# How can I prove the general formula for Taylor Series?

I am trying to prove the general form for practice and I am having some trouble with it. The question in my book is "Suppose f is infinitely differentiable on an open interval I containing the point x = c, and there exists a constant M > 0 such that |$f^{(n)} (x)$| $\leq$ M for all x $\in$ I and all n $\geq$ 0. Show that $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!} (x-c)^n$$ My thoughts were to use the definition of a power series $$\sum a_n (x-c)^n$$ and then to make the assumption that the power series converges with $|x-c| < M$ then to let $f(x)$ to be equal to our power series than evaluating each n-th derivative at $x=c$ than to make the statement that if we repeat this process infinitely many times we get the desired form of $a_n = \frac{f^{(n)}(c)}{n!}$

This is what I am unsure about because I am not sure if this would be strong enough to prove the Taylor's form especially with trying to get the form of $a_n$ any suggestions would be appreciated. Cheers

Your approach is a reasonable one to gain intuition. However, it tacitly assumes that one can differentiate the power series term by term as if it were a finite sum. The fact that it is not a finite sum creates a need to apply a more rigorous way forward. To that end we proceed.

Note that from Taylor's Theorem, we can write for a $n$ times differentiable function at $x=c$

$$f(x)=\sum_{k=0}^n\frac{f^{(k)}(c)}{k!}(x-c)^k+R_n(x,c) \tag1$$

where the remainder $R_n(x,c)=o(|x-c|^n)$ as $x\to c$.

From $(1)$ we have $R_n(x,c)=f(x)-\sum_{k=0}^n\frac{f^{(k)}(c)}{k!}(x-c)^k$,

If for $x\in (c-a,a+c)$, $f$ and all of its derivatives exist, and if $x_0\in (c-a,c+a)$, then $f$ is represented by its Taylor series at $x_0$ if and only if

$$\lim_{n\to \infty}R_n(x,c)=0$$

There are a variety of explicit forms for the remainder term and estimates thereof. The mean-value form of the remainder,

$$R_n(x,c)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-c)^{n+1}$$

for some $\xi\in(c,x)$. If all derivatives of $f$ are bounded above with $|f^{(k)}|\le M$ for $|x-c|< a$, then

$$|R_n(x,c)|\le M\frac{a^{n+1}}{(n+1)!}$$

and the Taylor series converges uniformly to $f$. In such as case, we can write

$$f(x)=\sum_{k=0}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k$$

for $x\in (c-a,c+a)$

• Thank you for your answer, I see where I was mis-thinking I was thinking that the limit of $R_n (x,c)$ going to zero for n going to infinity showed that it converged to the Taylor polynomial of n-th degree not the taylor series. Mar 17 '17 at 12:02