Example: $$ f(x)=\frac{1}{1+x} \qquad x\neq-1 $$ $$ f(x)=1-x+x^2-x^3+x^4-x^5+\;... \qquad |x| < 1 $$

Why Taylor series does not converge for all x in the domain of the function?

  • $\begingroup$ What do you mean it doesn't? It does converge. Check by yourself with $x = 1/2$ for example. $\endgroup$ – user88595 Mar 13 '14 at 10:39
  • $\begingroup$ Is your question: "why doesn't the taylor series converge everywhere?" $\endgroup$ – Joshua Pepper Mar 13 '14 at 10:40
  • $\begingroup$ @user88595 he meant why doesn't it work, say, 2. $\endgroup$ – Guy Mar 13 '14 at 10:40
  • $\begingroup$ @JoshuaPepper yes. "Why Taylor series does not converge for any x in the domain of the function?" That should mean every I think. $\endgroup$ – Guy Mar 13 '14 at 10:40
  • $\begingroup$ @Sabyasachi : Ah ok. Got confused by the domain he was talking about. $\endgroup$ – user88595 Mar 13 '14 at 10:41

This series is obtained from Newton's generalization of the binomial theorem which is only applicable for $|x|\lt 1$. This can be also obtained from the infinte geometric series, as

$$\lim_{n\to\infty}1-x+x^2-\cdots+(-1)^nx^n = \lim_{n\to\infty}\frac{1-x^{n+1}}{1+x}$$

Clearly, this only converges if $|x|\lt1$

Although it is useful to note, that Gottfried Leibniz admitted that the series,

$$1-2+4-8+\cdots = ?$$

Could either yield positive or negative infinty depending on how you subtract, and therefore neither can be correct, and the actual value should be finite. To quote:

"Now normally nature chooses the middle if neither of the two is permitted, or rather if it cannot be determined which of the two is permitted, and the whole is equal to a finite quantity."

He associated it with $\frac13$ which now happens to be the accepted convergent representation for this divergent sum, and can be otained by substituting $x=2$ in the Taylor series for $\frac{1}{1+x}$. So this formula isn't entirely inapplicable. Citation


It's actually surprising that the Taylor series converges to the function anywhere other than $x=0$ -- functions that are equal to their Taylor series are actually rather rare.

Of course, most of the functions people actually study do have this property of being "analytic", so it's easy not to get a sense of how special this property is.

It turns out there is a pleasing geometric description of what the radius of convergence of a Taylor series will be: if you look at the complex plane, your analytic function may have some singularities. The Taylor series around any point will converge in a disk around that point whose radius is as large as it can possibly be without including any of those singularities.

Your particular $f$ has a singularity at $x = -1$, and that is its only singularity in the complex plane. Thus, the Taylor series around $x=0$ will have a radius of convergence equal to $1$. Its Taylor series around $ x = 41$ will have a radius of converge equal to $42$. Its Taylor series around $x = \mathbf{i}$ will have radius of convergence $\sqrt{2}$, and so forth.


A quick answer is imagine $|x| > 1$, then $x^n \to \pm \infty$ as $n\to \infty$.

So your series is adding and subtracting larger and larger numbers at every step. How could this possibly converge?

A more mathematical reason is that this is the same as a well known formula : $$\dfrac{1}{1-x} = 1+x+x^2+x^3+...\qquad |x| < 1$$ multiplying both sides by $1-x$ you obtain : $$1 = (1+x+x^2+x^3+...+x^n)(1-x) = 1-x+x-x^2+x^2 - ... -x^{n-1}+x^{n+1} - x^n$$ Most terms cancel and you get : $$1 =\lim_{n\to\infty} 1-x^n$$ This only holds for $|x|<1$.

  • $\begingroup$ I know this. My question is, why it's like this? the taylor series of a function has the same nth derivative as the function, how come their values are not same for some x outside radius of convergence. I've expected taylor series converge for all x in the domain of the function. $\endgroup$ – kptlronyttcn Mar 13 '14 at 12:12

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