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The function $f(x)=\log\left(\frac{1+x}{1-x}\right)$ log satisfies the equation |
$f (x + 2)− 2 f (x −1)+ f (x) = 0$ $f (x)+ f (x +1)= f (x) f (x +1)$ $f(x_1)f(x_2)=f(x_1+x_2)$ $f(x_1)+f(x_2)=f\left(\frac{x_1+x_2}{1+x_1x_2}\right)$ |
$f(x_1)+f(x_2)=f\left(\frac{x_1+x_2}{1+x_1x_2}\right)$ |
$f(x_1)+f(x_2)=\log\frac{1+x_1}{1-x_1}+\log\frac{1+x_2}{1-x_2}=f\left(\frac{x_1+x_2}{1+x_1x_2}\right)$ |
