0

Let's say we have some function defined over an interval and we write down its Taylor or Maclaurin series around some internal point there.

If the series converges for example via ratio test (in that same interval), can we claim its sum is f(x)?

I think we can because the series converges iff the residual term tends to zero. But if that happens, then the sum is f(x). Right?

But in one very good textbook I see this:

"Our derivation of the binomial series shows only that it is generated by (1 + x)^m and converges for |x|<1. The derivation does not show that the series converges to (1 + x)^m. It does, but we leave the proof to Exercise 58."

?!

I don't understand this. Why do we have to prove something additional here?

Sorry for the bad formatting, I am typing on my phone.

Edit: See please my latest comments

peter.petrov
  • 12,963
  • 1
    No, a series converging is different from converging to the original function. See my answer in What's the difference between the "remainder" and "radius of convergence" for Taylor series that converge for all x? – peek-a-boo Jan 01 '22 at 19:04
  • But it is not any series, it is the Taylor series written from the same function. – peter.petrov Jan 01 '22 at 19:05
  • my response is still the same. Read that answer very carefully. As an explicit counter-example, the function $f(x)=e^{-1/x^2}$ if $x\neq 0$ and $f(0)=0$ is a $C^{\infty}$ function on $\Bbb{R}$, with $f^{(n)}(0)=0$ for all $n\geq 0$. So, the Taylor series about the origin is $T_{f,0}(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n=0$, i.e the series has infinite radius of convergence, yet for $x\neq 0$, $T_{f,0}(x)=0\neq f(x)$. EDIT: LOL I realized I gave this exact counterexample in that answer, so you you didn't even read my link prior to commenting. – peek-a-boo Jan 01 '22 at 19:07
  • I didn't but I will. I have been thinking on this for quite a while now. Thanks – peter.petrov Jan 01 '22 at 19:11
  • See the same example from a year earlier. A much easier example is the absolute value function: expand on $x_0 < 0$ and get $-x$ with infinite radius of convergence, expand on $x_0 > 0$ and get $x$ with infinite radius of convergence -- in neither case did you get a Taylor series that agrees with the function on its entire interval of convergence. – Eric Towers Jan 01 '22 at 19:18
  • But what is the flaw in my reasoning? Isn't the convergence of the series equivalent to the residual term tending to zero?! And that residual term tending to zero means the difference between f(x) and the Taylor polynomial of x goes to zero as n goes to infinity ? And doesn't that mean that the sum of the Taylor series is f(x). I just need to understand the flaw in my reasoning. – peter.petrov Jan 01 '22 at 19:28
  • If I may add... Are there some conditions for f(x) under which what I claim is true? For example if f(x) is differentiable infinitely many times maybe?! – peter.petrov Jan 01 '22 at 19:38
  • but just because the series converges (i.e $\lim\limits_{n\to\infty}T_{n,a,f}(x)$ exists), doesn't mean the limit of remainders is zero. That's the error you're making. You should read that answer, and these comments several times over (it's a non-intuitive fact so it takes time). Your latest comment asks about "if $f$ is differentiable infinitely many times maybe", and my previous comment already tells you this is not a strong enough condition. – peek-a-boo Jan 01 '22 at 19:41
  • I am reading them and yes, will reread them. OK, good that I at least have the point that is flawed now. Thanks again. – peter.petrov Jan 01 '22 at 19:45
  • Aaah, so if the residual tends to zero 0 the sum is f(x). But the residual may well tend to 5 let's say, then the series is again convergent of course, but the sum is of course not fx). Right? It will be (f(x) - 5) – peter.petrov Jan 01 '22 at 19:48
  • yes that's correct. – peek-a-boo Jan 01 '22 at 19:51
  • Thank you very much – peter.petrov Jan 01 '22 at 19:52

0 Answers0