# Power series

• Feb 6th 2010, 09:04 PM
srobrien
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:

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

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

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
• Feb 6th 2010, 09:08 PM
Prove It
Quote:

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:

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

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

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

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

Now replace $\displaystyle x$ with $\displaystyle -2t$ and $\displaystyle 4t$ as required...
• Feb 6th 2010, 09:20 PM
srobrien
Hi thanks for the help.

R
• Feb 7th 2010, 11:29 AM
srobrien
Ok, gave the first part a shot today, don't know if im right so feedback would be gratefully received.

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

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

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

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

First 4 terms of $\displaystyle \cos4t$:

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

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

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

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