Finding an expression and proof for $\sqrt{1-4x}$

Question

This is similar to $$\frac{1}{\sqrt{1-4x}}=\sum_{n\geq0}{2n\choose n}x^n$$

However I want to find an expression the same way for $$\sqrt{1-4x}$$ rather than $$\frac{1}{\sqrt{1-4x}}$$

Here's my thoughts so far:

$$(1-4x)^\frac{1}{2}=\sum_{n\geq0}{\frac{1}{2}\choose n}(-4)^nx^n$$ but this is where I get stuck.

Answer 1

Notice that $$\frac{d}{dx}\sqrt{1-4x}=\frac{-2}{\sqrt{1-4x}}$$ So now, $$\frac{-1}{2}\frac{-2}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n$$ and integrating with respect to $x$ yields $$\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1}=-\frac{1}{2}\sqrt{1-4x} +C $$ so we get that $$\sqrt{1-4x}+C=-2\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1} $$ Letting $x=\frac{1}{4}$, we find that $$C=-2\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^{n+1}(n+1)}=-\frac{1}{2}\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^n(n+1)}$$

Putting it all together, $$\begin{align}\sqrt{1-4x}&=\frac{1}{2}\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^n(n+1)} - 2\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1}\\&=\frac{1}{2}\sum_{n\geq 0} \binom{2n}{n}\frac{1-(4x)^{n+1}}{4^n(n+1)}\end{align}$$

Answer 2

There are many alternatives to your expression. One is $$\sqrt{1-4x}=\sum_{n\geq0}\frac1{1-2n}{2n\choose n}x^n$$ at least for $|x|< \frac14$

Answer 3

Hint:

Use Binomial on $\ \ \ \displaystyle \sqrt {1-4x}. \ $

You would find an expression for this as - $\displaystyle \sqrt {1-4x} = -2\sum_{n\ge 0}\frac1n\binom{2(n-1)}{n-1}x^n$.

Then find the constant term in this expansion i.e at $ n = 0.$

As the above summation isn't valid for $ n = 0.$ So, use this technique:

$\displaystyle \sqrt {1-4x} \ |_{x=0} = -2\sum_{n\ge 0}\frac1n\binom{2(n-1)}{n-1}x^n + k $.

FINAL RESULT:

$$\displaystyle \sqrt {1-4x} = 1-2\sum_{n\ge 1}\frac1n\binom{2(n-1)}{n-1}x^n$$

Answer 4

If the representation \begin{align*} \frac{1}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n \end{align*} for $|x|<\frac{1}{4}$ is supposed to be known, we can also use

\begin{align*} \sqrt{1-4x}&=\frac{1-4x}{\sqrt{1-4x}}\\ &=\sum_{n \geq 0}\binom{2n}{n}x^n-4\sum_{n\geq 0}\binom{2n}{n}x^{n+1}\\ &=1+\sum_{n\geq 1}\left(\binom{2n}{n}-4\binom{2n-2}{n-1}\right)x^n \end{align*}