008 >>12991689 >>12991714 >>12993975 >>12993978

Let's see who is actually smart on sci, find a general solution to this functional differential equation.

f ''(x) = f(f(x)) and there are general solutions

Anonymous

Anonymous Sat 17 Apr 2021 14:20:00 No.12989837 Report

Quoted By: >>12989895

>>12989790
no one on sci is smart enough lol

Anonymous

Anonymous Sat 17 Apr 2021 14:30:10 No.12989861 Report

Quoted By: >>12989895

>>12989790
Is this what OP means by this differential equation?

In other words find a function whose 2nd derivative is equal to the same function evaluated at all the points that make that function up.

For example f(f(x)) = ? when f(x) = x^2
is for x=2 ----> f(f(2)) ----> f(2) = 4 ----> f(4) = 16 and this has to be equal to its 2nd derivative

Anonymous

Anonymous Sat 17 Apr 2021 14:41:51 No.12989895 Report

Quoted By: >>12989923

>>12989861
>>12989861
If f(x) = x^2 then f(f(x)) = (x^2)^2 = x^4. Note that as (x^4)'' = (4x^3)' = 12 x^2 =/= x^4, f(x) = x^2 isn't a solution.

You're looking for a class of functions where evaluating it twice is the same as taking the 2nd derivative.

>>12989837
You'll have to wait for more than 17 minutes to get someone to do your homework for you.

Anonymous

Anonymous Sat 17 Apr 2021 14:53:10 No.12989923 Report

Quoted By:

>>12989895
the solutions must be complex valued, right?

Anonymous

Anonymous Sat 17 Apr 2021 15:04:18 No.12989952 Report

Quoted By: >>12989995

>>12989790
Is there no math for solving functional equations other than guessing? Has it been proven that there's no general methods for solving them? Does functional analysis has anything to do with it?

Anonymous

Anonymous Sat 17 Apr 2021 15:13:12 No.12989979 Report

Quoted By:

>>12989790
f(x) if second differential = 1st nested

Anonymous

Anonymous Sat 17 Apr 2021 15:18:22 No.12989995 Report

Quoted By: >>12990038

>>12989952
you need what mathematicians call a ansatz.

Anonymous

Anonymous Sat 17 Apr 2021 15:24:14 No.12990008 Report

Quoted By: >>12990082

>>12989790
do u guys want me to give you a hint?

Anonymous

Anonymous Sat 17 Apr 2021 15:35:11 No.12990038 Report

Quoted By: >>12990443

>>12989995
>ansatz
midwit cope in other words

Anonymous

Anonymous Sat 17 Apr 2021 15:50:38 No.12990082 Report

Quoted By: >>12991693 >>12991798

>>12990008
solutions must be of the form f(x)=a * x ^ b

a and b are both complex

Anonymous

Anonymous Sat 17 Apr 2021 17:25:23 No.12990424 Report

Quoted By: >>12991652

there was a stack exchange about this

Anonymous

Anonymous Sat 17 Apr 2021 17:28:29 No.12990443 Report

Quoted By:

>>12990038
Reminder that picking an ansatz or checking-and-guessing ARE, in fact, logically rigorous

Anonymous

Anonymous Sat 17 Apr 2021 21:25:49 No.12991652 Report

Quoted By:

>>12990424
will stack exchange solve the navier stokes problem?

Anonymous

Anonymous Sat 17 Apr 2021 21:36:13 No.12991689 Report

Quoted By: >>12993711

>>12989790
f(x) = 0 :^)

Anonymous

Anonymous Sat 17 Apr 2021 21:36:45 No.12991693 Report

Quoted By: >>12991826

>>12990082
So, ab(b-1)x^(b-2) = a(ax^b)^b

No general solutions, sorry OP.

Anonymous

Anonymous Sat 17 Apr 2021 21:40:51 No.12991714 Report

Quoted By:

>>12989790
this is an interesting question because you are basically asking when a second order recursion it a second order derivative. logistic equation comes to mind. I could probably figure this out given enough time but... I got stuff to do. do your own homework.

Anonymous

Anonymous Sat 17 Apr 2021 21:58:34 No.12991798 Report

Quoted By:

>>12990082
Then you're fucking retarded op, because not even f'(x) = f(x) is of that form, and that's just e^x.

Anonymous

Anonymous Sat 17 Apr 2021 22:05:26 No.12991826 Report

Quoted By:

>>12991693
a^b/b(b-1) = x^b^b/x^(b-2) = x^(b^b-b+2)

x = (a^b/(b(b-1)))^-(b^b-b+2)

Where's my Field Medal?

Anonymous

Anonymous Sat 17 Apr 2021 22:07:49 No.12991835 Report

Quoted By:

f(x)=0

Anonymous

Anonymous Sun 18 Apr 2021 05:58:14 No.12993711 Report

Quoted By:

>>12991689
Based trivial solution

Anonymous

Anonymous Sun 18 Apr 2021 07:25:14 No.12993950 Report

Quoted By: >>12993968

$<br />f(x)=(-2)^{\frac{1+i\sqrt7}{4}}x^{\frac{1-i\sqrt7}{2}}<br />$

Anonymous

Anonymous Sun 18 Apr 2021 07:34:32 No.12993968 Report

Quoted By: >>12993974

>>12993950
christ alive how did i fuck that up so badly
$\frac{1}{\sqrt{6}x^2}$

Anonymous

Anonymous Sun 18 Apr 2021 07:37:40 No.12993974 Report

Quoted By:

>>12993968
no wait i think the first one was right, exponents still get me bros

Anonymous

Anonymous Sun 18 Apr 2021 07:38:49 No.12993975 Report

Quoted By:

>>12989790
the identity function f(x) = x

Anonymous

Anonymous Sun 18 Apr 2021 07:39:35 No.12993978 Report

Quoted By: >>12993983

>>12989790
Something of the form 1/x2

Anonymous

Anonymous Sun 18 Apr 2021 07:41:54 No.12993983 Report

Quoted By:

>>12993978
good guess, but then f(f(x)) would have an x^4 term but the derivative would be x^-4

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