How can we find the doubling time of a curve?

Lets say I have a graph, how can I find the doubling time of the curve?

Alpha = [1.05 ];  % Initial conditions 
x0 = 100; % Initial conditions 
figure 
n = 1:30;
x = Alpha.^n*x0;
subplot(2,2,1);
plot(n,log2(x));
title('Figure 1A-Alpha=1.05'),xlabel('X-Axis','fontweight','bold'), ylabel('Y-Axis','fontweight','bold') 

Note: Doubling time refers to the days it takes for the values to double. Doubling time should stay the same throughout a single curve

Answer 1

[KSSV]

There is a formula for doubling time, if (n1,x1) and (n2,x2) are two data points selected, then doubling time Td is given by:

In your case: Follow the below:

Alpha = [1.05 ];  % Initial conditions 
x0 = 100; % Initial conditions 
figure 
n = 1:30;
x = Alpha.^n*x0;
plot(n,log2(x));
title('Figure 1A-Alpha=1.05'),xlabel('X-Axis','fontweight','bold'), ylabel('Y-Axis','fontweight','bold') 

% pick any two quantities randomly 
idx = [5 15]; 
x1 = x(idx(1)) ; x2 = x(idx(2)) ;
n1 = n(idx(1)) ; n2 = n(idx(2)) ; 
% doubling time 
Td = (n2-n1)*log(2)/log(x2/x1)

The doubling time is found to be 14.207. Change the values of (n1,q1) and (n2,q2) you will get the same value for the data picked.

[SAS]

Thanks KSSV, but can you explain about idx[5 15] and x1 = x(idx(1)) ; x2 = x(idx(2)) ;

is it indexing the values? mean x(idx(1)) means 5th value of x?

[KSSV]

Yes you are right. x(idx(1)) means x(5) and x(idx(2)) means x(15). You can give different indices in idx and check.

[SAS]

okay. One more thing pls,

I have another part of qtn saying Show why the doubling time of curves can be computed as 1/(log2 (alpha)).

[KSSV]

YOu have

x = Alpha^n*x0

Substitue the above formula in the given formula and simplify, you should get the given formula.

[SAS]

Substitute value of Alpha as

clc

clear all

Alpha1 = 1.05; % Initial conditions

x0 = 100; % Initial conditions

n=1:30

xa = Alpha1.^n*x0;

x = (1/log2(Alpha1)).^n*x0

figure

plot(n,x)

title('Figure 1A-Alpha=1/log2(1.05)')

figure

plot(n,log2(xa));

title('Figure 1B-Alpha=1.05');

Used Apha as 1/log2(1.05).

But the results are deviating too much.

[KSSV]

1/log2(Alpha) is another formula for doubling time. How you can replace this with Alpha?

[SAS]

Sorry man... I just got it.. it's formula for exponential growth.. Thanks once again... (Y)