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Could someone please help me solve for ε in terms of [itex]\delta[/itex] ?

  1. Dec 7, 2011 #1
    As part of my problem I need the following condition to hold:
    [itex]\frac{2^{n(H+\epsilon)}}{\epsilon}:=\delta^{-\theta}[/itex] for some [itex]\epsilon, \delta[/itex] and [itex]\theta[/itex] all in (0,1).
    Now, I would like to rearrange the equation (solve for [itex]\epsilon[itex/] in terms of the rest of the parameters) so as to have the condition be represented as:
    [itex]\epsilon = ...[/itex]

    So, I played around with it a little bit, to get:

    [itex]\epsilon-\frac{1}{n}\log \epsilon = \frac{-t}{n}\log \delta-H[/itex]

    I wonder if there is a closed form solution for [itex]\epsilon[/itex] now?

    Thanks,

    - A.
     
  2. jcsd
  3. Dec 7, 2011 #2

    SammyS

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    Hello azal. Welcome to PF.

    I fixed a bad tag in the quoted post.

    It's usual to solve for δ in terms of ε, not the other way around.

    I'm pretty sure there's no way to solve for ε in closed form.
     
    Last edited: Dec 7, 2011
  4. Dec 7, 2011 #3
    Hi Sammy,

    Thanks for your response.
    This is not an [itex]\epsilon,\delta[/itex] (limit) proof, although my notation may suggest it is.

    I guess I'll have to change the conditions then.

    Thanks again,
    - Azal.
     
  5. Dec 7, 2011 #4

    SammyS

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    In that case, the way to solve for ε is numerically.
     
  6. Dec 8, 2011 #5
    Thanks.
     
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