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Take some analytic function, $f(x)$, that goes from $-\infty$ to $\infty$, with a finite number of points such that $\frac{df}{dx}=0$. You can divide the y axis into intervals, where the boundary between each interval is the y value at a critical points (see graph).

Within each interval, there are an odd number of real inverse functions, $g_n(y)$. Number them in ascending order. Then, for every function I've tried, $\sum_n (-1)^n g_n(y)=ag_m(y)+b$ for some a,b and m that is constant across the interval. Does anyone know why this would be the case?

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It's not true in general. What is true is this. If $f(x) = a_n x^n + \ldots + a_0$ is a polynomial with $n$ odd, then in any interval where there are $n$ real inverse functions, the sum of those is constant, namely $-a_{n-1}/a_n$. So if you happened to take $n=3$, you would get $-g_1(y) + g_2(y) - g_3(y) = a_2/a_3 + 2 g_2(y)$.

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That does explain the n=3 case, but how about, say, n=5 in the middle region where there are 5 gs and you can't just use the fact that the sum is constant? –  palmtree3000 Aug 4 '12 at 1:25
Try e.g. $f(x) = x (x - 1) (x - 3) (x - 7) (x - 10)$. What do you get for $\sum_{j=1}^5 g_j(y)$? –  Robert Israel Aug 4 '12 at 1:54
g(0)=0,1,3,10. At other points we want to solve $x(x-1)(x-3)(x-7)(x-10)-y=0$, which seems nontrivial. –  palmtree3000 Aug 4 '12 at 17:02
It should be noted, however, that $x(x−1)(x−3)(x−7)(x−10)$ does have this property, according to my numerical inversion... –  palmtree3000 Aug 4 '12 at 17:25
Nevermind, I think that my function looked linear, but actually wasn't. Thanks for your help. –  palmtree3000 Aug 4 '12 at 18:30

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