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This is a problem I was trying to solve for a while with no succeed. Show that the equation $\cos (a_1x) + \cdots + \cos (a_nx) = 0$ has at least one solution in $[0,\frac {\pi}{a_1}]$, where $0 < a_1 < \cdots < a_n$.

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2 Answers 2

up vote 8 down vote accepted

By a simple scaling argument, it is clear we only need to prove the assertion for the special case $a_1 = 1$. Let us assume this is the case. Notice for any $\alpha > 0$, we have

$$\int_0^\pi \sin(x) \cos(\alpha x) dx = \begin{cases} 0, & \alpha = 1\\ \frac{1 + \cos(\pi\alpha)}{1-\alpha^2}, & \alpha\ne 1 \end{cases}$$

Let $\;\displaystyle \varphi(x) = \sum_{k=1}^{n} \cos(a_k x)\;$, above integrals implies

$$\int_0^\pi \sin(x)\varphi(x) dx = \sum_{k=2}^n \frac{1+\cos(\pi a_k)}{1-a_k^2} \le 0\tag{*1}$$ because $a_1 = 1$ and $a_k > 1$ for $k \ge 2$.

Notice $\sin(x) > 0$ over $(0,\pi)$ and $\varphi(x) > 0$ near $0$. If $\varphi(x)$ doesn't vanish for any $x \in (0,\pi)$, then $\varphi(x) > 0$ over the whole $(0,\pi)$. This will imply the integral in LHS of $(*1)$ is strictly positive. This is a contradiction and hence $\varphi(x)$ must has a root in $(0,\pi)$.

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Very, very nice! –  Stephen Montgomery-Smith Dec 30 '13 at 5:38
    
Nice. May I ask how did you reach the ideal !? –  Robert M Dec 30 '13 at 6:30
    
@RobertM It is more or less by trial and error. My first attempt is to use some form of comparison theorem, e.g Sturm comparison theorem, but this leads me nowhere. I then tried another common way for this sort of problem. Namely, scale the target function by a positive factor and show the integral is non-positive. Since we are dealing with trigonometric functions and the length of the interval is $\pi$, the most natural choice is to scale the function by $\sin x$. I'm glad it simply works. –  achille hui Dec 30 '13 at 11:34

This isn't a pretty proof, but here's my shot.

Assume $n=2$. With that, the following is true:

$$ \cos(\frac{1}{3}\pi) + \cos(\frac{2}{3}\pi) = 0 $$

Assume $x = \pi$. With this, the range for $x$ is now $[0,3\pi]$, which encompasses $\pi$.

As for even $n$ the cosine values have to be symmetrical by pairs, $a_1$ will be lower than $1/3$ as $n$ increases. As such, $x$ can have the value of $\pi$ for $n>2$. For $n$ the following is true:

$$ \cos(\frac{1}{n+1}\pi) + ... + \cos(\frac{n-1}{n+1}\pi) = 0 $$

because this is a sum of pairs that cancel themselves. And as proved earlier, $x$ can have the value of $\pi$. Note that there can exist infinite forms, as long as the pairs

$$ \cos(a_1x) + \cos(\pi-a_1x) $$

exist, as they will cancel themselves out. For odd $n$, an other member must be added, that will always be $\cos(\frac{1}{2}\pi) = 0$

If $x$ is contained between $[0,\frac{\pi}{a_1}]$ then $a_1x$ is contained between $[0,\pi]$, which contains the range for the 1st quadrant. As the first term in all these equations exists in the first quadrant, the reasoning is valid for all $a_1>0$.

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