# Prove that $ax = \cos(\pi\cdot x)$ has exactly one solution

Prove that $ax = \cos(\pi\cdot x)$ has exactly one solution when $0 \le x \le 1$. a is any positive real number.

I can solve this question fine by drawing $\cos(\pi\cdot x)$ out but it's considered informal. I need help on a written proof.

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This can be done by showing that $cos(\pi x)$ is strictly decreasing on $(0,1)$, i.e. $\frac{d}{dx}\cos(\pi x) < 0$, as if $\cos(\pi a) = \cos(\pi b)$ for $a < b$ then by the mean value theorem we have $\frac{d}{dx}\cos(\pi c) = 0$ for some $a < c < b$. $\frac{d}{dx}\cos(\pi x) = -\sin(\pi x)$, which you can show is less than $0$ on $(0,1)$ by interpreting $\sin(\pi x)$ it as the ratio of the opposite side of a triangle to the hypotenuse.

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"equivalent" is a strong word... – Aryabhata May 1 '11 at 7:16
@Moron: Sorry. It's very close though, as it must be at least non-increasing, but I'll change it. – Alex Becker May 1 '11 at 7:18
$\cos (\pi x)$ is decreasing in $(0,1)$, not increasing (derivative is $-\pi \ \sin(\pi x)$ which is negative in $(0,1)$). Also, the function could oscillate wildly yet have only one intersection point, so I would not call "equivalent" close. +1 though :-) – Aryabhata May 1 '11 at 7:20
The "you can show" part means that you should work that part out for yourself. – Alex Becker May 1 '11 at 7:31
Alex: Care to address the remarks above, by Aryabhata and me? As far as I am concerned, the present version of your post is far from being satisfying. – Did Aug 15 '11 at 15:44

Hint: on the interval $[0,1]$, the function $x\mapsto ax$ is increasing from $0$ to $a$ and the function $x\mapsto\cos(\pi x)$ is decreasing from $1$ to $-1$.

Validity check: one should be able to prove that the unique solution in $[0,1]$ is in $(0,1/2)$ and that every point in this interval can be made solution for a suitable value of the slope parameter $a$.

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A sketch is fully persuasive. However, the problem is intended to test your control over certain tools. So it is difficult to give an appropriate answer unless one knows the context in which the problem arises. I will assume that this context is a somewhat theoretically minded calculus course.

We will use a standard trick. When we are interested in the relation between two functions $g(x)$ and $h(x)$, it is often useful to examine the behaviour of their difference, that is, to study $f(x)$, where $f(x)=g(x)-h(x)$.

Let $f(x)=ax-\cos x$. We wish to show that $f(x)=0$ for exactly one value of $x$ between $0$ and $\pi$.

Note that $f(0)=-1$ and $f(\pi)=a\pi +1$. Thus $f(0)$ is negative, and (since $a$ is positive), $f(\pi)$ is positive. Since $f$ is continuous in our interval, it follows by the Intermediate Value Theorem that $f(c)=0$ for some $c$ between $0$ and $\pi$. (Indeed, as $x$ ranges over our interval, $f(x)$ takes on all values between $-1$ and $a\pi+1$.)

Note that we have not proved that $f$ is continuous in our interval, we have just asserted it. And of course we took it for granted that $\cos(0)=1$ and $\cos(\pi)=-1$. In the context of this problem, you are almost certainly not expected to prove continuity. But asserting it is important, since it shows that you know that one needs certain conditions on $f$ in order to be able to use the Intermediate Value Theorem.

Finally, we show that there cannot be more than one one value of $x$ in our interval such that $f(x)=0$. Here you are likely expected to use the derivative.

Note that $f'(x)=a+\sin x$. Between $0$ and $\pi$, $\sin x \ge 0$, so between $0$ and $\pi$, $f'(x)>0$. It follows that $f$ is an increasing function in our interval. So in our interval, $f(x)$ can be $0$ for at most one value of $x$.

We have proved that $f(x)=0$ for at least one $x$ in our interval, and also for at most one $x$, so $f(x)=0$ for exactly one value of $x$ in our interval.

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