2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Robert
    Mar 17 '17 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.