Polynomials

Recall that a polynomial of degree n is an expresion of the form

NiMsJiomSSRhX25HNiIiIiIpSSJ4R0YmSSJuR0YmRidGJyomLUkjYV9HRiY2IywmRipGJ0YnISIiRicpRilGL0YnRic=... NiMsKComSSRhXzJHNiIiIiIqJEkieEdGJiIiI0YnRicqJkkkYV8xR0YmRidGKUYnRidJJGFfMEdGJkYn where a_n is not zero.

Find a polynomial that passes through the points (1,6), (2,4).

First way: Using the quation of the line

slope:=(6-4)/(1-2);

y=6-slope*(x-1);

Second way: Start with y=m*x+b and find m and b solving a system of equations.

L:=x->m*x+b;

solve({6=L(1),4=L(2)},{m,b});

Third way: use the package CurveFitting

with(CurveFitting): PolynomialInterpolation([[1,6],[2,4]],z);

We know how to find the polynomial passing through two points.

Plot the points (1,-1), (2,-2) and (3,-3). Find a polynomial that passes through all of them.

plot([[1,-1],[2,-2],[3,-3]],style=point,color=blue);

p:=x-> a*x^2+b*x+c;

solve({p(1)=-1,p(2)=-2,p(3)=-3},{a,b,c});

Why does one get a=0?

Using the fact that the three points are on a line, one can proceed as in the first case.

with(CurveFitting): PolynomialInterpolation([[1,-1],[2,-2],[3,-3]],z);

Find a polynomial that passes through the points (1,6), (2,4) and (3,5)

plot([[1,6],[2,4],[3,5]],style=point,color=blue);

Let's try with the polynomial of degree 2, a*x^2+b*x+c

p:=x-> a*x^2+b*x+c;

solve({p(1)=6,p(2)=4,p(3)=5},{a,b,c});

plots[display](plot([[1,6],[2,4],[3,5]],style=point,color=blue),plot((3/2)*x^2-(13/2)*x+11,x=0..7));

EXERCISE: What happens if you try to find a polynomial of degree 3 passing through the points (1,6), (2,4) and (3,5)?

q:=x->d*x^3+a*x^2+b*x+c;

solve({q(1)=6,q(2)=4,q(3)=5},{a,b,c,d});

with(CurveFitting):

PolynomialInterpolation([[1,6],[2,4],[3,5]],x);

Now, consider the following data.

data:=[seq([i,i/2],i=1..10)];

Plot the points of data and find a polynomial passing through all of them.

plot(data,style=point);

with(CurveFitting):

S:=PolynomialInterpolation(data,x);

Consider now the following data. Plot data and data2 together.

data2:=[seq([i,i/2],i=1..7),[8,3.4],[9,4.7],[10,5]];

plot({data, data2},style=point, symbol=[box,diamond],color=[red,blue]);

Find a polynomial passing through the points of data2.

T:=PolynomialInterpolation(data2,x);

Plot the polynomial of data adn the polynomial of data2 in the same graph. Compare the results.

plots[display](plot(data2,style=point),plot({T,S},x=0..11,y=-5..10,thickness=3));

Exercise: Find a polynomial that interpolates the data (1,-5), (2,-8.3),(3,-11),(4,-14) and (5,-17) by defining first and structure of points and then using PolynomialInterpolation. Plot the results.

data3:=[[1,-5],[2,-8.5],[3,-10.8],[4,-14.21],[5,-17]];

r:=PolynomialInterpolation(data3,x);

Repeat the above procedure with the data [[1,-5],[2,-8],[3,-11],[4,-14]]. Compare both results by making the graph of both functions together. Find a value of x for which the value of the two functions differs significatively.

data4:=[[1,-5],[2,-8],[3,-11],[4,-14]];

s:=PolynomialInterpolation(data4,x);

plots[display](plot(data3,style=point,color=blue),plot({r,s},x=0..6));

Consider the following data.

data6:=[[1,1],[2,1.5],[3,3]];

plots[display](plot(data6,x=-1..5,y=-1..5,style=point,symbol=box,color=green));

Find a linear function that approximate the three points of data6. (this involves some guessing). Meassure the error using the square of the vertical distances.

L:=x->x/2+0.5;

Find the linar function that best approximate these points by the method of least squares. Plot the result wiht the points.

Line:=CurveFitting[LeastSquares](data6,x);

plots[display](plot(data6,style=point,symbol=box, color=green),plot([Line(x)],x=0..4,y=0..10));

Here is the graph of the line, with the segments representing the distances whose squares we added.

plots[display](plot(data6,style=point,symbol=box, color=green),plot([Line(x),L(x)],x=0..4,y=0..10,color=[red,blue]),seq(plot([data6[i][1],t,t=data6[i][2]..L(data6[i][1])],thickness=3),i=1..3));

Let us plot all the lines in the same graph.

plots[display](plot(data6,style=point,symbol=box, color=green),plot([Line(x),L(x)],x=0..4,y=0..10,color=[red,blue,orange,gold,black,yellow,green]));

Define the error function.

E:=(data6,M,B)->sum((data6[i][1]*M+B-data6[i][2])^2,i=1..3);

expand(E(data6,M,B));

Using the error function, find the error with the line you guessed.

E(data6,2,1);

plot3d(E(data6,M,B),M=-1..5,B=-1..4,view=0..5,axes=boxed,style=patchcontour);

Find the minimum of the error funcion.

solve({diff(E(data6,M,B),B)=0 ,diff(E(data6,M,B),M)=0},{M,B});

Find the line that best approximate data6 by least squares method.

line:=CurveFitting[LeastSquares](data6,x);

But one can consider another error. Write, x=n*y+c

E2:=(data6,n,c)->sum((data6[i][1]-(n*data6[i][2]+c))^2,i=1..3);

expand(E2(data6,M,B));

solve({diff(E2(data6,M,B),B)=0 ,diff(E2(data6,M,B),M)=0},{M,B});