Question

Write the Maclaurin series for $\frac{1}{\sqrt{1+x}}$in$\sum$ form using the binomial coefficient notation. Then find a formula for the binomial coefficients in terms ofn as we did in Example $2$ above

Short answer

The Maclaurin series is $\left(\frac{-1}{2}\right)=\frac{(-1{)}^n(2n-1)!!}{\left(2n\right)!!}$

Step-by-step solution

Given Information

Thebinomial series.

Definition of the binomial series.

The Taylor series for the function given by is the binomial series, where is an arbitrary complex number.

Prove the statement.

The binomial series states that $(1+x{)}^p=\sum _{n=0}^{\infty }\left(\begin{array}{l}p\\ n\end{array}\right)x^n$

$(1+x{)}^{-1/2}=\sum _{n=0}^{\infty }\left(\begin{array}{c}-1\\ n\end{array}\right)x^n$

$=1-\frac{1}{2}x+\frac{\left(\frac{-1}{2}\right)\left(\frac{-1}{2}-1\right)}{2!}{x}^2+\frac{\left(\frac{-1}{2}\right)\left(\frac{-1}{2}-1\right)\left(\frac{-1}{2}-2\right)}{3!}{x}^3+\frac{\left(\frac{-1}{2}\right)\left(\frac{-1}{2}-1\right)\left(\frac{-1}{2}-2\right)\left(\frac{-1}{2}-3\right)}{4!}{x}^4+\dots$

$=1-\frac{1}{2}x+\frac{\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)}{2!}{x}^2+\frac{\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\left(\frac{-5}{2}\right)}{3!}{x}^3+\frac{\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\left(\frac{-5}{2}\right)\left(\frac{-7}{2}\right)}{4!}{x}^4+\dots$

$=1-\frac{1}{2}x+\frac{3}{8}{x}^2-\frac{5}{16}{x}^3+\frac{35}{128}{x}^4+\dots$

The formula states that $\left(\begin{array}{l}p\\ n\end{array}\right)=\frac{p(p-1)(p-2)\dots (p-n+1)}{n!}$

For $n\ge 2$, the formula becomes as follows.

$\left(\begin{array}{l}\frac{-1}{2}\\ n\end{array}\right)=\frac{\left(\frac{-1}{2}\right)\left(\frac{-1}{2}-1\right)\left(\frac{-1}{2}-2\right)\left(\frac{-1}{2}-3\right)\left(\frac{-1}{2}-(n-1)\right)}{n!}$

$=\frac{\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\left(\frac{-5}{2}\right)\left(\frac{-7}{2}\right)\left(\frac{1}{2}-n\right)}{n!}$

$\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right)=\frac{(-1{)}^{n}1·3·5·7\dots (2n-1)}{2^{n}n!}$

$\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right)=\frac{(-1{)}^n(2n-1)!!}{\left(2n\right)!!}$

The statement has been proven.