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Math Help - Power series expansion proofs for the differential of sinx and cosx

  1. #1
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    Power series expansion proofs for the differential of sinx and cosx

    Let sin x and cos x be defined by their power series expansions

    how can we use term by term differentiation to show that

    d/dxsin x = cos x and

    d/dxcos x = − sin x.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by maximus101 View Post
    Let sin x and cos x be defined by their power series expansions

    how can we use term by term differentiation to show that

    d/dxsin x = cos x and

    d/dxcos x = − sin x.
    Is...

    \displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} (1)

    ... so that...

    \displaystyle \frac{d}{dx} \sin x = \sum_{n=0}^{\infty} (-1)^{n} (2n+1)\ \frac{x^{2n}}{(2n+1)!}=

    \displaystyle =  \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!} = \cos x (2)

    The same procedure is valid for \cos x...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    Is...

    \displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} (1)

    ... so that...

    \displaystyle \frac{d}{dx} \sin x = \sum_{n=0}^{\infty} (-1)^{n} (2n+1)\ \frac{x^{2n}}{(2n+1)!}=

    \displaystyle =  \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!} = \cos x (2)

    The same procedure is valid for \cos x...

    Kind regards

    \chi \sigma
    I think this question is entirely circular reasoning.

    You require knowing the derivatives of \displaystyle \sin{x} and \displaystyle \cos{x} in order to generate their series expansions in the first place...
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  4. #4
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    No it's not circular.

    Note that he differentiate the power series (given hypothesis) and identified that series as cosine.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    No it's not circular.

    Note that he differentiate the power series (given hypothesis) and identified that series as cosine.
    The first sentence of the question is "Let \displaystyle \sin{x} and \displaystyle \cos{x} be defined by their power series expansions..."

    How is one supposed to generate the power series expansions without already knowing the derivatives of \displaystyle \sin{x} and \displaystyle \cos{x}?
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  6. #6
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    Quote Originally Posted by Prove It View Post
    The first sentence of the question is "Let \displaystyle \sin{x} and \displaystyle \cos{x} be defined by their power series expansions..."

    How is one supposed to generate the power series expansions without already knowing the derivatives of \displaystyle \sin{x} and
    \displaystyle \cos{x}?
    Exactly the point!

    Let sin x and cos x be defined by their power series expansions

    So IF you were to define the sine & cosine functions by what we know their power series are. Guess what!! No surprise that \displaystyle {{d}\over{dx}}\,\sin(x)=\cos(x)\quad \text{  and  }\quad {{d}\over{dx}}\,\cos(x)=-\sin(x)

    It's not so much circular reasoning as it is an exercise to help students verify that the power series have the same properties as the functions they represent - indeed they are identical.
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