Results 1 to 11 of 11

Math Help - series convergens mn1 q1

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    series convergens mn1 q1

    a_{n+1}=\frac{4a_{n}-3}{a_{n}}
    a_{1}=2
    A.
    prove that for n \geq1
    1 \leq a_{n}\leq3
    ?
    B.
    prove that \mbox{\ensuremath{a_{n}}} converges
    and find the limit
    ?
    Last edited by transgalactic; February 18th 2011 at 11:09 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44
    it's a sequence not a serie.
    Here are some hints:
    A) Induction using the fact that the function defined by is increasing (because f'(x) is postive) then induction using
    B) is increasing and bounded (is that the word because i don't study maths in english ) so it convergences then solve f(x)=x to compute the limit.
    Hard luck .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i only started to learn it
    i tried to look on google for similar
    solved questions but i dint find any.

    do you have any web site which explain the theory of this subject
    ?
    and some solved similar questions
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44
    First of all , i wanna know if you understood my hints , if you didn't i can write the complete solution .
    I actually have some similar problems in french and arabic , wait some minutes so that i can translate and post them
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44

    we consider the function f such that
    1)show that
    2) show that for every positive integer n :
    3)Study the monotony of then deduce that it converges.
    4) compute it's limit .

    Hard luck .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    cant understand the first condition
    you say that the values of F output is a part of its range
    ?

    on convergence we have lagrange test cauchy test

    you didnt say any thing about it?
    Last edited by transgalactic; February 18th 2011 at 03:15 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44
    We don't have to use Lagrange or Cauchy , those are simple sequences.
    Let's stay with your problem:
    A) we will use mathematical induction:
    for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
    we assume that it's true for n and let's prove it for n+1
    since 1 \leq a_{n}\leq3 and f is increasing we have
    End of induction .
    B) let's prove that (a_{n}) is increasing :
    we have now proved that it's increasing .
    Now and since it's bounded , it converges.
    As i said , to compute the limit , solve the equation f(x)=x
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by Tarask View Post

    we consider the function f such that
    1)show that
    2) show that for every positive integer n :
    3)Study the monotony of then deduce that it converges.
    4) compute it's limit .

    Hard luck .
    The 'initial value' u_{0} is not specified and that isn't a minor detail. The 'recursive relation' can be written as...

    \displaystyle \Delta_{n}= u_{n+1}-u_{n}= u^{2}_{n} - \frac{u_{n}}{4}= f(u_{n}) (1)

    The function f(*) is represented here...



    There is only one 'attractive fixed point' at x_{0}=0 and that means that, if the sequence converges, it converges to 0. In particular the sequence converges monotonically for -\frac{3}{4} \le u_{0} < \frac{1}{4}, converges 'with oscillation' for -1 < u_{0} < -\frac{3}{4} and diverges for u_{0}< -1 and u_{0}> \frac{1}{4}...

    Kind regards

    \chi \sigma
    Last edited by chisigma; March 2nd 2011 at 09:28 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44
    Actually u_0=1/5 .
    Sorry i was in a hurry.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    Quote Originally Posted by Tarask View Post
    We don't have to use Lagrange or Cauchy , those are simple sequences.
    Let's stay with your problem:
    A) we will use mathematical induction:
    for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
    we assume that it's true for n and let's prove it for n+1
    since 1 \leq a_{n}\leq3 and f is increasing we have
    End of induction .
    B) let's prove that (a_{n}) is increasing :
    we have now proved that it's increasing .
    Now and since it's bounded , it converges.
    As i said , to compute the limit , solve the equation f(x)=x
    Quote Originally Posted by Tarask View Post
    We don't have to use Lagrange or Cauchy , those are simple sequences.
    Let's stay with your problem:
    A) we will use mathematical induction:
    for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
    we assume that it's true for n and let's prove it for n+1
    since 1 \leq a_{n}\leq3 and f is increasing we have
    End of induction .
    because a_n is between 3 and 1 the numerator is negative
    and denominator is positive so the whole thing is negative

    not positive
    B) let's prove that (a_{n}) is increasing :
    we have now proved that it's increasing .
    Now and since it's bounded , it converges.
    As i said , to compute the limit , solve the equation f(x)=x
    Quote Originally Posted by Tarask View Post
    We don't have to use Lagrange or Cauchy , those are simple sequences.
    Let's stay with your problem:
    A) we will use mathematical induction:
    for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
    we assume that it's true for n and let's prove it for n+1
    since 1 \leq a_{n}\leq3 and f is increasing we have
    End of induction .
    because a_n is between 3 and 1 the numerator is negative
    and denominator is positive so the whole thing is negative

    not positive
    B) let's prove that (a_{n}) is increasing :
    we have now proved that it's increasing .
    Now and since it's bounded , it converges.
    As i said , to compute the limit , solve the equation f(x)=x
    the numerator is negative
    denominator is positive so the whole thing is negative
    not possitive
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    so to find the limit we put L instead of a_n and a_n+1 and compute the roots?
    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: 1
    Last Post: May 5th 2008, 09:44 PM
  5. Urgent Question: Convergens of series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 13th 2006, 04:45 PM

Search Tags


/mathhelpforum @mathhelpforum