Results 1 to 3 of 3

Math Help - Series

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    2

    Series

    Write the 1st 6 terms of the power series for, along with the general power series using summation notation.

    y = e^{x}

    y = e^{-x}

    y = e^{-x^3}

    --------

    y = e^{x} \implies 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + \cdots + \sum_{n=0}^{\infty} \frac{x^n}{n!}

    And then the next I am assuming is the same as above but alternating signs?

    No idea for the 3rd.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by alikation View Post
    Write the 1st 6 terms of the power series for, along with the general power series using summation notation.

    y = e^{x}

    y = e^{-x}

    y = e^{-x^3}

    --------

    y = e^{x} \implies 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + \cdots + \sum_{n=0}^{\infty} \frac{x^n}{n!}

    And then the next I am assuming is the same as above but alternating signs?

    No idea for the 3rd.
    Ok for the first one and the second.

    For the third one, just replace x in the series of e^x by -x^3, just like you should have done for the second one :

    y=e^{-x} = 1+(-x)+(-x)^2/2+..



    y=e^{-x^3}=1+( ???) + (???)^2/2+..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by alikation View Post
    Write the 1st 6 terms of the power series for, along with the general power series using summation notation.

    y = e^{x}

    y = e^{-x}

    y = e^{-x^3}

    --------

    y = e^{x} \implies 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + \cdots + \sum_{n=0}^{\infty} \frac{x^n}{n!}

    And then the next I am assuming is the same as above but alternating signs?

    No idea for the 3rd.
    You're right about the second. For the third, wherever you see x, replace it with -x^3. So you will get
    1 - x^3 + \frac{(-x^3)^2}{2!} + \frac {(-x^3)^3}{3!} + ...

    which is equal to
    1 - x^3 + \frac{x^6}{2!} - \frac{x^9}{3!} + ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 02:12 AM
  2. Replies: 3
    Last Post: September 29th 2010, 07:11 AM
  3. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  4. Replies: 2
    Last Post: December 1st 2009, 01:45 PM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum