1. ## maclaurin expansion

Hi there, I am trying to find the Maclaurin expansion of 3rd degree of the function $\displaystyle f(x)=\sqrt[3]{1+3\sin{x}}$ but I get strange result ..
I need to ask what am I doing wrong ..
I started with
$\displaystyle (1+z)^{\frac{1}{3}}=1+z+\frac{\frac{1}{3}(-\frac{2}{3})}{2!}z^2+\frac{\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})}{3!}z^3+R(z)$
where for $\displaystyle R(z)$ there holds $\displaystyle R(z)=o(z^3)$ ...

then I substituted $\displaystyle z=3\sin{x}$ and I got

$\displaystyle (1+3\sin{x})^{\frac{1}{3}}=1+3\sin{x}-\frac{1}{9}(3\sin{x})^2+\frac{5}{3^4}(3\sin{x})^3+ R(3\sin{x})$

where

$\displaystyle R(3\sin{x})=o(x^3)$ since $\displaystyle \lim_{x\to 0}\frac{R(3\sin{x})}{x^3}=\lim_{x\to 0}\frac{27\sin^3{x}}{x^3}\frac{R(3\sin{x})}{27\sin ^3{x}}=27\lim_{z\to 0}\frac{R(z)}{z^3}=0$ ..

after that I used the Maclaurin expansion of $\displaystyle \sin{x}$ and I got

$\displaystyle (1+3\sin{x})^{\frac{1}{3}}=1+3\sin{x}-\sin^2{x}+\frac{5}{3}\sin^3{x}+R(3\sin{x})=$
$\displaystyle =1+3\bigg(x-\frac{x^3}{3!}+o(x^3)\bigg)-(x+o(x^2))^2 +\frac{5}{3}(x+o(x^2))^3+o(x^3)=$
$\displaystyle = 1+3x-x^2+\frac{7}{6}x^3+o(x^3)$

but the result should be $\displaystyle 1+x-x^2+\frac{3}{2}x^3$

where's the mistake then?
thanks for any help

2. ## Re: maclaurin expansion

[QUOTE=alteraus;794113]Hi there, I am trying to find the Maclaurin expansion of 3rd degree of the function $\displaystyle f(x)=\sqrt[3]{1+3\sin{x}}$ but I get strange result ..
I need to ask what am I doing wrong ..
I started with
$\displaystyle (1+z)^{\frac{1}{3}}=1+z+\frac{\frac{1}{3}(-\frac{2}{3})}{2!}z^2+\frac{\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})}{3!}z^3+R(z)$
where for $\displaystyle R(z)$ there holds $\displaystyle R(z)=o(z^3)$ ...
This is, at least, one problem: the first two terms should be $\displaystyle 1+ \frac{1}{3}z$, not $\displaystyle 1+ x$.

then I substituted $\displaystyle z=3\sin{x}$ and I got

$\displaystyle (1+3\sin{x})^{\frac{1}{3}}=1+3\sin{x}-\frac{1}{9}(3\sin{x})^2+\frac{5}{3^4}(3\sin{x})^3+ R(3\sin{x})$

where

$\displaystyle R(3\sin{x})=o(x^3)$ since $\displaystyle \lim_{x\to 0}\frac{R(3\sin{x})}{x^3}=\lim_{x\to 0}\frac{27\sin^3{x}}{x^3}\frac{R(3\sin{x})}{27\sin ^3{x}}=27\lim_{z\to 0}\frac{R(z)}{z^3}=0$ ..

after that I used the Maclaurin expansion of $\displaystyle \sin{x}$ and I got

$\displaystyle (1+3\sin{x})^{\frac{1}{3}}=1+3\sin{x}-\sin^2{x}+\frac{5}{3}\sin^3{x}+R(3\sin{x})=$
$\displaystyle =1+3\bigg(x-\frac{x^3}{3!}+o(x^3)\bigg)-(x+o(x^2))^2 +\frac{5}{3}(x+o(x^2))^3+o(x^3)=$
$\displaystyle = 1+3x-x^2+\frac{7}{6}x^3+o(x^3)$

but the result should be $\displaystyle 1+x-x^2+\frac{3}{2}x^3$

where's the mistake then?
thanks for any help

3. ## Re: maclaurin expansion

ooh,
so at least I found out there was a keying mistake in my book,
after using correct formula it goes well by procedure I used and I got correct result

thank you