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Did i developed these teylor series correctly

  1. Feb 14, 2009 #1
    A.
    [tex]
    sin^2 x=0+x^2-\frac{x^4}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{4n+2}}+O(4n+2)
    [/tex]
    B.
    [tex]
    xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^{n+1}}{n!}+O(n+1)
    [/tex]
    C.
    [tex]
    xsin^3 x=0+x^4-\frac{x^6}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{5n+6}}+O(5n+4)
    [/tex]
     
    Last edited: Feb 14, 2009
  2. jcsd
  3. Feb 15, 2009 #2
    A:
    [tex]
    cos 2x=cos^2x-sin^2x=1-2sin^2x\\
    [/tex]
    [tex]
    f(x)=sin^2x=\frac{1-cos 2x}{2}\\
    [/tex]
    [tex]
    f(0)=0\\
    [/tex]
    [tex]
    f'(x)=\frac{1-cos 2x}{2}=cos 2xsin2x=\frac{sin4x}{2}\\
    [/tex]
    [tex]
    f'(0)=0\\
    [/tex]
    [tex]
    f''(x)=2cos4x\\
    [/tex]
    [tex]
    f^{(3)}=-8sin4x\\
    [/tex]
    [tex]
    f^{(4)}=-32cos4x\\
    [/tex]
    i know to to use each derivative to build a member out of it.
    how to defne the n'th member??

    regarding b:
    [tex]
    xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^{n+1}}{n!}+O(n+1)
    [/tex]
    i am sure that its correct because i took the series for e^x and multiplied eah member by x.
    so the only error that could be is with the remainder .
    i thought that if we multiply the remainder by x we add 1 to it
    where is my mistake??

    regarding C:
    [tex]
    sin3x=3sinx-4sin^3x\\
    [/tex]
    [tex]
    f(x)=sin^3x=\frac{3sinx-sin3x}{4}\\
    [/tex]
    [tex]
    f(0)=0\\
    [/tex]
    [tex]
    f'(x)=\frac{3cosx-\frac{cos3x}{3}}{4}\\
    [/tex]
    [tex]
    f'(0)=0
    [/tex]
    [tex]
    f''(x)=\frac{-3sinx-\frac{sin3x}{9}}{4}\\
    [/tex]
    [tex]
    f''(0)=0
    [/tex]
    for every member i get 0
    ??

    did i solved A,B correctly
    where is my mistake for C

    ??
     
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