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Thread: did i developed these teylor series correctly..

  1. #1
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    did i developed these teylor series correctly..

    A.
    $\displaystyle
    sin^2 x=0+x^2-\frac{x^4}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{4n+2}}+O(4n+2)
    $
    B.
    $\displaystyle
    xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^ {n+1}}{n!}+O(n+1)
    $
    C.
    $\displaystyle
    xsin^3 x=0+x^4-\frac{x^6}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{5n+6}}+O(5n+4)
    $
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    A.
    $\displaystyle
    sin^2 x=0+x^2-\frac{x^4}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{4n+2}}+O(4n+2)
    $
    Remember that $\displaystyle 2\sin^2 x = 1 - \cos 2x \implies 2\sin^2 x = 1 - \sum_{n=0}^{\infty} \frac{(2x)^{2n}(-1)^n}{(2n+1)!}$.

    $\displaystyle
    xsin^3 x=0+x^4-\frac{x^6}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{5n+6}}+O(5n+4)
    $
    This one is similar but you need an identity for $\displaystyle \sin^3x$ like in the first one.
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  3. #3
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    A:
    $\displaystyle
    cos 2x=cos^2x-sin^2x=1-2sin^2x\\
    $
    $\displaystyle
    f(x)=sin^2x=\frac{1-cos 2x}{2}\\
    $
    $\displaystyle
    f(0)=0\\
    $
    $\displaystyle
    f'(x)=\frac{1-cos 2x}{2}=cos 2xsin2x=\frac{sin4x}{2}\\
    $
    $\displaystyle
    f'(0)=0\\
    $
    $\displaystyle
    f''(x)=2cos4x\\
    $
    $\displaystyle
    f^{(3)}=-8sin4x\\
    $
    $\displaystyle
    f^{(4)}=-32cos4x\\
    $
    i know to to use each derivative to build a member out of it.
    how to defne the n'th member??

    regarding b:
    $\displaystyle
    xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^ {n+1}}{n!}+O(n+1)
    $
    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??
    Last edited by transgalactic; Feb 14th 2009 at 10:05 PM.
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  4. #4
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    regarding C:
    $\displaystyle
    sin3x=3sinx-4sin^3x\\
    $
    $\displaystyle
    f(x)=sin^3x=\frac{3sinx-sin3x}{4}\\
    $
    $\displaystyle
    f(0)=0\\
    $
    $\displaystyle
    f'(x)=\frac{3cosx-\frac{cos3x}{3}}{4}\\
    $
    $\displaystyle
    f'(0)=0
    $
    $\displaystyle
    f''(x)=\frac{-3sinx-\frac{sin3x}{9}}{4}\\
    $
    $\displaystyle
    f''(0)=0
    $
    for every member i get 0
    ??
    Last edited by transgalactic; Feb 15th 2009 at 03:31 AM.
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  5. #5
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    can you tell me if i solved correctly A,B
    ad where is my mistake in C??
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    regarding C:
    $\displaystyle
    sin3x=3sinx-4sin^3x\\
    $
    $\displaystyle
    f(x)=sin^3x=\frac{3sinx-sin3x}{4}\\
    $
    $\displaystyle
    f(0)=0\\
    $
    $\displaystyle
    f'(x)=\frac{3cosx-\frac{cos3x}{3}}{4}\\
    $
    $\displaystyle
    f'(0)=0
    $
    $\displaystyle
    f''(x)=\frac{-3sinx-\frac{sin3x}{9}}{4}\\
    $
    $\displaystyle
    f''(0)=0
    $
    for every member i get 0
    ??
    You stopped too soon! Since sin(x) starts with x, the lowest power of sin^3(x) is x^3.
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  7. #7
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    it doesnt matter
    both sinx and sin3x transform into cosx and cos3x and vise versa
    in both cases i get zero for every derivative
    where did i stop early
    i transformed power 3 function into a much easier to differentiate
    power 1 function.
    whats the problem??

    what about A and B
    are they ok??
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