Seek help from MATLAB for remote programs

First of all, the front must be a t vector.

length(t) represents the number of t elements (t vector) for i=2:length(t).

2. First assign i to enter the loop, 3 to i, and so on.

The i-2 value enters the cycle, and the dep tpe==1i is assigned to 2 to calculate the output value.

a_coeff*(t(i))^2+b_coeff*t(i)+c_coeff;Sentence meaning: a coeff multiplication t(2) (indicating that the second element point of the t vector must be noted), flat plus b coeff multiplication t(2) surface, I talk about calculating the knot, assigning the second element point to output, similar to the c language pointer.

The program points are chaotic and judgment, and the two situations enter the same loop, so they figure out the logical relationship.

First of all, your delta=pi 2000*a is written incorrectly, it should be: delta=pi (2000*a), otherwise you will calculate the right and wrong numbers.

Secondly, the delta value range of the drawing is (pi 2000, pi 10), but your delta=pi (2000*a) is not in this range, of course, it cannot be drawn.

All you need to do is change the delta range, for example, to:

delta=linspace(pi/(2000*a),pi/10,1000);

clc;

a=0;fa=-inf;

b=3;fb=inf;

while b-a>eps*b

x=(a+b)/2;

fx=x^3-2*x-5;

if fx==0

breakelseif sign(fx)==sign(fa)a=x;

elseb=x;

endend

disp(x)

It's a little wrong to write, so I've changed it with you. The purpose of this problem is to find the weight of a DAO zero point of the function between 0 and 3 (the function has several zero points, but we are only concerned with this one between 0 and 3 for now). The structure is a cyclic approximation.

For example, in the first loop, x=, bring x into f(x) and find that it is less than zero, so we know that the zero point is between 3. In the second cycle, x=, bring x into f(x) to see whether it is greater than or less than zero, move one of the two points of a and b according to the result of greater than less than zero, and finally end the cycle when a and b are almost the same size. You're right, it's a dichotomy approximation.

It's not better to use abaqus and ansys directly, so why bother to go far.

Let the length, width and height be He Yan x, y, z respectively, and the problem is to find the minimum value of f(x, y, z)= under the condition xyz=512.

The Lagrangian function is f(x,y,z)=.

According to fx=0, fy=0, fz=0, we get this x=y=z. Substituting xyz-512=0 gives x=y=z=8.

You only count 2 sides, and the real cavity has 4 sides.

r2=find(f>t);This sentence is problematic because according to your program, when min(f)=0, t may be equal to max(f), then the result of r2 is an empty data, so it will be wrong.

If you don't understand your algorithm, check t=(min(f(:)max(f(:)max(f(:) 2; Is there a problem.