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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?

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  • $\begingroup$ would it be: $1+x=1+(ist-2nd+3rd...)(1st-2nd+3rd...)$ $\endgroup$
    – 123
    Sep 21 '14 at 22:07
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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

$$\left(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 ..\right)^2=\underbrace{ 1^2+(2\times 1\times\frac12 x)}_{=1+x}+\underbrace{(\frac12 x)^2-2\times \frac18x^2}_{=0}\\\underbrace{-2\times \frac12x\frac18x^2+2\times\frac1{16}x^3}_{=0}+\cdots$$

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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?

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  • $\begingroup$ If you identify $c=-x^2/8$, yes. $\endgroup$
    – Dmoreno
    Sep 21 '14 at 22:14
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$$ \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + \dots \right)^2 = \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + O(x^4) \right)^2 =$$ $$= 1+ \dfrac{1}{4}x^2 + \dfrac{1}{64}x^4 + \dfrac{1}{256} x^6+1-\dfrac{1}{4}x^2 - \dfrac{1}{8}x^3 + \dfrac{1}{8}x^3 +\dfrac{1}{16}x^4 - \dfrac{1}{64}x^5 + O(x^4) =$$ $$=1+x + O(x^4)$$

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  • $\begingroup$ Where does the $x$ come from? $\endgroup$
    – johnaton
    Sep 21 '14 at 22:14
  • $\begingroup$ $x$ has to be considered as a real number in a (narrow) neighbourhood of $0$. It's the same $x$ in your question. $\endgroup$
    – Crostul
    Sep 21 '14 at 22:16
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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.

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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 : $$[t^2]_3=x^2(\frac 14 -\frac 18 x)$$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\root{1 + x}\equiv \sum_{k\ =\ 0}^{\infty}{1/2 \choose k}x^{k}}$.

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

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