Roots of $ 16 x^5 - 20 x^3 + 5x + 1 = 0 $

Question

The following is from Edexcel further mathematics Core Pure Book 2 A Level Mixed Exercise 1 Question 9 part b:

9 a Use De Moivre's Theorem to show that

$$ \cos 5\theta \equiv 16 \cos^5 \theta - 20\cos^3 \theta + 5\cos \theta. $$

b Hence find all solutions to the equation

$$ 16 x^5 - 20 x^3 + 5x + 1 = 0 $$

giving your answers to $3$ decimal places where necessary.

Part $(a)$ was fine. Part $(b)$ I also got the correct answer. However, I find the logic I'm using to be dubious (although I suspect it is the same logic that Edexcel considers correct).

My answer to b:

When making trigonometric substitutions, I have been told that you should use a bijective function as the substitution: this helps determine the maximum domain on which you can choose the substitution. So let $x=\cos \theta,\quad 0\leq \theta \leq \pi.$ (Note that this is that largest domain on which the substitution is bijective). Using the identity from part $(a),$ the equation $ 16 x^5 - 20 x^3 + 5x + 1 = 0\ $ is equivalent to $\ \cos 5\theta = -1.\ $ Since we defined $0\leq \theta \leq \pi,\ $ it follows that $ 0\leq 5\theta \leq 5\pi.\ $ Thus we have to solve $\ \cos 5\theta = -1\ $ in the interval $ 0\leq 5\theta \leq 5\pi.\ $ This has solutions:

$$\ 5\theta = \pi, 3\pi, 5\pi,\ \implies \theta = \frac{\pi}{5}, \frac{3\pi}{5}, \frac{5\pi}{5},$$

$$\implies x=\cos\theta = \cos\frac{\pi}{5},\ \cos\frac{3\pi}{5},\ \cos\frac{5\pi}{5} = 0.809, -0.309, -1. $$

$$$$

The part I find dubious is that we have only looked for roots of the equation $ 16 x^5 - 20 x^3 + 5x + 1 = 0 $ that are between $-1$ and $1$. We haven't determined if there are roots of this equation $<-1$ or $>1.$ So we don't know yet if we have found all roots of the equation. Am I right about this?

Answer 1

You should be slightly more careful stating the solution, but the general idea is correct. What does $x=\cos \frac{\pi}{5}$, $\cos \frac{3\pi}{5}, \cdots$ mean? As stated, we need to solve $\cos 5\theta=-1$. And the original polynomial ($16x^5-20x^3+5x=-1$) is of fifth degree, so has at most $5$ real roots.

We could say $x=\cos \frac{(2k+1)}{5}\pi$, where $k \in \{0,1,2,3,4\}$. This gives us $5$ real values of $x$ which are possible roots of $16x^5-20x^3+5x=-1$. If you are allowed to use a calculator, then it would be easy to check (typing things like '$16x^5-20x^3+5x+1$ where x=cos(9*pi/5)' into Wolfram Alpha) that these are (I mean $x=\cos \frac{(2k+1)}{5}\pi$) indeed the roots of the equation.

Answer 2

You are mistaken in thinking that you have only looked for roots in the interval from $-1$ to $1$. You haven’t.

It is clear from the working that the only roots that exist just happen to be in this interval.

In case you are wondering why you only have three values, whereas you would expect five, it so happens that two of these roots, namely $\cos\frac{\pi}{5}$ and $\cos\frac{3\pi}{5}$ are double roots. And that is because if you continue to write down more solutions, $$\cos\frac{\pi}{5}=\cos\frac{9\pi}{5}$$ and $$\cos\frac{3\pi}{5}=\cos\frac{7\pi}{5}$$

Answer 3

Adam Rubinson, do not worry, indeed the equation $\,16x^5-20x^3+5x+1=0\;$ cannot have real roots in $\,(-\infty,-1)\cup(1,+\infty)\,.$

Proof:

If the equation $\,16x^5-20x^3+5x+1=0\,$ had a real root $\,x_0\in(-\infty,-1)\cup(1,+\infty)\,,\,$ then $\;|x_0|>1\;$ and

