Integration yields:

This yields an approximate error of 4%.

I tried using my Newton’s Divided Differences excel spreadsheet to fit a polynomial to the same function and the results are below:



Using the coefficients found above for the polynomial, I graphed the real function and the Newton’s Divided Difference Polynomial on http://www.coolmath.com and the results on the interval [0,1] are below:

Evaluating the integrals of these two functions on the interval  [0,0.75] yields results which produce approximately a 3% error. The error is less than the Lagrange method’s error over a smaller interval. This is just one particular example and there are many other functions that could be evaluated in a number of different ways. Each function is unique and there is no “one size fits all” method that will always produce the most accurate results.

This provides us with a weighted sum of function values. We can note that this method fails to work when a function has a singularity. A singularity is a point where the function is not well defined or cannot be differentiated. For example, this method would not work for functions such as f(x) = 1/x or   f(x) =|x|.

It utilizes the equation for the area of a trapezoid. A summation is done to provide total area under the function:

Will a polynomial fit give us an accurate evaluation of an integral?

• Trapezoidal Rule
• Gauss Quadrature

Sometimes, depending on the function, integrating can be a difficult thing to do. When an integral becomes difficult and messy, it may be easier to approximate a polynomial using any of the previously mentioned methods. Polynomials are easy to integrate and evaluate and therefore are preferred. Next, we will test the limits of polynomial interpolation to be used in solving difficult integrals.

A cubic spline essentially derives a 3^rd order polynomial for each interval between two data points and has the general equation:

In order to provide a smoother fit, the first and second derivatives at the interpolating points must be equal. Some conditions can be applied at the outermost points of a spline interpolating function. The derivatives at these points are set equal to zero so that they can extend as straight lines and provide a natural spline.

The actual derivation of the cubic spline can get complicated. It is difficult to satisfy iterative conditions because you never know where you are ending up. Since a spline interpolates on both ends, we need it to match the point to the left and to match the function to the right.

Iterative conditions:

The coefficients can be solved for in a messy algebraic system. Some of the results yielded from this extensive derivation are:

We can note that although these conditions get us no further to solving the system, if they hold true then we can transfer this condition on coefficients. Since we have already solved a & d, we just need to solve the system to get c. That can then be used to solve for b. Now that all the coefficients are known, the spline interpolation method is ready to be used.

1. To select N+1 points, split the semi-circle into N arcs of equal length

2. Project the arcs onto the x-axis, giving the following formula for each Chebyshev point xj:

clear

%Melissa Cabral, Peter Gagnon, Andrew Rodriguez

N=5;

xp=linspace(0,1,N+1); % equidistant points

x=xp; %this should be changed

M=N+1;

f = exp(xp); %function value

for k=1:M

%Lagrange Interpolation

for j=1:N+1

L(j,k)=1;

for i=1:N+1

if i~=j

L(j,k)=L(j,k).*(x(k)-xp(i))/(xp(j)-xp(i));

end

end

end

%now build the interpolating polynomial

P(k)=0;

for j=1:N+1

P(k)=L(j,k).*f(j)

end

end

You can verify just by looking at the function that the first term will be equal to f(x1) at x1 and equal to zero at x2 and x3. The second and third terms will have this same characteristic.

Second degree polynomials as well as higher order degree polynomials can be represented as the summation:

From this equation, you can see that the weighted average at each function value has been simplified and represented as Li(x) as seen below:

In this equation, n represents the number of data points you are interpolating.

The Lagrange Polynomial is definitely a more elegant interpolating method. However, elegant ≠ efficient. A drawback to this method is that if you need to change an interpolation point, you need to re-evaluate the entire polynomial. It does not build upon itself like the Newton’s Divided Difference Method. Newton’s Divided Differences seems like it is the more practical method for computational purposes.

Another drawback to this method and any method using equally spaced points is that the interpolating polynomial will tend to oscillate between the data points. As the order of the polynomial gets higher, and the number of data points increases, this oscillation can worsen.

An example of these effects can be seen below: (Graphs taken from Holistic Numerical Methods Institute)

If you choose equidistant points to interpolate your polynomial, you can get results that oscillate between points as seen in the figure above. However, if you choose points that are spaced further apart in the regions where the fit is good and choose points that are closer together in the regions where the oscillation is occurring you will notice a higher degree of accuracy.

This brings us to our next sub-topic: Chebshev points and Gauss-Lobatto points.

%Melissa Cabral, Peter Gagnon, Andrew Rodriguez

%interpolation point info

N=32;

%xp=linspace(-1,1,N+1); %equidistant points

%for j=1:N+1

%xp(j)=-cos(pi*(j-1)/N); %Chebyshev Gauss Lobatto points

%end

for j=1:N+1

xp(j)=-cos(pi*(2*j-1)/(2*N+2)); %Chebyshev Gauss points

end

f=exp(xp); %function value

%evaluate polynomial at these points

M=50;

x=linspace(-5.0,5.0,M);

ff=exp(x);

%Newton Divided Differences

for j=1:N

D(1,j)=(f(j+1)-f(j))/(xp(j+1)-xp(j+1)-xp(j));

end

for k=1:N-1

for j=1:N-k

D(k+1,j)=(D(k,j+1)-D(k,j))/(xp(j+k+1)-xp(j));

end

end

%build the basis polynomials

Pol(:,1)=(x(:)-xp(1))

for j=1:N-1

Pol(:,j+1)=Pol(:,j).*(x(:)-xp(j+1))

end

%Build the polynomial

P=f(1)+0.0*x;

for j=1:N

P(:)=P(:)+D(j,1).*Pol(:,j)

end

%plotting stuff

plot(x,P)

hold on

plot(x,ff,‘r’)

plot(xp,f,‘gp’)

In general, if we have n data points, the (n-1)^th order polynomial is:

(Click on equation for a better view)

This equation comes from intense algebraic manipulations that help us build this table of coefficients. It is often seen as an efficient method even though it may seem cumbersome. The divided differences build upon one another; therefore you only have to do the work once. Once the coefficients have been calculated, you can add additional points without having to re-calculate the previous points. It is not necessarily the most elegant method, however elegance is not the top priority when defining accuracy of an interpolation.

An example of Newton’s Divided Differences:

will be interpolated by a 3^rd order polynomial using the points:

To do repetitive calculations of this nature, I like to set up excel spreadsheets. The following table represents the divided differences:

Please note:

This excel file that I created can be used to calculate the divided differences of any function just by changing the input values. I will attach the table so that you may see the formulas and try plugging in some formulas of your own!

I also set it up so that it can evaluate both the original function and the interpolating polynomial at specific points. This is useful because by evaluating these two functions at specific points and comparing them, a relative error can be calculated. The relative error on my spreadsheet is set to calculate automatically once the input data is changed!

This gives us an interpolating polynomial of the form:

(Click on equation for a better view)

I used graphing software on http://www.coolmath.com to create a graphical depiction of our interpolating polynomial versus the actual function ln(x). Below are the results: