I'm going through a proof right now and am having trouble figuring out the math behind one line. It says:
$$\sum_{i=0}^\infty i^2p^{i-1} = \sum_{i=0}^\infty i(\frac{d}{dp}p^i) $$
I know this question is vague but can anybody explain why this is the case?
Answer
First of all, the summation notation is unnecessary and irrelevant. Remove it and we get
$$i^2p^{i-1}=i\big(\frac{d}{dp}p^i\big)$$
Okay. Recall the formula
$$\frac{d}{dx}x^k=kx^{k-1}, k\ne0$$
Then we have that
$$\frac{d}{dp}p^{i}=ip^{i-1}, i\ne0$$
and so, by substitution, we have
$$i\big(\frac{d}{dp}p^i\big)=i\big(\frac{d}{dp}p^i\big)$$
$$i\big(\frac{d}{dp}p^i\big)=i(ip^{i-1})$$
$$i\big(\frac{d}{dp}p^i\big)=i^2p^{i-1}$$
Does that answer your question?
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