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Taylor polynomialneed help bad!

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    find the taylor polynomial f4 for f(x)=sin(2x) and a=pi/4

    2. Relevant equations

    3. The attempt at a solution

    so replace x with 2x?
    you get ((-1)^n)(2x)^(2n+1)/(2n+1)!)

    is this right?
  2. jcsd
  3. Apr 21, 2010 #2
    Its a good thing to present it the right way...

    You know the formula right?

    [tex]\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} \cdot (x-a)^{n} = \sum_{n=0}^{\infty} \frac{f^{(4)}(\frac{\pi}{4})}{4!} \cdot (x-\frac{\pi}{4})^{4}[/tex]

    which you simply expand as show in your Calculus book which then inturn gives the right answer!
  4. Apr 21, 2010 #3
    i see i have to center it at pi/4 but this is ((-1)^n)(2x)^(2n+1)/(2n+1)!) centered at 0 so i have to derive it sin(2x) using center pi/4 right

    thanks man
  5. Apr 21, 2010 #4
    you welcome :) Generally do you have the Edwards and Penny Calculus Bible? You find the whole definition and example in there :)
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