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Math Help - Tunes

  1. #1
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    Tunes

    Bob cranked up his sound system. Initially, the volume is set to v>0.

    At the end of each second, the control circuitry performs two adjustments:

    1.) it resets the volume to the square root of its value

    then,

    2.) it increases the reset volume by the intial volume, v.

    Performing these two operations always increases the volume:

    initial volume: v(0) = v

    volume after first adjustment:


    v(1) = v(0) + sqr(v(0)) = v + sqr(v)


    volume after second adjustment:


    v(2) = v(1) + sqr(v(1))
    = v(0) + sqr(v(0) + sqr(v(0))
    = v + sqr(v + sqr(v))


    volume after third adjustment:


    v(3) = v + sqr(v + sqr(v + sqr(v)))

    etc., and obviously v = v(0) < v(1) < v(2) < v(3) < ...

    If this process continues indefinately, will the volume converge to a finite value? if so, what value?
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  2. #2
    MHF Contributor
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    Talking

    For the "infinitely-nested" radical expression, you have the following:

    . . . . . x\, =\, v\, +\, \sqrt{v\, +\, \sqrt{v\, +\, \sqrt{v\, +\, ...}}}

    By substitution, you can get:

    . . . . . x\, =\, v\, +\, \sqrt{x}

    Then:

    . . . . . x^2\, -\, 2xv\, +\, v^2\, =\, x

    . . . . . x^2\, -\, (2v\, +\, 1)x\, +\, v^2\, =\, 0

    Applying the Quadratic Formula, you get:

    . . . . . x\, =\, \frac{(2v\, +\, 1)\, + \, \sqrt{4v\, +\, 1}}{2}

    ...because v > 0, so x > 0. How does the above compare with the following?

    . . . . . \frac{2v\, +\, 2\, +\, \sqrt{4v\, +\, 4}}{2}

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  3. #3
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    x = 2v + 2 + √v+1
    Last edited by dxdy; April 5th 2009 at 01:07 PM.
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  4. #4
    MHF Contributor
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    Talking

    I'm sorry, but I don't understand what you are doing...?

    Please reply with a clear listing of your work and reasoning, explaining what you are attempting to accomplish. When you reply, please also answer the question about the relationship, if any, between the final expression and the "x=" expression, as this may give some direction regarding the bounds of x.

    Thank you!
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