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

[azal]

As part of my problem I need the following condition to hold:
\frac{2^{n(H+\epsilon)}}{\epsilon}:=\delta^{-\theta} for some \epsilon, \delta and \theta all in (0,1).
Now, I would like to rearrange the equation (solve for \epsilon[itex/] in terms of the rest of the parameters) so as to have the condition be represented as: <br /> \epsilon = ...<br /> <br /> So, I played around with it a little bit, to get:<br /> <br /> \epsilon-\frac{1}{n}\log \epsilon = \frac{-t}{n}\log \delta-H<br /> <br /> I wonder if there is a closed form solution for \epsilon now?<br /> <br /> Thanks,<br /> <br /> - A.

 

[SammyS]

As part of my problem I need the following condition to hold:

\displaystyle \frac{2^{n(H+\epsilon)}}{\epsilon}:=\delta^{-\theta} for some \epsilon, \delta and \theta all in (0,1).

Now, I would like to rearrange the equation (solve for \epsilon in terms of the rest of the parameters) so as to have the condition be represented as:
\epsilon = ...

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

\epsilon-\frac{1}{n}\log \epsilon = \frac{-t}{n}\log \delta-H

I wonder if there is a closed form solution for \epsilon now?

Thanks,

- A.

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.

 

[azal]

Hi Sammy,

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

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

Thanks again,
- Azal.

 

[SammyS]

Hi Sammy,

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

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

Thanks again,
- Azal.

In that case, the way to solve for ε is numerically.

 

[azal]

Thanks.