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Solution to puzzle 148: Power series

Skip restatement of puzzle.Find the power series (expanded about x = 0) for Square root of ((1+x)/(1-x)).

We write Square root of ((1+x)/(1-x)) as (1 + x)(1 − x2)−½, and expand the latter term as a binomial series.  We have

(1-x^2)^(-1/2) = 1 - x^2/2 + (-1/2)*(-3/2)x^4/2! + ...

The coefficient of the general term, x2n, is given by 1*3*5*...*(2n-1)/(2^n * n!).

The numerator of this expression is a double factorial: (2n − 1)!!.
The denominator may then be written in a similar form by "absorbing" the factors of 2n into n!, giving (2n)!!.

(Alternatively, we could multiply numerator and denominator by 2n n!, and simplify the expression, obtaining C(2n,n)/4^n .)

Therefore, the power series expansion of Square root of ((1+x)/(1-x)) = Sum over nonnegative n of ((2n-1)!!/(2n)!!) (x^2n + x^(2n+1))..

Source: To be announced

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