Taylor Series — Question 6

Question 6 – Solution

Step 1: Expand both exponentials.

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad e^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{n!}.$$ When the series are added, odd-power coefficients cancel and even-power coefficients double.

Step 2: Divide by two.

$$\boxed{\cosh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}},\qquad P_6(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}.$$

Step 3: State convergence.

Both exponential series converge absolutely for every real $x$, so their sum and scalar multiple do also. Thus the Maclaurin series for $\cosh x$ has infinite radius and represents the function on $(-\infty,\infty)$.