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Does a Taylor series always converge to its generating function? Can you please explain?

Also, I've encountered an exercise in my Math book.

What is the Taylor series generated by a function $ f = f(x)$ at a point $x = a$? What information do you need about f to construct the series? Give an example.

Thank you very much in advance! I'll be forever thankful!

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In general: no! A real valued function is called analytic if it is given by a convergent power series. (A Taylor series being an example of a power series.) The canonical counter-example is $f(x) = e^{-1/x^2}.$ You find that $f(x) \to 0$ as $x \to 0$ and, moreover, all of its derivatives tend to zero as $x \to 0$. The Taylor series of $f(x) = e^{-1/x^2}$ at $x=0$ is identically zero, while the function is not. Such a function is called a flat function. In fact, analytic functions form a very "small" subset of functions.

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Thank you very much! I understood it prefectly the way you explained it. – Grozav Alex Ioan Nov 19 '12 at 20:56
@GrozavAlexIoan You're very welcome! I'm really glad I could help. – Fly by Night Nov 20 '12 at 16:09

The Taylor series of a function $f(x)$ around $x=a$ does not necessarily converge anywhere except at $x=a$ itself, and if it converges the value at $x$ is not necessarily $f(a)$. What you need for it to converge to $f(x)$ for $|x-a| < r$ is that $f$ is analytic in the disk $|x-a| < r$ in the complex plane.

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No it doesn't, for expample take $$f:\mathbb R \to \mathbb R$$ $$ f(x) = 0~ \forall x \leq 0$$ and $$f(x) = e^{\frac 1 {x^2}}~\forall x > 0$$. You can show $f\in C^\infty(\mathbb R,\mathbb R)$ and $f^{(n)}(0)= 0 \forall n \in \mathbb N$. So the Taylorseries expansion in $0$ is the 0-function, but this doesn't describe $f$ for $x > 0$.

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As for the second part of your question, to construct the Taylor's series you need to know the value of the function and all of its derivatives at $x=a$.

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Thank you very much! Was hoping someone would answer that. – Grozav Alex Ioan Nov 19 '12 at 21:53

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