Question

Question: 55-58 = Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

55. \(y = {e^{ - {x^2}}}cosx\)

Short answer

The first three non-zero terms is \(1 - \frac{3}{2}{x^2} + \frac{{25}}{{24}}{x^4} - \ldots \)

Step-by-step solution

Step 1: To solve the Maclaurin series

We know the following Maclaurin series :

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

Step 2: To multiply the power series

Now we can multiply these power series like polynomials to get:

\({e^{ - {x^2}}} \times \cos x{\rm{ }} = \left( {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n}}}}{{n!}}} \right) \times \left( {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n}}}}{{(2n)!}}} \right)\)\( = \left( {1 - {x^2} + \frac{1}{2}{x^4} - \frac{1}{6}{x^6} + \ldots } \right) \times \left( {1 - \frac{1}{2}{x^2} + \frac{1}{{24}}{x^4} - \ldots } \right)\)\( = 1 - \frac{3}{2}{x^2} + \frac{{25}}{{24}}{x^4} - \ldots \)