Coding Assignment 2: Taylor Series

Problem 1-Calculating a Planet's Space Coordinates

To calculate a planet’s space coordinates, we have to solve the function 𝑓(𝑥) = 𝑥 − 1 − 0.5 sin 𝑥

Let the base point be 𝑎 = 𝑥i = 𝜋/2 on the interval [0, 𝜋]. Determine the highest-order Taylor series expansion resulting in a maximum error of 0.015 on the specified interval. The error is equal to the absolute value of the difference between the given function and the specific Taylor series expansion.

Pseudocode

1) Establish parameters

2) Find true value

3) Create a loop for the Taylor Series approximation

4) Find true error after each iteration

5) If stopping criteria is met, break loop

6) Calculate error between each iteration and true value

*It is important to note that this method was attempted, but then the code was written without a loop. Each iteration was done by hand and inserted into the next Taylor Series Approximation.

Analytical Solution

The analytical solution is equivalent to the true value that the Taylor Series is approximating. The analytical solution is found by plugging in the points in the x interval [0,pi] into the equation for f(x). Here, I decided to estimate the planet's space coordinates at x1=pi as a sample calculation.

Numerical Solution

The numerical solution represents the Taylor Series approximation of the planet's space coordinates at point x1. Before approximating this value, the following intermediate calculations were performed.

Next, a Taylor Series was used to approximate the coordinates at x1=pi with a absolute true error stopping criteria of 0.015. This stopping criteria indicates when the Taylor Series approximated coordinates close enough to the true coordinates of the planet. The 0 order Taylor Series was performed first and the corresponding absolute true error was 2.0708. Since this error is greater than the stopping criteria, the Taylor Series was iterated again, this time to the first order. This process was repeated until the fourth order, where the stopping criteria was met. The Taylor Series approximations and the absolute true error of them is shown below.

Results

Despite the code for my plot including axis labels, a title, and a legend, it would not include these into the graph of the results. Nothing I tried seemed to work and this was the only graph that would display from the code. The Taylor Series approximation was closer to the true value using higher order derivatives. The results of the Taylor Series approximations are shown below.

Error Analysis

As a higher order derivative was used in the Taylor Series approximation, the error between the true and approximate value decreased. It only took four iterations for the error to be less than the stopping criteria of 0.015. It is important to note that the The absolute error of the Taylor Series approximations are shown below.

The absolute true error of the Taylor Series approximations will never be exactly zero. When the order of the Taylor Series is smaller, truncation error is the cause of most of the overall error. As the order increases, the truncation error decreases but the round-off error increases. There is a point in the series where the truncation and round-off error are equal. Then, for successive orders of the Taylor Series approximation, round-off error dominates the overall error.

Discussion

The approximation was more accurate with each successive order of the Taylor Series. In this experiment, it only took 4 orders to reach the error stopping criteria. If an infinite number of terms was calculated, the Taylor Series would give an exact approximation. Overall, Taylor Series is an effective method for approximating the value of a function when used properly.

MATLAB Code

% 17 February 2020% Cameron King%% The purpose of this code is to approximate f(x)=1-x-0.5sin(x) using a% Taylor series. The initial value of x0=pi/2 will be used to estimate the% value of f(pi) with an absolute true error stopping criteria of 0.015.
%% Analytical Solution % I am going to be approximating the value of the function at pi using pi/2% with all the numbers within the interval.x=0:0.2:pi; % x-interval given in the problem with an arbitrary step size of 0.2true=x-1-(0.5*sin(x)); %% Numerical Solution % The Taylor Series for orders 0 through 4 are performed. The Taylor Series% approximation is stopped at fourth order because the error stopping% criteria is met after the fourth order.o0=(pi/2)-(1)-(0.5*sin(pi/2)); % pi/2 is the initial value given in the problemo1=0.07079+(1-0.5*cos(pi/2))*(pi/2); % first ordero2=1.6416+((0.5*sin(pi/2)/2)*(pi/2)^2); % second ordero3=2.2584+((0.5*cos(pi/2)/6)*(pi/2)^3); % third ordero4=2.2584+((-0.5*sin(pi/2)/24)*(pi/2)^4); % fourth order %% Absolute Error% The error is calculated using the formulas below. The problem indicates % the error is the "absolute value of the difference between the given % function and the specific Taylor series expansion.error0=abs(true-o0);error1=abs(true-o1);error2=abs(true-o2);error3=abs(true-o3);error4=abs(true-o4); %% Plot of Taylor Series Approximations and Analytical Solution % Plot of numerical and analytical solutionsplot(x,o0,"b-o")hold onplot(x,o1,"r-o*")plot(x,o2,"b-o")plot(x,o3,"g-o")plot(x,o4,"m-o")plot(x,true,"-")xlabel("x-values")ylabel("f(x)")legend("0 order","1st order","2nd order","3rd order","4th order","true value")title("Taylor Series Approximations")hold off
%% Error Plotplot(x,e0,"c-")hold onplot(x,e1,"g-")plot(x,e2,"r-")plot(x,e3,"b-")plot(x,e4,"m-")plot(x,true,"k-")xlabel("x")ylabel("f(x)")legend("0 order","1st order","2nd order","3rd order","4th order","true value")title("Absolute Error")hold off

