Results 1 to 8 of 8

Math Help - Sum of a series

  1. #1
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1

    Sum of a series

    Find the sum of the series:

    \sum^{\infty}_{n=0} (-1)^n \frac{2^{n+1}x^{6n}}{n!}

    = \sum^{\infty}_{n=0} \frac{(-2^nx^6)^n}{n!}

    = e^{-2^nx^6}

    Where am I wrong? Do I need an index shift because of the 2^{n+1} ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by mollymcf2009 View Post
    Find the sum of the series:

    \sum^{\infty}_{n=0} (-1)^n \frac{2^{n+1}x^{6n}}{n!}

    = \sum^{\infty}_{n=0} \frac{(-2^nx^6)^n}{n!}

    = e^{-2^nx^6}

    Where am I wrong? Do I need an index shift because of the 2^{n+1} ?

    2 \sum^{\infty}_{n=0} (-1)^n \frac{2^n x^{6n}}{n!}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    Also you need

    (-2*x^6)^n

    not (-2^n x^6)^n

    you get 2e^(-2x^6)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by skeeter View Post
    2 \sum^{\infty}_{n=0} (-1)^n \frac{2^n x^{6n}}{n!}
    Ok, here is what I did, but I'm not sure it's right:

    2 \sum^{\infty}_{n=0} \frac{(-2x^6)^n}{n!}

    When do I multiply that 2 back in, do I just put it in front of the e or does it get multiplied back into the argument before I take the e?

    e^{-4x^6} or is it 2e^{-2x^6}

    So sum is:

    \frac{1}{4x^6} or \frac{1}{x^6}

    Also, am I supposed to have an x in my answer? That doesn't seem like an "answer".

    Thanks!!!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    No the sum of the series is period.

    Since sum(x^n/n!) = e^x and you are substituting (-2x^6) for x

    The sum is a Taylor Series so it better have an x in it.

    What it can't have is n which is what you're summing over.

    not sure where you're getting 1/x^6 or 1/(4*x^6) from but they have nothing to do with the series.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    the sum of the original series is 2e^{-2x^6}

    where/how/why are you getting \frac{1}{4x^6} or \frac{1}{x^6} ?

    is there some other info you haven't posted?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by Calculus26 View Post
    No the sum of the series is period.

    Since sum(x^n/n!) = e^x and you are substituting (-2x^6) for x

    The sum is a Taylor Series so it better have an x in it.

    What it can't have is n which is what you're summing over.

    not sure where you're getting 1/x^6 or 1/(4*x^6) from but they have nothing to do with the series.
    Ugggg...totally see what I did...lol!

    My brain is doing:

    e^{-ln(4x^6)}

    Now see where I'm getting my crazy answer?

    Lordy, lordy....it's gonna be a long night...

    Thanks y'all!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    You're alright-- I thought as much but coudn't figure out why you were taking logarithms .

    Stream of consciousness often finds its way into my calculations--thats why I never became an engineer-- at least with math things don't blow up
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 01:12 AM
  2. Replies: 3
    Last Post: September 29th 2010, 06:11 AM
  3. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  4. Replies: 2
    Last Post: December 1st 2009, 12:45 PM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum