# Strange functional equation: $f(x)+f(\cos(x))=x$

BACKGROUND: A while ago, I became obsessed for a period of time with the following functional equation: $$f(x)+f(\cos(x))=x$$ I am only considering the unique real analytic solution to this functional equation (it is indeed unique... all derivatives of $f$ at $w\approx 0.739$, the Dottie Number, can be determined by differentiating this equation repeatedly). Here is a graph of the function: I had already given up a long time ago when I accidentally came across my notes on this problem today, and I figured I would put it one MSE to see if anyone could find anything interesting that I missed.

WHAT I FOUND: I am almost absolutely certain that there is no nice closed-form for this function, so I resorted to finding special values and other neater functional equations. So far, I have found the following special values of $f$ and its derivatives: $$f(w)=w/2$$ $$f'(w)=\frac{1}{1-\sqrt{1-w^2}}$$ $$f''(w)=\frac{1}{2-w^2}\frac{w}{1-\sqrt{1-w^2}}$$ $$f'(0)=1$$ $$f'(\pi/2)=2$$ $$f'(-\pi/2)=0$$ Here are some functional equations I have found for $f$: $$f(x+2\pi)-f(x)=2\pi$$ $$f(x+\pi)-f(x)=2\cos(x)+\pi$$ $$f(x)-f(-x)=2x$$ And here is a series representation for $f$, where $\cos^{\circ n}$ represents the cosine function composed $n$ times: $$f(x)=\frac{w}{2}+\sum_{n=0}^\infty \big(\cos^{\circ 2n}(x)-\cos^{\circ 2n+1}(x)\big)$$

QUESTIONS: This is a very open-ended question. I have a few unproven conjectures or particular unanswered questions about this function, but I really just want to see what interesting properties (especially special values, zeroes, maxima or minima, inflection points, and functional or differential equations) people can find.

An example of one of my conjectures: by looking at graphs, I have conjectured that $f(x+a)-f(x)$ is a sinusoid or a sum of sinusoids for all $a$. It is easy to show that this must be periodic, but I would like to prove or disprove that it can be expressed as a sum of sinusoids. In particular, can we find an expression for $f(x+\pi/2)-f(x)$ as a sum of sinusoids, possibly involving other mathematical constants like $w$? Here is a graph of $f(x+\pi/2)-f(x)$: I appreciate any contributions!

• The $g(x)=\sin(x)^2+x-\cos(x)$ seems to be pretty close approximation to $f(x)$ near origin. Also matches lots of your identities. – Sil Aug 18 '18 at 18:13
• @Sil Hmm, that is interesting... it matches very closely. – Franklin Pezzuti Dyer Aug 18 '18 at 18:19
• The function $h(x) = f(x) - x$ is $2\pi$-periodic and even. As you are only interested in continuous solutions, it can therefore be written as a cosine-series, $$h(x) = \sum_{n=0}^{\infty} c_n \cos(nx)$$ According to @Sil's comment, the first terms are $$\sin(x)^2-\cos(x) = \left( \frac12 - \frac12\cos(2x) \right) - \cos(x) = \frac12 - \cos(x) - \frac12\cos(2x).$$ – md2perpe Aug 18 '18 at 23:39
• Have you looked at $f$ in the frequency domain? – Jack M Aug 19 '18 at 0:53
• Maybe Fourier can be used to express $f(x+\pi/2)-f(x)$ as a sum of sines. – Gonzalo Benavides Aug 19 '18 at 1:13

