Find the Taylor series expansion of e^x about x = 0.

Step-by-step solution

1. 1

   Write down the template you are filling in. The Taylor series of $f$ about $x=0$ is

   $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}$

   so the entire job is to compute the derivatives of $f(x)=e^{x}$ at the single point $0$.

2. 2

   Differentiate repeatedly. The exponential is the fixed point of differentiation:

   $f'(x)=e^{x},\quad f''(x)=e^{x},\quad\ldots,\quad f^{(n)}(x)=e^{x}\ \text{for every }n\ge 0$

   This is the whole reason the expansion comes out so clean — there is no pattern of signs or growing coefficients to track.

3. 3

   Evaluate at the centre. Since $e^{0}=1$,

   $f^{(n)}(0)=1\qquad\text{for all }n\ge 0$

4. 4

   Substitute back into the template. Every numerator is $1$, so the coefficient of $x^{n}$ is just $\dfrac{1}{n!}$:

   $e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots$

5. 5

   Check the radius of convergence. By the ratio test on consecutive terms,

   $\left|\frac{x^{n+1}/(n+1)!}{x^{n}/n!}\right|=\frac{|x|}{n+1}\longrightarrow 0$

   for every fixed $x$, so the series converges absolutely on all of $\mathbb{R}$ (radius $R=\infty$).

6. 6

   Sanity-check one value. At $x=1$ the partial sums $1,\ 2,\ 2.5,\ 2.6\overline{6},\ 2.708\overline{3},\ 2.71\overline{6}$ climb toward $e=2.71828\ldots$, matching the claim $e^{1}=\sum 1/n!$.