1

Consider some function $f(k) = \frac 1 {a_1 a_2 \cdots a_k}$ so that the number of terms in the denominator changes with $k$. My question is there anyway to discuss its derivative, if so, how would one go about beginning to differentiate this? Here's my attempt at making any sense of it.

$$f(k) = \frac 1 {a_1 a_2 \cdots a_k}$$ $$\ln(f(k)) = -\ln({a_1 a_2 \cdots a_k)}$$ $$\frac {f'(k)}{f(k)}=-\frac{1}{a_1 a_2 \cdots a_k} \frac d {dk}(a_1 a_2 \cdots a_k)$$ $$f'(k)=-\frac d {dk}(a_1a_2\cdots a_k)$$ $$f'(k)=-\frac d {dk}\frac{1}{f(k)}$$ I'm not entirely sure what to do with this? Any thoughts or opinions?

Michael Hardy
  • 1
Ryan Goulden
  • 803
  • Your function isn't a function of a real variable, so you'll need make clear what you mean by derivative. Do you mean the finite difference $f(k+1)-f(k)$? – Hayden Jan 15 '17 at 21:40
  • What do you mean it's not a function of a real variable?? The input of the function would be in the naturals? – Ryan Goulden Jan 15 '17 at 21:41
  • Being a function of a real variable means you can input any real number, or at least all reals in an open interval. The derivative is defined by $\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}} = f'(x)$. $\mathbb{N}$ is discrete, so while this limit will exist, it won't be unique, and can be literally anything you want. – Hayden Jan 15 '17 at 21:43
  • The way you describe $f$, it is a sequence. What would it mean to have a non-natural number of $a_i$ terms? (What @Hayden says in his second comment..) – MathematicianByMistake Jan 15 '17 at 21:45
  • Yeah, it is indeed a sequence. And Yes, I was just thinking about having like a "half" of a term seems kind of meaningless. – Ryan Goulden Jan 15 '17 at 21:57

1 Answers1

0

The number of terms in the denominator is an integer but derivatives are concerned with infinitesimal changes at a point. Recall the definition of the derivative at a point $k$

$$f'(k) = \lim_{h \to 0} \frac{f(k+h)-f(k)}{h}$$

The limit does not make sense since $f$ only applies to natural numbers. Therefore the derivative isn't meaningful.

Hugh
  • 2,341
  • Hmm, I see. Then how can we say exactly that $$\frac{d}{dx} \sum_{n=1}^{x}\frac{1}{a^n}=\frac{{a^{-x}}ln(a)}{a-1}$$

    Do you first find the partial sum formula then differentiate?

    My problem comes from trying to differentiate this $$\prod_{l=1}^{k}\frac{1}{a_l}$$

     – Ryan Goulden Jan 15 '17 at 21:52
  • I guess those are a bit fundamentally different. – Ryan Goulden Jan 15 '17 at 21:55
  • @RyanGoulden : Proper MathJax notation is $\ln(a)$, not $ln(a)$. It is coded as \ln a. This not only prevents italicization but also results in proper spacing in things like $a\ln b$ and $a\ln(b). \qquad$ – Michael Hardy Jan 15 '17 at 22:08
  • Oh, thank you, I didn't know. My apologies. – Ryan Goulden Jan 15 '17 at 22:10
  • @RyanGoulden That expression for the derivative is got from getting a partial sum and taking the derivative. That is not strictly valid mathematically because the partial sum is only valid when $x$ is a natural number, if you treat it as true you may reach some contradiction. However it can still be useful in something like an optimization problem where we use derivatives to find a solution iteratively. It works for that sort of thing because the derivative is being used as a tool rather than being taken as a mathematical fact. – Hugh Jan 16 '17 at 00:45