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If there any way to expand the following?

$$\left(\sum_{i=1}^nx_i\right)^{\frac{1}{2}}$$

and more generally, a way to expand

$$\left(\sum_{i=1}^nx_i\right)^{\frac{p}{q}}$$

where $\gcd(p,q) = 1$

More closed to my original problem, is there any formula for: $$\left(\sum_{i=1}^nx_i\right)^{\frac{1}{2}} - \sum_{i=1}^nx_i^{\frac{1}{2}}$$

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  • $\begingroup$ What do you mean by "formula"? $\endgroup$
    – Igor Rivin
    Jun 2, 2012 at 19:30
  • $\begingroup$ This really does not see to be of the appropriate level. Voting to close. I am sure stackexchange would be more receptive. $\endgroup$
    – Igor Rivin
    Jun 2, 2012 at 19:46

1 Answer 1

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I think, you can use binomial theorem for rational exponents inductively for expanding $(\sum_{i=1}^{n}x_i)^\frac{1}{2}$. I am not very sure whether one can use multinomial theorem for rational exponents.

When you expand $(1+x)^{\frac{1}{2}}$, It has infinitely many terms, so to your other question, It is very less likely to have any formula, as one of the term in the expansion has infinitely many terms while $\sum_{i=1}^{n}x_i^{\frac{1}{2}}$ has only finitely many.

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