Question

I need some help with G-5b in problem set 1. How do you show that only one term is not zero? I managed to prove to myself that if u=x^p then u^(p)=p!, but I need to understand where the zeros come from when using the Leibniz formula.

Answer 1

You may see the correct pdf. There is a small mistake in the provided solution.

Answer 2

The correction is minor. It has to do only with the chosen order of coefficients and it does not affect the result. If you have any questions just ask me.

Answer 3

You've already proven to yourself that the p'th derivative of x^p is p!. Note that p! is a constant, so if you take the next derivative, you get zero:

\[u = x^p => u^{(p)} = p! => u^{(p+1)} = 0.\]

Once you have that, you'll notice that all of the early terms in the sequence have a derivative of u that is much higher than p (starting with u^(p+q) and going down), and all of the late terms have a similarly high derivative of v. The only time this is not true is when we are at the qth term in the sequence, when we have: \[u^{(p)}v^{(q)}\]. Once you work out the value of the binomial coefficient for that term, and cancel out, you're left with (p+q)! as the answer.