$16x_0^5-20x_0^3+5x_0+1=0\;\;,$

$x_0\left[\,4\!\left(x_0^2-1\right)\!\left(4x_0^2-1\right)+1\,\right]=-1\;\;,$

$\big|x_0\big|\!\cdot\!\big|\,4\!\left(x_0^2-1\right)\!\left(4x_0^2-1\right)+1\,\big|=1\;\;,$

which is impossible because

$|x_0|>1\;\;$ and $\;\;\big|\,4\!\left(x_0^2-1\right)\!\left(4x_0^2-1\right)+1\,\big|>1\,.$

Answer 4

I will give a high level discussion of my perspective on this type of equation-solving and what is and isn't logically valid. My discussion will not necessarily be suited for the average A-level student (and I agree with what you say - from my recollection of taking A-levels, I think the exam board likely doesn't actually expect much justification or logical rigour here).

In general, in the process of "solving an equation $f(x) = 0$" (ie finding an explicit description for the set $X$ of $x$ such that $f(x) = 0$), one technique is to repeatedly "do something to both sides" of the equality and apply algebraic identities, ending up with some equation $g(x) = 0$ which is easier to analyse. Say we know that $X'$ is exactly the set of $x$ such that $g(x) = 0$. Since we manipulated the original equation by a series of "$\implies$" steps, we know that $X \subseteq X'$. The step that students often get wrong is jumping the gun and assuming that $X = X'$. But to be fully rigorous at this stage, one must prove this - ie check that $X' \subseteq X$. It's quite possible that if you've been overly zealous in your manipulations, you've overshot and $X$ might be a proper subset of $X'$. If $X'$ is still a nice enough set, you can still find $X$ by just taking the subset of $X'$ of $x$ that do satisfy $f(x) = 0$.

The classic example is an equation like $\sqrt{|x + 1|} = x$. If you square both sides, you get $|x + 1| = x^2$, which you can solve to see $x = \tfrac 12 (1 \pm \sqrt 5)$. But of course one of these is not actually a solution.

In this case, the template of the solution is quite different. We wish to solve $f(x) = 0$. The way we do this is to consider some function $c: [0, \pi] \to [-1, 1]$, and instead we look for solutions to $f(c(\theta)) = 0$. If $\Theta$ is the set of such $\theta$, then $\{c(\theta) : \theta \in \Theta\} \subseteq X$. However, it is possible that it's a proper subset. This means that there may be cases where we carry out this argument and find a small number of solutions, but one cannot conclude immediately that those are all the solutions. If you're aware of this, then there's no more logical peril than in the other template I describe - you just have to be careful to verify that you've not undershot and have indeed found all the solutions.

In the case of your particular question, one can verify that (with multiplicities) the three roots obtained are all the roots of your equation, by using the theorem that a quintic has at most $5$ roots with multiplicity. Since what you've written doesn't include any justification that you have the full set of solutions, I would agree that you don't know yet if you've found all the roots to the equation. But if you do this verification, I would say that solution is on logically solid ground. However, it does get a bit "lucky", and there are similar problems for which that technique would fail to solve the problem.

(Incidentally, there are all sorts of ways to verify that you have actually got all the roots. Angelo's answer gives one way to see it. You can also look at derivatives. I think a nice way is to argue that for $k$ a tiny bit less than $1$, the equation $16x^{5}-20x^{3}+5x+k = 0$ does have $5$ distinct real roots, and as we take the limit $k \to 1$, two of those pairs of roots merge together. Hence those must be repeated roots.)

Here is an example of such a problem: say we're trying to solve $16x^{5}-20x^{3}+5x+2 = 0$. Suppose $x = \cos \theta$ for $\theta \in [0, \pi]$. Then by the nice identity we proved, we have $\cos 5\theta + 2 = 0$, so $\cos 5\theta = -2$. This is absurd, because for $\theta \in \Bbb R$, we have $\cos \theta \ge -1$. Therefore there are no solutions obtained by this method. However, there is in fact a real solution to this equation, which follows from the fact that this polynomial has odd degree. This is frustrating!

We can actually make a small change to the second template to make it more powerful. If the function $c$ is surjective onto the domain of $f$, then actually we're guaranteed to have $\{c(\theta) : \theta \in \Theta\} = X$ (exercise for the reader). This is a very nice guarantee, particularly if the equation $f(c(\theta)) = 0$ is easy to work with.

