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Using Taylor's theorem

  1. Jul 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Using Taylor's theorem, prove that
    [tex]x-\frac{x^3}{6}<sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}[/tex] for x>0.

    2. Relevant equations



    3. The attempt at a solution
    I know what is Taylor theorem but i haven't done some basic questions on it, i mean the easy ones. I have no idea on how should i start here.
     
  2. jcsd
  3. Jul 1, 2012 #2

    Curious3141

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    Start by stating the Taylor series for sin x.

    Recognise it's an alternating series. What can you say about the behaviour of the partial sums as you alternately add and subtract terms, given that x is positive?
     
  4. Jul 1, 2012 #3
    Series for sin x:
    [tex]sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}........[/tex]

    The terms are getting smaller and smaller. For a positive x, i notice that
    [tex]x-\frac{x^3}{3!}<x-\frac{x^3}{3!}+\frac{x^5}{5!}[/tex]
    But i can't find my way to the answer, there's a sin x between.
     
  5. Jul 1, 2012 #4

    Curious3141

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    In a convergent alternating series, the partial sums oscillate around the total sum.

    So [itex]x[/itex] overestimates [itex]\sin x[/itex] because the next term in the series is negative.

    And [itex]x-\frac{x^3}{3!}[/itex] underestimates [itex]\sin x[/itex] because the next term in the series is positive.

    Can you complete the rest of the simple proof?
     
  6. Jul 1, 2012 #5
    I have never heard about "convergent alternating series" but i understand what you are saying.
    As the partial sums oscillate, [itex]x-\frac{x^3}{3!}+\frac{x^5}{5!}[/itex] overestimates sin x. So i think i have reached the answer, thanks Curious for the help! :smile:

    I will have to check about the convergent series, do you have any good link or should i refer Wikipedia?
     
  7. Jul 1, 2012 #6

    Curious3141

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    A brief google search gave me this document: http://www.millersville.edu/~bikenaga/calculus/altser/altser.pdf [Broken]

    The second half of page 3 is most relevant to what you're looking for.
     
    Last edited by a moderator: May 6, 2017
  8. Jul 1, 2012 #7
    Thank you once again Curious! :smile:
     
    Last edited by a moderator: May 6, 2017
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