Question

Find the Maclaurin series of f (by any method) and its radius of convergence. Graph f and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and f?

38. \(f(x) = {e^{ - {x^2}}} + cosx\)

Short answer

\(f(x) = \sum\limits_{}^{} {{{( - 1)}^n}{x^{2n}}\left( {\frac{1}{{n!}} + \frac{1}{{(2n)!}}} \right)} \,\;\;R = \infty \)

Step-by-step solution

Step 1: The Maclaurin series

We know the following Maclaurin series:

\(\begin{aligned}{l}\cos x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{x^{2n}}}}{{(2n)!}}} \\{e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \\{e^{ - {x^2}}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - {x^2})}^n}}}{{n!}} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{x^{2n}}}}{{n!}}} } \end{aligned}\)

Step 2: The series and explanation

We also know that the addition of the two series will be convergent on the intersection of their intervals of convergence. However, since both series converge for all x, their addition will also converge for all x. This is the series:

\(\begin{aligned}{l}{e^{ - {x^2}}} + \cos x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{x^{2n}}}}{{n!}}} + \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{x^{2n}}}}{{(2n)!}}} \\ = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}{x^{2n}}} \left( {\frac{1}{{n!}} + \frac{1}{{(2n)!}}} \right)\end{aligned}\)

NOTE : We can change the expression \(\frac{1}{{n!}} + \frac{1}{{(2n)!}}\) to different form, but this seemed like the simplest form.

Step 3: The examination of the graph

Now let’s examine the graphs of the function f(x) and the first four Taylor polynomials. Notice that the polynomials are starting to converge towards the graph as n grows. For an even n, the polynomials are above the graph, while polynomials are below the graph for an odd n.