Understanding extended binomial theorem for coefficient in square roots

Question

In the question, compute the coefficient of $x^{7}$ in $\sqrt{1-7x}$

How come $\sqrt{1-7x}$ can result in a coefficient for $x^{7}$?

Is there some generating function that I don’t see? Or does it it work for any function?

The answer is $$ \left(\frac{1}{_2 C_7} \right) \cdot -7^{7} $$

I know the algorithm, but I don’t see how it works for square roots? As far as I understand, once you have your generating function you plug stuff into the extended binomial theorem assuming that there is a composite function. Then you multiply the combination by the coefficient from the original generating function.

Hence the answer, $ \left(\frac{1}{_2 C_7} \right) \cdot -7^{7} $

Answer 1

Letting $[x^n]\,f(x)$ denote the coefficient of $x^n$ in the Maclaurin series of $f(x)$, we have $$\begin{eqnarray*}[x^7]\sqrt{1-7x}&=&7^7\cdot [x^7]\sqrt{1-x}= 7^6\cdot[x^6]\frac{d}{dx}\sqrt{1-x}\\&=&-\frac{7^6}{2}[x^6]\frac{1}{\sqrt{1-x}}=\color{red}{-\frac{7^6}{2\cdot 4^6}\binom{12}{6}}\end{eqnarray*} $$ due to the well-known $\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n$.

Answer 2

The generalized binomial theorem states that, for $|x| < 1$, we have $(1+x)^a =\sum_{n=0}^{\infty} \binom{a}{n} x^n $ where $\binom{a}{n} =\dfrac{\prod_{k=0}^{n-1} (a-k)}{n!} $.

For $a = \frac12$ this is $\binom{\frac12}{n} =\dfrac{\prod_{k=0}^{n-1} (\frac12-k)}{n!} =\dfrac{ (\frac12)(-\frac12)(-\frac32)...(-\frac12-n+1)}{n!} $.

The first few successive values are

$ \binom{\frac12}{1} =\dfrac{\frac12}{1} =\frac12\\ \binom{\frac12}{2} =\dfrac{\frac12(-\frac12)}{2!} =-\frac18\\ \binom{\frac12}{3} =\dfrac{\frac12(-\frac12)(-\frac32)}{3!} =\frac1{16}\\ \binom{\frac12}{4} =\dfrac{\frac12(-\frac12)(-\frac32)(-\frac52)}{4!} =-\frac{5}{128}\\ ...\\ \binom{\frac12}{7} =\dfrac{\frac12(-\frac12)(-\frac32)(-\frac52)(-\frac72)(-\frac92)(-\frac{11}{2})}{7!} =\frac{33}{2048}\\ $

Putting $-7x$ for $x$ we get $(1-7x)^{1/2} =\sum_{n=0}^{\infty} \binom{\frac12}{n} (-7x)^n =\sum_{n=0}^{\infty} \binom{\frac12}{n} (-7)^nx^n $ so the coefficient of $x^7$ is $\binom{\frac12}{7}(-1)^77^7 =-\frac{33}{2048}7^7 =-\frac{27176919}{2048}. $

Note that, for $n \ge 2$,

$\begin{array}\\ \binom{\frac12}{n} &=\dfrac{\prod_{k=0}^{n-1} (\frac12-k)}{n!}\\ &=\dfrac12\dfrac{\prod_{k=1}^{n-1} (\frac12-k)}{n!}\\ &=\dfrac12\dfrac{\prod_{k=1}^{n-1} (1-2k)}{2^{n-1}n!}\\ &=\dfrac{(-1)^{n-1}}{2}\dfrac{\prod_{k=1}^{n-1} (2k-1)}{2^{n-1}n!}\\ &=\dfrac{(-1)^{n-1}}{2}\dfrac{\prod_{k=1}^{n-1} ((2k-1)(2k))}{2^{n-1}n!\prod_{k=1}^{n-1}(2k)}\\ &=\dfrac{(-1)^{n-1}}{2}\dfrac{(2n-2)!}{2^{n-1}n!2^{n-1}(n-1)!}\\ &=\dfrac{(-1)^{n-1}(2n-2)!}{2^{2n-1}n!(n-1)!}\\ &=\dfrac{(-1)^{n-1}(2n)!}{(2n-1)2^{2n}n!^2}\\ &=\dfrac{(-1)^{n-1}}{(2n-1)4^{n}}\binom{2n}{n}\\ \end{array} $