Taylor series at $a=1$

Question

I have to find Taylor series at $a=1$ for

$ f(x)=\begin{cases} \frac{e^{x}-e}{x-1},\quad &\text{if } x\ne1\\ e,\quad &\text{if } x=1\\ \end{cases} $

I haven't found Taylor series for such functions before and I also don't know how to find n-th derivative of this function. Can anyone help me with this?

Answer 1

Using the series expansion of $e^{x-1}$ we get $f(x)=e \frac {e^{x-1}-1}{x-1}=e\sum\limits_{k=1}^{\infty} \frac {(x-1)^{k}} {(x-1)k!} =\sum\limits_{k=1}^{\infty}\frac { e{(x-1)^{k-1}}} {k!}=\sum\limits_{k=0}^{\infty}\frac { e{(x-1)^{k}}} {(k+1)!}$

Answer 2

First I compute

$$f'(x)=\dfrac{e^x-f(x)}{x-1}.$$ And

$$ f''(x)=\dfrac{e^x-2f'(x)}{x-1}.$$

So that I find a pattern

$$ f'''(x)=\dfrac{[e^x-2f''(x)](x-1)-[e^x-2f'(x)]}{(x-1)^2}=\dfrac{[e^x-2f''(x)](x-1)-f''(x)(x-1)}{(x-1)^2}=\dfrac{e^x-3f''(x)}{x-1}.$$

And I can give a general formul for the $n$-th derivative

$$ f^{(n)}(x)=\dfrac{e^x-nf^{n-1}(x)}{x-1}.$$

In the Taylor series, this terms appear evaluated at $x=1$. Then $f(1)=e$ but $f''(1)=0$, $f'''(1)=0$ etc so we have to make sense of $\dfrac{0}{0}$ limits. We can use Hôpital for this. Then

$$ \lim_{x\rightarrow 1}f^{(n)}(x)=\lim_{x\rightarrow 1}\dfrac{e^x-nf^{n-1}(x)}{x-1}=\dfrac{0}{0}\rightarrow \lim_{x\rightarrow 1} [e^x-nf^{(n)}(x)]=e-n \lim_{x\rightarrow 1} f^{(n)}(x)$$

Now, since all this limits must tend to $0$ (differentiable),

$$e-n \lim_{x\rightarrow 1} f^{(n)}(x)=0 \Rightarrow \lim_{x\rightarrow 1} f^{(n)}(x)=\dfrac{e}{n} \quad n\geq 1.$$

And I get for the Taylor series

$$f(x)= \begin{cases} e+\displaystyle\sum_{n=1}^{\infty} \dfrac{e}{n! }\dfrac{(x-1)^n}{n}\quad x\neq 1\\\\ e \quad x=1 \end{cases}.$$

Answer 3

With $y:=x-1$,$$\frac{e^{x-1+1}-e}{x-1}=e\frac{e^y-1}y=e\sum_{k=1}^\infty \frac{y^k}{k!y}=e\sum_{k=1}^\infty \frac{(x-1)^{k-1}}{k!}.$$