4

The expansion would be $\sum_{n=0}^\infty$$\frac{1}{2}\choose n $$x^{6n}$

How to evaluate binomial coefficient with rational numbers? If $\frac{1}{2}\choose n $=$2n\choose n $$\times \frac{(-1)^{n+1}}{2^{2n}(2n-1)}$ what would be the expression for binomial coefficient if rational number is $\frac{3}{2}$ instead of $\frac{1}{2}$?

This is one example of Macclaurin expansion for the above expression: $$1+\frac{1}{2x^6}+r(\frac{1}{x^{11}})$$

Can someone show the steps for expanding the expression, and how to generate remain?

Bhaskar Vashishth
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user300045
  • 3,449

3 Answers3

3

The generalised binomial coefficient $\,\dbinom{\tfrac12}n$ is simply: $$\frac{\cfrac12\Bigl(\cfrac12-1\Bigr)\Bigl(\cfrac12-2\Bigr)\dots\Bigl(\cfrac12-n+1\Bigr)}{n\,!}.$$

Bernard
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2

If you want an alternative representation, one can simply recall that

$$\frac{d^n(1+x)^{1/2}}{dx^n}=\frac{(-1)^{n+1}(2n-3)!!}{2^n}(1+x)^{-(2n-1)/2}$$

for $n\ge 2$, where $(2n-3)!!=1\cdot3\cdot5\cdots (2n-3)$ is the double factorical. Then, we have

$$(1+x^6)^{1/2}=1+\frac{x^6}{2}+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}(2n-3)!!}{2^nn!}x^{6n}\tag 1$$

Note that we can write the double factorial in terms of single factorials as

$$\begin{align} (2n-3)!!&=\frac{(2n)!}{(2n)(2n-1)(2n-2)(2n-4)\cdots 2}\\\\ &=\frac{(2n)!}{(2n-1)2^nn!} \tag 2 \end{align}$$

Substituting $(2)$ into $(1)$ gives the expansion

$$\begin{align} (1+x^6)^{1/2}&=1+\frac{x^6}{2}+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}(2n)!}{4^n(2n-1)(n!)^2}x^{6n}\\\\ &=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2n)!}{4^n(2n-1)(n!)^2}x^{6n} \end{align}$$

NOTE:

Another way of writing the generalized binomial coefficient for $\binom{1/2}{n}$ is thus

$$\binom{1/2}{n}=\frac{(-1)^{n+1}(2n)!}{4^n(2n-1)(n!)^2}$$

Mark Viola
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0

When $n$ and $m$ are positive integers, the binomial coefficient $\binom{n}{m}$ is defined as: $$ \binom{n}{m}=\frac{n!}{m!(n-m)!}\tag{1}$$ hence by replacing $a!$ with $\Gamma(a+1)$ it is natural to consider the extended definition: $$\binom{n}{m}=\frac{\Gamma(n+1)}{\Gamma(m+1)\,\Gamma(n-m+1)}\tag{2}$$ from which it follows that: $$\binom{\frac{1}{2}}{n}=\frac{\Gamma(3/2)}{n!\Gamma(3/2-n)}=\frac{\frac{1}{2}\cdot\left(-\frac{1}{2}\right)\cdot\ldots\cdot\left(\frac{3}{2}-n\right)}{n!}=\frac{1}{4^n}\binom{2n}{n}\frac{(-1)^{n+1}}{2n-1}\tag{3}$$ as you stated. Since: $$ \frac{\binom{n+1}{m}}{\binom{n}{m}}=\frac{n+1}{n+1-m}\tag{4}$$ it follows that: $$ \binom{\frac{3}{2}}{n}=\frac{\frac{3}{2}}{\frac{3}{2}-n}\binom{\frac{1}{2}}{n}=\frac{3}{4^n}\binom{2n}{n}\frac{(-1)^n}{(2n-1)(2n-3)}.\tag{5}$$

Jack D'Aurizio
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