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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/   
    
Associated Topics:
High School Functions
High School Sequences, Series

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