Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$

Show this equation holds by squaring both sides and comparing terms up to $x^3$.

I wonder, how can I square the right hand side? 
 A: Notice that
$$(a+b+c+d)^2=\underbrace{a^2+b^2+c^2+d^2}_{\text{the sum of square of all terms}}+\underbrace{2ab+2ac+2ad+2bc+2bd+2cd}_{\text{the sum of the double products of the terms taken 2 by 2 }}$$
so we find
$$\left(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 ..\right)^2=\underbrace{ 1^2+(2\times 1\times\frac12 x)}_{=1+x}+\underbrace{(\frac12 x)^2-2\times \frac18x^2}_{=0}\\\underbrace{-2\times \frac12x\frac18x^2+2\times\frac1{16}x^3}_{=0}+\cdots$$
A: Hint:
\begin{align}
(a + b + c + \ldots )^2  & =  (a+b+c+\ldots) \times (a+b+c+\ldots)   \\
& = a^2 + ba + ca + \ldots + a b + b^2 + cb + \ldots ac + bc +c ^2 + \ldots 
\end{align}
Can you take it from here?
A: $$ \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + \dots \right)^2 = \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + O(x^4) \right)^2  =$$
$$= 1+ \dfrac{1}{4}x^2 + \dfrac{1}{64}x^4 + \dfrac{1}{256} x^6+1-\dfrac{1}{4}x^2 - \dfrac{1}{8}x^3 +  \dfrac{1}{8}x^3 +\dfrac{1}{16}x^4 - \dfrac{1}{64}x^5 + O(x^4) =$$
$$=1+x  + O(x^4)$$
A: You can put $t=\frac 12 x-\frac 18 x^2 + \frac 1{16}x^3$ then, you get:
$$\sqrt{\sqrt{1+x}}=\sqrt{1+t}=1+ \frac 12 t--\frac 18 [t^2]_3 + \frac 1{16}[t^3]_3 ... $$
where $[P(x)]_3$ for a polynom $$P(x)=a_0+a_1 x +a_2x^2+a_3x^3+a_4x^4 ... +a_n x^n $$ is:
$$[P(x)]_3=a_0+a_1x+a_2x^2+a_3x^3.$$
For example , since: $t=x(\frac 12 -\frac 18 x + \frac 1{16}x^2)$  we have :
$$t^2=x^2(\frac 12 -\frac 18 x + \frac 1{16}x^2)^2$$ then : $$[t^2]_3=x^2(\frac 14 -\frac 18 x)$$
A: When you square the right-hand side, you will get another (infinite) polynomial $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots$ . Now go through the coefficients $a_i$ and ask yourself which terms in the product $\big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big) \big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big)$ will contribute to $a_i$.
For example, there is only one product that contributes to $a_0$, namely $1 \cdot 1$, since the degree of $x$ in every other term is at least 1. Next, to get $a_1$, you can multiply $1$ from the first factor with ${1 \over 2} x$ from the second factor, or ${1 \over 2} x$ from the first and $1$ from the second. Thus $a_1 = 1\cdot {1 \over 2} + {1 \over 2} \cdot 1 = 1$. Continue similarly for the rest.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\root{1 + x}\equiv \sum_{k\ =\ 0}^{\infty}{1/2 \choose k}x^{k}}$.

\begin{align}
1 + x&=\pars{\root{1 + x}}^{2}
=\sum_{k\ =\ 0}^{\infty}{1/2 \choose k}x^{k}
\sum_{j\ =\ 0}^{\infty}{1/2 \choose j}x^{j}
\\[5mm]&=\sum_{k\ =\ 0}^{\infty}{1/2 \choose k}x^{k}
\sum_{j\ =\ 0}^{\infty}{1/2 \choose j}x^{j}
\sum_{n\ =\ 0}^{\infty}\delta_{n,k\ +\ j}
=\sum_{n\ =\ 0}^{\infty}x^{n}\bracks{%
\sum_{k\ =\ 0}^{\infty}\sum_{j\ =\ 0}^{\infty}{1/2 \choose k}
{1/2 \choose j}\delta_{j,n\ -\ k}}
\\[5mm]&=\sum_{n\ =\ 0}^{\infty}x^{n}\bracks{%
\color{#00f}{\sum_{k\ =\ 0}^{\infty}{1/2 \choose k}{1/2 \choose n - k}}}
\end{align}
  We have to prove the coefficient of $\ds{x^{n}}$
  $\ds{\pars{~\color{#00f}{\mbox{the blue expression}}~}}$ is equal to $\ds{1}$ when
  $\ds{n = 0,1}$ and, otherwise, it vanishes out.

However,
\begin{align}
&\color{#00f}{\sum_{k\ =\ 0}^{\infty}{1/2 \choose k}{1/2 \choose n - k}}
=\sum_{k\ =\ 0}^{\infty}{1/2 \choose k}\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{1/2}
\over z^{n - k + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{1/2}
\over z^{n + 1}}\sum_{k\ =\ 0}^{\infty}{1/2 \choose k}z^{k}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{1/2}
\over z^{n + 1}}\pars{1 + z}^{1/2}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 + z \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
={1 \choose n}=
\left\lbrace\begin{array}{lcrcl}
1 & \mbox{if} & n & = & 0
\\
1 & \mbox{if} & n & = & 1
\\
0&&&& \mbox{otherwise}
\end{array}\right.
\end{align}

Then,
  $$
1 + x = \sum_{n\ =\ 0}^{\infty}x^{n}{1 \choose n}= 1 + x
$$

