Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - maclaurin expansion

  1. #1
    Junior Member
    Joined
    May 2013
    From
    Czech Republic
    Posts
    28
    Thanks
    1

    maclaurin expansion

    Hi there, I am trying to find the Maclaurin expansion of 3rd degree of the function 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
    (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 R(z) there holds R(z)=o(z^3) ...

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

    (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

    R(3\sin{x})=o(x^3) since \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 \sin{x} and I got

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

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

    where's the mistake then?
    thanks for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470

    Re: maclaurin expansion

    [QUOTE=alteraus;794113]Hi there, I am trying to find the Maclaurin expansion of 3rd degree of the function 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
    (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 R(z) there holds R(z)=o(z^3) ...
    This is, at least, one problem: the first two terms should be 1+ \frac{1}{3}z, not 1+ x.

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

    (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

    R(3\sin{x})=o(x^3) since \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 \sin{x} and I got

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

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

    where's the mistake then?
    thanks for any help
    Thanks from alteraus
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2013
    From
    Czech Republic
    Posts
    28
    Thanks
    1

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. maclaurin expansion
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 8th 2011, 07:40 AM
  2. Maclaurin expansion
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 8th 2010, 05:40 PM
  3. Replies: 6
    Last Post: May 1st 2009, 11:37 AM
  4. Maclaurin expansion of e^x
    Posted in the Calculus Forum
    Replies: 8
    Last Post: August 1st 2008, 09:18 PM
  5. Maclaurin expansion
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 16th 2007, 07:53 PM

Search Tags


/mathhelpforum @mathhelpforum