Determining n in sigma ($\Sigma_{x=0}^n$) Reffered here by https://mathoverflow.net/questions/41750/determining-n-in-sigma-x0n
I'm not entirely sure if this question falls under MathOverflow but neither of my Calculus AP  teachers in high school could help me with this:
Given $\Sigma_{x=0}^n {f(x)\over2}$ and the output of the summation, how would you find $n$? I've learned how to determine the $n$ given an arithmetic or geometric sequence, but not for an arbitrary function.
Specifically, when $f(x) = 40 + 6\sqrt{x}$.
12 Oct 2010. Edit:
It seems like I need to explain the entire situation for finding $n$, the number of trapezoids, for trapezoidal rule. It started on a simple review question for Calc AP and a TI-83 program that my calc teacher gave to me to solve the definite integral with trapezoidal rules. Aiming to major in Computer Science, I took it a bit further and completely took apart the program resulting in my original question on StackOverflow: https://stackoverflow.com/questions/3886899/determining-the-input-of-a-function-given-an-output-calculus-involved
Since there were tumbleweeds for a response, I took it as a personal challenge to reverse engineer the trapezoidal program into an algebraic form with my notes found on my forum: http://www.zerozaku.com/viewtopic.php?f=19&t=6041
After reverse engineering the code into some algebra, I derived the formula: 
$$TrapRule(A, B, N) = {(B-A)\over N}({F(A)\over2}+\sum_{k=1}^NF(k)+{F(B)\over2})$$
Given the values of A and B are constant for the definite integral, I should be able to isolate and solve for $N$. The problem, however, was determining $N$ in $sum_{k=1}^N$ and I came to the conclusion that it was an issue that I called recursive complexity because it was impossible to determine without recursively adding for the summation.
Eventually, I found MathOverflow and they referred me here. I was hoping only to get help on the issue for a summation because its beyond my skill as a high school student. Now that others have proposed other solutions for my dilemma, I guess I can throw out my thesis Dx
Thanks for the help though, I'll definitely be returning for more.

An approximation to $\displaystyle \sum_{k=1}^{n} \sqrt{k}$ can be found here (on this very site): How closely can we estimate $\sum_{i=0}^n \sqrt{i}$
It was shown that $\displaystyle \frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2} -\frac{2}{3} < \sum_{k=1}^{n} \sqrt{k} < \frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2}$
Thus if $\displaystyle S = \sum_{k=0}^{n} f(x)/2$, then 
$$\sum_{k=0}^{n} \sqrt{n} = \frac{S - 10n(n+1)}{3}$$
Using the approximation above
$$\frac{2n\sqrt{n}}{3} + \frac{\sqrt{n}}{2} = \frac{S - 10n(n+1)}{3}$$
This is a fourth degree equation in $\sqrt{n}$ which can be solved exactly (closed formula) in terms of $S$ and would give you a value of $n$ which is close to $n$. 
My guess is that taking the integer part of the square of an appropriate root will be sufficient to give $n$ or $n-1$ (and so a formula might exist, after all!)
If you are not looking for a formula, but a procedure, you can always try binary search.
Hope that helps.
