Number of Real Roots of Polynomial w/ Monotonic Behaviour

[chaoseverlasting]

Homework Statement

Let f(x) be a pllynomial of degree n, an odd positive integer, and has monotonic behaviour then the number of real roots of the equation
f(x)+ f(2x) + f(3x)...+ f(nx)= n(n+1)/2

Homework Equations

The Attempt at a Solution

This seems like the summation of 1+2+3+...+n, and that would be a root of the equation, but I don't know if that's the only one.

 

[AKG]

Let f(x) be a pllynomial of degree n, an odd positive integer, and has monotonic behaviour then the number of real roots of the equation
f(x)+ f(2x) + f(3x)...+ f(nx)= n(n+1)/2

You didn't finish asking a question. Do you mean that you have to show that the number of solutions of

f(x) + ... + f(nx) = 0

is n(n+1)/2, or are you asked to find the number of solutions of the equation

f(x)+ f(2x) + f(3x)...+ f(nx)= n(n+1)/2?

 

[HallsofIvy]

Not to mention: what do you mean by "has monotonic behavior"? Do you mean that it is monotonic for all x? If so then f(x)= 0 has exactly one root.

 

[chaoseverlasting]

You have to find the number of roots of the equation. And it is monotonic for all x. How did you get f(x)=0 has only one root?

 

[AKG]

So you have to find the number of x-values such that

f(x) + ... + f(nx) = n(n+1)/2

Right? f(x)=0 has only one root because f is monotonic, so once it crosses the x-axis once, it will never cross again. Of course, it must cross the x-axis at least once because f is an odd-degree polynomial.

 

[HallsofIvy]

I would recommend looking at simple cases first. It is easy to show that if f(x)= ax+ b, there is exactly one root to that equation, but what if f(x)= x³ or variations on that?

 

[chaoseverlasting]

Since f(x) is monotonic for all x, then it has only one solution. I.e, it only cuts the line y=0 at one point. Let f(x) be a function of the type $$f(x)=a_nx^n+a_{n-1}x^{n-1}...$$
Then f(x)+f(2x)+f(3x)+...+f(nx) will also be a function of the type $$F(x)=A_nx^n+A_{n-1}x^{n-1}...$$ which will also be monotonic, and hence have only one solution.

Is that right?

 

[AKG]

$x \mapsto f(x)$ is either monotonic increasing or monotonic decreasing. Without loss of generality, suppose it is monotonic increasing. Then what can you say about the functions $x \mapsto f(kx)$ where k is some positive integer?

 

[chaoseverlasting]

f(kx) is also monotonically increasing or decreasing. Graphically, if k<1, then the graph expands, and if k>1, then the graph contracts, but the graph remains basically the same.

 

[AKG]

Right, and it is monotonic in the same way, i.e. if f(x) is increasing, then f(kx) is increasing (for k > 0) and if f(x) is decreasing then f(kx) is decreasing (for k > 0). Now you know that f(x), f(2x), ..., f(nx) are all monotonic, and either ALL strictly increasing or ALL strictly decreasing. What can you say about the sum f(x) + ... + f(nx)?

 

[chaoseverlasting]

f(x)+...+f(nx) is also monotonic. Thats why it can only intersect the x-axis at one point. Hence, f(x)+...+f(nx) has only one root.

 

[AKG]

Okay, but you're not done yet. You actually want to find the number of roots of f(x) + ... + f(nx) - n(n+1)/2. f(x) + ... + f(nx) is monotonic, and -n(n-1)/2 is just a constant, so what can you say about their sum? Also, e^x is monotonic, but it has no zeroes. Both e^x and any odd-degree polynomial, so what more do you need to say about f(x), and then about f(x) + ... + f(nx), and then about f(x) + ... + f(nx) - n(n-1)/2 in order to count the roots?