Taylor Series — Question 1

Question 1 – Solution

Step 1: Compute the coefficients.

Every derivative of $f(x)=e^x$ is $e^x$, so $f^{(n)}(0)=1$. Hence $$P_N(x)=\sum_{n=0}^{N}\frac{x^n}{n!},\qquad \boxed{e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}}.$$

Step 2: Bound the remainder.

Taylor’s theorem gives $$R_N(x)=\frac{e^c x^{N+1}}{(N+1)!}$$ for some $c$ between $0$ and $x$. If $|x|\le1$, then $c\le1$, so $$\boxed{|R_N(x)|\le\frac{e|x|^{N+1}}{(N+1)!}\le\frac{e}{(N+1)!}}.$$

Step 3: Justify equality with the function.

For any fixed real $x$, $e^c\le e^{|x|}$ and $e^{|x|}|x|^{N+1}/(N+1)!\to0$ by the Ratio Test. Thus $R_N(x)\to0$ for every $x$, proving the series represents $e^x$ on all of $\mathbb R$.