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Lecture 11

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Lecture 11

Chap. 15

- 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

- 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

- 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.

- Linear
- Piecewise-linear
- Polynomial
- Cubic Spline
- Interpolation via Gaussian Processes
- Other

- Accuracy of interpolated value
- Amount of work to obtain an interpolated value
- Amount of initial data needed
- …

- From http://en.wikipedia.org/wiki/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.

- 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).

- 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 n-1 that touches all of the points
- Can be computationally complex to find it

- Piecewise – polynomial
- Low degree polynomials for intervalsbetween 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)

- 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

- 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)

- 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)

- answer_y = spline(x, y, target_x)

- 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

- 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”.

- Degree of polynomial is the largest exponent
- Polynomial in x:
- f(x) = ax n + bxn-1 + cxn-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.

- 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.

- 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.

- 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)

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

- 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))

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));

- 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 first-order. They take differences of adjacent terms.

- 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?

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)’})