/sci/ - Science & Math

Previous Thread >>10603810

If you write x as cos(a), then you get [math] f(\cos{2a}) = 2\cos(a)f(\cos(a)) [/math]. Using this repeatedly, [math] f(\cos{2^na}) = 2^n f(\cos(a)) \prod_{k = 0}^{n-1} \cos(2^k a) [/math]. No idea what to do after this though.

thinly veiled homework thread

>>10607483
cos(a) only takes values in [0,1] so this would only prove the case for that range.

>>10607599
>cos(a) only takes values in [0,1]
never go full retard

If we assume f(x) is cos(x) aren't we losing generality?

Seems like a bad way to "solve" it.

>>10607620
Why would you assume f(x) to be cos(x)

>>10607175
Trivial

I don’t even see why this would be true

>>10607782
cope

>>10607604
lmao that's a hard yikes from me

>>10607604
woops meant [-1,1]. Didnt read the "for x in [-1,1]" part.

Anyway, solved it. Just compute f'(x) and show that it's zero and find a point where f(x) = 0

>>10608228
why is f differentiable?

>>10607175
I put more time on this one than I should.
Anyway, I think a solution is as follows:
First replace x by -x in the starting equation to see that f is odd
f(-x)=-f(x) (at 0 take limit).

Start from this observation:
>>10607483
To warm up take [math] a=\pi/(2^n+1)[/math].
This number comes from the equation [math]2^na=\pi-a[/math].
Then [math] -f(cos(a))=f(cos(a))2^ncos(a)...cos(2^{n-1}a)[/math].
But [math]2^{n-1}a<\pi/2[/math] so the factors are positive,
so the only posibility is [math]f(cos(a))=0[/math].
The idea now is to see that [math]f(cos(ka))=0[/math],
for k odd.
As before from
[math]2^nka=k\pi-a[/math],
[math]-f(cos(a))=f(cos(a))2^ncos(ka)...cos(2^{n-1}ka)[/math].
It turns out that the sign of cos(ka)...cos(2^{n-1}ka) is always positive.

This is pretty ugly to see: first by reasons of parity if r<n-1 then
cos(k2^ra) has the same sign as cos(k2^r/2^n*pi)
("can replace the denominator in a by 2^n").
Next the set of exponents r such that cos(k2^r/2^n*pi) depends on the
expression of k in base 2. If k=(1 blocks)(0-blocks)(1-blocks)...(1-blocks),
then at the start and end of each 1 block (except the last one) there will be
an negative exponent. For the last add no extra exponents if this block has lenght 1, and two extra (including r=n-1) if it has lenght more than one. In any case this is even so we are done.

This numbers [math]cos(k/(2^n+1)\pi)[/math] are dense so this finishes the thing.

>>10608290
>Next the set of exponents r such that cos(k2^r/2^n*pi) depends on the
I mean the set of exponents r such that cos(k2^r/2^n*pi) is negative.

>>10607175
f(2x^2-1) is a even function of x, if the equality is to hold, then 2x*f(x) must also be even, and so f(x) is odd.
The fourier transform of the left side is 4*F(f(x)/|x|) and of the right side is 2*F(x*f(x)) where F(f) is the fourier transform of F. if f != 0, those are clearly different, for the left one is a odd function of x and the right one is a even function of x. So the respectives fourier transforms are purely imaginary and purely real, respectively. Thus f is 0.
I probably messed something up

>>10608572
Because we are only interested in the interval [-1,1] we can replace f by a function f* with supp(f*) = [-1,1] and the required transforms then will exist

>>10607863
I didn’t say it wasn’t true, baka

Aren't cirno problems supposed to be the easy ones?