# B How to prove this approximation?

1. Feb 3, 2017

### Kumar8434

I've arrived at it not by using some mainstream mathematics. I'm looking for a proof which involves some widely-known mathematics. I'm sorry if I'm using my own notation, but it's the only way to make the expression compact.
The notation is:
$$log^n_xy$$: For log with the base x applied n times to y. For example, $$log^3y=log(log(log(y))$$ all with the same base.
The approximation is:$$\frac{\log_ax_2-\log_ax_1}{\log^{n+1}_ax_2-\log^{n+1}_ax_1}*\frac{1}{((ln(a))^n}\approx \prod_{i=1}^nlog_a^ix_1$$, when $$\frac{\log_{a}^{n}x_2}{\log_{a}^{n}x_1}\approx 1$$
1. Is it correct? 2. How can it be proved?

2. Feb 3, 2017

### Staff: Mentor

What is the domain of x?

It can't be the Reals right because at some point the log argument will be negative and the log of a negative number is undefined.

What would be an interesting thing to do is to write a matlab program to plot log(log...(x)) on the complex plane to see how it behaves?
Code (Matlab M):

n=100
z=3
for i=1:n
z=log(z)
zz(i)=z
end
plot(zz,'*')

3. Feb 3, 2017

### Staff: Mentor

Should work, as long as the n-times repeated logarithm is well-defined. How I would try to prove it: First reduce everything to the natural logarithm, then replace ln(xi) by yi - you don't use the value without logarithm anyway. Then invert both sides of the equation (1/...), and interpret it as derivative. Induction can be useful.

4. Feb 3, 2017

### Staff: Mentor

In my plot it shows it converges to a non-zero complex number of $( 0.3181 + 1.3372i )$ after 100 iterations.

5. Feb 3, 2017

### Staff: Mentor

I was assuming we don't make n too large, so everything stays real.

6. Feb 3, 2017

### stevendaryl

Staff Emeritus
So here's how I would go about it. Let's just assume that the base is $e$. Let's assume that $x_1 = x$ and $x_2 = x + \delta x$. Let's assume that $x$ is so big that $x > 0, log(x) > 0, log(log(x)) > 0, ...log^n(x) > 0$. (Up to a maximum value of $n$; when $n$ gets really large, $log^n(x)$ becomes negative.)

Then to first order, $log(x+\delta x) = log(x(1+ \frac{\delta x}{x})) = log(x) + log(1+ \frac{\delta x}{x}) \approx log(x) + \frac{\delta x}{x}$
Then $log^2(x+\delta x) \approx log(log(x) + \frac{\delta x}{x}) \approx log(log(x)) + \frac{\delta x}{x log(x)}$
Etc.

So $log^j(x+\delta x) \approx log^j(x) + \frac{\delta x}{x \cdot log(x) \cdot log(log(x)) \cdot ... log^{j-1}(x)}$

I think your fact follows from that.

7. Feb 3, 2017

### Staff: Mentor

The last equation is nearly identical to the equation in the first post, just rearranged a bit and with different labels.

8. Feb 3, 2017

### Kumar8434

I
I'm sorry, I'm not very good in maths. How can expressions like $log^nx$ be reduced to the natural logarithm. If we apply change of base formula to the deepest logarithm, then keep applying change of base formula until we get into the outermost logarithm, then wouldn't that make the problem very clumsy?

Last edited by a moderator: Feb 3, 2017
9. Feb 3, 2017

### Staff: Mentor

All the inner changes should be negligible within the given approximation.

10. Feb 3, 2017

### Kumar8434

@steven Daryl: That equation is not at all identical to mine. In your equation the requirement of the approximation is that x1 and x2 shpuld be close. The requirement of my approximation is that the log applied n times to x2 should be close to log applied n times to x1. Since even with numbers of large differences, if we continually apply log to bpth of them, then it's a good chance that the resulting numbersthat we get will be close for some value of n. Then my approximation can be applied in that case.

My approximation gives somewhat closer values.

Last edited: Feb 3, 2017
11. Feb 3, 2017

### Staff: Mentor

That's what I found with the numeric approximation that any starting x will approach the same complex number value.

Hence the ratio of the two will be 1.

12. Feb 3, 2017

### Kumar8434

I didn't understand that.

13. Feb 3, 2017

### Kumar8434

Try computin $\prod_{j=1}^nlog^nlogx_1$ My approximation will give somewhat closer values. I don't think my approximation follows from your last equation.

Last edited by a moderator: Feb 3, 2017
14. Feb 3, 2017

### stevendaryl

Staff Emeritus
It's the same: What I wrote was:

$log^j(x + \delta x) \approx log^j(x) + \frac{\delta x}{x \cdot log(x) \cdot ... \cdot log^{j-1}(x)}$

Rearranging gives:
$\frac{\frac{\delta x}{x}}{log^j(x + \delta x) - log^j(x)} \approx log(x) \cdot ... \cdot log^{j-1}(x)$

Since $\frac{\delta x}{x} \approx log(x+\delta x) - log(x)$, we have:

$\frac{log(x+\delta x) - log(x)}{log^j(x + \delta x) - log^j(x)} \approx log(x) \cdot ... \cdot log^{j-1}(x) = \Pi_{k=1}^{k=(j-1)} log^k(x)$

Now if you let $x_1 = x$ and $x_2 = x+ \delta x$, and let $j = n+1$, it becomes:
$\frac{log(x_2) - log(x_1)}{log^{n+1}(x_2) - log^{n+1}(x_1)} \approx \Pi_{k=1}^{k=n} log^k(x_1)$, which is what you wrote.

So the two are the same in the limit as $x_2 \rightarrow x_1$.

15. Feb 3, 2017

### stevendaryl

Staff Emeritus
They are the same, except that I'm using the additional approximation $log(x_2) - log(x_1) \approx \frac{x_2 - x_1}{x_1}$

16. Feb 3, 2017

### MAGNIBORO

so the aproximation works if $\frac{x_1}{x_2} \approx 1$
because $log(1+ \frac{\delta x}{x}) \approx \frac{\delta x}{x}$
This is not a little limited?

17. Feb 3, 2017

### stevendaryl

Staff Emeritus
Yes, it's limited to the case where $x_2$ is not very far from $x_1$. The formula obviously breaks down if $x_2 = 1,000,000,000$ and $x_1 = 0.0000001$

18. Feb 3, 2017

### micromass

Staff Emeritus
Isn't this just a trivial application of the mean value theorem?

19. Feb 3, 2017

### Staff: Mentor

$$\log_a (\log_a x) = \frac 1 {\ln a} \ln \left(\frac 1 {\ln a} \ln(x)\right) = \frac 1 {\ln a} \left( \ln(\ln(x)) - \ln (\ln a)\right) \approx \frac 1 {\ln a} \ln(\ln(x))$$
for x >> aa. Same for larger log chains.

20. Feb 3, 2017

### Kumar8434

The reason you can't use this additional approximation in your proof is that the requirement of this additional approximation is that x2 should be closer to x1. Try using x2=10^7 and x1=10^6 in $\frac{x_2-x_1}{x_1}\approx lnx_2-lnx_1$ and you won't get even a single digit right. But you can use my approximation for x2=10^7 and x1=10^6 because the IMPORTANT thing there is that $ln^nx_2$ should be closer to $ln^nx_1$ which is true for n=3.