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998244353 | 7 months ago
You can represent any function f: [-pi, pi] -> R as an infinite sum
f(x) = sum_(k = 0 to infinity) (a_k sin(kx) + b_k cos(kx))
for some coefficients a_k and b_k as long as f is sufficiently nice (I don't remember the exact condition, sorry).This is very useful, but the functions sin(x), sin(2x), ... , cos(x), cos(2x), ... don't constitute a basis in the formal sense I mentioned above as you need an infinite sum to represent most functions. It is still often called a basis though.
seanhunter|7 months ago
The convergence criteria for Fourier series vary depending on how strongly you need convergence but I think basically if a function is differentiable on the interval you care about then the Fourier series provably converges on that interval and otherwise if it has jump discontinuities and that sort of thing, then depending on whether it is square-integrable or a bunch of other properties) you can prove weaker forms of convergence (absolute, pointwise etc).
To address your comment I don’t see why an infinite sum prevents something being a basis. In fact I would specifically say that can’t be true because then there would never be a basis for any infinite-dimensional space- any time you want to take an inner product in such a space you need an infinite sum, and you need such an inner product to construct the basis. A sibling comment pointed me in the direction of a Hilbert basis, which seems to be what I was thinking of.