1. ## Power series

Hi, can anyone help with this question, I can't get me head around power series.

A sound wave is given by the function:

$f(t) = 0.5e^{-2t} \cos 4t$

Write down the first four terms of the power series expansions of $e^{–2t}$and $cos 4t$

Determine the cubic (up to and including the third power of t) approximation of f(t) and calculate the accurate and approximate values of f(0.02) giving your results correct to 3 decimal places.

Thanks

2. Originally Posted by srobrien
Hi, can anyone help with this question, I can't get me head around power series.

A sound wave is given by the function:

$f(t) = 0.5e^{-2t} \cos 4t$

Write down the first four terms of the power series expansions of $e^{–2t}$and $cos 4t$

Determine the cubic (up to and including the third power of t) approximation of f(t) and calculate the accurate and approximate values of f(0.02) giving your results correct to 3 decimal places.

Thanks
Are you allowed to take some things for granted?

Like $e^x = \sum_{k = 0}^{\infty}\frac{x^k}{k!}$

and $\cos{x} = \sum_{k = 0}^{\infty}\frac{(-1)^kx^{2k}}{(2k)!}$?

Now replace $x$ with $-2t$ and $4t$ as required...

3. Hi thanks for the help.

R

4. Ok, gave the first part a shot today, don't know if im right so feedback would be gratefully received.

First 4 terms of $e^{-2t}$:

$e^{-2t} = 1 + (-2t) + \frac{(-2t)^2}{2!} + \frac{(-2t)^3}{3!} + ...$

$e^{-2t} = 1 -2t + \frac{(4t)^2}{2!} - \frac{(8t)^3}{3!} + ...$

$e^{-2t} = 1 -2t + 2t^2 - \frac{(8t)^3}{3!} + ...$

First 4 terms of $\cos4t$:

$\cos4t = 1 - \frac{(4t)^2}{2!} + \frac{(4t)^4}{4!} - \frac{(4t)^6}{6!} + ...$

$\cos4t = 1 - \frac{(16t)^2}{2!} + \frac{(256t)^4}{4!} - \frac{(4096t)^6}{6!} + ...$

$\cos4t = 1 - (8t)^2 + \frac{(64t)^4}{3!} - \frac{(4096t)^6}{6!} + ...$

Am I anywhere close? If so where is next from here?
Sorry about the basic questions, this subject was briefly covered by a module and didn't really go into much depth.

Thanks again