Results 1 to 3 of 3

Thread: Sequences

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    36

    Sequences

    1a) Define $\displaystyle (x_n)$ as being the sequence:

    $\displaystyle x_1 = 3$

    $\displaystyle x_{n+1} = \frac{1}{2}\cdot (x_n + \frac{3}{x_n})$

    Prove $\displaystyle x_n$ converges and find the lim.

    b) Let $\displaystyle b > 1$ and define $\displaystyle (x_n)$ as being the sequence:

    $\displaystyle x_1 = b$

    $\displaystyle x_(n+1) = \frac{1}{2}\cdot (x_n + \frac{b}{x_n})$

    Prove $\displaystyle x_n$ converges and find the lim.

    c) Suppose you're on an island with only a solar-powered very basic calculator. Use the result from part b to approximate $\displaystyle sqrt(17)$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    1) a
    We'll prove that the sequance is low bounded by $\displaystyle \sqrt{3}$
    $\displaystyle x_1=3>\sqrt{3}$

    $\displaystyle \displaystyle x_{n+1}=\frac{x_n+\frac{3}{x_n}}{2}\geq\sqrt{x_n\c dot\frac{3}{x_n}}=\sqrt{3}$ (I used AM-GM inequality).

    Now, we'll prove that the sequence is descrescent.
    $\displaystyle x_2=2\Rightarrow x_1>x_2$.
    Suppose that $\displaystyle x_{n-1}>x_n$.
    Then, $\displaystyle x_n-x_{n+1}=x_n-\frac{x_n^2+3}{2x_n}=\frac{x_n^2-3}{2x_n}>0\Rightarrow x_n>x_{n+1}$ (I used the fact that the sequence is bounded).

    So, the sequence is convergent. Let $\displaystyle \lim_{n\to\infty}x_n=x$.
    Applying the limit in the recurrency relation we have

    $\displaystyle \displaystyle x=\frac{1}{2}\left(x+\frac{3}{x}\right)\Rightarrow x^2=3\Rightarrow x=\sqrt{3}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, alikation0!

    I recognized the function . . . and "eyeballed" the answers.


    b) Let $\displaystyle b > 1$ and define $\displaystyle (x_n)$ as being the sequence:

    $\displaystyle x_1 = b$

    $\displaystyle x_(n+1) = \frac{1}{2}\cdot (x_n + \frac{b}{x_n})$

    Prove $\displaystyle x_n$ converges and find the lim.

    c) Suppose you're on an island with only a solar-powered very basic calculator.
    Use the result from part b to approximate $\displaystyle \sqrt(17)$

    I'm old enough to remember those "very basic calculators".
    They were the size of a TI-30 (but an inch thick), weighed a pound,
    . . used a 9-volt battery (used up quickly by those red LEDs)
    .. cost about $100 and did only basic arithmetic.

    Back then, we learned many tricks for approximating more complex answers.
    . . And one of them was square roots.


    Suppose we want $\displaystyle \sqrt{N}$.

    Make a first approximation, $\displaystyle a_1.$

    If we're very very lucky, $\displaystyle a_1$ is exact.
    . . Then the two factors of $\displaystyle N$ are equal: .$\displaystyle a_1 \,=\,\frac{N}{a_1}$

    Most likely, the two factors are not equal.
    . . Obviously, one is too small and the other too large.

    Then a better approximation is the average of these two factors.

    So we will calculate: .$\displaystyle a_2 \:=\:\frac{a_1 + \frac{N}{a_1}}{2} \;=\;\frac{1}{2}\left(a_1 + \frac{N}{a_1}\right)$


    Hence, $\displaystyle a_2$ is a better approximation of $\displaystyle \sqrt{N}.$

    Then: .$\displaystyle a_3 \:=\:\frac{1}{2}\left(a_2 + \frac{N}{a_2}\right)$ is an even better approximation.

    And that is the source of that recursive function: .$\displaystyle a_{n+1} \;=\;\frac{1}{2}\left(a_n + \frac{N}{a_n}\right)$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Actually, I used this form: .$\displaystyle a_{n+1} \;=\;\frac{a_n^2 + N}{2a_n} $

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence in sequences of sequences
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Oct 19th 2010, 07:28 AM
  2. Sequences and the sequences' arithmetics
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 6th 2010, 09:31 PM
  3. Monotone sequences and Cauchy sequences
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Mar 21st 2009, 08:59 PM
  4. Sequences Q3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 9th 2009, 05:08 AM
  5. Replies: 5
    Last Post: Jan 16th 2008, 04:51 PM

Search Tags


/mathhelpforum @mathhelpforum