Problem 2- Finite Difference Approximations

Consider the function 𝑓(𝑥) = 𝑥^3 − 2𝑥 + 4 on the interval [−2,2] with ℎ = 0.25. Use the forward, backward, and centered finite difference approximations for the first and second derivatives to illustrate graphically which approximation is most accurate. Graph all three first derivative finite difference approximations along with the theoretical, and do the same for the second derivative as well.

Pseudocode

1) Establish parameters h and x

2) Find the first derivative and second derivative analytical solutions:

For loop

equation for f'(x)

equation for f''(x)

end

3) Find the first derivative and second derivative numerical solutions for forward, backward, and center approximation methods:

For loop

Forward first derivative approx. equation

Backward first derivative approx. equation

Centered first derivative approx. equation

Forward second derivative approx. equation

Backward second derivative approx. equation

Centerd second derivative approx. equation

end

4) Plot first derivative approximations with analytical value

plot(x, forward first derivative)

plot(x, backward first derivative)

plot(x, centered first derivative)

plot(x, analytical first derivative)

Include legend, xlabel, ylabel, title

5) Plot second derivative approximations with analytical value

plot(x, forward second derivative)

plot(x, backward second derivative)

plot(x, centered second derivative)

plot(x, analytical second derivative)

Include legend, xlabel, ylabel, title

6) Calculate error between each iteration and true value:

abs((true-approx)/true)*100% for each first and second derivative approximation

7) Plot error of the first derivative approximation

plot(x, forward first derivative error)

plot(x, backward first derivative error)

plot(x, centered first derivative error)

plot(x, analytical first derivative error)

Include legend, xlabel, ylabel, title

8) Plot error of the second derivative approximation

plot(x, forward second derivative error)

plot(x, backward second derivative error)

plot(x, centered second derivative error)

plot(x, analytical second derivative error)

Include legend, xlabel, ylabel, title

Analytical Solution

The analytical solution represents the true values of the first and second derivatives at each step over the interval. The analytical solution for the first derivative is found by taking the derivative of f(x) and inserting every value between the interval [-2,2] that is a multiple of the step size (0.25). The general analytical solution for the second derivative was found by taking the derivative of f'(x) and inserting the same x-values into it.

The function and its first and second derivatives is shown below.

A sample calculation when x=1 is shown below. This solution represents one of the range of 17 solutions.

Numerical Solution

First, the derivative of the function was estimated using forward, backward, and centered approximations at each step over the interval. The equations of these formulas can be seen in the MATLAB code below. Next, the second derivative of the function was approximated using all three methods as well at each x-value over the interval. Then, each array was compared to the true value found in the analytical solution using a true error calculation.

Results

The three first derivative approximations are parabolic over the interval. All approximations and the analytical value are the same at the point x=0. However, as the x-values stray from zero, the forward and backward approximations differentiate from the analytical value. The forward approximation estimates the first derivative to be higher than the true derivative for x-values greater than zero and lower for x-values less than zero. The backward approximation estimates the first derivative to be lower than the true derivative for x-values greater than zero and greater than for x-values less than zero. The centered approximation technique predicted the analytical solution perfectly for each x-value.

The second derivative approximations correctly resemble lines. The graphs should be linear because the original function is a cubic function and the second derivative of a cubic function is a linear function. Similar to the first derivative approximation, the centered approximation method estimates the exact second derivative. The forward approximation over-estimates the second derivative, while the backward approximation under-estimates the second derivative. The difference between the forward and backward methods and the true value remains the same because the slope of the lines is the same.

Error Analysis

The percent relative error was calculated between the approximations and the analytical solutions. The formula for percent relative error is:

The error between the first and second derivative approximations is shown in the two graphs below.

In the first derivative percent relative error graph, the error is smallest for the centered finite difference approximation. The error for the forward and backward approximations is mirrored and the average error along the interval is the same for both approximation methods.

In the second derivative percent relative error graph, the error is zero for the centered finite difference approximation. The error for the forward and backward approximations is the same, which is shown be the overlap of every point along the x-interval. Additionally, the error for the forward and backward approximations is highest at x=0, or at the middle of the interval. As the x-value moves from zero, the error decreases.

Discussion

The centered finite difference approximation correctly predicted the analytical value for the first and second derivatives. In most cases, the centered approximation will give the true derivative. Even so, the centered approximation method is more effective than the forward and backward approximation methods in most situations. This is because the centered approximation method uses both the values of the function one step size above and below the approximated value. In contrast, the forward approximation method just uses the value of the function one step size above the value being approximated. Lastly, the backward approximation method uses the analytical value of the function at a point one step size below the value being approximated.

