The MATLAB program is desperate for answers

The result of my run is: beq =

ub =optimization =

z = I don't get the same error as you, this should be a problem with the matlab version, I'm using matlab2010a, you can try it.

Hello landlord.

The procedure is as follows: t=linspace(0,2*pi,50); % is evenly taken by 50 points in the interval [0:2].

y1=sin(2*; y2=3*cos(t+;%y1=sin(2*; y2=3cos(t+;

I hope it will be helpful to the landlord, and I hope the landlord will adopt it, and welcome to communicate.

The subject gives that this question is actually a question of the shortest distance from the point to the curve. This question can be answered in the MATLAB environment like this:

In the first step, let the coordinates of the point on the curve y=1 x be (x,y), then according to the formula for the distance between the two points.

can be obtained. d=sqrt((x-2) 2+(y-1) 2)=sqrt((x-2) 2+(1 x-1) 2)=sqrt(x 2+1 x 2-4x-2 x+5) Then, we use the MATLAB language to customize the function, ie.

d=sqrt(x^2+1/x^2-4*x-2/x+5);

In the second step, the derivative of x of the function is taken and equal to 0, i.e.

eq=diff(d)=0

In the third step, use the vpasolve function to find the value of x in the first quadrant.

The fourth step is to calculate the y-value, ie.

y=1 x step 5, calculate the d value, ie.

d=sqrt(x 2+1 x 2-4*x-2 x+5) is calculated as follows.

The ** of this MATLAB problem can be completed with a combination of the following functions.

Function one, polynomial curve fitting function polyfit() function two, polynomial calculation function polyval().

Solution: 1. Create an array of years from 2016 to 2022 x2, create a profit array from 2016 to 2022 y3, and use the polyfit() function to find the coefficient of y=p1x+p2.

4. Using the polyval() function, the profit in 2023 and 2024.

After programming, we run it and get the following result.

The subject's programming problem can be solved in this way.

1) Find the mean and standard deviation of y.

The average value of y is calculated using the mean() function; The standard deviation of y is calculated using the std() function; Namely.

y_mean=mean(y)

y_std=std(y)

2) Use the cubic spline interpolation method to find the value of y every time x changes.

The interpolation calculation is performed using the interp1() function, i.e., xi=:15

yi=interp1(x,y,xi,'spline') %'spline'Cubic spline interpolation.

3) Use the cubic polynomial to fit the curve, find the polynomial, and use the polynomial to find the value of y when x=[6,8,10].

The polynomial coefficients are calculated using the polyfit() function, i.e., p=polyfit(x,y,3).

Use the polyval() function to calculate the y value corresponding to x, i.e., x=6; y1=polyval(p,x)

Running results and**.

When you answer, check the English first.

The limit problem of the recursive series given by the subject can be solved according to the following ideas.

1. According to the recursive relationship of x(n+2)=2x(n+1)+x(n), the value is evaluated in a loop.

2. Use the x(n+1) x(n) limit relation to determine whether x(n+1) x(n) is close to x(n) x(n-1), and if not, the cycle continues.

3. When x(n+1) x(n) x(n) x(n-1), its value is the limit value we require.

4. Solve the result.

limit(na)x(n+1) x(n)=its exact solution is 1+sqrt(2).

%%% Q1.

Colon generation method.

x=0:(2*pi)/9:2*pi

Linear fixed-point method.

x=linspace(0,2*pi,10)%% Q2.

a=rand(3,5)

num=8;% assumes that 8 is the last digit of the student number.

i1=find(a>num* %single subscript[i,j]=ind2sub(size(a),i1); The single subscript identifier of an element is converted into the full subscript identifier corresponding to the element in the matrix.

i2=[i j] % full subscript.

% Q3.

cleara=rand(3,3)

e=diag(a) %extract the diagonal elements of matrix a, b=diag(e) %, generate diagonal elements, and the diagonal elements are the diagonal matrix of the elements in e, c=a-b %%, the diagonal elements are all 0, and the rest of the elements are the same as the corresponding a-array elements.