I want to calculate the following sum in Haskell: (m + i)^n from i = m to n. So far I have thought of this:
sum2017 m n
|m > n = 0
|otherwise = (c + m)^n + sum2017 (m+1) n
where c = m
but the problem is that c changes every time due to getting assigned a new value from recursive calls
You can stuff the actual recursion in a local function, keeping the c binding outside:
sum2017 m = go m
where go μ n
| μ > n = 0
| otherwise = (c + μ)^n + go (μ+1) n
c = m
...of course, you may then omit the c = m entirely and just use (m + μ)^n in the recursion.
This specific example can also easily be done without any manual recursion at all, like
sum2017 m n = sum [(m+μ)^n | μ<-[m..n]]
In Haskell, you want to avoid explicit recursions as much as possible.
So what do you do instead? You use looping function that does it for you ;)
sum2017 m n = sum $ ((^ n) . (+ m)) <$> [m .. n]
Or you could use lambda functions instead of this pointfree: (\x -> (c + x) ^ n)
If you decide to have a local function as leftaroundabout suggested, makes it recurse on itself, not on the global function:
sum2017 m = go m
where go μ n
| μ > n = 0
| otherwise = (c + μ)^n + go (μ+1) n
c = m
Because otherwise, it can get out of control/memory pretty fast
(^ n) . (+ m).)