Taylor Series — Question 5

Question 5 – Solution

Step 1: Start with the parent series.

$$e^u=\sum_{n=0}^{\infty}\frac{u^n}{n!}\qquad(u\in\mathbb R).$$ Set $u=2x$: $$\boxed{e^{2x}=\sum_{n=0}^{\infty}\frac{(2x)^n}{n!}=\sum_{n=0}^{\infty}\frac{2^n x^n}{n!}}.$$

Step 2: Extract the requested polynomial.

$$P_4(x)=1+2x+2x^2+\frac43x^3+\frac23x^4.$$

Step 3: Check convergence.

For $a_n=(2x)^n/n!$, $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{2|x|}{n+1}\longrightarrow0$$ for every real $x$. Hence the radius is infinite and the series represents $e^{2x}$ on all of $\mathbb R$.