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What is the derivative for the following function $$\frac{d}{dw} \sum_{i=1}^{\lfloor\frac{K_o}{w} \rfloor} i^{-\gamma} \left(1-(1-w)e^{-a w}\right),\quad w \in \mathbb{R}^+ \text{ and } K_o,\gamma, \text{ and } a \text{ are positive constants} $$

M. Emara
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1 Answers1

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As one could see, the problem is $$U(w):=\left\lfloor\frac{K_0}{w}\right\rfloor$$ since it is not a fixed number. So, let us examine the following cases:

  1. $w>K_0$. Then, we have that $\frac{K_0}{w}<1$, so $U(w)=0$ and the result is trivially $0$.
  2. $K_0\geq w>\frac{K_0}{2}$. Then, we have that $1\leq\frac{K_0}{w}<2$, so $U(w)=1$ and the result is $$e^{-aw}+a(1-w)e^{-aw}=(1+a-w)e^{-aw}$$
  3. $\frac{K_0}{2}\geq w>\frac{K_0}{3}$. Then, we have that $2\leq\frac{K_0}{w}<3$, so $U(w)=2$ and we have to differentiate: $$\sum_{i=1}^2i^{-\gamma}(1-(1-w)e^{-aw})=\left(1-(1-w)e^{-aw}\right)\sum_{i=1}^2i^{-\gamma}$$ Which is: $$(1+a-w)e^{-aw}\left(1+\frac{1}{2^\gamma}\right)$$
  4. The general case $\frac{K_0}{n}\geq w>\frac{K_0}{n+1}$ where $U(w)=n$, so the all-wanted derivative is - this is simple to calculate: $$(1+a-w)e^{-aw}\left(1+\frac{1}{2^\gamma}+\frac{1}{3^\gamma}+\dots+\frac{1}{n^\gamma}\right)$$

Also note that the given function may not be differentiable at the points $w_n=\frac{K_0}{n}$, for the most values of $a,K_0,\gamma$.

Vassilis Markos
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  • Thank you. According to that the derivative for a summation with respect to the an integer upper limit, $$(\frac{d}{dn}\sum_{i=0}^{n}f(n)g(i) = \frac{d}{dn}f(n) \sum_{i=1}^{n}g(i), \quad n \in \mathbb{N}^{+} \text{ and } f: \mathbb{N}^{+} \to \mathbb{R}^{+})$$. Is it correct? – M. Emara Jan 27 '18 at 19:13
  • How do you define differntiation on $\mathbb{N}$. Note that, in the above case $w$ is a usual real variable, it just happens that the upper limit of summation is not fixed. – Vassilis Markos Jan 27 '18 at 19:19
  • sorry I mean for an integer $n$. Is the derivative is carried for the function $f(n)$ without affecting the summation over the $g(i)$ function? – M. Emara Jan 27 '18 at 19:28
  • Again, how is differentiation defined on $\mathbb{Z}$? – Vassilis Markos Jan 27 '18 at 19:34
  • @ Βασίλης, like in the following link – M. Emara Jan 27 '18 at 19:45
  • Wait, that's another thing. I'm asking what is the notion of $\frac{d}{dn}$ where $n$ is an integer. In which way do you differentiate over integers or naturals? How can you take limits? Differentiating a sum with a variable upper limit can not in general be handled as above, though. – Vassilis Markos Jan 27 '18 at 19:48