Results 1 to 7 of 7

Math Help - {a_n}\to A and define {b_n}

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    {a_n}\to A and define {b_n}

    \{a_n\}_{n=1}^{\infty}\to A and define b_n=\frac{a_n+a_{n+1}}{2}, \ \ \forall n

    Prove \{b_n\}_{n=1}^{\infty}\to A

    Is it as simple as b_n=\frac{A+A}{2}=A

    Therefore, \{b_n\}_{n=1}^{\infty}\to A
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: {a_n}\to A and define {b_n}

    Quote Originally Posted by dwsmith View Post
    \{a_n\}_{n=1}^{\infty}\to A and define b_n=\frac{a_n+a_{n+1}}{2}, \ \ \forall n

    Prove \{b_n\}_{n=1}^{\infty}\to A

    Is it as simple as b_n=\frac{A+A}{2}=A

    Therefore, \{b_n\}_{n=1}^{\infty}\to A
    You need a little more rigor.

    Since a_n \to A Pick N \in \mathbb{N} such that for all n > N

    |a_n-A|< \frac{\epsilon}{2}

    Now what is

    |b_n-A|=\bigg| \frac{a_n+a_{n+1}}{2}-\frac{2A}{2}\bigg|=...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Re: {a_n}\to A and define {b_n}

    I don't know why you picked \frac{\epsilon}{2} and I don't know where to go from the =\cdots.

    Can you elaborate further.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,390
    Thanks
    1476
    Awards
    1

    Re: {a_n}\to A and define {b_n}

    Quote Originally Posted by dwsmith View Post
    I don't know why you picked \frac{\epsilon}{2} and I don't know where to go from the =\cdots.
    Actually we could pick just {\epsilon} if you note that
    \left| {\frac{{a_n  + a_{n + 1} }}{2} - A} \right| \leqslant \frac{1}{2}\left| {a_n  - A} \right| + \frac{1}{2}\left| {a_{n + 1}  - A} \right|~.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: {a_n}\to A and define {b_n}

    Quote Originally Posted by dwsmith View Post
    I don't know why you picked \frac{\epsilon}{2} and I don't know where to go from the =\cdots.

    Can you elaborate further.
    you can rewrite this difference as the sum of two differences, and use a version of the triagle inequality.....

    the "idea" behind this is that any convergent series is Cauchy, so for large enough n, two successive terms are as close as we like (this means something in terms of epsilon), so the "average" of the two successive terms (for large enough n) is going to be close to the limit as well.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Re: {a_n}\to A and define {b_n}

    Quote Originally Posted by Plato View Post
    Actually we could pick just {\epsilon} if you note that
    \left| {\frac{{a_n  + a_{n + 1} }}{2} - A} \right| \leqslant \frac{1}{2}\left| {a_n  - A} \right| + \frac{1}{2}\left| {a_{n + 1}  - A} \right|~.
    Ok, so from that, we can say:

    \frac{1}{2}|a_n-A|<\epsilon\Rightarrow |a_n-A|<2\epsilon\Rightarrow |a_n-A|<\epsilon_2

    Therefore, \{a_n\}\to A and the same argument for a_{n+1}.

    From this, we can then use the definition of b_n with both sequences being A, correct?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: {a_n}\to A and define {b_n}

    since {an}-->A, for any ε > 0, there IS some N with |an - A| < ε for all n > N. so 1/2 of that is less than ε/2.

    if n > N, surely n+1 > N as well, thus |a(n+1) - A| < ε, too. there is no need for a "secondary" epsilon:

    \left|\frac{a_n + a_{n+1}}{2} - A}\right|\leq\frac{1}{2}|a_n-A|+\frac{1}{2}|a_{n+1}-A| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Define each of the following.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 8th 2013, 03:19 PM
  2. define x ~ y
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 30th 2011, 01:10 PM
  3. Replies: 1
    Last Post: March 13th 2010, 12:13 PM
  4. Define gcd(a,b,c)
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 18th 2009, 06:00 AM
  5. define a map
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 27th 2008, 03:37 PM

/mathhelpforum @mathhelpforum