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Taylor series

  1. May 3, 2009 #1
    Find P5(x), the 5th order Taylor series, of sin (x) about x = 0. Hence find the 4th
    order Taylor series for x sin (2x) about x = 0.

    In this question why is it required to find the 5th order taylor series of sin(x) to find the 4th order taylor series of xsin(2x)?
     
  2. jcsd
  3. May 3, 2009 #2

    Mark44

    Staff: Mentor

    After you have P5(x), it's a simple matter to find the power series for xsin(2x). "Hence" in this problem just means "after that," I believe.
     
  4. May 4, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Mark44, I believe Ry122's question was specifically why it would be necessary to find the 5th order polynomial of one in order to find the 4th order polynomial of the other. And, in fact, it is not necessary. Multiplying an nth order polynomial by x gives, of course, an n+1 order polynomial.

    It would be sufficient to find the 3rd order Taylor's polynomial for sin(x) in order to find the 4th order Taylor's polynomial for x sin(2x). It would have made sense if you were looking for the 4th order Taylor's polynomial of sin(2x)/x.
     
  5. May 4, 2009 #4

    Mark44

    Staff: Mentor

    My answer was a little oblique. I don't think it is necessary to find the 5th order Taylor polynomial. My comment about "hence" was intended to convey my belief that this word was not used in it usual mathematics sense of "it therefore follow that..."
     
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