Desarrollo en serie de Taylor al cuadrado

Question

Considera la siguiente expansión $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$

Demuestra que esta ecuación se cumple elevando al cuadrado ambos lados y comparando los términos hasta $x^3$.

Me pregunto, ¿cómo puedo elevar al cuadrado el lado derecho?

Answer 1

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

\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} Hay que demostrar que el coeficiente de $\ds{x^{n}}$ $\ds{\pars{~\color{#00f}{\mbox{la expresión azul}}~}}$ es igual a $\ds{1}$ cuando $\ds{n = 0,1}$ y, de lo contrario, se anula.

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}

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