# How does the Taylor Series converge at all points for certain functions

The way my professor defined Taylor polynomials is: the $$n^{th}$$ degree Taylor polynomial $$p(x)$$ of $$f(x)$$ is a polynomial that satisfies $$\lim_{x\to 0}{f(x)-p(x) \over x^n} = 0$$. This is actually the little-o notation $$o(x^n)$$, which means $$(f(x)-p(x)) \ll x^n$$ as $$x$$ approaches $$0$$. From this I have got the intuition that Taylor Polynomials work only for $$|x| < 1$$ because $$x^n$$ gets smaller as $$n$$ gets bigger only when $$|x| < 1$$. And the textbook seemed to agree with my intuition, because the textbook says “Taylor polynomial near the origin” (probably implying $$|x| < 1$$).

Since Taylor Series is basically Taylor polynomial with $$n\to\infty$$, I intuitively thought that the Taylor Series would also only converge to the function it represents in the interval $$(-1, 1)$$.

For example, in the case of $$1\over1-x$$, it is well known that the Taylor series only converges at $$|x| < 1$$.

However, all of a sudden, the textbook says that the Taylor series of $$\cos x$$ converges for all real $$x$$. It confused me because previously I thought the Taylor series would only work for $$|x|<1$$. Now, I know that the Taylor Series is defined like this: $$f(x) = Tf(x) \Leftrightarrow \lim_{n\to\infty}R_{n}f(x) = 0$$

And I know how to get the maximum of Taylor Remainder for $$\cos x$$ using Taylor's Theorem, and I know that the limit of that Taylor Remainder is $$0$$ for all real $$x$$, which makes the Taylor Series of $$cosx$$ converge to $$\cos x$$, pointwise. However, I just can't get why my initial intuition is wrong (why taylor series converges for all $$x$$ for certain functions, like $$\cos x$$, also $$\sin x$$ and $$e^x$$, etc.)

• For the specific example $e^x$, the terms in its series are $x^n/n!$. Note that $n!$ grows much faster than $x^n$ as $n$ grows. That's one reason to believe the Taylor series of $e^x$ converges. Apr 24, 2020 at 18:10
• In my mind, the key idea of calculus is the local linear approximation $f(x) \approx f(a) + f'(a)(x - a)$, which is a good approximation when $x$ is near $a$. It is natural to ask, what if we approximate $f$ locally by a quadratic or cubic function rather than by a linear function. This leads to the idea of Taylor polynomial approximation, which is indeed initially intended to be a local approximation to $f$. But when we look at the remainder term in Taylor's theorem, there's something kind of amazing, surprising that happens which is that often it's small even when $x$ is far from $a$. Apr 24, 2020 at 18:11
• @trisct Yes that is true indeed. The limit of Taylor remainder for $e^x$ is also zero for all x, which further explains why the Taylor Series converges. But I still I can't see why my initial intuition is wrong.
– the
Apr 24, 2020 at 18:12
• @littleO So it just happened to be that the Taylor remainder's limit is zero even for x far away?
– the
Apr 24, 2020 at 18:14
• @linearAlg In my mind, yes. It is just one of the miracles of math, which makes us wonder why things have worked out more nicely than we deserved. Apr 24, 2020 at 18:16

Actually, things may go wrong in $$(-1,1)$$. For instance, the Taylor series centered at $$0$$ of $$f(x)=\frac1{1-nx}$$ only converges to $$f(x)$$ on $$\left(-\frac1n,\frac1n\right)$$. And if$$f(x)=\begin{cases}e^{-1/x^2}&\text{ if }x\ne0\\0&\text{ if }x=0,\end{cases}$$then the Taylor series of $$f$$ only converges to $$f(x)$$ if $$x=0$$.
On the other hand, yes, Taylor series centered at $$0$$ are made to converge to $$f(x)$$ near $$0$$. But that's no reason to expect that they don't converge to $$f(x)$$ when $$x$$ is way from $$0$$. That would be like expecting that a non-constant power series $$a_0+a_1x+a_2x^2+\cdots$$ takes larger and larger values as the distance from $$x$$ to $$0$$. That happens often, but $$1-\frac1{2!}x^2+\frac1{4!}x^4-\cdots=\cos(x)$$, which is bounded.
In other words, you have gone from $$a\implies b$$ to $$\tilde a\implies \tilde b,$$ which you can see to be clearly false, identically. That is, it is not necessarily true for all $$a,\,b.$$
Since you already know why the series for entire functions like $$\cos x$$ converges everywhere (as you explain towards the end of your post), you should now see where your original intuition (I would say erroneous belief) misled you.
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