Question #347682

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

1
Expert's answer
2022-06-12T14:57:03-0400
g(y)=y1y,y1g(y)=\dfrac{y}{1-y}, y\not=1

g(y)=y1(y),y1g(-y)=\dfrac{-y}{1-(-y)}, y\not=-1

12[g(y)+g(y)]=12[y1y+y1+y]\dfrac{1}{2}[g(y)+g(-y)]=\dfrac{1}{2}[\dfrac{y}{1-y}+\dfrac{-y}{1+y}]

=12(y+y2y+y21y2)=y21y2=\dfrac{1}{2}(\dfrac{y+y^2-y+y^2}{1-y^2})=\dfrac{y^2}{1-y^2}

=g(y2),y±1=g(y^2), y\not=\pm1

Therefore


12[g(y)+g(y)]=g(y2),y±1\dfrac{1}{2}[g(y)+g(-y)]=g(y^2), y\not=\pm1


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