Fourier cosine series expansion of $$f(x)=1,~~~ x\in (0,\pi)$$
Hint is "thought is better than calculation".
Asked
Active
Viewed 1,921 times
1
-
3You're tasked with expanding $f$ in terms of $1, \cos x, \cos 2x,\ldots$. – Cameron Williams Nov 30 '15 at 03:43
-
Correct, but I'm drawing a blank on what to do. I tried normal expansion methods but got nowhere. – MC989 Nov 30 '15 at 03:51
-
1The Fourier series is unique.. You need to represent the $1$ function as a linear combination of $1,\cos x,\cos 2x,\ldots$. What is the obvious way to do this? – Cameron Williams Nov 30 '15 at 03:53
-
So, do I just let the fourier series for cosine equal 1? – MC989 Nov 30 '15 at 04:01
1 Answers
1
So, do I just let the Fourier series for cosine equal $1$?
Yes, the cosine expansion of this function has just one nonzero term: $1$. The rest of coefficients are therefore zero.
This is something that always happens when you expand a vector in a basis which contains that vector. If your basis of $\mathbb{R}^2$ is $\{(2,3), (1,-5)\}$, and you want to expand the vector $v=(1,-5)$ in this basis, the expansion is just that: $$(1,-5) = 0\cdot (2,3)+1\cdot (1,-5)$$
Indeed, this particular choice of coefficients evidently works, and since the coefficients are unique, any other method would produce same ones.