# Proving d^l[(x-1)^l]/dx^l = l!

1. Dec 20, 2011

### RK455

1. The problem statement, all variables and given/known data

Sorry I can't do this in latex, the PC I'm on doesn't display it and I can't confidently reel it off without checking so I'll have to do this in an ugly way! :/

Prove:
d^n/dx^n [(x-1)^n] = n!
For integer n

2. Relevant equations

I'm trying induction but it's never been my strong point, I'm not convinced I actually understand how induction works, whenever I try and get my head around it, it seems circular..

3. The attempt at a solution

Testing when n=0 gives (x-1)^0 = 1 = 0! so it holds when n=0

Now assuming true for n=k gives:
d^k/dx^k [(x-1)^k] = k!

If this holds for n=k, that it holds when n=k+1 must be implied by n=k being true:
d^(k+1)/dx^(k+1) [x-1]^(k+1) = d^k/dx^k(d/dx((x-1)^(k+1))) separating the differentiation operator
= d^k/dx^k((k+1)(x-1)^k) evaluating the inner derivative
= (k+1)d^k/dx^k((x-1)^k) bringing the constant (k+1) outside the derivative
= (k+1) k! using the result I'm trying to prove
= (k+1)!

Sorry, that's hideous.. is it OK?

2. Dec 20, 2011

### daveb

Yes, it is hideous, but (unless my eyes glazed over something important), your induction works just fine!

3. Dec 20, 2011

### RK455

OK thanks :)

A further question:
This proof is actually part of my solution to Q20 part C on this paper:
http://www.maths.cam.ac.uk/undergrad/nst-pastpapers/2009/PaperNST_IA_2.pdf [Broken]

I've managed to do the proof OK and get the required result but it seems a bit odd to me. I'm not sure why P(1) = 1 when if we sub x=0 into v, we get 0^l for all values of l greater than 0.

If v(1) = 0 for all positive l, then surely P(1) = 0 as well since P(1) is simply the derivative of v(1) l times multiplied by a constant factor?
Thanks

Edit: That was dumb of me... just realised we're taking the 'l'th derivative of the function v at x=1.. just because v(1) = 0, doesn't mean its derivatives are.

Thanks again

Last edited by a moderator: May 5, 2017
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