Chapter 8.7, Problem 55E

To determine

To Find: First three nonzero terms in the Maclaurin series

Answer to Problem 55E

First three nonzero terms

  $e^{-x^2}\cos x\approx 1-\frac{3x^2}{2}+\frac{25x^4}{24}$

Explanation of Solution

Given information:

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for function

  $y=e^{-x^2}\cos x$

Concept Used:

Maclaurin series for cosx

  $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\mathrm{..}$

Maclaurin series for $e^x$

  $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x3}{3!}\mathrm{..}$

Calculation:

Maclaurin series of

  $y=e^{-x^2}\cos x$

UsingtheMaclaurin series for $\cos x$ and $e^x$

  $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x3}{3!}\mathrm{..}$

Then

  $e^{-x^2}=1-\frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}\mathrm{..}$

And

  $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^5}{6!}\mathrm{..}$

Multiply these expressions, collecting like terns just as for polynomials

  $e^{-x^2}\cos x=1-\frac{x^2}{1!}+\frac{x^4}{4!}-\frac{x^2}{1!}+\frac{x^4}{2!}+\frac{x^4}{2!}$

First three nonzero terms

  $e^{-x^2}\cos x\approx 1-\frac{3x^2}{2}+\frac{25x^4}{24}$