1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Use the calculus rule $\frac{d}{dx} x^i = i \frac{d}{dx} x^{i-1}$ with $x=p$. $\endgroup$ – angryavian Jun 9 '17 at 22:46
4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Thanks, that makes sense! Math rocks! $\endgroup$ – MarksCode Jun 9 '17 at 23:02
  • $\begingroup$ Haha! Indeed it does. $\endgroup$ – Frpzzd Jun 9 '17 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.