B How to prove this approximation?

[stevendaryl]

I've already proved under what circumstances would it work, that's why I've written the circumstance in my first post.

Your first post asks

1. Is it correct? 2. How can it be proved?

which to me seems to be asking for a proof, not giving a proof.

 

[stevendaryl]

Okay, I am officially done with this thread.

 

[Kumar8434]

Your first post asks
which to me seems to be asking for a proof, not giving a proof.

I can't give that proof here because it is not allowed here.

 

[mfb]

I can't give that proof here because it is not allowed here.

What?

Did you try the approach I suggested in post 3? It is not too different from the one stevendaryl posted, but it allows larger differences between the numbers.

 

[stevendaryl]

I can't give that proof here because it is not allowed here.

As @mfb pointed out, your formula and mine are actually IDENTICAL.

I have a formula saying: log^n(x_2) \approx log^n(x_1) + \frac{(x_2 - x_1)}{x_1} \frac{1}{\Pi_{j=1}^n log^j(x_1)}

To get your formula, let x_1 = log(y_1) and let x_2 = log(y_2). This gives:

log^n(log(y_2)) \approx log^n(log(y_1)) + \frac{(log(y_2) - log(y_1))}{log(y_1)} \frac{1}{\Pi_{j=1}^n log^j(log(y_1))}

Which can be written as:

log^{n+1}(y_2) \approx log^{n+1}(y_1) + (log(y_2) - log(y_1))\frac{1}{log(y_1) \Pi_{j=1}^n log^{j+1}(y_1)}
= log^{n+1}(y_1) + (log(y_2) - log(y_1))\frac{1}{\Pi_{j=1}^{n+1} log^{j}(y_1)}

This is your approximation, rearranged, with the change of variables
x_2 \Rightarrow y_2
x_1 \Rightarrow x_2
n \Rightarrow n+1

 

[mfb]

@stevendaryl: The formula is a good approximation even in a range where the approximations you used are no longer valid. Proving that doesn't work with your approach. Which is correct, but not general enough.

 

[stevendaryl]

@stevendaryl: The formula is a good approximation even in a range where the approximations you used are no longer valid. Proving that doesn't work with your approach. Which is correct, but not general enough.

His formula is the same as mine, after substituting x_2 \Rightarrow log(x_2), x_1 \Rightarrow log(x_1), n \Rightarrow n+1. So if my formula is good as long as \frac{x_2}{x_1} < K, then the replacement would extend that range to \frac{x_2}{x_1} < (x_1)^K.

 

[mfb]

That is still not sufficient to cover the range where the original formula is a good approximation.