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In response to my article on Taking Square Roots, I received email asking about a power series (a series like e^x=1+x+x^2/2!+x^3/3!+...) for taking square roots. I thought I had once derived such a series, but I could not find it. But, I got an idea, and here it is. How about the Binomial Theorem, which says:
(x+y)^n = x^n + nx^(n-1)y + n(n-1)y^2/2! + n(n-1)(n-2)y^3/3! ...
If we make x=1 and n=1/2 (square root), we get:
(1+y)^(1/2) = 1 + y/2 - y^2/4(2!) + 3y^3/8(3!) - 3(5)y^4/16(4!) + 3(5)(7)y^5/32(5!) ...
If we want to find the square root of some number z, then z=1+y, which means that y=z-1. But we find that y must be in the range -1<y<1, so 0<z<2, this series only works when z is between 0 and 2, not inclusive. We could use another series for z>=2. Let's find the series for (1+y)^(-1/2):
(1/(1+y))^(1/2) = 1 - y/2 + 3y^2/4(2!) + 3(5)y^3/8(3!) - 3(5)(7)y^4/16(4!) ...
Here, you can use z=(1/(1+y)), which makes y=1/z-1. And again -1<y<1, which makes z>1/2. So, with this series, we can find the square root of anything greater than 1/2. Actually, it's slow for z>10.
Note: y^2 means y squared. y^(1/2) is the square root of y. I could have tidied up the equations with superscipts. But the equations would be messy anyway, in HTML. So, please forgive me for the notation.