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

  • Thread starter RK455
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In summary, the homework statement is to prove that if v(1) = 0 for all positive l, then P(1) = 0. The attempted solution is to test when n=0 and find that x-1=1. This gives d^0/dx^0 [(x-1)^0] = 1. Assuming this holds for n=k, we can conclude that d^k/dx^k [(x-1)^k] = k. Finally, if k=n, we can conclude that d^(k+1)/dx^(k+1) [x-1]^(k+1) = d^k/dx^k(
  • #1
RK455
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Homework Statement



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

Homework 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..

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?
 
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  • #2
Yes, it is hideous, but (unless my eyes glazed over something important), your induction works just fine!
 
  • #3
daveb said:
Yes, it is hideous, but (unless my eyes glazed over something important), your induction works just fine!

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 realized 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
 
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1. What is the definition of d^l[(x-1)^l]/dx^l?

The notation d^l[(x-1)^l]/dx^l represents the second derivative of the function (x-1)^l. It is the rate of change of the rate of change of the function with respect to x.

2. How do you prove d^l[(x-1)^l]/dx^l = l?

To prove d^l[(x-1)^l]/dx^l = l, we can use the definition of the second derivative and apply the power rule. This will result in d^l[(x-1)^l]/dx^l = l*(l-1)*(x-1)^(l-2). Since (l-1)*(x-1)^(l-2) is a constant, we can simplify it to d^l[(x-1)^l]/dx^l = l.

3. What is the significance of proving d^l[(x-1)^l]/dx^l = l?

Proving d^l[(x-1)^l]/dx^l = l demonstrates the relationship between the rate of change and the rate of change of the rate of change of a function. This is useful in understanding the behavior of functions and their derivatives.

4. Can d^l[(x-1)^l]/dx^l be equal to a different constant besides l?

No, d^l[(x-1)^l]/dx^l can only be equal to l. This is because the second derivative of a function is dependent on the function itself, and in this case, the function is (x-1)^l. Therefore, the constant in the second derivative must also be l.

5. How does proving d^l[(x-1)^l]/dx^l = l relate to the power rule for derivatives?

The power rule for derivatives states that the derivative of x^n is nx^(n-1). In the case of proving d^l[(x-1)^l]/dx^l = l, we are essentially applying the power rule twice. This is because the first derivative of (x-1)^l would result in l*(x-1)^(l-1), and the second derivative would then result in l*(l-1)*(x-1)^(l-2), which simplifies to l.

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