Finding terms for maclaurin series

[dyy]

Problem- Find the terms through x^5 in the Maclaurin series for f(x)=e^(-x) * cos(x)

After trying to find f'(x), f''(x), f'''(x), and so on, the derivatives seemed overly complex. I heard that it is possible to use known Maclaurin series (e.g. e^(-x) and cos(x) separately) and then perform multiplications, divisions, etc. How would I combine them?

Thanks!

 

[Opalg]

Problem- Find the terms through x^5 in the Maclaurin series for f(x)=e^(-x) * cos(x)

After trying to find f'(x), f''(x), f'''(x), and so on, the derivatives seemed overly complex. I heard that it is possible to use known Maclaurin series (e.g. e^(-x) and cos(x) separately) and then perform multiplications, divisions, etc.

That is definitely the best method to use. Given that \(\displaystyle e^{-x} = 1 - x + x^2/2! - x^3/3! + x^4/4! - x^5/5! + \ldots\) and \(\displaystyle \cos x = 1 - x^2/2! + x^4/4! - \ldots\), multiply the two series to get \(\displaystyle e^{-x}\cos x = (1 - x + x^2/2! - x^3/3! + x^4/4! - x^5/5! + \ldots)(1 - x^2/2! + x^4/4! - \ldots)\). There's a theorem that says it's okay to multiply these "infinite brackets" in the usual way (given that each series is absolutely convergent). So just multiply each term in the first bracket by each term in the second one, ignoring any powers greater than x^5.

 

[mathcracker]

Or, you can try to find the Taylor Expansion of the complex function:

\(\displaystyle f(z) = e^{iz}\)

and then look for the real part of the expansion. Typically, complex analysis provides the easiest solutions....

 

[morson]

and then look for the real part of the expansion.

Isn't that just the Taylor series for \(\displaystyle cos(z)\)..

 

[soroban]

It can be multiplied like this . . .

\(\displaystyle \L\begin{array}{ccccccccccccccc}\;e^{-x} & \;=\; & 1& - & x & + & \frac{x^2}{2!} & - & \frac{x^3}{3!} & + & \frac{x^4}{4!} & - & \frac{x^5}{5!}& + & \cdots \\
\cos x & \;=\; & 1 & - & \frac{x^2}{2!} & + & \frac{x^4}{4!} & - & \cdots & & & & & & \end{array}\)
. . . . . . . . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
. . . . . . . . . .\(\displaystyle \L\begin{array}{ccccccccccccc}1 & - & x & + & \frac{x^2}{2!} & - & \frac{x^3}{3!} & + & \frac{x^4}{4!} & - & \frac{x^5}{5!} & + & \cdots \\

& & & - & \frac{x^2}{2!} & + & \frac{x^3}{2!} & - & \frac{x^4}{2!2!} & + & \frac{x^5}{2!3!} & + & \cdots \\
& & & & & & & & \frac{x^4}{4!} & - & \frac{x^5}{4!} & + & \cdots\end{array}\)
. . . . . . . . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

. . . . . . . . . .\(\displaystyle \L\begin{array}{ccccccccccccc}1 & - & x & \qquad\qquad & \qquad & + & \frac{x^3}{3} & - & \frac{x^4}{6} & + & \frac{x^5}{30} & + & \cdots\end{array}\)