# Functional equations (F) -

1. $f(x) + f(cos(x)) = x$ [ Orignal equation ]
2. $f(-x)+f(cos(x))+x=0$ [ From (F$1$) ]
3. $f(x)+f(-x)+2f(cos(x))=0$ [ From (F$1$) and (F$2$) ]
4. $f(x)-f(-x)=2x$ [ From (F$1$) and (F$2$), Found by OP ]
5. $f(x+\pi)-f(x)=2cos(x)+\pi$ [ From F$1$, Found by OP ]
6. $f(cos^{-1}(x))+f(x)=cos^{-1}(x)$ [ From F$1$ ]
7. $f(x)=f(x+2n\pi)-2n\pi, n\in \mathbb{Z}$ [ Derived below, D$1$ ]
8. $f(x)=x+\{ f(x+(2n+1)\pi ) + (x+(2n+1)\pi )\}, n\in \mathbb{Z}$ [ Derived below, D$2$ ]
9. $f'(x)+f'(-x)=2$ [ From F$4$ ]
10. $f''(x)=f''(-x)$ [ From F$9$ ]
11. $f'(x)-sin(x)\cdot f'(cos(x))=1 \Leftrightarrow f'(cos(x)={{(f'(x)-1)}\over {sinx}}$ [ From F$1$ ]
12. $f''(x)+sin^{2}(x)\cdot f''(cos(x))=cos(x)\cdot f'(cos(x)) ={{(f'(x)-1)}\over {sin(x)}}\cdot cos(x)$ [ From F$11$ ] $\Leftrightarrow sin(x)\cdot f''(x)-cos(x)\cdot f'(x)+sin^{3}(x)\cdot f''(cos(x))+cos(x)=0$

# Special values and their relations (V) -

## (Derived from various F) -

1. $f(w)=w/2$ [ Found by OP ]
2. $f'(w)=\frac{1}{1-\sqrt{1-w^2}}$ [ Found by OP ]
3. $f''(w)=\frac{1}{2-w^2}\frac{w}{1-\sqrt{1-w^2}}$ [ Found by OP ]
4. $f'(0)=1$ [ Found by OP ]
5. $f'(\pi /2)=2$ [ Found by OP ]
6. $f'(-\pi /2)=0$ [ Found by OP ]
7. $f(0)+f(1)=0$
8. $f(\pi )-\pi = f(0)+2$
9. $f(2\pi )- f(\pi)=\pi -2$
10. $f(2\pi)=f(0)+2\pi =2\pi - f(1)$

# Solutions of functional equations (S) -

I shall limit myself to solutions of F$1$ and F$4$.

1. ## Solution of F$4$ -

From F$10$, we see that any even function, $e(x)=f''(x)$, would lead to -

$f(x)=\int (\int e(x) dx) dx +c_1\cdot x + c_2$, the required solution.

Since, integration of an even function is odd and vice-verse the double-integral above yields an even function, giving - $f(x)=e(x)+c_1\cdot x + c_2$.

We may generalise the above to -

$f(x) = \Sigma (c_i \cdot e_i(x)) + x + c$, where $e_i$ is an even function

Example-

$f_n(x)= c_0 . (\Sigma_{i=1}^{n}a_ix^{2i})(1+\vert x \vert ) +x +c$

Further, since the modulus of an odd function is an even function, we may define-

$E(x)=\Sigma (e_i(x))) + \Sigma (\vert o_i(x)\vert )$, where $o_i$ are odd functions, $e_i$ are even functions except those obtained from odd functions by taking their modulus and the summation is over all possible functions.This shall yield -

$f(x) = E(x) + c_1\cdot x +c$

# Derivation (D)-

1. Using, $f(cos(x)=x-f(x) and cos (x+2n\pi)=cos(x)$

We have, $x-f(x) = (x+2n\pi ) - f(x+2n\pi )$

The result follows from simplification.

1. Using, $f(cos(x)=x-f(x) and cos (x+(2n+1)\pi)=-cos(x)$

We have, $x-f(x) = f(x+(2n+1)\pi ) - (x+(2n+1)\pi )$

The result follows from simplification.

• I am not sure F11 is correct. – user254433 Sep 4 '18 at 19:28
• @user254433 Corrected, of course, i shall now have to find some other method to solve F$1$. – The Long Night Sep 5 '18 at 14:42
• @user254433 Thanks for pointing it out, by the way :D – The Long Night Sep 5 '18 at 15:16