Taylor Series — Question 7

Question 7 – Solution

Step 1: Multiply the exponential series.

$$xe^x=x\sum_{n=0}^{\infty}\frac{x^n}{n!}=\boxed{\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}}.$$ Multiplication by $x$ raises every exponent by one but leaves the coefficient denominator tied to the original index.

Step 2: Reindex if powers $x^k$ are preferred.

Set $k=n+1$: $$xe^x=\boxed{\sum_{k=1}^{\infty}\frac{x^k}{(k-1)!}}.$$ The first five nonzero terms are $$x+x^2+\frac{x^3}{2}+\frac{x^4}{6}+\frac{x^5}{24}.$$

Step 3: State convergence.

Multiplying an everywhere-convergent series by $x$ does not change its infinite radius. Therefore the identity holds for every real $x$.