# Help me improve my question

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.

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 $a,b,c,d,e,y$

• 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 ... Jan 29 '15 at 21:11
• 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. Jan 30 '15 at 15:52

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 $\mathbb Z_n \simeq \{0,\ldots, n-1\}$?