# Thread: Factorials

1. ## 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

2. 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

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

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

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.

3. im trying to find the derivative of cos x using the series
$\displaystyle cos x = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!)$

i worked out that each term could be given by

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

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

would it be possible just to differentiate each term ie $\displaystyle (x^2/2!)$ becomes $\displaystyle x$ and repeat for each term, how would u state that you are assuming something is continuing though.

4. Originally Posted by mrQwerty
im trying to find the derivative of cos x using the series
$\displaystyle cos x = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!)$

i worked out that each term could be given by

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

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

would it be possible just to differentiate each term ie $\displaystyle (x^2/2!)$ becomes $\displaystyle x$ 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

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

then by differentiating we get $\displaystyle (\cos(x))'=-sin(x)=\sum_{n=1}^{\infty}\frac{2n(-1)^nx^{2n-1}}{(2n)!}$

5. sorry to be a pain, but ive never seen

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

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.

6. Originally Posted by mrQwerty
sorry to be a pain, but ive never seen

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

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

7. thank you, but how would you be able to help me get from

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

to the series you showed me.

thanks again your help is really appreciated