1
$\begingroup$

I asked a question, and several people stated that it is not clear enough. So I fixed all what they been asking me to, but I did not got a feedback if my fix is satisfiable.

Please tell me if the question is clear enough or should I add additional modifications?

My goal is to find solution for:

$$f'(n)=\begin{cases} a &\text{if $n \bmod c=0$ or in the range of $\{c-d,c-1\}$}\\ b &\text{otherwise} \end{cases}$$ $$\frac{e}{n}=y - \sum_{0}^{n}{f'(n)}$$

  • $a,b,c,d,e,y$ are known positive integers

I want to get $n$ representation with $a,b,c,d,e,y$

$\endgroup$
  • $\begingroup$ For me it is too unusual to have the derivative ($f'$) in such a functional definition so I just skipped and assume there are much more clever people around who'll have an idea how to deal with this. Maybe the question is ok, maybe not: I don't see it immediately but also see no obvious problem ... $\endgroup$ – Gottfried Helms Jan 29 '15 at 21:11
  • $\begingroup$ The notation is very problematic. Especially the sum is probably meant to be $$\sum_{i=0}^n f'(i) \ne \sum_{?=0}^n f'(n) = (n+1)f'(n)$$ Also, a "range" is usually an interval while $\{c-d, c-1\}$ is a set. Furthermore if you give such functional equations, you should give the domain of your function $f$. Is it defined on $\mathbb N$ or on $\mathbb R$ or on $\mathbb Z_n \simeq \{0,\ldots, n-1\}$? As-is it is unclear what you are asking. $\endgroup$ – AlexR Jan 30 '15 at 15:52
2
$\begingroup$

The notation is very problematic. Especially the sum is probably meant to be $$\sum_{i=0}^n f'(i) \ne \sum_{?=0}^n f'(n) = (n+1)f'(n)$$ Also, a "range" is usually an interval while $\{c-d, c-1\}$ is a set. Furthermore if you give such functional equations, you should give the domain of your function $f$. Is it defined on $\mathbb N$ or on $\mathbb R$ or on $\mathbb Z_n \simeq \{0,\ldots, n-1\}$?
As-is it is unclear what you are asking. Try to provide complete information. Everything you are given, including the context under wich the question arose, can be helpful to understand the questions intent and to provide a good and useful answer.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .