Functional equations (F) -
- $f(x) + f(cos(x)) = x$ [ Orignal equation ]
- $f(-x)+f(cos(x))+x=0$ [ From (F$1$) ]
- $f(x)+f(-x)+2f(cos(x))=0$ [ From (F$1$) and (F$2$) ]
- $f(x)-f(-x)=2x$ [ From (F$1$) and (F$2$), Found by OP ]
- $f(x+\pi)-f(x)=2cos(x)+\pi$ [ From F$1$, Found by OP ]
- $f(cos^{-1}(x))+f(x)=cos^{-1}(x)$ [ From F$1$ ]
- $f(x)=f(x+2n\pi)-2n\pi, n\in \mathbb{Z}$ [ Derived below, D$1$ ]
- $f(x)=x+\{ f(x+(2n+1)\pi ) + (x+(2n+1)\pi )\}, n\in \mathbb{Z} $ [ Derived below, D$2$ ]
- $f'(x)+f'(-x)=2$ [ From F$4$ ]
- $f''(x)=f''(-x)$ [ From F$9$ ]
- $f'(x)-sin(x)\cdot f'(cos(x))=1 \Leftrightarrow f'(cos(x)={{(f'(x)-1)}\over {sinx}}$ [ From F$1$ ]
- $f''(x)+sin^{2}(x)\cdot f''(cos(x))=cos(x)\cdot f'(cos(x)) ={{(f'(x)-1)}\over {sin(x)}}\cdot cos(x)$ [ 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) -
- $f(w)=w/2$ [ Found by OP ]
- $f'(w)=\frac{1}{1-\sqrt{1-w^2}}$ [ Found by OP ]
- $f''(w)=\frac{1}{2-w^2}\frac{w}{1-\sqrt{1-w^2}}$ [ Found by OP ]
- $f'(0)=1$ [ Found by OP ]
- $f'(\pi /2)=2$ [ Found by OP ]
- $f'(-\pi /2)=0$ [ Found by OP ]
- $f(0)+f(1)=0$
- $f(\pi )-\pi = f(0)+2$
- $f(2\pi )- f(\pi)=\pi -2$
- $f(2\pi)=f(0)+2\pi =2\pi - f(1)$
Solutions of functional equations (S) -
I shall limit myself to solutions of F$1$ and F$4$.
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)-
- 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.
- 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.
Best Answer
We have $$f(x) = f\left(\dfrac{x}2\right)\cos\left(\dfrac{x}2\right) = f\left(\dfrac{x}4\right)\cos\left(\dfrac{x}4\right)\cos\left(\dfrac{x}2\right)$$ Hence, we have $$f(x) = f\left(\dfrac{x}{2^{n}}\right) \prod_{k=1}^n \cos\left(\dfrac{x}{2^k}\right) = \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right)\dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n} \right)}$$ Hence, we have that $$f(x) = \lim_{n \to \infty} \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right) \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \lim_{n \to \infty}f\left(\dfrac{x}{2^n}\right) \lim_{n \to \infty} \dfrac1{2^n} \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \dfrac{\sin(x)}{x} \lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right)$$ If we assume that $\lim_{x \to 0} f(x)$ exists at the origin, we then have that $$\lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right) = c$$ from which we obtain that $$f(x) = c\cdot \dfrac{\sin(x)}x$$