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Given $$f(x)=a_0+a_1\cos x+a_2\cos2x+\cdots+a_n\cos nx$$ where $a_0,a_1,a_2,\cdots,a_n$ are non-zero reals such that $a_n>|a_0|+|a_1|+\cdots+|a_n-1|$ then how many roots of $f(x)=0$ are there in $0\le{x}\le{2\pi}$? I have tried to check the conditions for lower power equations. For $a_0+a_1\cos x=0$ there are two roots and for $a_0+a_1\cos x+a_2\cos 2x=0$ I am getting four roots. After this I tried to use induction on the equation but could not find any possible way to solve the problem. So how can we approach this type of problem?

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  • $\begingroup$ I would guess that you could use Rouche's theorem to show that your polynomial has the same number of roots as $a_n\cos(nx)$. This would be $2n$ and would agree with the small examples you computed $\endgroup$
    – Sam Ballas
    Oct 6, 2023 at 11:36
  • $\begingroup$ I am very sorry @SamBallas but I don’t know Rouche’s theorem. This problem is from a high school contest of mathematics in India where we cannot use such theorem which is not taught. We have to use elementary ideas to solve the given problem. But thank you for hint , I got to know something new. $\endgroup$
    – Sillyasker
    Oct 6, 2023 at 11:53
  • $\begingroup$ I see, well perhaps that is not the best strategy then. $\endgroup$
    – Sam Ballas
    Oct 6, 2023 at 11:55
  • $\begingroup$ See math.stackexchange.com/questions/2266985/… This is not the same coefficient conditions, but similar constructions also work here. $\endgroup$ Oct 6, 2023 at 11:56

1 Answer 1

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Part 1: $f(x)$ has at least $2n$ roots in $[0,2\pi]$.

Just evaluate $f(x)$ at $x = \frac{k\pi}{n}$ for $k=0,\ldots,2n.$ Then $$ f\left(\frac{k\pi}{n}\right) = \sum_{m=0}^{n-1} a_m\cos\left(\frac{km\pi}{n}\right) + a_n\cos\left(k\pi\right) = \sum_{m=0}^{n-1} a_m\cos\left(\frac{km\pi}{n}\right) + a_n (-1)^k $$ Because of $a_n > \sum_{m=0}^{n-1}|a_m|,$ the sign of $f(k\pi/n)$ is determined by $(-1)^k.$ This means that the sign of $f(x)$ changes at least $2n$ times between $0$ and $2\pi,$ which in turn means that there must be at least $2n$ roots.

Part 2: $f(x)$ has at most $2n$ roots in $[0,2\pi]$.

Note that $x=0$, $x=\pi$ and $x=2\pi$ cannot be roots, because $|\cos nx|=1$ in that case and $a_n\cos nx$ consequently dominates the other terms of the sum.

We set $x=\arccos u$ and use the ordinary convention that the result of $\arccos$ is in $[0,\pi]$. Then $$ f(\arccos u) = \sum_{m=0}^{n} a_m\cos\left(m\arccos u\right) =\sum_{m=0}^{n} a_m T_m(u) =: g_1(u) $$ with the Chebyshev polynomials of the first kind $T_m(u).$ Therefore, $g_1(u)$ is a polynomial of degree $n$ and can have at most $n$ roots which in turn means that $f(x)$ can have at most $n$ roots in $(0,\pi).$

Likewise, you can define $x=\pi+\arccos v$ and you get $$ f(\pi+\arccos v) = \sum_{m=0}^{n} a_m\cos\left(m\pi+m\arccos v\right) =\sum_{m=0}^{n} (-1)^m a_m T_m(v) =: g_2(v) $$ $g_2(v)$ is also a polynomial of degree $n$ and can have at most $n$ roots which in turn means that $f(x)$ can have at most $n$ roots in $(\pi,2\pi).$

Put all this together, and you get that $f$ has exactly $n$ roots in $(0,\pi)$ and exactly $n$ roots in $(\pi,2\pi)$.

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