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Maclaurin series

  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-9-14_11-35-29.png

    Note - I do not know why there is a .5 after the ampere. I think it is an error and I have asked my lecturer to clarify.
    2. Relevant equations
    upload_2016-9-14_11-36-20.png

    3. The attempt at a solution
    f(t)=sint2 f(0)=sin(0)2=0
    f'(t)=2sintcost f'(0)=sin2(0)=0
    f''(t)=2cos2t f''(0)=2cos2(0)=2
    f'''(t)=-4sin2t f'''(0)=-4sin2(0)=0
    fiv(t)=-8cos2t fiv(0)=-8cos(0)=-8

    .... until four non-zero terms

    Am I on the right track?
     
  2. jcsd
  3. Sep 14, 2016 #2

    Dr Transport

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    Doing your differentials incorrectly....

    [itex] i = \sin(t^2) [/itex], the first derivative is [itex] 2t\cdot\cos(t^2)[/itex]
     
  4. Sep 14, 2016 #3
    Thanks Doc.

    The next derivative will then be:

    upload_2016-9-14_15-44-20.png

    right?

    I have put in the 0 value up until the eighth term and I am still getting zero as the answer.
    Only f''(t), f'''(t) and fvi have yielded non-zero terms thus far.
     
  5. Sep 14, 2016 #4

    Ray Vickson

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    If you want the first four nonzero terms in the Maclauren expansion of ##f(t) = \sin(t^2)## you will need the first 14 derivatives of ##f##; that is, you will need
    $$\left(\frac{d}{dt}\right)^k f(0), \; k=1, 2, \ldots, 14.$$
     
  6. Sep 14, 2016 #5
    Hi Ray

    I got another non-zero term at the fourth non zero term at the 10th derivative.
     
  7. Sep 14, 2016 #6
    upload_2016-9-14_16-46-6.png

    Plugging in zero values for x I get 60480
     
  8. Sep 14, 2016 #7

    Ray Vickson

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    You will get another one at the 14th derivative.

    It must be obvious by now that the way I did it was using some other approach!
     
  9. Sep 14, 2016 #8
    (d/dt) = 2t⋅cos(t2) ?
     
  10. Sep 14, 2016 #9

    pasmith

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    You can obtain the first four terms of the series for [itex]\sin(t^2)[/itex] simply by obtaining the first four terms in the series for [itex]\sin x[/itex] and substituting [itex]x = t^2[/itex].

    Only if you have a more complicated function than [itex]at^n[/itex] as the argument, such as [itex]\sin(t + 3)[/itex] or [itex]\sin(e^t)[/itex], will calculating derivatives compete with "write down the series for [itex]\sin(x)[/itex] and substitute for [itex]x[/itex]" in efficiency.
     
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