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Math Help - Power series

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by Jaffa View Post
    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?

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

    \cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + ...
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  3. #3
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    Hello, Jaffa!

    You're forgetting the factorials . . .


    Find the first four terms for these series:

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

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

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

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

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



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

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

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

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

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  4. #4
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    Hello Soroban
    Thanks for the help, but I am not sure what you mean by factorials
    could you please explain
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