Results 1 to 5 of 5
Like Tree3Thanks
  • 2 Post By romsek
  • 1 Post By Vinod

Thread: Need help with Maclaurin Series problem

  1. #1
    Member
    Joined
    Sep 2017
    From
    nj
    Posts
    179
    Thanks
    1

    Need help with Maclaurin Series problem

    Need help with Maclaurin Series problem-capture.png

    How can i recognize the function?

    If the first problem (a) is

    f(x) = summation of x^(3n+2) startinng from n=0, what is the function then? I know that x^n is geometric series function, but im unable to conclude the function
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,242
    Thanks
    2665

    Re: Need help with Maclaurin Series problem

    I toyed with common series a bit and came up with

    $\sum \limits_{k=0}^\infty ~(-1)^k x^{3k-1} = \dfrac{x^2}{1+x^3}$

    Look at the common Maclaurin series templates.

    You may have to use a power of $x$ rather than just $x$

    You may have to adjust things by dividing or multiplying by $x$

    Use Wolfram Alpha Series function to test out your hypotheses.

    Series[f(x), {x, 0, n}]

    will give you $n$ terms of the series of $f(x)$ about $x=0$
    Last edited by romsek; Apr 8th 2018 at 10:03 AM.
    Thanks from topsquark and HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Vinod's Avatar
    Joined
    Sep 2011
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    319
    Thanks
    5

    Re: Need help with Maclaurin Series problem

    Hello, $\displaystyle\sum_{k=0}^{\infty}(-1)^k x^{3k+2}=\frac{x^2}{1+x^3}$. Error in your post was rectified
    Last edited by Vinod; Apr 26th 2018 at 08:40 PM.
    Thanks from romsek
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,926
    Thanks
    3098

    Re: Need help with Maclaurin Series problem

    $\displaystyle \sum_{n=0}^\infty x^{3n+2}= \sum_{n=0}^\infty (x^3)^nx^2= x^2\sum_{n=0}^\infty (x^3)^n$ which is now a geometric series in $\displaystyle x^3$. Since $\displaystyle \sum_{n= 0}^\infty r^n= \frac{1}{1- r}$ the sum of this series is $\displaystyle x^2\frac{1}{1- x^3}= \frac{x^2}{1- x^3}$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Vinod's Avatar
    Joined
    Sep 2011
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    319
    Thanks
    5

    Re: Need help with Maclaurin Series problem

    Hello, in the summation,you forgot to enter $(-1)^n$.Then you will get $\frac{x^2}{1+x^3}$ posted in #2 of this thread.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. maclaurin series problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 8th 2010, 04:18 PM
  2. Maclaurin series problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 28th 2009, 11:58 PM
  3. Maclaurin Series Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 27th 2009, 09:30 PM
  4. MacLAurin Series problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 26th 2009, 11:10 PM
  5. Maclaurin Series Problem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Apr 20th 2008, 05:36 PM

/mathhelpforum @mathhelpforum