# Power series

• Mar 18th 2009, 03:14 PM
Jaffa
Power series
I am trying to work out the first four terms for these series

i)e^-2t I get 1 -2t+4t^2/2 -8t^3/3

ii)cos4t Iget = 1-8t^2/2 +256t4/4

Is this right, or can you explain where I am going wrong
• Mar 18th 2009, 03:27 PM
skeeter
Quote:

Originally Posted by Jaffa
I am trying to work out the first four terms for these series

i)e^-2t I get 1 -2t+4t^2/2 -8t^3/3

ii)cos4t Iget = 1-8t^2/2 +256t4/4

Is this right, or can you explain where I am going wrong

factorials?

$\displaystyle e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + ...$

$\displaystyle \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + ...$
• Mar 18th 2009, 04:00 PM
Soroban
Hello, Jaffa!

You're forgetting the factorials . . .

Quote:

Find the first four terms for these series:

. . . $\displaystyle (1)\;e^{-2t} \qquad (2)\;\cos4t$

(1) Series: .$\displaystyle e^x \:=\:1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \hdots$

We have: .$\displaystyle e^{\text{-}2t} \;=\;1 + (\text{-}2t) + \frac{(\text{-}2t)^2}{2!} + \frac{(\text{-}2t)^3}{3!} + \hdots$

. . . . . . . . . . $\displaystyle = \;1 - 2t + \frac{4t^2}{2} - \frac{\text{-}8t^3}{6} + \hdots$

. . . . . . . . . . $\displaystyle = \;1 - 2t +2t^2 - \frac{4}{3}t^3 + \hdots$

(2) Series: .$\displaystyle \cos x \;=\;1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \hdots$

We have: .$\displaystyle \cos(4t) \;=\;1 - \frac{(4t)^2}{2!} + \frac{(4t)^4}{4!} - \frac{(4t)^6}{6!} + \hdots$

. . . . . . . . . . . . .$\displaystyle = \;1 - \frac{16t^2}{2} + \frac{256t^4}{24} - \frac{4096t^6}{720} + \hdots$

. . . . . . . . . . . . .$\displaystyle = \;1 - 8t^2 + \frac{32}{3}t^4 - \frac{256}{45}t^6 + \hdots$

• Mar 18th 2009, 04:13 PM
Jaffa
Hello Soroban
Thanks for the help, but I am not sure what you mean by factorials