Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By Prove It

Math Help - Taylor series kinda stuck

  1. #1
    Newbie
    Joined
    Aug 2013
    From
    Quebec
    Posts
    5

    Taylor series kinda stuck

    Hello there :P first of all i love that site

    been trying to learn math before going back to school so you guys are a good way to learn and love math


    so i do have a trouble here with a taylor series... been doing them for quite a while but that one i can't get halfway before seeing mistake appearing



    f(x)= sin(x/2) i could use any point i wanted i took a= pi for fun... now i can't get aways with it

    anyone got patience to show me how i can't get the convergence information?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,784
    Thanks
    1569

    Re: Taylor series kinda stuck

    Use the Ratio Test
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2013
    From
    Quebec
    Posts
    5

    Re: Taylor series kinda stuck

    I do understand the ratio test everywhere but with taylor and Mclaurin...

    I'll give you an example that i didn't post here because i already did 2 post today haha :P

    i have that Maclaurin question with the fonction F(x)=ln(1-2x)

    So i do the Maclaurin as best as i can and get the general fonction ... then i apply myself to `the ratio test and its gives me as answer 1... how with a one do i get my intervalle and lenght of r?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,784
    Thanks
    1569

    Re: Taylor series kinda stuck

    You need to understand that a Taylor or MacLaurin Series is a FUNCTION, it will give DIFFERENT series for different values of x. So it makes sense that you would need to know what values of x this function will give a series that is convergent. So in most cases, you should not be getting a single number, rather, an INTERVAL or SET of x values that you know will give a convergent series.

    Because we have no guarantee that the value of x we choose will give a positive term series (in fact, in most cases, it won't), that means we have to test for ABSOLUTE CONVERGENCE. We can say for certain that the series is absolutely convergent when \displaystyle \begin{align*} \lim_{n \to \infty} \frac{\left| a_{n + 1} \right| }{\left| a_n \right| } < 1 \end{align*}.

    So you need to first get an expression for \displaystyle \begin{align*} \frac{ \left| a_{n + 1} \right| }{ \left| a_{n} \right| }  \end{align*}, see what happens when you make \displaystyle \begin{align*} n \to \infty \end{align*}. You will get an expression in terms of x. Then you need to solve for the values of x which will make this expression less than 1. This will ensure convergence.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stuck with a Taylor series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 16th 2012, 09:59 AM
  2. Getting stuck on series - can't develop Taylor series.
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: October 5th 2010, 09:32 AM
  3. ap's...kinda stuck
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 7th 2010, 01:04 PM
  4. Replies: 9
    Last Post: April 3rd 2008, 06:50 PM
  5. Kinda stuck with this one
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: August 27th 2007, 05:00 PM

Search Tags


/mathhelpforum @mathhelpforum