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Math Help - Factorials

  1. #1
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    Factorials

    hey, trying to do a question, but don't know how to differentiate factorials, how would u differentiate (x-2)! any help would be great thanks
    Last edited by mrQwerty; April 29th 2008 at 12:27 AM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by mrQwerty View Post
    hey, trying to do a question, but don't know how to differentiate factorials, how would u differentiate (2x-2)! any help would be great thanks
    You would need to use the gamma function

    \Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt=(x-1)!

    \Gamma(2x-1)=\int_{0}^{\infty}t^{2x-2}e^{-t}dt=(2x-2)!

    Why do you need to take the derivative of a factorial?

    From here you could take the derivative of an integral I guess.

    I hope this helps.
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  3. #3
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    im trying to find the derivative of cos x using the series
    cos x = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!)

    i worked out that each term could be given by

    Tn = n^(2n-2)/(2n-2) * (-1)^(n+1)

    although now that i think about it, thats not going to help is it.

    would it be possible just to differentiate each term ie (x^2/2!) becomes x and repeat for each term, how would u state that you are assuming something is continuing though.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mrQwerty View Post
    im trying to find the derivative of cos x using the series
    cos x = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!)

    i worked out that each term could be given by

    Tn = n^(2n-2)/(2n-2) * (-1)^(n+1)

    although now that i think about it, thats not going to help is it.

    would it be possible just to differentiate each term ie (x^2/2!) becomes x and repeat for each term, how would u state that you are assuming something is continuing though.
    An alternative way would be listing cosine of x in its explicit form

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

    then by differentiating we get (\cos(x))'=-sin(x)=\sum_{n=1}^{\infty}\frac{2n(-1)^nx^{2n-1}}{(2n)!}
    Last edited by Mathstud28; April 28th 2008 at 08:25 PM.
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  5. #5
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    sorry to be a pain, but ive never seen

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

    also my question asks specifically to use the series to confirm the derivative of cosx is sinx. so im guessing that i have to use the series.


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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mrQwerty View Post
    sorry to be a pain, but ive never seen

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

    also my question asks specifically to use the series to confirm the derivative of cosx is sinx. so im guessing that i have to use the series.


    Two things...the derivative of cos(x) is -sin(x) and secondly that is the series cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!} is exactly the series you are talking about?

    Dont believe me put in values for n,1,2,3,4 and you will get the same series that you listed above
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  7. #7
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    thank you, but how would you be able to help me get from

    Tn = x^(2n-2)/(2n-2) * (-1)^(n+1)

    to the series you showed me.

    thanks again your help is really appreciated
    Last edited by mrQwerty; April 29th 2008 at 12:16 AM.
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