Results 1 to 9 of 9

Math Help - interval of convergence

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    17

    interval of convergence

    What is the interval of convergence?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Here.
    Attached Thumbnails Attached Thumbnails interval of convergence-picture14.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Nitz456 View Post
    What is the interval of convergence?

    we proceed by the ratio test ( i hope you know it, if not, say so)

    let An = (n^2 * x^2n)/2^2n

    then by the ratio test, we want

    lim{n-->infinity} |A(n+1)/An| < 1

    lim{n-->infinity} |A(n+1)/An|
    = lim{n-->infinity} |[(n+1)^2 * x^(2n + 2)]/2^(2n + 2) * (2^2n)/(n^2 * x^2n)|
    = lim{n-->infinity} |(2^2n)/2^(2n + 2) * [(n+1)^2]/n^2 * x^(2n + 2)/x^2n|
    = lim{n-->infinity} |2^-2 * (n^2 + 2n + 1)/n^2 * x|
    = (1/4)|x|

    for (1/4)|x| < 1
    we must have |x| < 4

    so -4 < x < 4

    so our interval of convergence is (-4,4)

    now we check the end points to see if that should be a closed interval

    plug in -4 and then 4 for x and see if we have convergence

    Note: E is summation

    E (n^2 * (4)^2n)/2^2n = E (n^2 * 2^4n)/2^2n = E n^2 * 2^2n

    this diverges as n--> infinity

    a similar thing happens when x = -4

    so the interval of convergence is (-4,4)

    hmm, i got a different answer from TPH, wonder where i went wrong

    oh, i see! i put x^(2n + 2)/x^2n = x (see the blue writing above), but it was supposed to be x^2. ok.

    so just follow TPH's way, it's shorter--and correct
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Jhevon View Post
    we proceed by the ratio test ( i hope you know it, if not, say so)
    If you want to use the ratio test you need the fact that non of the terms are zero. That is, it fails for x=0. Hence you need to add "if x!=0 we can use the ratio test ....".

    And then consider the case when x=0 (but that case is trivial).

    hmm, i got a different answer from TPH, wonder where i went wrong
    I think you forgot a square somewhere.
    You get that,
    |x|<4
    But it should be,
    |x^2|<4
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ThePerfectHacker View Post
    If you want to use the ratio test you need the fact that non of the terms are zero. That is, it fails for x=0. Hence you need to add "if x!=0 we can use the ratio test ....".

    And then consider the case when x=0 (but that case is trivial).


    I think you forgot a square somewhere.
    You get that,
    |x|<4
    But it should be,
    |x^2|<4
    yeah, i already caught that, thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ThePerfectHacker View Post
    If you want to use the ratio test you need the fact that non of the terms are zero. That is, it fails for x=0. Hence you need to add "if x!=0 we can use the ratio test ....".

    And then consider the case when x=0 (but that case is trivial).
    and thanks for this, i forgot that little detail.

    and this is the day before my exam! oh boy!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2006
    Posts
    17
    Thanks kindly folks! I used the ratio test answer because we've not yet learned of the root test.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Nitz456 View Post
    Thanks kindly folks! I used the ratio test answer because we've not yet learned of the root test.
    make sure you correct my error though, i have not corrected it
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Nov 2006
    Posts
    17
    mos def, sir.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Interval of convergence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: September 13th 2011, 01:01 AM
  2. Interval of convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 25th 2010, 04:08 AM
  3. Replies: 1
    Last Post: May 13th 2010, 01:20 PM
  4. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  5. Interval of convergence.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 19th 2009, 03:14 PM

Search Tags


/mathhelpforum @mathhelpforum