Why $f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)\, x^n}{n!} $?

Question

Let $f:\mathbb R\longrightarrow \mathbb R$ a function that is $\mathcal C^\infty(\mathbb R) $. Suppose $$\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}\,x^k$$ converge and its radius is $R$. Why $$f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}\,x^k$$ for all $x\in ]-R,R[$ ?

I know that for a fixed $n$, $$f(x)=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}\,x^k+o(x^n),$$ in a neighborhood of $0$, but a priori it won't be correct for $x$ too large (take for example the sinus or exponential, we have $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!},$$ for all $x\in \mathbb R$, not only in a neighborhood of $O$. So why can we extend the formula to all $x\in\mathbb R$ ?

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Ok, let forget $e^{-1/x^2}$. Consider for example $\sin(x)$. Why $$\sin(x)=\sum_{k=0}^\infty \frac{(-1)^{n+1}}{(2n)!}x^{2n},$$ for all $x\in\mathbb R$ whereas $$\sin(x)=\sum_{k=0}^n \frac{(-1)^{n+1}x^{2k}}{(2k)!}+o(x^{2n})$$ in a neighborhood of $0$ only.

Answer 1

This is false. For example, if $f(x)=\exp(-1/x^2)$ for $x\neq 0$ and $f(0)=0$, then $f$ is $C^\infty$ with $f^{(k)}(0)=0$ for all $k$. In particular, $$\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}\,x^k$$ converges for all $x$ since every term is $0$. However, it is not true that $$f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}\,x^k$$ for any $x$ besides $x=0$.

(It happens to be true for certain functions like $f(x)=e^x$, but you must prove it using particular properties of those functions. How you prove it in the case $f(x)=e^x$ depends on what your definition of $e^x$ is; often $e^x$ is simply defined to be $\sum\frac{1}{k!}x^k$ so it is true by definition.)

Answer 2

This is not true. The function $$ f(x)=\cases{0& if $x=0$\\e^{-1/x^2}& otherwise} $$ is $C^{\infty}$, and for any $k\in\Bbb N$ we have $f^{(k)}(0)=0$, so the series converges for all $x$, but only converges to $f(x)$ for $x=0$.

Answer 3

This is true only for what is called analytic functions (sums of power series).

It happens that for functions of a complex variable, being differentiable (in the complex sence) implies infinitely differentiable and even analytic, whereas for functions of a real variable, infinitely differentiable does not imply analytic. The function $\mathrm e^{-\tfrac1{x^2}}$ mentioned in other answers is an example of a $\mathcal C^\infty$ function which is not analytic.

Answer 4

It depends on the size of the radius.

For some $\sum_{k=0}^{\infty}a_k\,x^k$ you will have $r=\lim_{k\to\infty}|\frac{a_k}{a_{k+1}}|$.

Calculating the radius for $e^x$ provides $r=\infty$.

But then consider the function $f(x)=\frac{1}{1-x}$. Its MacLaurin series $\sum_{k=0}^{\infty}x^k$ has a radius of convergence of $r=1$. But that given fraction function clearly is well-defined beyond too.

Just plot the graph of that latter function and you would understand that the series would be well-defined even for $x\in [-1,1)$.

--- rk