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Taylor series question

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Taylor's theorem can be stated f(a+x)=f(a)+xf'(a)+(1/2!)(x^2)f''(a)+...+(1/n!)(x^n)Rn
    where Rn=fn(a+y), 0≤y≤x
    Use this form of Taylor's theorem to find an expansion of sin(a+x) in powers of x, and show that in this case, mod([itex]\frac{x^n Rn}{n!}[/itex])[itex]\rightarrow[/itex]0 as n[itex]\rightarrow[/itex][itex]\infty[/itex] for all x.

    2. Relevant equations



    3. The attempt at a solution

    sin(a+x)=sin(a)+xcos(a)-[itex]\frac{1}{2!}[/itex]x^2sin(a)-[itex]\frac{1}{3!}[/itex]x^3cos(a)...

    I don't know how to prove the next bit. Also, I don't understand why Rn=fn(a+y) rather than Rn=fn(A). Any help would be great.
     
    Last edited: Jan 8, 2012
  2. jcsd
  3. Jan 8, 2012 #2
    I'm still stuck on this. Any help would be brilliant!
     
  4. Jan 8, 2012 #3

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    Hi Lucy Yeats! :smile:

    ##R_n=f^{(n)}(a+y)## is part of the remainder term.
    Taylor's theorem states that there is such an y so that the remainder term is equal to the sum of the remaining terms in the series.

    You already have the expansion for sin(a+x), although perhaps you should find a generic formula for the terms.

    Furthermore you would need to find the limit of the remainder term.
    Can you think of an upper and a lower bound for Rn?
     
    Last edited: Jan 8, 2012
  5. Jan 8, 2012 #4
    Hello again!

    The question says 0≤y≤x, so maybe fn(a)≤Rn≤fn(a+x)?
     
  6. Jan 8, 2012 #5

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    That would only be true if fn would be a monotonically increasing function.
    But sines and cosines are notorious for it that they are not.

    What do you think ##f^{(n)}(u)## looks like, knowing that f(u)=sin(u)?
     
  7. Jan 8, 2012 #6
    fn(u) is always between 1 and -1?
     
  8. Jan 8, 2012 #7

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    How did you come to that idea?
     
  9. Jan 8, 2012 #8
    Because the gradient is always sin, cos, -sin, or -cos, which have ranges between -1 and 1.
     
  10. Jan 8, 2012 #9

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    Right!

    So...
     
  11. Jan 8, 2012 #10
    Can I have another hint? I really can't see the next step.
     
  12. Jan 8, 2012 #11

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    You are supposed to find the limit of ##|\frac{x^n R_n}{n!}|##.
    What do you know and what can you say about this limit?
     
    Last edited: Jan 8, 2012
  13. Jan 9, 2012 #12
    So mod(Rn) is between -1 and 1, so I thought about the x^n/n! part. The limit will only be zero if x is less than one. But x could be greater than 1, so I'm confused.
     
  14. Jan 9, 2012 #13

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    So suppose x>1, say x=100.
    What will x^n/n! be for large(r) values of n?
     
  15. Jan 9, 2012 #14
    So if x=100, 100^n>n!
     
  16. Jan 9, 2012 #15

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    How do you know that 100^n>n! ?
    Is that true for every n?
     
  17. Jan 9, 2012 #16
    Is that ony true if n<x?
     
  18. Jan 9, 2012 #17

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    Let's pick a smaller value for x, say x=3.

    Then x^3 = 3x3x3 > 1x2x3 = n!
    x^4 = 3x3x3x3 > 1x2x3x4 = n!
    What do you get for n=5, 6, 7, 10, 100?
     
  19. Jan 10, 2012 #18

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    Hey Lucy!

    Did you give up on this thread?
    That would be a pity!
     
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