Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor polynomial question

  1. Dec 9, 2007 #1
    Find the Taylor polynomial of degree 10 about x=0 for f(x)=sin2x (show all work)

    This is what i have:

    M10= f(0)+f[tex]\hat{}[/tex]1(0)x+f[tex]\hat{}[/tex]2(0)x[tex]\hat{}[/tex]2/2!+f[tex]\hat{}[/tex]3(0)x[tex]\hat{}[/tex]3/3!+....+f[tex]\hat{}[/tex]10(0)x[tex]\hat{}[/tex]10/10!

    f(x)=sin2x
    f(0)=sin2(0)=0

    f[tex]\hat{}[/tex]1(x)=2cos2x
    f[tex]\hat{}[/tex]1(0)=2cos2(0)=2

    f[tex]\hat{}[/tex]2(x)=-4sin2x
    f[tex]\hat{}[/tex]2(0)=-4sin2(0)=0

    f[tex]\hat{}[/tex]3(x)=-8cos2x
    f[tex]\hat{}[/tex]3(0)=-8cos2(0)=-8

    f[tex]\hat{}[/tex]4(x)=16sin2x
    f[tex]\hat{}[/tex]4(0)=16sin2(0)=0

    f[tex]\hat{}[/tex]5(x)=32cos2x
    f[tex]\hat{}[/tex]5(0)=32cos2(0)=32

    f[tex]\hat{}[/tex]6(x)=64sin2x
    f[tex]\hat{}[/tex]6(0)=64sin2(0)=0

    f[tex]\hat{}[/tex]7(x)=128cos2x
    f[tex]\hat{}[/tex]7(0)=128cos2(0)=128

    f[tex]\hat{}[/tex]8(x)=256sin2x
    f[tex]\hat{}[/tex]8(0)=256sin2(0)=0

    f[tex]\hat{}[/tex]9(x)=412cos2x
    f[tex]\hat{}[/tex]9(0)=412cos2(0)=412

    f[tex]\hat{}[/tex]10(x)=824sin2x
    f[tex]\hat{}[/tex]10(0)=824sin2(0)=0


    So, M10=0+2x+0-8x[tex]\hat{}[/tex]3/3!+0+32x[tex]\hat{}[/tex]5/5!+0+128x[tex]\hat{}[/tex]7/7!+0+412x[tex]\hat{}[/tex]9/9!+0

    Therefore, 2x-8x[tex]\hat{}[/tex]3/3!+32x[tex]\hat{}[/tex]5/5!+128x[tex]\hat{}[/tex]7/7!+412x[tex]\hat{}[/tex]9/9!
     
  2. jcsd
  3. Dec 9, 2007 #2
    Find the Taylor polynomial of degree 10 about x=0 for f(x)=sin2x (show all work)

    This is what i have:

    M10= f(0)+f[tex]\hat{}[/tex]1(0)x+f[tex]\hat{}[/tex]2(0)x[tex]\hat{}[/tex]2/2!+f[tex]\hat{}[/tex]3(0)x[tex]\hat{}[/tex]3/3!+....+f[tex]\hat{}[/tex]10(0)x[tex]\hat{}[/tex]10/10!

    f(x)=sin2x
    f(0)=sin2(0)=0

    f[tex]\hat{}[/tex]1(x)=2cos2x
    f[tex]\hat{}[/tex]1(0)=2cos2(0)=2

    f[tex]\hat{}[/tex]2(x)=-4sin2x
    f[tex]\hat{}[/tex]2(0)=-4sin2(0)=0

    f[tex]\hat{}[/tex]3(x)=-8cos2x
    f[tex]\hat{}[/tex]3(0)=-8cos2(0)=-8

    f[tex]\hat{}[/tex]4(x)=16sin2x
    f[tex]\hat{}[/tex]4(0)=16sin2(0)=0

    f[tex]\hat{}[/tex]5(x)=32cos2x
    f[tex]\hat{}[/tex]5(0)=32cos2(0)=32

    f[tex]\hat{}[/tex]6(x)=64sin2x
    f[tex]\hat{}[/tex]6(0)=64sin2(0)=0

    f[tex]\hat{}[/tex]7(x)=128cos2x
    f[tex]\hat{}[/tex]7(0)=128cos2(0)=128

    f[tex]\hat{}[/tex]8(x)=256sin2x
    f[tex]\hat{}[/tex]8(0)=256sin2(0)=0

    f[tex]\hat{}[/tex]9(x)=412cos2x
    f[tex]\hat{}[/tex]9(0)=412cos2(0)=412

    f[tex]\hat{}[/tex]10(x)=824sin2x
    f[tex]\hat{}[/tex]10(0)=824sin2(0)=0


    So, M10=0+2x+0-8x[tex]\hat{}[/tex]3/3!+0+32x[tex]\hat{}[/tex]5/5!+0+128x[tex]\hat{}[/tex]7/7!+0+412x[tex]\hat{}[/tex]9/9!+0

    Therefore, 2x-8x[tex]\hat{}[/tex]3/3!+32x[tex]\hat{}[/tex]5/5!+128x[tex]\hat{}[/tex]7/7!+412x[tex]\hat{}[/tex]9/9!

    does this look right? anyone?
     
  4. Dec 9, 2007 #3

    rock.freak667

    User Avatar
    Homework Helper

    well your differentials of cos(2x) are wrong as
    [tex]\frac{d}{dx}(cos2x)=-2sin2x[/tex]

    but if I were you, I would just find sinx around x=0 and after you find it, replace all the x's with 2x
     
    Last edited: Dec 9, 2007
  5. Dec 10, 2007 #4
    It was looking right until the end...

    Sixth derivative - you forgot a negative sign
    Ninth derivative - 512, not 412
    Tenth derivative - the sign probably stays the same after double corrections, 512*2=1024

    Other than those small errors, you are golden!
     
  6. Dec 10, 2007 #5
    Please see my reply on your other double post.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Taylor polynomial question
  1. Taylor Polynomials (Replies: 2)

  2. Taylor Polynomials (Replies: 2)

  3. Taylor polynomials (Replies: 1)

Loading...