# Is there no formula for $\cos(x^2)$?

I was wondering if there was a "formula" or an "identity" for $\cos(x^2)$, as there is for $\cos(2x)$. My question is closely related to this one, which was only asking for $\cos(ab)$. For instance $$\cos(x^2)= \frac{\cos^3(x)+\sin^3(x)}{2\cos^2(x)+2016}$$ could be such a formula.

More precisely, I would like to know if the function $f : x \mapsto \cos(x^2)$ belongs to $F = \Bbb R(\cos, \sin)$.

Here, the space $F$ denotes all the rational functions of the form $$x \mapsto \frac{P(\cos(x), \sin(x))}{Q(\cos(x), \sin(x))}$$ where $P, Q \in \Bbb R[X,Y]$ and $Q(\cos(x), \sin(x))≠0$ for all real numbers $x$. (Notice that $x \mapsto \cos(nx)$ belongs to $F$ (if $n≥1$). It can be proved by induction on $n$.)

My guess is no and here is why : thanks to partial fraction decomposition, every rational function has a primitive. Doing some tangent half-angle substitutions, one can see that every $g \in F$ has an explicit primitive (if I'm not mistaken).

But my function $f : x \mapsto \cos(x^2)$ above has no elementary primitive. I think this can be shown thanks to Liouville's theorem.

Does my reasoning is correct ? Do you have any easier argument (or counterargument) ?

Any comment will be appreciated !

• This answer can be relevant ; the antiderivative of a rational function like $1/(x^5-x-1)$ can involve algebraic numbers which are not expressible by radicals, however. Dec 25, 2016 at 13:05
• Related (in view of the answers below): math.stackexchange.com/questions/282644 Nov 27, 2018 at 15:40

There's a simpler reason: Every rational function in $\cos x, \sin x$ has (not necessarily minimal) period $2 \pi$, but $\cos(x^2)$ is not periodic.

Much easier: any rational function of $\sin$, $\cos$ will be periodic. Your function isn't.

• Thank you ! My idea was indeed too complicated… ! Feb 4, 2016 at 15:23

One can use binomial expansion in combination with the complex extension of trig functions:

$$\cos(xy)=\frac{e^{xyi}+e^{-xyi}}{2}=\frac{a^{xy}+a^{-xy}}2$$

Using $a=e^i$ for simplicity.

We also have:

$$(a+a^{-1})^n=\sum_{i=0}^{\infty}\frac{n!a^{n-i}a^{-i}}{i!(n-i)!}=\sum_{i=0}^{\infty}\frac{n!a^{n-2i}}{i!(n-2i)!}$$

Which is obtained by binomial expansion.

We also have:

$$(a+a^{-1})^n=(a^{-1}+a)^n=\sum_{j=0}^{\infty}\frac{n!a^{2j-n}}{j!(n-j)!}$$

And, combining the two, we get:

$$(a+a^{-1})^n=\frac{\sum_{i=0}^{\infty}\frac{n!a^{n-2i}}{i!(n-i)!}+\sum_{j=0}^{\infty}\frac{n!a^{2j-n}}{j!(n-j)!}}2=\frac12\sum_{i=0}^{\infty}\frac{n!}{i!(n-i)!}(a^{n-2i}+a^{-(n-2i)})$$

If we have $\cos(n)=\frac{a^n+a^{-n}}2$, then we have

$$(2\cos(n))^k=(a^n+a^{-n})^k=\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}\frac{a^{n(k-2i)}+a^{-n(k-2i)}}2$$

Furthermore, the far right of the last equation can be simplified back into the form of cosine:

$$\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}\frac{a^{n(k-i)}+a^{-n(k-i)}}2=\sum_{i=0}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

Thus, we can see that for $\cos(ny)$, it simply the first of the many terms in $\cos^n(y)$ and we may rewrite the summation formula as:

$$(2\cos(n))^k=\cos(nk)+\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

And rearranging terms, we get:

$$\cos(nk)=2^k\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

This becomes explicit formulas for $n=0,1,2,3,\dots$

I note that there is no way by which you may reduce the above formula without the knowledge that $n,k\in\mathbb{Z}$.

Also, it is quite difficult to produce the formulas for, per say, $\cos(10x)$ because as you proceed to do so, you will notice that it requires knowledge of $\cos(8x),\cos(6x),\cos(4x),\dots$, which you can eventually solve, starting with $\cos(2x)$ (it comes out to be the well known double angle formula), using this to find, $\cos(4x)$, use that to find $\cos(6x)$, etc. all the way to $\cos(10x)$.

Notably, this can be easier than Chebyshev Polynomials because it only requires that you know the odd/even formulas less than the one you are trying to solve. (due to $-2i$)

But this is the closest I may give to you for the formula of $\cos(xy)$, $x,y\in\mathbb{R}$.

It is also true for $x,y\in\mathbb{C}$.

Then, use $x=y$ to get $\cos(xx)=\cos(x^2)$

$$\cos(x^2)=\cos^x(x))-\sum_{i=1}^{\infty}\frac{x!}{i!(x-i)!}(\cos(x(x-2i)))$$

It is definitely not periodic in the $2\pi i$ sense, but it is a formula you can use.

Obviously, I don't recommend using it because it is complicated.

• Thank you, your approach is interesting. As pointed out in this comment, there are some typos. At the end, when you write $x!/i!(x-i)!$, it can be useful to notice that this means $x(x-1)\cdots (x-i+1)/i!$ (see here) Feb 17, 2016 at 9:58
• @Watson Thank you. And I do know that. Feb 17, 2016 at 12:30
• @Watson oh so long ago... Jan 17, 2017 at 17:19

$e^{xi} = \cos(x) + i \sin(x)$

$e^{x^2i} = (\cos(x) + i \sin(x))^x$

$e^{-x^2i} = (\cos(x) + i \sin(x))^{-x}$

$$\cos(x^2) = \frac {e^{x^2i} + e^{-x^2i}}{2} = \frac {1}{2}[(\cos(x) + i \sin(x))^x + (\cos(x) + i \sin(x))^{-x}]$$

$$\cos(x^2) = \frac {e^{x^2i} + e^{-x^2i}}{2} = \frac {1}{2}[(\cos(x) + \sqrt {\cos^2(x) - 1})^x + (\cos(x) - \sqrt {\cos^2(x) - 1})^{x}]$$