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If there is a function in the form $y=\cos(x)+a\cos(bx)$ do there exists real number values for $a$ and $b$ such that for every real number value for $x$ there is either a positive number value for $y$ or a $0$ value for $y$?

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  • $\begingroup$ You should rule out $b=0$ and $|b|=1$, as these make different (and easy) questions. $\endgroup$ – Yves Daoust Apr 8 '17 at 10:23
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Let $f(x) = \cos(x)+a\cos(bx)$, and suppose $f(x) \ge 0$ for all $x \in \mathbb{R}$.

First suppose $b=0$. Then $f(x) = \cos(x) + a,$ which has range $[a-1,\infty).$

Thus, for $b=0$, we get the solution pairs $(a,b) = (t,0),\,$ for any $t \ge 1.$

Next, suppose $b \ne 0$.

If $|a| > 1$, then when $a\cos(bx)$ realizes a value of $-a$, $f(x)$ will be negative, contradiction.

Similarly, if $|a| < 1$, then when $\cos(x)$ realizes a value of $-1$, $f(x)$ will be negative, contradiction.

It follows that $|a| = 1,$ hence either

\begin{align*} f(x) &= \cos(x) + \cos(bx)\\[4pt] &\,\text{or}\\[4pt] f(x) &= \cos(x) - \cos(bx)\\[4pt] \end{align*}

First suppose $f(x) = \cos(x) + \cos(bx)$.

Then when one of the summands realizes a value of $-1$, the other must realize a value of $+1$ to compensate. But that means each of the summands has a period which is a multiple of the other. It follows that periods are equal, hence $b = \pm 1$. But $\cos\,$ is an even function hence, $b = \pm 1$ implies $\cos(bx) = \cos(x)$. But then $f(x) = 2\cos(x)$, contradiction, since then $f(\pi) < 0$.

Next suppose $f(x) = \cos(x) - \cos(bx)$.

If $x_0$ is such that $\cos(x_0) = -1$, then $\cos(bx_0)$ must also be $-1$, else $f(x_0) < 0$.

It follows that the period of $\cos(bx)$ divides the period of $\cos(x)$.

Similarly, if $x_0$ is such that $\cos(bx_0) = 1$, $\cos(x_0)$ must also be $1$, else $f(x_0) < 0$.

It follows that the period of $\cos(x)$ divides the period of $\cos(bx)$.

Therefore the periods must be equal, hence $b = \pm 1$. But then, since $\cos$ is an even function, $b = \pm 1$ implies $\cos(bx) = \cos(x)$, hence $f(x)$ is identically $0$.

This gives the solution pairs $(a,b) = (-1,\pm 1).$

To summarize, we have the solution pairs

\begin{align*} (a,b) &= (-1,1)\\[4pt] (a,b) &= (-1,-1)\\[4pt] (a,b) &= (t,0),\,\text{for any $t \ge 1$}\\[4pt] \end{align*}

and those are the only solutions.

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  • $\begingroup$ Mh, isn't $(2,0)$ also a solution ? $\endgroup$ – Yves Daoust Apr 8 '17 at 10:05
  • $\begingroup$ Ah, yes. I guess I the key is to deal with the case $b=0$ first. I'll fix it -- thanks. $\endgroup$ – quasi Apr 8 '17 at 10:32
  • $\begingroup$ As I wrote in a comment, $b=0$ should not really be considered as the expression $\cos x+a$ is of a different nature. Same for $\cos x+a\cos\pm x$. $\endgroup$ – Yves Daoust Apr 8 '17 at 10:38
  • $\begingroup$ Well, the problem statement made no such restrictions, so for completeness, I dealt with all possible cases (although, as you noted, in my prior version, I was a little careless). $\endgroup$ – quasi Apr 8 '17 at 11:11
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Just set $b=0$ and $a=10$, then $$ y = \cos (x) + a \cos (bx) = \cos (x) + 10 \cos (0) =\cos (x)+10\geq -1 +10 =9>0 $$ for all real $x $.

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Let $f(x)=\cos(x)+a\cos(bx)$. We will show:

If $f$ is non negative then $f\equiv 0$ (*)

First, assume that $b$ is rational, say $b=p/q$ (in lowest terms). Then

$$\int_0^{q\pi}f(x)dx=\int_0^{q\pi}[\cos(x)+a\cos(bx)]dx=0$$ which is implies (*).

If $b$ is irrational, let $\epsilon>0$. See Dirichlet's approximation theorem, if needed. There are integer numbers $p,q$ with $q>\max\{1/\sqrt\epsilon,|a/b|\}$, such that $$\left|b-\frac pq\right|<\frac1{q^2}$$

Then, using that $|\sin'|\le 1$ and by the mean value theorem, $$\begin{align} \left|\int_0^{q\pi}f(x)dx\right|&=\left|\int_0^{q\pi}[\cos(x)+a\cos(bx)]dx\right|\\ &=\left|\frac ab\sin(bq\pi)\right|\\ &=\left|\frac ab[\sin(bq\pi)-\sin(p\pi)]\right|\\ &\le\left|\frac ab(bq-p)\right|\pi<\frac \pi q<\pi\sqrt\epsilon \end{align}$$

So, if $f$ were non negative, then $$\sup f<\frac{\pi\sqrt\epsilon}{q\pi}<\epsilon$$ and $f\equiv 0$.

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  • $\begingroup$ For $b$ irrational, $y(\pi/2+k\pi)=a\cos(b(\pi/2+k\pi))$ necessarily takes both signs. $\endgroup$ – Yves Daoust Apr 8 '17 at 9:39
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We assume $b\ne0$ and $|b|\ne 1$ for (hopefully) obvious reasons.

If there is a solution,

$$I(t):=\int_0^t(\cos x+a\cos bx)\,dx=\sin t+\frac ab\sin bt$$ must be increasing, and as it is bounded, must converge. This is not possible.

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Assuming $b \not = 0$.

If you allow $y = 0$, then yes. Otherwise, no:

$\int cos(x) + a cos(bx) dx = sin (x) + (a/b)sin (bx)$, integrate again to get $- cos(x) - (a / b^2) cos(bx)$. Keep repeating this to get the integral $sin(x) + a / b^n sin(bx).$ If $b \not = 1$, then either $a / b^n$ is very very small, in which case $sin(x) $ dominates, or it is very big, in which case $a / b^n sin(bx)$ dominates. In either of these cases,the integral will have regions on which it is positive or negative, meaning the integrand had to have regions in which it was positive or negative, and inducting back we see that $cos(x) + a cos(bx)$ had to be positive in some places, negative in others.

In the case $b = 1$, we have just $(a + 1) cos(x)$...

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