Expansion of (x+y)^(1/2)Date: 06/07/99 at 05:54:31 From: Dheera Subject: Expansion of (x+y)^(1/2) Is there a way to expand (x+y)^(1/2)? And if so, how is it derived? Date: 06/07/99 at 13:08:22 From: Doctor Rob Subject: Re: Expansion of (x+y)^(1/2) Thanks for writing to Ask Dr. Math! Yes, there is, and it's due to Sir Isaac Newton. Just use the Binomial Theorem: (x+y)^n = x^n + n/1!*x^(n-1)*y + n*(n-1)/2!*x^(n-2)*y^2 + n*(n-1)*(n-2)/3!*x^(n-3)*y^3 + ..., (x+y)^(1/2) = x^(1/2) + (1/2)/1!*x^(-1/2)*y + (1/2)*(-1/2)/2!*x^(-3/2)*y^2 + (1/2)*(-1/2)(-3/2)/3!*x^(-5/2)*y^3 + ..., = x^(1/2)*[1 + (1/2)/1!*(y/x) + (1/2)*(-1/2)/2!*(y/x)^2 + (1/2)*(-1/2)(-3/2)/3!*(y/x)^3 + ...]. The difference here is that the sum on the right doesn't have just n+1 terms, but infinitely many. This is called an infinite series. If y/x <= 1, that is, y <= x, the part in [...] converges, and the formula makes sense. If, instead, y >= x, then swap the roles of x and y, and again, you can succeed. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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