Lecture 11
Chap. 15
Outline-1
• 15.1 Interpolation
– 15.1.1 Linear Interpolation
– 15.1.2 Cubic Spline Interpolation
– 15.1.3 Extrapolation
• 15.2 Curve Fitting
– 15.2.1 Linear Regression
– 15.2.2 Polynomial Regression
Outline-2
• 15.3 Numerical Integration
– 15.3.1 Determination of the Complete Integral
– 15.3.2 Continuous Integration Problems
• 15.4 Numerical Differentiation
– 15.4.1 Difference Expressions
– 15.4.2 MATLAB Implementation
Interpolation
• We have at least two points with independent
(think, horizontal axis) and dependent (think,
vertical axis) components.
• We have another independent component,
whose value is in between the smallest and
the greatest of the previous independent
components.
• We would like to know the corresponding
dependent component.
Methods of Estimating the Result of
Interpolation
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Linear
Piecewise-linear
Polynomial
Cubic Spline
Interpolation via Gaussian Processes
Other
Quality Considerations
• Accuracy of interpolated value
• Amount of work to obtain an interpolated
value
• Amount of initial data needed
• …
Example Interpolation
• From
http://en.wikipedia.org/wiki/Interpolation
Linear Interpolation
• If the initial points all lie on a line (y = mx + b)
then use the equation of the line on newX.
• If there are only two initial points, they define
a line.
Piecewise Linear
• Use adjacent pairs of initial points to define
lines.
• With the point whose value is to be estimated
by interpolation, choose the relevant defined
line. Y = m(i) x(i) + b(i).
Polynomial
• If we had a polynomial for y in terms of x,
– then, with an x to be interpolated,
– we would evaluate the polynomial
• So, what’s left is to obtain a polynomial
– For n points, there is one polynomial of degree n1 that touches all of the points
– Can be computationally complex to find it
Spline
• Piecewise – polynomial
– Low degree polynomials for intervals
between furnished points
– Connection from one interval to the next smooth
• Smaller error than linear interpolation (if
linear is not perfect)
• Less work than perfect polynomial (if “real”
polynomial is of higher degree than that of
spline)
Cubic Spline
• The curve between each pair of points is a
third-degree polynomial. There is a parameter
t that varies from 0 to 1, and the distance
from the starting point to the ending point
goes from 0 to 100%.
• X = at 3 + b t 2 + c t + d
• Y = et 3 + f t 2 + g t + h
• So, need to know the values of a, b, …h
How to … Linear
• X = [ the independent axis values]
• Y = [the dependent axis values]
• Only works when X, Y pairs are all on the same
line.
• answer_y = interp1(x,y, target_x)
How to…Piecewise Linear
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•
•
•
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X = [ the independent axis values]
Y = [the dependent axis values]
whichInterval = 0;
for i = 1:length(X)
if target_x > X(i)
– whichInterval = whichInterval+1;
•
•
•
•
end
Interval_x = [ x(whichInterval) x(whichInterval+1)]
Interval_y = [ y(whichInterval) y(whichInterval+1)]
answer_y = interp1(Interval_x, Interval_y, target_x)
How to…Cubic Spline
• answer_y = spline(x, y, target_x)
15.1.3 Extrapolation
• This is where one tries to guess beyond the
limits of the available data.
• Be aware that error in this process can be
significant.
• Cannot use interp1
• Can use spline
15.2 Curve Fitting
• In polynomial interpolation we observed that for
n points there is a polynomial of degree n-1 that
includes all of the points.
• Suppose we believe that the points are not
perfect, as in measured data.
• Suppose we choose a degree of polynomial less
than n, and then choose some coefficients for the
polynomial, such that the difference between the
measured data and the resulting polynomial is
“minimized”.
Polynomial Curve Fitting-1
• Degree of polynomial is the largest exponent
• Polynomial in x:
– f(x) = ax n + bx n-1 + cx n-2 … dx + e
– has degree n
– has n+1 coefficients
– so, with n+2 choices, the polynomial is
determined
• Choose/guess the degree from information
about the problem.
Polynomial Curve Fitting -2
• What constitutes a “good” fit, for a particular
choice of degree, n?
• Least squares is one kind of measure of how good
the fit is. If for each point x,y, we take the
difference between the y of the point and the y
of the polynomial curve, square the difference
(yields a positive number) and add up all those
squared differences, we get a measure of
goodness of fit called least squares.
Linear Regression
• If in curve fitting, we choose the degree n to be 1,
we are fitting by linear regression.
• To get MATLAB to determine the coefficients, we
call polyfit with n=1, which looks like
polyfit(x,y,1)
• Polyfit returns the coefficients in linearly
decreasing order of power of the variable in the
polynomial. If n = 1, f(x)= ax+b, and the first
coefficient returned is a, and the second is b.
Polynomial Regression
• Suppose we choose a degree of polynomial,
called n, > 1.
• There will be more coefficients.
• For example, n=3, the number of coefficients
will be 4.
• coef = polyfit (x, y, n)
• Now that coef is set, it can be used.
• New_y = polyval(coef, new_y)
More degrees of freedom
yields closer fit
function viewFits()
x = 0:5;
fine_x = 0:.1:5;
y = [0 20 60 68 77 110];
for order = 2:5
y2=polyval(polyfit(x,y,order), fine_x);
subplot(2,2,order-1)
plot (x,y,'o', fine_x, y2)
axis([-1 7 -20 120])
ttl = sprintf('Degree %d Polynomial Fit',...
order);
title(ttl)
xlabel('Time (sec)')
ylabel('Temperature (degrees F)')
end
15.3 Numerical Integration
• Using piecewise linear technique
– Add areas from trapezoidal rule
– KT = (b – a)/2n)(f(a) + 2f(x1) + 2f(x2)… + 2f(xn-1) + f(b))
• Using piecewise parabolic technique
– Add areas from Simpson’s rule
– KS = (b – a)/2n)(1/3)(f(a) + 4f(x1) + 2f(x2) + 4f(x3)+ … +
2f(xn-2) + 4f(xn-1) + f(b))
Numerical Integration Code
function K = trapezoid(v,a,b)
%h = trapezoid(v,a,b)
K=(b-a)*(v(1)+v(end)+...
2*sum(v(2:end-1)))/(2*(length(v)-1));
function K = simpson(v,a,b)
%h = simpson(v,a,b)
K=(b-a)*(v(1)+v(end)+…
4*sum(v(2:2:end-1))+…
2*sum(v(3:2:end-2)))/ (3*length(v)-1));
Numerical Differentiation
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Forward difference:
f’(xk) = (f(x k ) – f( x k-1 ))/(x k – x k-1)
Backward difference:
f’(xk) = (f(x k+1 ) – f( x k ))/(x k+1 – x k)
Centered difference:
f’(xk) = (f(x k+1 ) – f( x k-1 ))/(xk+1 – x k-1)
• Both forward and backward differences are firstorder. They take differences of adjacent terms.
MATLAB function diff
• function diff returns a vector of differences
between adjacent terms
• v(2)-v(1), v(3)-v(2), …
• Which of forward, backward is this? Could it
be used for either one?
• Suppose we wanted centered. Could we use
diff? Would we need to prepare by a step?
Forward Difference Code
x=-7:0.1:9;
f = polyval([0.0333,-0.3,-1.3333,16,0,-187.2,0],x);
plot(x,f)
hold on
df = diff(f)./diff(x)
plot(x(2:end),df, ‘g’)
grid
legend({‘f(x)’,’f ‘’(x)’})