# Number of Real Roots of Polynomial w/ Monotonic Behaviour

• chaoseverlasting
In summary, the conversation discusses finding the number of real roots for an equation involving a polynomial of degree n, an odd positive integer, with monotonic behavior. It is mentioned that f(x)=0 has only one root due to its monotonicity, and further exploration is done to determine the number of roots for f(x) + ... + f(nx) = n(n+1)/2. It is concluded that the sum of monotonic functions is also monotonic, and the equation will have only one root due to its monotonicity. However, to find the number of roots for f(x) + ... + f(nx) - n(n+1)/2, it is necessary to consider the constant term and the behavior of the

## 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

## 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.

chaoseverlasting said:
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?

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.

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?

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.

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?

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?

$x \mapsto f(x)$ 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?

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.

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)?

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.

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, ex is monotonic, but it has no zeroes. Both ex 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?

## 1. How do you determine the number of real roots of a polynomial with monotonic behavior?

The number of real roots of a polynomial with monotonic behavior can be determined by analyzing the end behavior of the polynomial and its degree. If the degree is odd, the polynomial will have at least one real root. If the degree is even, the polynomial may have either 0 or 2 real roots. Additionally, the number of sign changes in the coefficients can give an upper bound on the number of real roots.

## 2. Can a polynomial with monotonic behavior have imaginary roots?

No, a polynomial with monotonic behavior can only have real roots. This is because monotonic behavior means the polynomial is either always increasing or always decreasing, and therefore cannot cross the x-axis and have imaginary roots.

## 3. What is the relationship between monotonic behavior and the number of real roots?

The relationship between monotonic behavior and the number of real roots is that a polynomial with monotonic behavior will have a specific number of real roots based on its degree. As mentioned before, the degree and end behavior can give an indication of the number of real roots, but this is not always the case. In some cases, a polynomial with monotonic behavior may have more real roots than what is predicted by these factors.

## 4. How does the leading coefficient affect the number of real roots of a polynomial with monotonic behavior?

The leading coefficient of a polynomial with monotonic behavior does not affect the number of real roots. As long as the polynomial has monotonic behavior, the number of real roots will be determined by its degree and end behavior, not the leading coefficient.

## 5. Are there any exceptions to the relationship between monotonic behavior and the number of real roots?

Yes, there can be exceptions to the relationship between monotonic behavior and the number of real roots. This can occur when the polynomial has multiple roots or when the polynomial has a repeating root. In these cases, the polynomial may have more real roots than what is predicted by the degree and end behavior.