# Express $f_n(x)=\cos{(n\arccos{x})}$ as a polynomial.

I had an interesting problem on an exam a few days ago in elementary calculus. It reads:

Show that for $n\geq 2,$ the function $f_n(x)=\cos{(n\arccos{x})}, \ x\in[-1,1]$ is a polynomial of degree $n$ and determine the coefficient for $x^n$.

I was not able to work this problem and after the exam I looked this up online and it seems as if this is related to the Chebyshev polynomials. But we have never covered these kinds of polynomials in this course. Is there other ways to do this?

What I tried to do in the exam was to compute $f_n(x)$ for $n=1,2...$ and see if I can find a pattern and formulate an induction hypothesis and then prove it. I got that

$$f_1(x)=\cos{(1\cdot \arccos{x})}=x,\\f_2(x)=\cos{(2\cdot \arccos{x})}=1-2\cos^2{(\arccos{x})}=1-2x^2\\ f_3(x)=\cos{(3\cdot \arccos{x})}=4\cos^3{(\arccos{x})}-3\cos{(\arccos{x})}=4x^3-3x$$

As you can see it quickly becomes ugly and there is no pattern to be seen. So, to formulate a hypothesis for even numbers $n=2k$ was not hard, but for odd numbers, $n=2k+1$ i could not do it.

Is this start a good one or is this totally wrong? Would this method work If I was a bit better at math? any other tips/tricks that only uses elementary calculus? We are not allowed to use expansions in this course.

• $\cos (nt) = \operatorname{Re} \bigl(\cos t + i\sin t\bigr)^n$. The powers of $\sin$ are all even in the real part, so using $\sin^2 t = 1 - \cos^2 t$ we can write it as a polynomial on $\cos t$. Set $t = \arccos x$. Commented Dec 22, 2017 at 13:33
• The course is called real introductory calculus and we are not allowed to use complex analysis/numbers. But that was a nice method though. Commented Dec 22, 2017 at 13:46
• I don't like that you are "not allowed" to use something. This is mathematics, not law. Commented Dec 22, 2017 at 14:19
• It asks you to show that it is a polynomial. Prove by induction "$f_n(x)$ is a polynomial of degree $n$". Do not try to find the exact form of $f_n(x)$, which is not asked for. Commented Dec 22, 2017 at 14:32
• @Parseval Please, if you are ok, you can accept the answer and set it as solved. Thanks!
– user
Commented Feb 1, 2018 at 22:32

The polynomials can be computed recursively: set $\:T_n(x)=\cos(n\arccos x)$ and start from the trigonometric identity: $$\cos(a-b)+\cos(a+b)=2\cos a\cos b.$$ It specialises as $$\cos(n-1)t+\cos(n+1)t=2\cos t\cos nt.$$ Setting $t=\arccos x$, you can read the above formula as $$T_{n+1}(x)=2x\,T_n(x)-T_{n-1}(x).$$

• But I don't understand... What is the coefficient for $x^n$ here? Setting $n=30$ I just get $$T_{31}(x)=2xT_{30}(x)-T_{29}(x).$$ I don't see any $x^n$ anywhere in that formula. Commented Dec 24, 2017 at 8:46
• @Parseval: similarly you have $T_{30}(x)=2xT_{29}(x)-T_{28}(x)$, and so on. To compute it explicitly, you have to compute $T_2, T_3,\dots,T_{30}$. However, you can prove by induction that $T_n$ is a polynomial of degree $n$, with leading coefficient $2^n$. Commented Dec 31, 2017 at 11:24

HINT

$$\cos{n\theta}=\cos^n{\theta}- \binom {n} {2}\cos^{n-2} \theta \cdot \sin^2 \theta+ \binom {n} {4}\cos^{n-4} \theta \cdot \sin^{4} \theta -\cdots$$

NOTE

The relation is obtained by binomial theorem and Euler's equality:

• $\ \cos \theta +i \sin \theta = e^{i \theta}$
• $\ (e^{i\theta})^n = (\cos\theta+i\sin\theta)^n$
• $\ (e^{i\theta})^n = e^{i(n\theta)} = \cos(n\theta)+i\sin(n\theta) = (\cos\theta+i\sin\theta)^n$

Substituting $t=\arccos(x)$, we have:

$$$\int f_n(x)=$$$

$$$=-\int (\sin(t)\cdot \cos(nt))=$$$

$$$=-\frac{1}{2}\int(\sin((n+1)t)+\sin((1-n)t)=$$$

$$$=\frac{1}{2}\left(\frac{\cos((n+1)\arccos(x))}{n+1}-\frac{\cos(\arccos((n-1)x))}{n-1}\right)=$$$

$$$=\frac{1}{2}\left(\frac{f_{n+1}}{n+1}-\frac{f_{n-1}}{n-1}\right)$$$

(passing from the second to the third equation we have used the prosthaphaeresis)

Now, equating the coefficients of $x^{n+1}$ in the LHS and RHS, we get:

$$$2 c_n=c_{n+1}$$$

$$$c_1=1$$$

$$$c_{n}=2^n$$$

(It doesn't match your computation since you switched the signes in f_2) Just for your knowledge, the pattern you were looking for is, in fact, not so "nice". It is:

$$$\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n}{2k} \left (x^2-1 \right )^k x^{n-2k}$$$

Hint:

With twice differentiating of $f_n(\cos x)=\cos nx$ we get $$f''_n(\cos x)\sin^2x=(\cos x-n^2)\cos nx$$ then with $\cos x=u$ solve DE $$(1-u^2)y''=(u-n^2)y$$ with series expansion $\displaystyle y=\sum_{n\geq0}a_nu^n$ where $f_n(u)=y$.

