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Find the least value

  1. Oct 3, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Least value of [itex] \left|z+\frac{1}{z}\right| if |z|\geq3 is [/itex]


    2. Relevant equations


    3. The attempt at a solution
    [itex] \left|z-\frac{-1}{z} \right| \geq |z|-\left| \frac{-1}{z} \right| [/itex]

    So the minimum value will be

    [itex]|z|- \frac{1}{|z|} [/itex]

    Now for minimum value
    |z| = 3 as it is the minimum value of |z|

    Substituting the value of |z| I get
    [itex]3- \frac{1}{3} [/itex]
    =[itex] \frac{8}{3}[/itex]

    But the correct answer is [itex] \frac{10}{3} [/itex]
     
  2. jcsd
  3. Oct 3, 2012 #2

    Ray Vickson

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    Your reasoning is wrong (or at least, incomplete---see below), but you have arrived at the correct answer. The minimum value is, indeed, 8/3, and the posted answer of 10/3 is incorrect. Hint: to show that 10/3 is incorrect, look for a value z = iy along the imaginary axis that gives z + 1/z = (8/3)i, so for that z we have |z + 1/z| = 8/3, which is less that the alleged minimum of 10/3.

    So, what did you do wrong? Well, your inequality
    [tex] \left|z-\frac{-1}{z} \right| \geq |z|-\left| \frac{-1}{z} \right| [/tex]
    is, indeed, true, but for some z it might be a strict inequality, so there is no guarantee that minimizing the right-hand-side will give an achievable minimum to the left-hand-side. I'll leave it up to you to see what else needs to be done to fix the argument.

    RGV
     
  4. Oct 4, 2012 #3

    utkarshakash

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    I can't make out any fault in my reasoning. Can you please bring it out?
     
  5. Oct 4, 2012 #4

    Ray Vickson

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    I have already explained why your reasoning is incomplete: you have some expression A that you want to minimize, and you know that A ≥ B for some other expression B. Minimizing B does not necessarily minimize A.

    For example, if f(x) = (x-1)2+ x4 and g(x) = (x-1)2, we certainly have f(x) ≥ g(x) for all x (and f = g for some x). However, the minimum of g(x) is = 0, and it occurs at x = 1, while the minimum of f(x) is = 0.2892734 and it occurs at x = 0.5897545 (as obtained using numerical methods). So, in this case, minimizing f by minimizing the simpler function g would fail. However, if I change f to f(x) = x2 + x4 and g(x) to g(x) = x2, we would get the correct answer by minimizing g instead of f.

    So, sometimes your method works and sometimes it fails. The issue is whether you can apply it in your particular problem.

    RGV
     
  6. Oct 6, 2012 #5

    utkarshakash

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    Hmmm...... that seems quite correct. Thanks
     
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