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F(x)=sin x, centered at x=pi/2 Question

  1. Jul 20, 2005 #1
    hi, im trying to get the Taylor Series for f(x)=sin x, centered at x=pi/2,
    but i am seem to be getting an incorrect taylor series, any help?
    Thanks

    f'(x) = cos(x)
    f(2)(x) = - sin(x)
    f(3)(x) = - cos(x)
    f(4)(x) = sin(x)
     
  2. jcsd
  3. Jul 20, 2005 #2
    Do you want to offset sin(o) = pi/2?
     
  4. Jul 21, 2005 #3
    Taylor series for f(x) centered about a=pi/2

    the a is :

    f(x) = sum[ f^(n)*(a)/n! * (x-a)^n ], n=0.....infinity
     
  5. Jul 21, 2005 #4

    lurflurf

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    The formula for taylor series centered about a is easily derived by integration by parts.
    [tex]\sum_{i=0}^{\infty} \frac{(x-a)^i}{i!}f^{(i)}(a)[/tex]
    a nice way to write the sine derivatives is
    [tex]\sin^{(n)}(x)= \sin(x+n\frac{\pi}{2})[/tex]
     
  6. Jul 21, 2005 #5
    You've listed the first four derivatives of sin(x). Evaluated at x = pi/2, the values of the first four derivatives are 0, -1, 0, 1 and the cycle repeats itself. What this tells you is that the 'even numbered' derivatives(ie. 2nd, 4th, 6th etc derivatives) are non-zero whilst the others are. Further, these even numbered derivatives alternate between -1 and 1. So it seems reasonable to suggest that the nth, where n is a natural number, derivative evaluated at x = pi/2 is given by [tex]\left( { - 1} \right)^{2n} [/tex].

    Edit: Nevermind the error message before. What I have said above alludes to a possible way to representing the nth derivative evaluated at x = pi/2. I haven't done the question myself, but if you try variations of what I suggested above you'll probably get the answer.

    Edit 2: You should probably just ignore what I said, it's a bit misleading now that I think about it.
     
    Last edited: Jul 21, 2005
  7. Jul 21, 2005 #6
    Thanks for the suggestions, i was google-ing this question, and the problem is that ALL solutions for this question are answered using : a is centered at 0.

    In other words, Maclaurin Series.

    When a != 0, it totally throws me off.

    :grumpy:
     
  8. Jul 23, 2005 #7
    replace x with x-pi/2
     
  9. Jul 26, 2005 #8
    life might be better if you center yourself at zero.. lol
     
  10. Jul 26, 2005 #9

    lurflurf

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    No this it is no harder. Plus this is nice if one wants to compute
    sin(pi/2+.00001) approximately.
     
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