## Fitting the q-exponential for decreasing function

### Spss

Analyze/Regression/Nonlinear
enter parameters in box as below for Excel
enter variable to be fitted in upper left box
enter equation Y*(1-(1-q)*binlogged/k)**(1/(1-q))
where binlogged (x, whatever) is the name of the binsize variable to which your data are keyed

### Excel

Tools/Solver
Minimize sum of squares (can be sum of sums)
the Ss are a row of squared differences
they refer to two different rows:
the raw data
the fitted data, from three parameters:
Y
k
q
Those values are initialized
Instruct solver to change those values
Add constraints to the parameters
Y>10
k>5
q>0

## Fitting power laws and the q-exponential for time series

### Power law trend: Spss

Analyze/Regression/Nonlinear
enter parameters in box as below for Excel
t0>10
k>5
c>0
enter variable to be fitted in upper left box
enter equation N0/(t0-year)**c
where year is the name of the actual variable for the year to which your data are keyed

### q-exponential trend: Spss

enter equation N0*(1-(1-q)*(t0-year)/k)**(1/(1-q))

### q1-q2-exponential trend: Spss

popgrowth.eps
popgrowth.pdf

Ignore the following

```From tsallis@santafe.edu Thu Aug  3 15:44:14 2006
Date: Sun, 08 May 2005 07:17:26 -0600
From: Constantino Tsallis
To: Douglas R. White
Subject: Re: monolog plot question

A tutorial on fitting q The fitting of q-laws as replacement for power-law fit: A social science tutorial,
Douglas R. White

Doug, yes, you can. Start with the data series which is the UPPER one among
those that you sent me in the Excel a few days ago. That one might be the

Do like this:

1) Let us call y(x) your data. Estimate with your eye the EXTRAPOLATED
value y(0) in the log-log plot that you sent me.

2) Calculate, for all x,

y(x) / y(0)

3) Then calculate, for all x,

z(x) = {[y(x) / y(0)]^(1-q) - 1} / (1-q)

4) Then plot in a linear-linear scale

z(x) versus x

for various values of q (say q=1.3, 1.4, 1.5, 1.6, 1.7, etc...).

The q that produces the "best straight" line provides you the estimate of q
that you are looking for.

5) The SLOPE of the linear-linear plot z versus x directly provides you (-
kappa). Then your data have been so fitted with

y(x) = y(0) exp_q (- x / kappa)

Comment (i): The "best straight line" can be found either by eye, or, more
systematically, by making a linear regression, and choosing that value of q
that gives the "linear correlation coefficient r" closest to unity.

Comment (ii): Once you have a quite good set [y(0), q], you can slightly
improve them. To do this you go back to your log-log representation that
you sent me in the Excel, and change slightly around your pair [y(0), q]
until you are satisfied the most.

Comment (iii) Stefan Thurner / Vienna and I are just now writing a paper
where we essentially follow the procedure that I described to you. I am
attaching the figure we get, so that you will have an illustration in front
of your eyes. The inset precisely shows the linear correlation coefficient
r versus q, which allows a quite precise determination of q for our problem.

Cheers and good luck!

Constantino

--------------------------

At 00:42 8/5/2005 -0700, you wrote:

>I note from the Malacrne and Mendes paper 2nd page right that using their
>generalized monolog plot that q can be estimated directly, independent of
>the other parameters.
>
>Constantino, given the error bars, would you advising using that method to
>estimate the q values in our city data?
>
>Doug ```