Let $\displaystyle f(x)$ be a polynomial of degree $\displaystyle n$, an odd positive integer, and has monotonic behaviour , then the number of real roots of the equation

$\displaystyle f(x)+f(2x)+......+f(nx) = \frac{1}{2} n(n+1)$ is equal to

Printable View

- Jan 8th 2011, 02:32 AMjacksNo. of real roots.
Let $\displaystyle f(x)$ be a polynomial of degree $\displaystyle n$, an odd positive integer, and has monotonic behaviour , then the number of real roots of the equation

$\displaystyle f(x)+f(2x)+......+f(nx) = \frac{1}{2} n(n+1)$ is equal to - Jan 8th 2011, 07:11 AMsnowtea
I think the point is that $\displaystyle f(x)$ is monotonic means that $\displaystyle f(kx)$ is monotonic in the same direction (increasing/decreasing).

The sum of of monotonic functions in the same direction is monotonic.

How many real roots can a monotonic function have?

E.g. how many times can a strictly increasing function intersect the x-axis? - Jan 8th 2011, 07:12 PMjacks