MATLAB Code

% 17 February 2020% Cameron King
% The purpose of this code is to approximate the first and second% derivatives of a given function f(x) using forward, backward, and% centered finite difference approximations. The three methods will be% compared to determine the most accurate approximation.
%% Set Parametersh=0.25; % interval lengthx=-2:0.25:2; % defined interval with step size of 0.25
%% Analytical Solution% f(x)=(x.^3)-(2.*x)+4; % the given function
% Loop to find the exact solution for the 1st and 2nd derivativefor j=1:17 df(j)=3*(x(j)^2)-2; % the 1st derivative of the function df2(j)=6*x(j); % the 2nd derivative of the function to the end
%% Numerical Solutionfor j=1:17 % First Derivative Finite Difference Approximations ff(j)=(f(x(j)+h)-f(x(j)))/h; % forward approximation fb(j)=(f(x(j))-f(x(j)-h))/h; % backward approximation fc(j)=(f(x(j)+h)-f(x(j)-h))/(2*h); % centered approximation % Second Derivative Finite Difference Approximations sf(j)=(f(x(j)+0.5)-(2*f(x(j)+0.25))+f(x(j)))/(0.25)^2; % forward approximation sb(j)=(f(x(j))-2*f(x(j)-0.25)+f(x(j)-0.5))/(0.25)^2; % backward approximation sc(j)=(f(x(j)+0.25)-2*f(x(j))+f(x(j)-0.25))/(0.25)^2; % centered approximationend
%% Plot of the first derivative approximation and the analytical valueplot(x,ff,'r-o') % 1st derivative forward approximationhold ongrid onplot(x,fb,'b-o') % 1st derivative backward approximationplot(x,fc,'k-o') % 1st derivative centered approximationplot(x,df,'m-*') % 1st derivative analytical (exact) valuelegend('Forward','Backward','Centered','Analytical')xlabel('X-values')ylabel ('1st Derivative of f(x)')title('First Derivate Approximations and Analytical Value')hold off
%% Plot of the second derivative approximation and the analytical valueplot(x,sf,'r-o'); % 2nd derivative forward approximationhold ongrid onplot(x,sb,'b-o'); % 2nd derivative backward approximationplot(x,sc,'k-o'); % 2nd derivative centered approximationplot(x,df2,'m-*'); % 2nd derivative analytical (exact) valuelegend('Forward','Backward','Centered','Analytical');xlabel('X-values');ylabel('2nd Derivative of f(x)');title('Second Derivate Approximations and Analytical Value');hold off

%% Error Analysis% Percent relative error of the between numerical and analytical solutions eff=abs((df-ff)./df)*100; % 1st derivative forward approximation efb=abs((df-fb)./df)*100; % 1st derivative backward approximation efc=abs((df-fc)./df)*100; % 1st derivative centered approximation esf=abs((df2-sf)./df2)*100; % 2nd derivative forward approximation esb=abs((df2-sb)./df2)*100; % 2nd derivative backward approximation esc=abs((df2-sc)./df2)*100; % 2nd derivative centered approximation
%% Plot Error% error of first derivative approximationsplot(x,eff,'r-o');hold ongrid onplot(x,efb,'b-o');plot(x,efc,'k-o');legend('Forward','Backward','Centered');xlabel('X-values');ylabel('Percent Relative Error between Approximations and Analytical Value');title('First Derivative Percent Relative Percent Error');hold off
%% Plot Error% error of second derivative approximationsplot(x,esf,'r-o');hold ongrid onplot(x,esb,'b--*');plot(x,esc,'k-o');legend('Forward','Backward','Centered');xlabel('X-values');ylabel ('Percent Relative Error between Approximations and Analytical Value');title ('Second Derivative Percent Relative Error');hold off
% Declared function that will always remain the samefunction y=f(x) % allows the function f(x) to be indexed in each loop y=x.^3-2.*x+4; % given f(x) function in the problemend

Peer Review Reflection

I learned how to solve these problems most by reviewing my partner, Max Album's code. For the first submission, I tried to solve the second problem using a for loop multiple times. I could not solve it, and I had to resort to plugging in every value by hand. Max's approach, however, allowed me to see how to implement a for loop with the proper syntax. The one step that I was missing in my original attempt was the function callback at the bottom of the code. This command sets the function of the problem as a constant throughout so that it never has to be stated again. This function command made indexing the function with the interval easier for me to solve.

I made several small mistakes that Max pointed out in the code review process. In several instances, I made syntax mistakes such as not including quotes around my plot parameters. I also had several missing parentheses that Max highlighted. Max also gave me helpful encouragement about what was correct about my code versus what needed to be changed. This made it much easier to make small changes rather than reworking the entire problems.

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