Compute $T^3_0f$ for sectional-defined function?

Question

Compute $T^3_0f$ for $f:(-1,1)\to \mathbb{R}, x\mapsto f(x)=\begin{cases}\ln \frac{1-x}{1+x} \, \quad x\neq 0 \\1 \quad \quad \, \,\quad x=0\end{cases}$

I'm a little confused, but shouldn't $[T_0^3f](x)=1$? Or is it a trick question?

Answer 1

The degree $3$ Taylor polynomial of $f$ centered at $0$ does not exist.

We are required to compute $$ f(0) + f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f'''(0)x^3 \text{.} $$ However, $f$ is not continuous at $x = 0$. In particular , $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \ln \frac{1-x}{1+x} = \ln 1 = 0 \neq 1 = f(0)$. Since $f$ is not continuous at $x = 0$, $f$ is not differentiable at $x = 0$, so $f'(0)$ does not exist (and also $f''(0)$ and $f'''(0)$ do not exist). (Consider that we want $\lim_{h \rightarrow 0} \frac{f(0+h) - f(0)}{h}$, where the numerator is approaching $0 - 1$, not zero, so the limit is $\pm \infty$, depending on the sign(s) of $f(0+h)$ which we have not investigated.) Therefore, the degree $3$ Taylor polynomial of $f$ centered at $0$ does not exist.

This can be corrected if there is an error in the specification of $f$. If instead of $f(0) = 1$, we replace that one piece of the piecewise definition with $f(0) = 0$, then $f$ is continuous and (thrice) differentable at $x = 0$, so the needed derivatives for the requested Taylor polynomial exist. (You should be able to find $f'(0) < 0$, $f''(0) = 0$, and $f'''(0) < 0$.)