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Taylor series

  1. Feb 21, 2012 #1
    Find the Taylor series for f(x) centered at a. Assume power series expansion.
    f(x) = sin(x) at a= pi/2

    My answer ended up being the summation from n=0 to infinity of ((x-pi/2)^n ) / n!
    And the radius of convergence is R=infinity

    This is right ?
     
  2. jcsd
  3. Feb 21, 2012 #2

    HallsofIvy

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    No, that is not right. (In fact it looks like a series for [itex]e^{x-\pi/2}[/itex]. If you show how you got that, we might be able to point out an error.
     
  4. Feb 21, 2012 #3
    Oh yea I followed the example in the book for e^x that's probly why

    So where do I start then? All I know about Taylor series stuff is:

    F(x)= summation [ (f^n)(a)*(x-a)^n ] / n!



    Ok well when I followed it in the book you have (f^n)(pi/2) = sin(pi/2) and then you put the a value which is pi/2 into the definition of a Taylor series so you get:

    Summation [ sin(pi/2)(x-pi/2)^n ] / n!
     
    Last edited: Feb 21, 2012
  5. Feb 21, 2012 #4

    Mark44

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    This is not true.
    f(x) = sin(x), f(pi/2) = 1
    f'(x) = cos(x), f'(pi/2) = 0
    f''(x) = -sin(x), f''(pi/2) = -1
    etc.
     
  6. Feb 21, 2012 #5
    Ok so how does it go (I don't have my book with me)
    It's like f(x) = f'(x)*(x-a) + f''(x)*(x-a)^2 + f'''(x)*(x-a)^3 + .... With the first part has 1! Under it, the second part has 2! Under it, and the third part had 4! Under it??

    So you just plug in that stuff? Idk
     
  7. Feb 21, 2012 #6

    Mark44

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    It goes like this:
    f(x) = f(a) + f'(a) * (x - a)/1! + f''(a) * (x - a)2/2! + f'''(a) * (x - a)3/3! + ... + f(n)(a) * (x - a)n/n! + ...
     
  8. Feb 22, 2012 #7
    Oh yea I think that's what I meant!

    So when you plug in values you get 1+0-1(((x-pi/2)^3)/2)+0+... right?
    So it has (-1)^n. I don't really know what else the series would be... Umm I get this:

    Summation n=0 to infinity: (-1)^n * ((x-pi/2)^(n+2)) / (n+2)!

    Does that work?
     
  9. Feb 22, 2012 #8

    Mark44

    Staff: Mentor

    No, your exponent on the third term is wrong, and you should show more terms. With too few terms, you won't be able to get the general term correctly.

    Make a table with the values for n, f(n)(x), and f(n)([itex]\pi/2[/itex]), similar to the one I did in post #4, but with more entries than I showed.
    No.
     
  10. Feb 23, 2012 #9
    oh, yea i meant this:

    1+0-1(((x-pi/2)^2)/2)+0+... i just had a typo.


    but okay, i did all the way up to f'''''''(x) and when you plug in the a value (pi/2) you get"
    1, 0, -1, 0, 1, 0, -1, 0

    i ended up getting: f(x) = 1 - (x-pi/2)2 / 2! + (x-pi/2)4 / 4! + (x-pi/2)6 / 6!

    so i dont know what the series would be. all i know is (-1)n should be there.
     
  11. Feb 23, 2012 #10

    Mark44

    Staff: Mentor

    That is the series (sort of), if you make a couple of corrections.
    1) You have a sign error on the last term you show.
    2) This is supposed to be an infinite series, so there is no last term. To indicate this, add "+ ..." after the last term you show.

    With these corrections, that is your Taylor series for f(x) = sin(x). Does the series have to be shown in closed form (i.e., as a summation)? If so, you need to be able to identify the general term of the series.

    Some things to note:
    Only even degree terms show up. IOW, the exponents are 0, 2, 4, 6, 8, ... This means that the exponent should be 2n.
    The terms alternate in sign. You can show this by having a factor of (-1)n or possibly (-1)n + 1.
     
  12. Feb 23, 2012 #11
    Yes I realize those corrections. Typo and the second is just me typing slower than im thinking.

    But it doesn't say if it had to be in closed form but I think that's how all the answers are so yea.

    Ok so te answer is just summation (-1)^n * ((x-pi/2)^(2n))/(2n)! Right?
     
    Last edited: Feb 23, 2012
  13. Mar 4, 2012 #12
    Is my series in the last post at the end the correct answer?
     
  14. Mar 4, 2012 #13

    SammyS

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    WolframAlpha says that's sin(x) .
     
  15. Mar 4, 2012 #14
    Cool awesome, that's what I wanted!
     
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