Easy bounty time!
Hello! I would like to get some help with the next problem:
I'm trying to learn how to develop the real function into the power series. After reading my book, i want to check if i understood correctly what is the process of the develpoing the given real function. So, this is how i understand what i need to do in order to develop the given real function $f$:
1) Check if the given function $f$ is infinitely diferentiable and where.
2) Chose the point $x_0$ in which we are going to develop the function.
3) Check if the given function $f$ is continuous with all of its derivatives, up to the $n$-th order, in some neighborhood of the point $x_0$. If this is fulfilled, we have that we can write $f(x) = P_n(x, x_0) + R_n(x)$.
4) Check if $\lim_{n \to \infty} R_n(x) = 0$.
5) Check if $f(x_0) = P_n(x, x_0) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n$. This means that we have to check the convergence of the Taylor series that we got and to calculate the sum of the series if the series is convergent.
6) If all conditions are fulfilled, than we can say that function $f$ can be developed into the power series $\sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n$ and we can call that function analytic.
Please, could you tell me if i understand this process corectly and if not where i make a mistake?
I don’t know whether there exists a canonical notion or a definition of a power series expansion $S(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ of a function $f:\Bbb R\to\Bbb R$ at a point $x_0\in\Bbb R$. So I guess that we need that $S(x)$ converges to $f(x)$ at least for $x=x_0$ and, in order to relate to $f$ not only a value of $S$ at a point $x_0$, but also its coefficients, a (at least pointwise) convergence of $S(x)$ to $f(x)$ for all $x$ from some open neighborhood $U$ of $x_0$. Remark that even the latter condition is formally weaker than analyticity of $f$ on $U$.
If a function $f$ is analytic then it is infinitely differentiable, but the converse is not generally true. A well-known example is a bump function $\lambda(x):\Bbb R\to\Bbb R$ such that $\lambda(x)=\exp\tfrac 1{x^2-1}$ for $|x|< 1$ and $\lambda(x)=0$, otherwise. This is so because $\lambda^{(n)}(0)=0$ for all $n\ge 0$ and a remark that if $f$ and $g$ are two analytic functions defined on the same connected open set $U$, and if there exists an element $x_0\in U$ such that $f^{(n)}(x_0) = g^{(n)}(x_0)$ for all $n\ge 0$ then $f(x) = g(x)$ for all $x\in U$.
For an analytic function, the coefficients $a_n$ can be computed as $a_n=\frac {f^{(n)}(x_0)}{n!}$, which means that every analytic function is locally represented by its Taylor series.