Is there a generalized way to find an inverse of a function defined as a sum (if the inverse exists)?

Question

Suppose I had a summation for a monotonic increasing function, like $f(n)=\sum_{k=1}^{n}k$. We already know this has a closed form solution in the form of $\frac{n(n+1)}{2}$. If you want to find $n$, you could use the quadratic formula. But, what if you didn't know the closed formula? What if all you knew was some $\sum_{k=1}^{n}g(k)$? Is there some way to invert this process to find $n$ like with $g^{-1}(n)$?

Answer 1

It's possible to find many inverses of a sum. for example, if a+b=10, then $1+9$, $2+8$, $3+7$, $4+6$, $5+5$ are all solutions to the equation and there's also many negative solutions like $(-1)+11$.

Good way to think about this, is that given 10 length sequence of successors, you split the sequence to two parts and the cut point is arbitrarily chosen, thus 0+10, 1+9, 2+8 etc are all good working solutions.

So steps to solve this kind of problems($a+b=c$):

• take $c$
• split it to successors $1+1+1+..+1$
• split the successors to many groups: $(1+1+1)+(1+1+1+1+1)$
• mark the groups as variables $a+b$

Now, to get more complexity, we should do $f(x)+f(y)+f(z)=d$, would become $$ d \mapsto (a+b+c) \mapsto (a,b,c) \mapsto (f^{-1}(a),f^{-1}(b),f^{-1}(c)) \mapsto (x=f^{-1}(a),y=f^{-1}(b),z=f^{-1}(c))$$ i.e. after finding $a,b,c$ for solutions to $a+b+c=d$, just apply inverse of the function.

Now if you have something like $f(1)+f(2)+f(3)=d$, then we need to use the previous logic, and add just ${x=1,y=2,z=3}$ to get $(f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3)$.