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I'm looking for an intuitive understanding instead of a formal proof. Thanks for the help.

Cameron Buie
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user73232
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    The series diverges for $x=-1$ and for $|x|>1$. How could you then use it to approximate anything? – Andrés E. Caicedo Apr 18 '13 at 14:58
  • @AndresCaicedo Sorry, I don't follow. What do you mean by diverge? Isn't a taylor series a polynomial? How can a polynomial series converge or diverge to a value? – user73232 Apr 18 '13 at 15:00
  • A Taylor series is an infinite series. A Taylor polynomial only considers the first few values of the Taylor series. They are different objects. – Andrés E. Caicedo Apr 18 '13 at 15:01
  • A Taylor series is a limit of Taylor polynomials. This series in particular converges on the interval $(-1,1]$ and diverges elsewhere. – Cameron Buie Apr 18 '13 at 15:02

2 Answers2

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Intuitively, you can think of the interval of convergence of the Taylor series of $f$ centred at $a$ as the largest symmetric interval centred at $a$ for which the resulting infinite series converges to the function $f$ - here, when I say symmetric, I am disregarding the endpoints as it is possible to have intervals of convergence of the form $[a - R, a + R)$ and $(a - R, a+ R]$.

Given the interval you are asking about, the Taylor series of $\ln(1+x)$ has been taken at the origin, so the corresponding interval of convergence, disregarding endpoints is of the form $(-R, R)$. As $\ln(1 + x)$ is only defined for $x \in (-1, \infty)$, we have $(-R, R) \subset (-1, \infty)$. The largest the interval $(-R, R)$ can be whilst satisfying this condition is $(-1, 1)$ - in fact, this is what the interval turns out to be.

However, we must check the endpoints. Well, as $-1 \notin (-1, \infty)$, it can't be in the interval of convergence, and by calculating the Taylor series, you can see that the infinite sum converges for $x = 1$ (it is the alternating harmonic series). So the interval of convergence for $f(x) = \ln(1+x)$ is $(-1, 1]$, that is, the Taylor series converges to $f(x)$ for $|x| < 1$ and $x = 1$.

Michael Albanese
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  • nice! but why must the interval of convergence for the Taylor series be syymetric around $a$? – Vincent Tjeng Apr 18 '13 at 15:15
  • Thanks, that really helped! – user73232 Apr 18 '13 at 15:22
  • @VincentTjeng: This is a fact about power series in general, you should be able to find a proof of it most real analysis texts. If you know of the ratio test, you may be able to convince yourself that it is true. – Michael Albanese Apr 18 '13 at 15:26
  • ok. would it be correct in general to say that, disregarding endpoints, if I have a function defined for $x \in (x_{min}, x_{max})$, then the interval of convergence for the Taylor series of $f$ centered at $a$ would always be the smaller of $(x_{min}, 2a-x_{min})$ and $(2a-x_{max}, x_{max})$? – Vincent Tjeng Apr 18 '13 at 15:29
  • No. If $f$ is defined on $\mathbb{R}$, the Taylor series of $f$ centred at $0$, sometimes called the Maclaurin series of $f$, could have interval of convergence ${0} = [0, 0]$. – Michael Albanese Apr 18 '13 at 15:32
  • I'll find a real analysis text about this question tomorrow, but just one last quick question - if it is known that for instance the Maclaurin series of $f$ converges for some $k \neq 0$, would my earlier suggestion hold? – Vincent Tjeng Apr 18 '13 at 15:39
  • What is $k$? Is $k \in (x_{min}, x_{max})$? – Michael Albanese Apr 18 '13 at 15:40
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    @Vincent: Consider the function $$f(x)=\frac1{1+x^2}.$$ This is defined for all $x\in\Bbb R,$ but the Maclaurin series $$\sum_{k=0}^\infty (-1)^kx^{2k}$$ only converges for $|x|<1.$ Ultimately, this has to do with the fact that the complex extension $f(z)=\frac1{1+z^2}$ is defined on the whole complex plane except for at $z=\pm i,$ so the disk of convergence of the corresponding complex power series is $|z|<1.$ – Cameron Buie Apr 18 '13 at 15:57
  • @MichaelAlbanese yes, I intended $k$ to be defined as you suggested. Cameron, thank you for the elaboration; it's far clearer for me now. I'll find a text on real analysis to understand the proof of that. – Vincent Tjeng Apr 19 '13 at 02:28
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Write $$\log(1+x)= \sum_{n=1}^ \infty (-1)^{n+1}\frac {x^{n}}{n}.$$ Now you can apply the ratio test to get the radius of convergence. Finally, you need to check convergence at the boundary points.

Michael Albanese
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Srijan
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