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

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somnath
somnath 11 May 2014
답변: Star Strider 11 May 2014
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

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Star Strider
Star Strider 11 May 2014
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).

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Image Analyst
Image Analyst 11 May 2014
편집: Image Analyst 11 May 2014
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?

lvn
lvn 11 May 2014
This is for a linear extrapolation:
interp1(x,y,4,'linear','extrap')
ans =
2.5000e+03

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Star Strider
Star Strider 11 May 2014
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.

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