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Math Help - Sequences

  1. #1
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    Sequences

    1a) Define (x_n) as being the sequence:

    x_1 = 3

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

    Prove x_n converges and find the lim.

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

    x_1 = b

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

    Prove 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 sqrt(17)
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) a
    We'll prove that the sequance is low bounded by \sqrt{3}
    x_1=3>\sqrt{3}

    \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.
    x_2=2\Rightarrow x_1>x_2.
    Suppose that x_{n-1}>x_n.
    Then, 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 \lim_{n\to\infty}x_n=x.
    Applying the limit in the recurrency relation we have

    \displaystyle x=\frac{1}{2}\left(x+\frac{3}{x}\right)\Rightarrow x^2=3\Rightarrow x=\sqrt{3}
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  3. #3
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    Hello, alikation0!

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


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

    x_1 = b

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

    Prove 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 \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 \sqrt{N}.

    Make a first approximation, a_1.

    If we're very very lucky, a_1 is exact.
    . . Then the two factors of N are equal: . 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: . a_2 \:=\:\frac{a_1 + \frac{N}{a_1}}{2} \;=\;\frac{1}{2}\left(a_1 + \frac{N}{a_1}\right)


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

    Then: . 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: . a_{n+1} \;=\;\frac{1}{2}\left(a_n + \frac{N}{a_n}\right)

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

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

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