So I have a test next week and I saw this question with no answer and I would like to some help.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ infinitely differentiable

and let $\sum _{n=0}^{\infty} a_nx^n$ the taylor series around $0$

and $f(1) = \sum _{n=0}^{\infty} a_n$

Which one of the following is correct?

  1. $\sum _{n=1}^{\infty} \frac{(-1)^na_n}{n}$ is convergent

  2. $\sum _{n=0}^{\infty} (-1)^na_n$ is convergent

  3. $\sum _{n=1}^{\infty} \frac{(-1)^na_n}{2^n}$ is absolute convergent

  4. $f(0.5) = \sum _{n=0}^{\infty} \frac{a_n}{2^n}$

So as far as I know, the Taylor series around $0$ is actually $\sum _{n=1}^{\infty} \frac{f^{n}(0)}{n!}x^n$

My question is this: why do they mention that $f(1) = \sum _{n=0}^{\infty} a_n$? Is it not obvious?

If not (and I am guessing it is not), how does it help me here ?

I would like (and be very happy) for some instruction here.

I took $f = e^x$ meaning the taylor series is $\sum _{n=1}^{\infty} \frac{1}{n!}x^n$

But I have no idea if e = $\sum _{n=1}^{\infty} \frac{1}{n!}$

And even though with this example both $1,2,3$ are true.

  • $\begingroup$ 2 is clearly wrong $\endgroup$
    – resgh
    Sep 3 '13 at 8:11
  • $\begingroup$ Finding an example with converges doesn't help you a lot. Finding an example which diverges does; it directly answers the corresponding question. Think about functions with "problems", e.g. arctan(x) $\endgroup$
    – MSalters
    Sep 3 '13 at 8:12
  • 1
    $\begingroup$ The point of the exercise seems to be that even when the Taylor series of a smooth function is convergent, it needn't converge to $f$. Hence it's not redundant to say $f(1)=\sum a_n$, nor is it obvious anywhere else (except that we know $\sum a_n (0.5)^n$ is convergent, as a property of power series). $\endgroup$ Sep 3 '13 at 8:17
  • $\begingroup$ With regard to your second question, the Taylor series for $e^x$ converges along the entire line, hence you are correct. This is a possible method for calculating $e$, in fact (although not the most efficient). $\endgroup$ Sep 3 '13 at 8:21
  1. Is incorrect. Take $a_n=\frac{(-1)^n}{\log n}$, and use this theorem of Borel's to find $f$. This fails to ensure that $f(1)=\sum a_n$, so we must correct it by taking, for example, the Cauchy function $g(x)=e^{-1/x^2}$, which has a vanishing McLauren series, and defining $h(x) = f(x) - e\left(f(1) - \sum a_n\right)g(x)$.

  2. Is incorrect, for the same counterexample as (1).

  3. This one is correct. We were given that $\sum a_n x^n$ converges at $x=1$, hence as a power series we know the radius of convergence $r\geq 1$, and the series converges absolutely on $(-1,1)$ (and uniformly on compacta), and in particular at $x=-\frac{1}{2}$.

  4. This is incorrect. The reason is that for a smooth function to have a convergent Taylor series is insufficient for that series to converge to the function itself. An application of the standard example of a smooth transition function which vanishes outside of $[0,1]$ such that $g(\frac{1}{2})=1$ will give an example of a smooth function whose Taylor series is zero, converging to the function outside $(0,1)$ (and in particular at $x=1$) but not in $(0,1)$.

As an aside, I'd like to emphasize that the question is designed to bring to light the fact that Taylor series needn't converge to the original functions. Viewed this way, note that we can quickly do away with $(2)$ on the basis that $(1)$ would then also be true (by Abel's test, for example), and keeping in mind the aforementioned point we should be suspicious of $(4)$. (I admit that $(1)$ is tricky--if not for the counterexample then for the fact that these coefficients are normalized derivatives of some smooth function.)

Also, as I mentioned in comments above, the Taylor series of $f(x)=e^x$ does converge to $f$ along the real line. Put differently, for all $x\in\mathbb{R}$ we have $e^x = \sum_{n\geq 0}\frac{x^n}{n!}$.


why do they mention that $f(1)=\sum_{n=0}^\infty a_n$ is it not obvious ?

No, that follows only if the Taylor series of $f$ converges to $f$. Which is not implied by your conditions...


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