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I read somewhere that the coefficient of $(1+x)^n$ is $1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots.$

For $n=-\frac{1}{2},$ the denominator becomes $2^n \cdot n!,$ while the numerator becomes $1 \cdot 3 \cdot \dots \cdot (2n-1) = \frac{(2n!)}{2^n \cdot n!}.$ The whole coefficient becomes $\frac{\binom{2n}{n}}{4^n},$ and the sign of the coefficient would be $(-1)^n.$

However, what happens when we are expanding $(1-x)^n.$ More specifically $(1-x)^{-\frac{1}{2}}.$ I have seen some solutions out there, but they gave differing answers with complicated processes. Is there a technique similar to the one above for it? It seems like there should, but just tweaking the above one would include raising $-1$ to a real, possibly negative, power.

Main question: What are the coefficients of $x^n$ for an integer $n$ in $(1-x)^{-\frac12}$?

Sub-question: Is there a technique similar to the one mentioned for $(1+x)^n$ to find the coefficients for $(1-x)^n$? I believe we cannot just plug in (-x) because it would involve raising $-1$ to any real power.

Mike Smith
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  • Just replace $x^k$ with $(-x)^k$ in the expansion of $(1+x)^n$ and you obtain the expansion of $(1-x)^n.$ Example $n=-1/2$ – Anne Bauval May 29 '23 at 11:36
  • I guess I am a bit confused by the intuitive reasoning behind the generalized definition of nCr for some real n. Is there a specific reason it was defined as such? How do we know it's "correct" to define it as such?

    We are taking it to the -1/2 power, so how does it even have terms with exponents of x greater than n?

     – Mike Smith May 29 '23 at 12:00
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    https://en.wikipedia.org/wiki/Binomial_series – Anne Bauval May 29 '23 at 12:06
  • Please focus your question on a specific point and clearly state it so that you can get the answer you want. – NoChance May 29 '23 at 12:11
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    @NoChance Does that help? – Mike Smith May 29 '23 at 12:50
  • Yes this is much better. As for the sub-question, Taylor Series comes to mind. Using n=1/2 looks tempting. – NoChance May 29 '23 at 12:54
  • The "subquestion" is unclear. Why would it "involve raising −1 to any real power"? The exponents of $x$ in $1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots$ are non-negative integers, hence so are the exponents of $x$ in $1 - nx + \frac{n(n-1)}{2!}x^2 - \cdots.$ Please, look again at the first comment above and the link. – Anne Bauval May 29 '23 at 13:03
  • Do you have a problem with the geometric series $(1-x)^{-1}=1+x+x^2+x^3+\dotsb$ having exponents going to infinity even though the power is $-1$? – Gerry Myerson May 29 '23 at 13:13
  • @AnneBauval sorry that was a typo – Mike Smith May 30 '23 at 19:09
  • @GerryMyerson now that I think of it, it does not make the most sense either. I know how to algebraicly solve for its value which just gave me some more assurance I guess. – Mike Smith May 30 '23 at 19:10

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