Answer to Question #347682 in Calculus for Stella

Question

If g(y) = y/(1-y), show that 1/2[g(y) + g(-y)]= g(y^2).

Answer 1

$$g(y)=\dfrac{y}{1-y}, y\not=1$$

$$g(-y)=\dfrac{-y}{1-(-y)}, y\not=-1$$

$$\dfrac{1}{2}[g(y)+g(-y)]=\dfrac{1}{2}[\dfrac{y}{1-y}+\dfrac{-y}{1+y}]$$

$$=\dfrac{1}{2}(\dfrac{y+y^2-y+y^2}{1-y^2})=\dfrac{y^2}{1-y^2}$$

$$=g(y^2), y\not=\pm1$$

Therefore

$$\dfrac{1}{2}[g(y)+g(-y)]=g(y^2), y\not=\pm1$$