How to extrapolate data from a matrix of x axis values and corresponding y axis values?

let x axis values be x=[.1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 1.7 1.8 1.9 2] and corresponding y axis values be y=[50 65 70 80 90 100 110 120 150 180 220 295 400 ];

 I want to find value of y corresponding to x = 4

[Star Strider]

Your data don’t make sense. The 13 x-values and 20 y-values don’t ‘correspond’ in any meaningful sense, unless they’re e-mailing each other.

It’s never a good idea to extrapolate more than a very short distance beyond the region-of-fit. You have no idea if the function y(x) even exists at y(4).

That said, my guess for y(4) = 3.1E+8 m/s. It’s as good as any. As for accuracy, I guarantee it to be good within ±1 attometre/s ( 99.99% confidence limits).

[somnath]

my mistake!! I reframed the question

Answer 1

[Image Analyst]

To extrapolate you must assume some kind of function, like a line or quadratic or something. Maybe you can use polyfit:

 x=[.1 .2 .3 .4 .5 .6 .7 .8 .9 1.7 1.8 1.9 2] 
 y=[50 65 70 80 90 100 110 120 150 180 220 295 400 580 1000 2400 5000 8900 15000 24000];
 plot(x, y(1:13), 'bd-', 'MarkerSize', 10)

 coeffs = polyfit(x, y(1:13), 2);
 xfit = linspace(x(1), x(end), 50);
 yfit = polyval(coeffs, xfit);
 hold on;
 plot(xfit, yfit, 'ro-', 'LineWidth', 2);
 grid on;

% Get value for x = 4:
yfit4 = polyval(coeffs, 4)

 % Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')

By the way, why does y have 20 elements and x have only 13?

[somnath]

my mistake!! I reframed the question

Answer 2

[lvn]

This is for a linear extrapolation:

interp1(x,y,4,'linear','extrap')

ans =

     2.5000e+03

Answer 3

[Star Strider]

The data are so discontinuous that extrapolating them is almost absurd. However a 5-th degree polynomial gives the best fit to the data:

x = [.1 .2 .3 .4 .5 .6 .7 .8 .9 1.7 1.8 1.9 2]; 
y = [50 65 70 80 90 100 110 120 150 180 220 295 400 ];

[p,S,mu] = polyfit(x,y,5);
xe = linspace(min(x),4);
[ye,dlta] = polyval(p,xe,S,mu);

figure(1)
plot(x,y, '*')
hold on
plot(xe,ye,'-b', xe,ye+dlta,'-r', xe,ye-dlta,'-r')
hold off
% axis([0  3    0  500])
grid

fprintf(1,'\n\tExtrapolated value at x = 4 is %.4f ± %.4f\n\n', ye(end), dlta(end))

and produces:

  Extrapolated value at x = 4 is 74066.6051 ± 13925.4417

Which considering its original exponential behaviour (with the 20-element y-vector), is entirely consistent.

This definitely shows the dangers of extrapolating so far from the region-of-fit with discontinuous data and a poorly-characterised function. A large number of functions will ‘fit’ to it, generating a large number of different extrapolations, and since we have no idea what the function actually does outside the region-of-fit (if it does anything; it may not be defined beyond the data given), they are all entirely wild guesses and all entirely without any valid mathematical or statistical support.