Factorials

**mrQwerty** · April 28th 2008, 04:35 PM

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

**TheEmptySet** · April 28th 2008, 06:13 PM

Originally Posted by mrQwerty

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.

**mrQwerty** · April 28th 2008, 07:59 PM

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.

**Mathstud28** · April 28th 2008, 08:05 PM

Originally Posted by mrQwerty

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

**mrQwerty** · April 28th 2008, 08:24 PM

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.

**Mathstud28** · April 28th 2008, 08:27 PM

Originally Posted by mrQwerty

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

**mrQwerty** · April 28th 2008, 08:45 PM

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

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