Linear Systems - 2 (W9)

- Q1 -

In this question, it is asked to find the coefficients of the regression line. Since it is a line, its degree is 1. In the code, we will define our x and y values, then we will use polyfit function.

clear all
x=[0,1,2];
y=[1,2,4];
c=polyfit(x,y,1)

Correct answer is a = 1.50 and b = 0.83, in this case, E.

- Q2 -

In this question, it is asked to find a specific value after building a parabola. Since it’s a parabola, its degree is 2. First, we’ll define the x and y values, then we will use the polyfit function, then we’ll use the polyval function in order to get the value of 3. Lastly, we’ll use the given formula to find the value.

clear all
x=[-2, -1.3, -1, -0.7, -0.4, -0.1];
y=[0.3, 0.5, 1.5, 1.3, 0.8, 0.1];
c=polyfit(x,y,2);
p3=polyval(c,3);
abs(p3-1.5)

Correct answer is 17.7300, in this case, C.

- Q3 -

We are asked to approximate the function f(x) = x*sin(x) using a parabola. To do this, we consider N=30 linearly spaced nodes between 0 and π/2. These nodes are used to create data points (x_i, f(x_i)). We need to find the coefficients of the parabola. Specifically, we are asked to determine the coefficient of the maximum degree term in the parabola.

clear all

n=30;
f=@(x) x.*sin(x);
x=linspace(0,1/2*pi);
y=f(x);
c=polyfit(x,y,2)
c(1)

Corrent answer is 0.3776, in this case, A.

- Q4 -

In this question, it is asked to find the coefficients of the straight line.Since it is a line, its degree is 1. We will define our x and y values, then we will use polyfit function.

clear all

x=[0.2, 1.14, 0.54, 0.87, 1.25, 2.36, 0.19, 0.54, 0.51, 0.33]; 
y=[1.25, 2.36, 2.58, 1.87, 2.68, 3.41, 0.65, 0.47, 1.36, 1.25]; 
c=polyfit(x,y,1)

Correct answer is 1.187, in this case, C.

- Q5 -

We are given data regarding the speed of a body at different points in time. Our x values are the time values, which are [0, 8, 18] seconds, and the corresponding y values are the speed values, which are [44, 43, 67] meters per second. The question asks us to approximate this data using a least square line, which is its degree 1. We need to determine the velocity of the body after 2 seconds by evaluating the line at that time point.

clear all

x=[0, 8, 18];
y=[44, 43, 67];
c=polyfit(x,y,1); 
polyval(c,2)

Correct answer is 4.25081967e+01, in this case, B.

- Q6 -

We are given a set of points (x, y) and we want to approximate these points using a second-degree exponential polynomial of the form p_e(x) = a * e^(2x) + b * e^x + c. The polynomial has three coefficients: ‘a’, ‘b’, and ‘c’. We need to determine the approximate value of the coefficient ‘a’ that provides the best least square approximation for the given set of points.

clear all

x=[0.34, 0.19, 0.25, 0.61, 0.47, 0.35, 0.83];
y=[0.58, 0.54, 0.91, 0.28, 0.75, 1.17, 0.38];
z=exp(x);
c=polyfit(z,y,2)
c(1)

Correct answer is -0.18, in this case, B.

- Q7 -

In this question, it is asked to find the coefficients of the regression line, which means its degree is 1. We will define our x and y values, then we will use polyfit function. Then, we will find the norm of the residual.

clear all
x=[0,1,2,3];
y=[1,2,4,8];
z=exp(x);
c=polyfit(z,y,1)
p=polyval(c,z);
res=norm(p-y)

Correct answer is a_0 = 9.65e-01, a_1 = 3.57e-01, and r = 5.33e-01, in this case, E.

- Q8 -

We are asked to approximate the function f(x) = x^2*log(1+x) using a polynomial p(x) of degree 4. We need to build the polynomial p(x) by fitting it to 22 nodes in the interval [0, 5]. The question specifically asks us to find the approximate value of p(2), which is the value of the polynomial at x = 2.

clear all

n=22;
f=@(x) x.^2.*log(1+x);
x=linspace(0,5,n);
y=f(x);
c=polyfit(x,y,4);
p=polyval(c,2)

Correct answer is 4.4056, in this case, D.