# Could someone please help me solve for ε in terms of $\delta$ ?

1. Dec 7, 2011

### 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 $\epsilon[itex/] in terms of the rest of the parameters) so as to have the condition be represented as: [itex]\epsilon = ...$

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.

2. Dec 7, 2011

### SammyS

Staff Emeritus
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
3. Dec 7, 2011

### 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.

4. Dec 7, 2011

### SammyS

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

5. Dec 8, 2011

### azal

Thanks.

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