MAT331, Feb 9th, 2010Wikis will be used for second project.LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=If a curve fits, wear it (Chapter 2 of the online book)EXERCISE: Suppose we measure the height of a baby, and we get the following results:When the baby is 1 day old, the height is 20.1 in.When the baby is 15 days old, the height is 20.6 in.When the baby is 20 days old, the height is 21.2 in.When the baby is 29 days old, the height is 21.9 in.JSFH1. Make a graph of the points determined by the above data2. What is the height of the baby at 25 days old?3. Find a polynomial passing through all the points.4. Find a line passing through all the points.5.1.babyPoints:=[[1,20.1],[15,20.6],[20,21.2],[29,21.9]];plot(babyPoints,style=point);2. From the given data, we cannot determine the high at 25 days old, but we say that there is a large probability that it is between 21.2 and 21.9. 3.p:=x->a*x^5+b*x^4+c*x^3+d*x^2+e*x+f;babyPoints[1];
babyPoints[1][1];
babyPoints[1][2];p(babyPoints[1][1])=babyPoints[1][2];
p(babyPoints[2][1])=babyPoints[2][2];equations:={seq( p(babyPoints[n][1])=babyPoints[n][2] , n=1..4 )};solve(equations,{a,b,c,d,e,f});There infinitely many polynomials of degree 5 passing through the four given points. Let's find a polynomial of degree 3 passing through the four points.q:=x->c*x^3+d*x^2+e*x+f;newequations:={seq( q(babyPoints[n][1])=babyPoints[n][2] , n=1..4 )};sol:=solve(newequations,{c,d,e,f});Now, let's obtain the polynomial,r:=subs( sol , q(x) ); graph1:=plot(r,x=-1..33,thickness=3);
graph2:=plot(babyPoints,style=point,color=blue,symbolsize=18,symbol=solidbox);graph1;
graph2;with(plots):display(graph1,graph2);with(CurveFitting);PolynomialInterpolation( babyPoints , y );4. There is no line passing through the four points because the points are not colinear.EXERCISE: Given the points, [1,-1/2], [2,1], [2.4,1] and [3,1/3] find a piecewise linear function with domain the interval [1,3] passing through all of them.newPoints:=[ [1,-1/2], [2,1], [2.4,1] , [3,1/3]];plot(newPoints,style=point,symbolsize=18,symbol=solidbox);We'll find the equation of the line passing through the newPoints[1] and newPoints[2](x-newPoints[1][1])/(newPoints[1][1]-newPoints[2][1])=(x-newPoints[1][2])/(newPoints[1][2]-newPoints[2][2]);We'll find all the equations.lines:=i->(newPoints[i][2]-newPoints[i+1][2])*(x-newPoints[i][1])/(newPoints[i][1]-newPoints[i+1][1])=(y-newPoints[i][2]);The first line is lines(1);The second line is lines(2);lines(3);with(CurveFitting);degree1:=Spline(newPoints,x,degree=1);Answer: The function we are looking for is defined by the formulaf(x)=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f(x)=1 for x between 2 and 2.4f(x)=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 for x between 2.4 and 3..degree2:=Spline(newPoints,x,degree=2);plot([degree2, degree1],x=0..4,thickness=3);EXERCISE: Is it possible to find a linear function passing through [1,-1/2], [2,1] and [3,1/3]? Why?EXERCISE: Find the linear function passing closest to the points [1,-1/2], [2,1] and [3,1/3]EXERCISE: Find the quadratic function passing closest to the points [1,-1/2], [2,1] and [3,1/3]EXERCISE: Find the quadratic function passing closest to the points [1,2], [2.4,1.5] , [3,1] and [-0.4,3]EXERCISE: Is it possible to find a linear function passing through the following points? Justify your answer.EXERCISE: Find the linear function passing closest to the points [1,-1/2], [2,1], [2.4,1] and [3,1/3] . You can use the graph below to study the distance. pts:=[ [1,-1/2], [2,1], [2.4,1] , [3,1/3] ];Note: The graph below works only if all the vertical blue segments have length larger than 0. L:=x->x+3;with(plots):display(plot(pts, x = 0 .. 5, style = point, symbol = box, color = green), plot([L(x)], x = 0 .. 5, color = red), seq(plot([pts[i][1], t, t = pts[i][2] .. L(pts[i][1])], thickness = 3, color = blue), i = 1 .. 3));EXERCISE: Consider the pointsnewpts := [seq([i, 1.5*i+1], i = -1 .. 5), [8, 1.5*8+.5], seq([i, 1.5*i+1], i = 9 .. 11)];a. Find a polynomial passing through those points.b. Fit the points with a line using the least squares method. Use the graph of the previous exercise to visualize the line you found, and see how "close" the line and the points are.c. Fit the points with a quadratic function using the least squares method.d. Find a piecewise linear function passing through all the points.e. Plot the points and all the functions together.f. Compare the results. In particular, compute the value of the function you found in each case when x=12.3 Fitting points to a circleHere is a program that produces a list of points approximately in a circle. You do not have to worry about the details of this program (but you are wellcome to do it if you want), just run it to obtain the points. Seed:=readlib(randomize)():
with(stats):
myrand := (a,b) -> evalf(a+(b-a)*(rand()/10^12)):
points_wanted := 50:
circlePts := proc()
local fuzz, a, b, r,s,x,i;
a:= myrand(-3,3);
b:= myrand(-3,3);
r:= myrand(1,2);
fuzz := [ random[normald[0,0.1]](points_wanted)];
x := [ random[uniform[a-r,a+r]](points_wanted)];
for i from 1 to points_wanted do if (myrand(-10,10) > 0) then s[i]:=1 else s[i]:=-1 fi:od:
return [seq([x[i],b+s[i]*sqrt(r^2-(x[i]-a)^2)+fuzz[i]],i=1..points_wanted)];
end:myCirclePts:=circlePts():