Let $t=\arccos x.$ Then $\sin t = \sqrt{1-x^2}/x$ and $\sin 2t = 2\sqrt{1-x^2}.$

We can prove by induction that $\sin nt = \mbox{poly}\times \sqrt{1-x^2}$ and $\cos nt =$ poly for each $n$. The (four) basis steps are above. Assume both identities are true for $n$. Then

$$\sin (n+2)t = \sin nt \cos 2t + \sin 2t \cos nt$$ $$= \mbox{(poly)}\sqrt{1-x^2}(2\cos t-1) +\sqrt{x^2-1}\mbox{(poly)}$$ $$= \mbox{(poly)}\sqrt{1-x^2}.$$

And then

$$\cos (n+2)t = \cos nt\cos 2t - \sin nt \sin 2t$$ $$=\mbox{(poly)}(2\cos t-1) - \left(\mbox{(poly)}\sqrt{1-x^2}\right)\sqrt{1-x^2}$$ $$= \mbox{(poly)}(2x^2-1) - \mbox{(poly)}(1-x^2) = \mbox{(poly)}.$$

Summary. We prove that $f_n$ is a polynomial of degree $n$ by showing that its $(n+1)$-th derivative vanishes.

We use the change of variable $x=\cos\theta$ and the chain rule $$\tag{1} \frac{d}{dx} = \frac{-1}{\sin \theta}\frac{d}{d\theta},$$ to show that $$\tag{2}\frac{d f_n}{dx}=\sum_{k=0}^{n-1} a_k f_k,$$ for some coefficients $a_k\in\mathbb R$. Iterating, we have that $\frac{d^n f_n}{dx^n}=af_0$ for some $a\in\mathbb R$. Since $f_0$ is a constant, we conclude that $\frac{d^{n+1}f_n}{dx^{n+1}}=0$.

Proof. Because of (1), $$\frac{d f_n}{dx} = n\frac{\sin n\theta }{\sin \theta}=n\frac{e^{in \theta}-e^{-in\theta} }{e^{i\theta}-e^{-i\theta}}.$$ Using the relation $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b^2+\dots + a^2 b^{n-2} + b^n)$ with $a=e^{i\theta}, b=e^{-i\theta}$, we rewrite the right-hand side as $$n(e^{i(n-1)\theta} +e^{i(n-3)\theta} +\ldots + e^{-i(n-1)\theta})= \begin{cases} \displaystyle n +2n\sum_{j=1}^{(n-1)/2} \cos(2j\theta), & n\text{ odd} \\ \displaystyle 2n\sum_{j=1}^{n/2}\cos((2j-1)\theta), & n\text{ even}.\end{cases}$$ With the substitution $x=\cos\theta$, the terms in the right hand side equal $f_{2j}(x)$ or $f_{2j-1}(x)$, thus (2) is proved. $\square$

Once demonstrated that $f_{n}(x)$ has the recursion formula as cleverly indicated by Bernard, and thus is a polynomial since $f_{0}(x)=1$ and $f_{1}(x)=x$, then we can look and find its zeros. \eqalign{ & f_n (x) = 0\quad \Rightarrow \quad n\,\arccos x = {\pi \over 2} + m\pi \quad \Rightarrow \cr & \Rightarrow \quad \,x = \cos \left( {{{1 + 2m} \over n}{\pi \over 2}} \right) \cr}

In the given interval for $x$ then $$x \in \left[ { - 1,1} \right]\quad \Rightarrow \quad 0 \le {{1 + 2m} \over n}{\pi \over 2} \le \pi \quad \Rightarrow \quad 0 \le m \le n - 1 \le {{2n - 1} \over 2}$$ which gives that the function has in fact $n$ distinct real zeros.

So, since $f_n (1) = 1$, we can express the polynomial as $$f_n (x) = {{\prod\limits_{0{\kern 1pt} \le \,m\, \le \,n - 1} {\left( {x - \cos \left( {{{1 + 2m} \over n}{\pi \over 2}} \right)} \right)} } \over {\prod\limits_{0{\kern 1pt} \le \,m\, \le \,n - 1} {\left( {1 - \cos \left( {{{1 + 2m} \over n}{\pi \over 2}} \right)} \right)} }}$$

From here, we can recover the coefficients of the power expansion of $f_n(x)$ by the Vieta's Rules, besides that by the mentioned recursion formula.

My friend used the De Moivre's formula to prove this.

First notice that (ps. I implicitly use De Moivre's) $$cos(n*arcosx) \\ = Re(cos(n*arccosx) + i * sin (n*arccosx) ) \\ = Re(e^{n*arccosx*i}) \\ = Re((e^{arccosx*i})^n) \\ = Re((cos(arccosx) + i * sin (arccosx) )^n)$$

also notice $$cos(arccosx) = x$$ and $$sin (arccosx)= \sqrt{1-x^2}$$

placing them in previous equation:

$$Re((x + i *\sqrt{1-x^2} )^n)$$

then use binomial expansion $$Re(\sum_{k=0}^{n} \binom{n}{k} x^{n-k} * (i \sqrt{1-x^2} )^n))$$

notice that the real part consists of exactly terms where k is even. Thus each $$\sqrt{1-x^2}$$ is raised to some multiple of 2.

Hence it follows that $$cos(n*arcosx)$$ is a polynomial