How do you binomially expand this?

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SUMMARY

The discussion focuses on the binomial expansion of the expression (x^2 + y^2)^0.5 using the generalized binomial theorem. The theorem states that (x + y)^\alpha can be expressed as a sum involving generalized binomial coefficients, defined as \(\left(\begin{array}{c}\alpha \\ i\end{array}\right) = \frac{\alpha(\alpha + 1)(\alpha - 2) \cdots (\alpha - i - 1)}{i!}\). For the case where \(\alpha = 1/2\), the expansion results in an infinite series, specifically (x^2 + y^2)^\alpha = \sum \left(\begin{array}{c}\alpha \\ i\end{array}\right)x^{2i} y^{2(\alpha - i)}.

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(x^2+y^2)^0.5
 
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You can look at the "generalized binomial theorem" here:
http://en.wikipedia.org/wiki/Binomial_series

Basically, it says that
[tex](x+ y)^\alpha= \sum \left(\begin{array}{c}\alpha \\ i\end{array}\right)x^i y^{\alpha- i}[/tex]
where
[tex]\left(\begin{array}{c}\alpha \\ i\end{array}\right)= \frac{\alpha(\alpha+ 1)(\alpha- 2)\cdot\cdot\cdot(\alpha- i-1)}{i!}[/tex]
is the "generalized binomial coefficient". If [itex]\alpha[/itex] is a positive integer, the "generalized binomial coefficient" is the usual binomial coefficient and is eventually 0 so the sum is finite. If [itex]\alpha[/itex] is not a positive integer (and for your problem, it is 1/2) the sum is an infinite series.

With [itex]x^2[/itex] and [itex]y^2[/itex] instead of x and y, it just becomes
[tex](x^2+ y^2)^\alpha= \sum \left(\begin{array}{c}\alpha \\ i\end{array}\right)x^{2i} y^{2(\alpha- i)}[/tex]
 

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