抽象函数

常用的几个抽象函数模型：
${\color{Red} ①f(x+y)=f(x)+f(y)\Leftrightarrow f(x)=kx;\qquad} $
${\color{Green} ②f(x+y)=f(x)+f(y)+c\Leftrightarrow f(x)=kx+b;\qquad}$
${\color{Red} ③f(xy)=f(x)+f(y)\Leftrightarrow f(x)=\log_{a}{\left | x \right | } ;\qquad} $
${\color{Green} ④f(x+y)=f(x)f(y)\Leftrightarrow y=a^x} $
${\color{Red}⑤ f(x+y)+f(x-y)=2f(x)f(y)} {\color{Green} \Leftrightarrow y=\cos \omega x} $
${\color{Red}⑥ f(x+y)+f(x-y)=f(x)f(y)} {\color{Green} \Leftrightarrow y=A\cos \omega x} $
${\color{Green} ㈦⑦f(x+y）=f(x)+f(y)+2xy\Leftrightarrow f(x)=x^2} $
${\color{Peach} ⑧f(x+y）=f(x)f(\cfrac{\pi}{2}-y)+f(y)f(\cfrac{\pi}{2}-x) \Longrightarrow f(x)=\sin x} $
${\color{Purple} ⑨f^2(x)-f^2(y)=f(x+y)f(x-y)\Leftrightarrow \sin^2x-\sin^2y=\sin (x+y)\sin (x-y)}$正弦平方差公式
请宝贝们，证明一下正弦平方差定理，在右边往左边证。

设$f(x)=A\cos \omega x\Rightarrow f(x+y)=A(\cos\omega x\cos \omega y-\sin \omega x\sin \omega y);\quad $
$f(x-y)=A(\cos\omega x\cos \omega y+\sin \omega x\sin \omega y)\Rightarrow 左边=f(x+y)+f(x-y)=2A\cos\omega x\cos \omega y$
$右边=f(x)f(y)=A^2\cos\omega x\cos \omega y\mapsto 2A=A^2,A=2$
$f(1)=2\cos \omega =1\Rightarrow \omega=\cfrac{\pi}{3} \Rightarrow f(x)=2\cos \cfrac{\pi}{3} x
$
$2(\cos \cfrac{\pi}{3} +\cos \cfrac{2\pi}{3}+\cos \cfrac{3\pi}{3}+\cos \cfrac{4\pi}{3})=-3$

构造：$f(x)=x^2\log \left | x \right | \qquad $
$f(xy)=(xy)^2\log \left | (xy) \right | =x^2y^2(\log \left | x \right |+\left |y \right |)=y^2f(x)+x^2f(y)$

同第一题$f(x)=2\cos \omega$

左边$＝\log xy+m=\log x+\log y+m;\quad $
右边$=f(x)+f(y)-1=\log x+m+\log y+m-1; \quad $
$m=2m-1\Rightarrow m=1$
$f(4)=\log_{a}{4} +1=2\Rightarrow a=4;\therefore f(x)=\log_{4}{x} +1$
$\therefore f(\cfrac{1}{2} )=\log_{4}{\cfrac{1}{2}} +1=\log_{2^2}{2^{-1}} +1=-\cfrac{1}{2}+1=\cfrac{1}{2}$

设$f(x)=\cos \omega x\because f(4)=\cos 4\omega =-1\Rightarrow \omega =\cfrac{\pi}{4} $
$\therefore f(x)=\cos \cfrac{\pi}{4} x$

$\sin x$

设$f(x)=ax^2+bx+c,故左边＝f(x)+f(y)=ax^2+bx+c+ay^2+by+c;\quad $
右边＝$f(x+y)-xy-1=a(x+y)^2+b(x+y)+c-xy-1=ax^2+ay^2+2axy+b(x+y)+c-xy-1；$
左边＝右边；$\begin{cases} c-1=2c\\2a-1=0\end{cases}$
$a=\cfrac{1}{2},c=-1,f(x) =\cfrac{1}{2}x^2+bx-1,f(1)=1,\Rightarrow b=\cfrac{3}{2}$
$f(x) =\cfrac{1}{2}x^2+\cfrac{3}{2}x-1$

设$f(x)=\sin \omega x,f(1)=\sin \omega =1 \Rightarrow \omega =\cfrac{\pi}{2}$
$f(2x+1)为偶函数，故f(2x+1)=f(-2x+1),即f(x)关于x=1对称；$
$f(0)=0,sin为奇函数，关于(2,0)对称，T=\cfrac{2\pi}{\omega}=4$