# Bounded minimum value problem (from an exam today in Australia).

A question appeared in today's exam in Victoria (Australia), in which a graphical calculator was also allowed to be used (just for context).
This question is causing some 'controversy', both in the interpretation of the wording, as well as the actual answer.

Consider the function $$g_a(x)=\sin(\frac{x}{a})+\cos(ax), a\in \mathbb{Z}^{+}. \text{ and } x\in \mathbb{R}$$. State the greatest possible minimum value of $$g_a$$ (1 mark).

(This is not necessarily word for word (except for the last sentence), but this was indeed the premise of the question)

For this, I stated that this value is $$-\sqrt2$$, occurring for $$a=1$$. Upon graphing, if we call the minimum value $$m$$, it becomes clear that $$-2. (The $$-2$$ can be seen as this only occurs upon the superposition of minima, which never occurs but does get approached).

Now, a solution posted online is that:

$$\lim_{a\to \infty}g_a(x)=\cos(ax) \\ \therefore \min(g_a)=-1$$

However, to me this limit is nonsensical as the value of $$x$$ for which the minimum occurs increases as $$a\to \infty$$ sufficiently so as for there to always exist a value of $$x$$ such that $$-2, even if this occurs for very large (negative or positive) values of $$x$$. This can be seen just by graphing on something like Desmos, where if zoomed out, you will see the value go below $$-1$$.

What is a mathematically rigorous way of showing this, seeing as this was given to Year 12s?

• In addition to your point, it is not even true that $\lim_{a \to \infty}g_a(x) = \cos ax$, as should be obvious because $a \to \infty$, and thus is not available to be used on the right hand side. It is true that $\sin \frac xa \to 0$, but $\cos ax$ oscillates between $-1$ and $1$ as $a$ increases, so the limit does not converge. Nov 4, 2021 at 20:34
• Here's a sketch of a rigorous argument. Let $a$ be large. Note that if $x_0=3\pi a/2$ then $\sin(x_0/a)=-1$. Show that if $x_0-1<x<x_0+1$ then $\sin(x/a)$ is close to $-1$. Show that $\cos(ax)$ goes through a full period for $x_0-1<x<x_0+1$, in particular, takes on the value $-1$ for some $x=x^*$ in that interval. Then $g_a(x^*)$ is close to $-2$. Nov 4, 2021 at 22:31

It looks like attacking this with the "standard tools" (derivatives to find extreme points) will not work for general $$a$$. For $$a=1$$ it boils down to solving

$$g'_1(x) =\cos(x) - \sin(x)=0,$$

which should be doable for a students, yielding the minimum of $$-\sqrt{2}$$ that you also found. But for $$a=2$$ it already becomes

$$g'_2(x) =\frac12\cos\left(\frac{x}2\right) - 2\sin(2x)=0,$$

which I wouldn't know how to attack except with expanding everything in terms of $$\tan\left(\frac{x}4\right)$$, yielding to (it looks like) a polynomial equation of degree $$8$$. This seems certainly well out what to expect from year $$12$$ students. And of course higher $$a$$ make this even more complicated.

If anyone has a good method to solve that, I'd be interested!

A nice method to "solve" the $$a=1$$ case without looking at derivatives is to consider that

$$g_1(x)^2=(\sin(x) + \cos(x))^2 = \sin^2(x) + \cos^2(x) + 2 \sin(x)\cos(x) = 1 + \sin(2x) \le 2\; \forall x \in \mathbb R,$$

implying $$|g_1(x)| \le \sqrt{2}$$ and finally $$g_1(x) \ge -\sqrt{2}$$. Together with $$g_1(\frac{5\pi}4)=-\frac12\sqrt{2} - \frac12\sqrt{2}= - \sqrt{2}$$ this proves that the minimum of $$g_1(x)$$ over the reals is really $$-\sqrt{2}$$.

To show that $$-\sqrt{2}$$ is indeed the greatest possible minimum, I'd like to re-frame the question a bit.

Consider

$$f_a(x):=g_a(ax) = \sin(x) + \cos(a^2x).$$

The map $$x \to ax$$ is a bijection from $$\mathbb R$$ onto itself. So the minimum value of $$g_a(x)$$ is also the minimum value of $$f_a(x)$$, it's just taken at a different argument (which we don't care about).

$$f_a(x)$$ has some advantages over $$g_a(x)$$ for understanding it (at least for me):

1. it is the sum of 2 parts, one of which is no longer dependent on $$a$$, and
2. it is periodic with a period not depending on $$a$$, namely $$2\pi$$.

This gives us a shot to prove that for $$a \ge 2$$, the minimum of $$f_a(x)$$ is at most $$-\frac12\sqrt{2} - 1 < -\sqrt{2}.$$

To do that, first fix $$a \ge 2$$ and consider the interval $$I=[\frac{5\pi}4, \frac{7\pi}4]$$. On that interval, $$\sin(x)$$ is always at or below $$-\frac12\sqrt{2}$$, at the endpoints of $$I$$ the sine function takes this value. That interval has length $$\frac\pi 2$$.

Also note that $$\cos(a^2x)$$ has a period $$p_a=\frac{2\pi}{a^2}$$. For $$a \ge 2$$, that means $$p_a \le \frac{2\pi}{2^2}=\frac\pi 2$$.

So $$\cos(a^2x)$$ goes over the full wave it represents in any interval of length $$\frac\pi 2$$, which is also the length of the Interval $$I$$ defined above. That means there is some point $$x_0 \in I$$ with $$\cos(a^2x_0)=-1$$.

That means, taken together with the property of $$I$$ regarding the sine function, that

$$f_a(x_0)=\sin(x_0) + \cos(a^2x_0) = \sin(x_0) - 1 \le -\frac12\sqrt{2} - 1 < -\sqrt{2}.$$

Since there is an argument to $$f_a$$ that produces a value less than $$-\sqrt{2}$$ (for $$a \ge 2$$), the minimum of $$f_a$$ must obviously also be less than $$-\sqrt{2}$$, finally proving that $$-\sqrt{2}$$, is actually the greatest minimum, take when $$a=1$$.

• Amazing, I was trying to go on a similar train of thought which was fruitless if working with $g_a$. Thank you very much! Nov 5, 2021 at 13:45