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Why can an arbitrary function be defined by a power series?
On Wikipedia:
http://en.wikipedia.org/wiki/Taylor_series#Derivation
It says that, "An arbitrary function may be defined by a power series" for showing that a taylor series of a function exists. Why can an arbitrary function be defined by a power series? I understand how this is true for polynomials, but not for functions like sin(x).
Thanks.
If a complex function is differentiable on an open set V and, for some z_0 ∈ V, there is r > 0 such that the closure of the disk D(z_0, r) is in V, then, for every z ∈ D(z_0, r), f is given by a power series about z. f(z) = f(z_0) + f'(z_0) z + f''(z_0)/2! z² + f''(z_0)/3! z³......
If a complex function is differentiable on an openset of the complex plane, then all of its derivatives exist on this set. That's whay Wilkipedia says an arbitrary function can be given by a power series. But when it says arbitrary, it doesn't mean any function at all, but just those which are differentiable, called analytic (or holomorphic) functions.
f(x) = sin(x) is differentiable on the entire complex plane. And indeed, for every complez z, sin(z) = z - z³/3! + z^5/5! - z^7/7!.....
On the real line, the existence of the 1st derivatives doesn't imply the existence of the other derivatives. But if all the derivatives exist on an open interva, then Taylor theorem applies. If the Lagrange Remainder converges to 0, on this interval, the f can indeed be given by a power series. Such functions are called real analytic.
It's not possible to get into details here, this subject is vast. I suggest you consult a book on Real or Complex Analysis.
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