Question on taylor series

Question

When we try to to find the Taylor series of let's say $e^x$ in $x=-1$ and of order up to $3$, I would think that we just find the Maclaurin series of $e^x$ and then instead of simply $x$ we put $x+1$.But my book as a first step does the following: $t=x+1$ so $x=t-1$ so $f (t − 1) = e^{(t−1)}$. I feel that this is pretty simple so not understanding it is pretty frustrating.What is going on here?

Answer 1

You could start with the Maclaurin series.

$e^x = 1 + x + \frac 12x^2 + \cdots$

Then plug $x+1$

$e^{x+1} = 1 + (x+1) + \frac 12(x+1)^2 + \cdots\\ e(e^{x}) = 1 + (x+1) + \frac 12(x+1)^2 + \cdots\\ e^{x} = e^{-1} + e^{-1}(x+1) + \frac {e^{-1}}2(x+1)^2 + \cdots\\ $

But for other functions, it is not so easy to factor out that $+1$ and you should start from scratch.

$a_n = \frac {f^{(n)}(-1)}{n!}$

Answer 2

Assume we want the expansion of $f(x)$ arround $x=a$. $f$ is defined arround $a$.

let $g(x)=f(x+a)$. $g$ is then defined arround $0$.

we expand $g(x)$ arround zero at order $n$ as

$$g(x)=P_n(x)+x^n\epsilon(x)$$ then we get the expansion of $f(x)$ arround $a$ using that

$$f(x)=g(x-a)=$$ $$P_n(x-a)+(x-a)^n\epsilon(x)$$ with $$\lim_{x\to a}\epsilon(x)=0$$

In your case $$f(x)=e^x,\;\; a=-1, \;\; g(x)=e^{x-1}$$

$$g(x)=e^{-1}\Bigl(1+x+\frac{x^2}{2}+\frac{x^3}{6}\Bigr)+x^3\epsilon(x)$$ therefore

$$f(x)=e^{-1}\Bigl(1+(x+1)+\frac{(x+1)^2}{2}+\frac{(x+1)^3}{6}\Bigr)+(x+1)^3\epsilon(x)$$

with $$\lim_{x\to -1}\epsilon(x)=0$$

Answer 3

We want to write $e^x$ in a series with $(x+1)^n$. To do this we can write $$ \begin{align} e^x &=\frac1e\cdot e^{x+1}\\ &=\frac1e\left(\sum_{n=0}^\infty\frac{(x+1)^n}{n!}\right)\\ &=\sum_{n=0}^\infty\frac{(x+1)^n}{e\,n!} \end{align} $$