সংযুক্তি: কোনো ফাংশনের টেইলর সিরিজ কনভার্জ করলেও সেটা ওই ফাংশনের মানের সমান নাও হতে পারে

It turns out a function $f$ can have a Taylor series that doesn't converge to $f$ anywhere, except at its center. (The value at the center point $a$ is just $f(a)$ so no convergence issues.) It's possible that in every neighborhood of the center, the Taylor series converges to the wrong function.

The classic example of such a function is $f(x)=\begin{cases}e^{-1/x},&x>0\\0,&x\le0\end{cases}$. (Its close relative $g(x)=\begin{cases}e^{-1/x^2},&x\ne0\\0,&x=0\end{cases}$ also works.) Certainly $f(x)\ne0$ for any $x>0$. Nevertheless, its Maclaurin series is just the constant function $0$, which is not equal to $f$ for any $x>0$ !!

Here's a graph of the function $f$:

Notice it looks pretty flat at $x=0$, which is intuitively how the Taylor series can just be the constant zero - All the derivatives of $f$ at $0$ are just zero!

Ready for the details and proof?

We will show that $f^{(n)}(0)=0$, so that its Maclaurin series is just $M(x)=0$, which is not equal to $f$ for any $x>0$.

Since $f(x)=0$ for $x<0$, then $f^{(n)}(x)=0$ for $x<0$, which implies that the left handed derivative at $0$ is: $$f^{(n)}_-(0)=\lim_{x\to0^-}\frac{f^{(n-1)}(x)-f^{(n-1)}(0)}{x-0}=0.$$ So to conclude that $f^{(n)}(0)=0$, we now need to show that $f^{(n)}_+(0)=0$.

We first use induction to show that for $x>0$, there exists some polynomial $p_n(t)$ such that $$f^{(n)}(x)=p_n(1/x)e^{-1/x}.$$ In other words, the $n$th derivative of $f$ is a finite sum of terms of the form $\frac{c}{x^k}e^{-1/x}$ for some constants $c$ and $k$ that may vary from term to term. (It can even be shown that $p_n$ is of degree $2n$.) Taking a few derivatives of $f$ will give you some motivation for this result.

Why is this fact useful? If we have $f^{(n)}(x)=p_n(1/x)e^{-1/x}$, then we take the limit as $x\to0^+$ to compute $f^{[n]}=0$ - We'll look more into these details after proving the fact.

This statement is true for $n=1$, since $$f'(x)=e^{-1/x}\cdot\frac{1}{x^2},$$ where $p_1(t)=t^2$ and $p_1\big(\frac{1}{x}\big)=\frac{1}{x^2}$.

Now we suppose that $f^{(k)}(x)=p_k(1/x)e^{-1/x}$ for some $n=k$. We differentiate once to get $$\begin{align*} f^{(k+1)}(x)=\frac{d}{dx}\left[f^{(k)}(x)\right]&=\frac{d}{dx}\left[p_k(1/x)e^{-1/x}\right] \\ f^{(k+1)}(x)&=p'_k(1/x)\cdot\frac{-1}{x^2}e^{-1/x}+p_k(1/x)e^{-1/x}\cdot\frac{1}{x^2} \\ f^{(k+1)}(x)&=\left[p'_k(1/x)\cdot\frac{-1}{x^2}+p_k(1/x)\cdot\frac{1}{x^2}\right]e^{-1/x}, \end{align*}$$ and we see that $p_{k+1}(1/x)=\left[p'_k(1/x)\cdot\frac{-1}{x^2}+p_k(1/x)\cdot\frac{1}{x^2}\right]$, which is still a polynomial in terms of $\frac{1}{x}$. Thus $f^{(k+1)}(x)=p_{k+1}e^{-1/x}$, so we can conclude that $f^{(n)}(x)=p_n(1/x)e^{-1/x}$ for all $n\in\mathbb{N}$.

Now using our formula for the $k$th derivative, we can compute $f^{[k]}_+(0)$.

Method 1: We have $$f^{[n]}(x)=p_n(1/x)e^{-1/x}.$$ Now we use the fact that $f^{[n]}_+(0)=\lim_{x\to0^+}f^{[n]}(x)$ when the limit exists. (Proof of this fact by using continuity of $f^{[n-1]}$ and mean value theorem.) Note that $$\lim_{x\to0^+}\frac{1}{x^k}e^{-1/x}=\lim_{t\to+\infty}t^ke^{-t}=0$$ for any $k\in\mathbb{N}$. Since $p_n(1/x)$ is just a bunch of powers of $\frac{1}{x}$ with constants, when we multiply by $e^{-1/x}$ we get the limit $$f^{[n]}_+(x)=\lim_{x\to0^+}p_n(1/x)e^{-1/x}=0.$$

Method 2 (induction): When $n=1$, then $$f'_+(0)=\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0^+}\frac{e^{-1/x}}{x}=\lim_{t\to\infty}te^{-t}=0,$$ where we used the equality $\lim_{x\to0^+}f\left(\frac{1}{x}\right)=\lim_{t\to\infty}f(t)$ (provided either limit exists).

Now we suppose that $f^{(k)}_+(0)=0$ for some $n=k$. By the definition of the derivative,

$$\begin{align*} f^{(k+1)}_+(0)&=\lim_{x\to0^+}\frac{f^{(k)}(x)-f^{(k)}(0)}{x-0} \\ &=\lim_{x\to0^+}\frac{f^{(k)}(x)}{x}\\ f^{(k+1)}_+(0)&=\lim_{x\to0^+}p_k(1/x)\cdot\frac{1}{x}\cdot e^{-1/x} \\ &=\lim_{t\to\infty}\frac{p_k(t)\cdot t}{e^t}=0, \end{align*}$$

since $e^t\to\infty$ faster than any power (or polynomial) of $t$. This can also be seen by repeated use of L'Hospital's Rule or other methods.

Thus $f^{(k+1)}_+(0)=0$, and so $f^{(n)}_+(0)=0$ for all $n\in\mathbb{N}$. Since we already know that $f^{(n)}_-(0)=0$, we can now conclude that $f^{(n)}(0)=0$.

Thus the Taylor series for $f$ centered at $x=0$ is $$M(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty0=0,$$ which is convergent $\forall\,x\in\mathbb{R}$. But since $f(x)\ne0$ for $x>0$, no open interval around $x=0$ exists such that $M(x)=0$ is equal to $f(x)$.