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Thursday, June 23, 2011

2011/057) Given |a | < 1 and |b| < 1


If 1+a+a^2+\cdots =x and 1+b+b^2+\cdots.=y then find 1+ab+a^2.b^2+.... in terms of x and y

Solution
we have
x = \dfrac{1}{1-a}
or (1-a) = \dfrac{1}{x}
or a = 1 – \dfrac{1}{x} = \dfrac{x-1}{x}
similarly
b = \dfrac{y-1}{y}
now as |a | and |b| both < 1 so |ab| < 1 and hence
1+ ab + a^2b^2 + a^3b^3 + \cdots= \dfrac{1}{1-ab} = \dfrac{1}{1- \frac{x-1}{x} * \frac{y-1}{y}}
= \dfrac{xy}{xy - (x - 1)(y - 1)}
= \dfrac{xy}{x + y - 1} 

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