Luckily for us, the function $\cos$ extends to $\Bbb C$, to give a surjective map $\Bbb C \to \Bbb C$. Moreover, the identity $\cos 5\theta \equiv 16 \cos^5 \theta - 20\cos^3 \theta + 5\cos \theta$ remains valid for $\theta \in \Bbb C$, and so do many familiar facts about the behaviour of $\cos$ (these facts do all need proving. If you're familiar with complex analysis, this all follows from general nonsense like Picard's theorem and the identity principle). So we will be able to use the same technique to find all complex roots of $16x^{5}-20x^{3}+5x+2 = 0$, from which we can work out the real roots if we like. I will write out this whole solution for the more advanced reader's benefit, but this is probably not something the average A-level student would be expected to follow.

Off we go, then. Let $\theta \in \Bbb C$. We wish to solve the equation $\cos 5\theta = -2$. Let's write $5\theta = u + iv$ with $u, v \in \Bbb R$.

Then by standard identities, we have $\cos(u + iv) = \cos u \cosh v + i \sin u \sinh v$, so we must have $\cos u \cosh v = -2$ and $\sin u \sinh v = 0$. From the latter equation it follows that $v = 0$ or $u = \pi n$ for some $n \in \Bbb Z$. But we already know that $v$ cannot be $0$.

Now the former equation says $(-1)^n \cosh v = -2$. Since $v$ is real, we have $\cosh v \ge 0$. So $n$ must be odd and we must have $\cosh v = 2$, and hence $v = \pm \operatorname{arccosh} 2$.

So, the general solution to the original equation is $x = \cos(\tfrac 15((2n + 1)\pi \pm i \operatorname{arccosh} 2))$ (in other words, $\Theta = \{\tfrac 15((2n + 1)\pi \pm i \operatorname{arccosh} 2) : n \in \Bbb Z\}$ and we are now looking at $\{\cos \theta : \theta \in \Theta\}$). Here the choice of sign for the $\pm$ gives the conjugate pairs of solutions to this quintic. You can work out the ugly formula in terms of $\cosh$ and $\sinh$ if you like.

Taking $n = 0$ gives the solutions $0.837242435715400 \pm 0.156613988262999i$.

Taking $n = 1$ gives the solutions $-0.319798153619534 \pm 0.253406756123210i$.

Taking $n = 3$ gives the real solution $-1.03488856419173$. Hurray!

Now we can argue that by periodicity and symmetry properties of $\cos$, this must be all of $\{\cos \theta : \theta \in \Theta\}$, and hence all of $X$. Alternatively, we already know that $|X| \le 5$ because $X$ is a quintic. So we've found all solutions.

Lastly - if your solution had used the modified template, with $\theta \in \Bbb C$, then you could actually be confident you had all the solutions without doing an extra verification, so long as you had justified well enough that you've found all possible $\theta$ (and so long as you accept that $\cos: \Bbb C \to \Bbb C$ is surjective). If you know enough about the complex cosine function, you would in fact have to change almost nothing about the argument, because the complex cosine function in general doesn't introduce "extra" preimages of numbers in $[-1, 1]$. But again, such knowledge is not expected of A-level students and it's not something they can assume in their answers to exam questions, as far as I know.

(In fact more generally, "$\cos x = \cos y$ iff $x - y \in 2\pi \Bbb Z$ or $x + y \in 2\pi \Bbb Z$" still holds for $x, y \in \Bbb C$. This is because $\cos x - \cos y = -2\sin(\tfrac 12(x + y))\sin(\tfrac 12(x - y))$, and $\sin z = 0$ iff $z \in \pi \Bbb Z$. You can see this last fact by directly working it out, or by some fun general nonsense again :) - if there was some $z_0 \notin \Bbb R$ with $\sin z_0 = 0$, then $2z_0$ would be a period of $\sin$, because $\sin(z + 2z_0) = \sin z \cos 2z_0 + \cos z \sin 2z_0 = \sin z (1 - 2 \sin^2 z_0) + 0 = \sin z$, but any doubly-periodic entire function must be constant.)