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 [itex]\epsilon[itex/] 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.

 

[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.