Question

(Fourier Series) if I take the Fourier Series of a constant function (not piecewise-defined or anything, just a constant function, like pi/4), does the integral I'd typically use to evaluate it become indefinite? @dan815

Answer 1

(Or otherwise how would I deal with the bounds presumably being -infty<x<infty, or what? @zepdrix

Answer 2

u wanna basically define a function with infinite period right

Answer 3

this is what fourier transforms is all about

Answer 4

i have a good example for you mendicant

Answer 5

\[F _{x}[1](k)=\int\limits_{-\infty}^{\infty}\exp[-2\pi ikx] dx \rightarrow \delta(k)\]

Answer 6

so, it is an infinity infinity type of integral, and that delta symbol is the dirac delta function

Answer 7

the exp{} essentially symbolizes e^[}

Answer 8

and thats it!

Answer 9

The terms of your Fourier seris will look like this, I don't think we should try to dress it up to be something fancier than it really is, all the other coefficients are 0. \[\Large 0 \sin(0x)+\frac{\pi}{4}\cos(0x)+0\sin(1x)+0\cos(1x)+\cdots\]

Answer 10

thats true xD

Answer 11

suppose i ask u to expand it as a sin series

Answer 12

It's an even function, so why would anyone expect you to represent it as an odd function?
\[\ \Large f(x) =C x^0\] satisfies
\[\ \Large f(x) = f(-x) \] is true for \[\ \Large f(-x) = C(-x)^0 =C(-1)^0x^0= f(x) \] Is that fair to say?

Answer 13

If f(x) only has even terms it will only be expressible in terms of cosine terms. This is just a general fact. If you want to express an even function in terms of odd functions it's actually very similar to trying to represent an odd number by only adding even numbers together... It's nonsense.

Answer 14

to torture you and try to build up the dirac function ofc

Answer 15

@Rizags can you elaborate on this function, i tried to integrate and was not successful
$$
F _{x}[1](k)=\int\limits_{-\infty}^{\infty}\exp[-2\pi ikx] dx \rightarrow \delta(k)

$$