/sci/ - Science & Math

[Anonymous No.8873946]

Let a function meets all reuirements to have a taylor seies and a fourier series. Then is there a way to equate the two series and derive one from the other?

Inb4 local and global handwaving argument.

[Anonymous No.8873951]

>Let a function meets all reuirements to have a taylor seies and a fourier series.
If a function has a Taylor series, that Taylor series is unique.
If a function has a Fourier series, that Fourier series is unique.
So if that function has both then of course you can equate them.

If you feel unsatisfied with that answer then maybe you should be more precise with what you mean when you say
>Let a function meets all reuirements [sic]

[Anonymous No.8873964]

>>8873951

Cool. Let me fix.

Let a function meet all reuirements to have a taylor seies and a fourier series. Then what is the derivation from one series to the other?

Taylor -> Fourier

Or

Fourier -> Taylor


Inb4 local and global handwaving argument

[Anonymous No.8873973]

>>8873946

Compute <x^n|e^inx>

[Anonymous No.8873984]

>>8873973

What does that even mean?

Do you know a book that explicitly states the derivation? If so, what is the name of the book?

[Anonymous No.8874013]

>>8873973

Bra ket notation.

Its just the inner product between two functions. Id help but i dont know.

[Anonymous No.8874022]

>>8873973

Computing a value doesnt answer the question of the derivation.

[Anonymous No.8874031]

>>8873984
Not that guy but off the top of my head, I know Kraut's "Fundamentals of Mathematical Physics" has a chapter on infinite series which discusses the bra-ket construction.

[Anonymous No.8874040]

>>8873973
Is it [math] \sum_k x^k \exp(ikx) [/math]?

[Anonymous No.8874049]

>>8874031

Thanks but i meant a book that shows the derivation between the two series, not the bra ket notation.

[Anonymous No.8874062]

>>8873946
Yes they should. As >>8873951 said they are unique.

So if you take the fourier series and expand out the sine's and cosine's in their own power series, you will ultimately get a power series for the whole function that should be the same as the taylor series. (Asumming all these series converge in the same nbhd and such)

[Anonymous No.8874072]

>>8874062

That was my original plan. Using brute force. But the fourier coefficients are coupled with the terms and makes that rather difficult.

I figured there is a more elegant approach. I may be wrong.

[Anonymous No.8874076]

>>8874072
It is not a calculation you actually want to do, it is just that you can prove that it can be done using the the properties of formal power series (again assuming everything converges nicely).

[Anonymous No.8874088]

>>8873946
It is simply change of basis

Going from Taylor to Fourier
1. Compute Fourier Series of [math]x^k[/math] for each [math]k[/math]
2. To find the coefficient of [math]e^{inx}[/math] in the Fourier Series simply collect the contributions of each term of the Taylor series to it

Going from Fourier to Taylor
1. Compute Taylor Series of [math]e^{ikx}[/math]
2. To find the coefficient of [math]x^n[/math] in the Taylor Series simply collect the contributions of each term of the Fourier series to it

[Anonymous No.8874112]

>>8874088

If you showed it here, i dont think you would use the word "simple" as you did.

[Anonymous No.8874124]

>>8874049
But that's what the bra-ket notation is, the inner product that it denotes is the coefficient of the Taylor series in terms of the Fourier series coefficients, and vice versa.

[Anonymous No.8874197]

>>8874112
f=[math]a_0+a_1*x+a_2*x^2+....[/math]
f=[math]b_0+b_1*e^{ix}+b_2*e^{2ix}....[/math]

[math]a_n=\sum_k <x^n|e^{ikx}>*b_k[/math]
[math]b_k=\sum_n <e^{ikx}|x^n>*a_n[/math]

[Anonymous No.8874207]

[math]
f=a_0+a_1*x+a_2*x^2+....
\\
f=b_0+b_1*e^{ix}+b_2*e^{2ix}....
\\
a_n=\sum_k <x^n|e^{ikx}>b_k
\\
b_k=\sum_n <e^{ikx}|x^n>a_n
[/math]

[Anonymous No.8874215]

>>8874112
It's not that bad

The [math]e^{inx}[/math] coefficient of [math]x^k[/math] is

[math]k \left( \frac{k-1}{in} \left( \frac{k-2}{in} \left( ... \left( \frac{1}{in} + m_1 \right) + ... \right) + m_{k-2} \right) + m_{k-1} \right) \frac{-1^{n+1}}{in} [/math]

where [math]m_k = \frac{1}{2\pi} \int_{-\pi}^{\pi}{x^k dx} [/math]

The [math]x^n[/math] coefficient of [math]e^{ikx}[/math] is trivial

[Anonymous No.8874225]

>>8873946
having a taylor series (C^inf) and being equal to the taylor series (analytic) are two different things

[Anonymous No.8874231]

>>8874207

[math] \color{YellowGreen}{>not \ \ using \ \ \langle \ \ \rangle} [/math]

[Anonymous No.8874384]

>>8874124
>Brain-let notation
Only Physicists could invent such a hack job of good mathematics

[Anonymous No.8874408]

>>8873946
you need to be more specific about what you mean by "having" a taylor series. for example, it's possible to construct a smooth function f such that the taylor series around x=0 is identically zero, but f(x)>0 for all x>0. so it's not possible to determine the fourier series just given the taylor coefficients at a point.

[Anonymous No.8874439]

>>8874408

You cant compute the fourier series for the function x^2?

[Anonymous No.8874443]

>>8874384

bra ket notation is just a nice way of expressing the interplay between vectors and one forms, which is a pretty necessary formality when you're dealing with complex vector spaces.

[Anonymous No.8874449]

>>8874439
huh?

[Anonymous No.8874501]

>>8873946
what is that image saying? What is A_l ?

[Anonymous No.8874506]

>>8874408
consider exp(-1/x^2)
even if it is smooth, it still has an essential singularity at 0. meaning, the taylor series is valid for |x|<0 so nowhere