How can I determine if the absolute value of a coefficient of x in each element of a matrix was less than 0.01 consider it zero?

For example I have the below matrix 2*2

a=[ e-8(X^2)+12X-e-6(X)+14, -10(X^2)+0.001X-e-6(X^3); X, 4]

I want to write the code so that it give me:

a=[ 12X+14, -10(X^2); X 4]

[Matt J]

Given what data as input? Do you have many pairs of (X,a) to which you could fit the coefficients?

[Walter Roberson]

could you confirm that this is a question symbolic expressions? If so I posted code for this about 4 years ago but it might be hard to locate. Might be easier to rewrite.

[Misi]

Actullay, the below is my code, I want to calculate the inverse of matrix V but the matlab cannot do that so I wanted to consider the small coefficient of px as a zero, maybe it help me to determine the inverse of V.

the line 59 in the following code cannot be calculated.

syms px

c11=158e9;
c22=15.51e9;
c33=15.51e9;
c44=3.2e9;
c66=4.4e9;
c12=5.64e9;
c13=5.64e9;
c23=7.21e9;
c55=4.4e9;
h=0.01;
m=1;
n=1;
a=1;
b=1;


 
c1=-c13/c33;
c2=c11-c13^2/c33;
c3=c12-c13*c23/c33;
c4=c22-c23^2/c33;
c5=-c23/c33;
c6=c66;
c10=1/c33;
a11=1/c55;
a22=1/c44;
z=m*pi/a;
e=n*pi/b;
t=px*z^2;


Buckmatrix=[0 0 0 a11 0 -z;
0 0 0 0 a22 -e;
0 0 0 z e -t;
-t+c2*z^2+c6*t^2 (c3+c6)*z*e c1*z 0 0 0;
(c3+c6)*z*e -t+c6*z+c4*e^2, c5*e 0 0 0;
    -c1*z, -c5*e, c10 0 0 0];

Buckmatrix=vpa(Buckmatrix,2);
[V,D] = eig(Buckmatrix);
Tf=V*exp(D*h)*inv(V);
FinalM=[Tf(3,1) Tf(3,2) Tf(3,6);Tf(4,1) Tf(4,2) Tf(4,6);Tf(5,1) Tf(5,2) Tf(5,6)];
Ans=det(FinalM);
Ans=vpa(Ans,8)

Answer 1

[Image Analyst]

Then just write it as a=[ 12X+14, -10(X^2); X 4] with whatever value you have for X.

Why can't you? Do you not know what the coefficients are? Like they're not hard coded in like you have but they're somehow contained in another matrix b so that

b = [ e-8, 12, -e-6,+14, -10,+0.001, -e-6; 1, 4]
a = [ b(1) * (X^2)+ b(2) * X + b(3) * X+b(4), b(5) * (X^2), b(6) * X + b(7) * (X^3); ...
    b(8) * X, b(10)]

so you can just use that a but zero out elements of b before using b, like

b = [ e-8, 12, -e-6,+14, -10,+0.001, -e-6; 1, 4] % Original list of coefficients.
b(b < 0.01) = 0; % Get rid of small coefficients.
X = 42; % Whatever.....
% Now compute a
a = [ b(1) * (X^2)+ b(2) * X + b(3) * X+b(4), b(5) * (X^2), b(6) * X + b(7) * (X^3); ...
    b(8) * X, b(10)]

[Misi]

Hi many thank for response, but it cannot be use for my code

actually the below is my code and the matlab cannot calulate the inverse of V matrix in line 59

syms px

c11=158e9;
c22=15.51e9;
c33=15.51e9;
c44=3.2e9;
c66=4.4e9;
c12=5.64e9;
c13=5.64e9;
c23=7.21e9;
c55=4.4e9;
h=0.01;
m=1;
n=1;
a=1;
b=1;

 
c1=-c13/c33;
c2=c11-c13^2/c33;
c3=c12-c13*c23/c33;
c4=c22-c23^2/c33;
c5=-c23/c33;
c6=c66;
c10=1/c33;
a11=1/c55;
a22=1/c44;
z=m*pi/a;
e=n*pi/b;
t=px*z^2;
Buckmatrix=[0 0 0 a11 0 -z;
0 0 0 0 a22 -e;
0 0 0 z e -t;
-t+c2*z^2+c6*t^2 (c3+c6)*z*e c1*z 0 0 0;
(c3+c6)*z*e -t+c6*z+c4*e^2, c5*e 0 0 0;
    -c1*z, -c5*e, c10 0 0 0];
Buckmatrix=vpa(Buckmatrix,2);
[V,D] = eig(Buckmatrix);
Tf=V*exp(D*h)*inv(V);
FinalM=[Tf(3,1) Tf(3,2) Tf(3,6);Tf(4,1) Tf(4,2) Tf(4,6);Tf(5,1) Tf(5,2) Tf(5,6)];
Ans=det(FinalM);
Ans=vpa(Ans,8)

[Sam Chak]

Can you introduce the if-else statement such that you have more control freedom to define the rules when a particular input is given?

Answer 2

[Walter Roberson]

sympref('floating', false);
tolerance = 0.01;
syms X
a=[  1e-8*(X^2)+12*X-1e-6*(X)+14, -10*(X^2)+0.001*X-1e-6*(X^3); X, 4]

a = 

a_new = mapSymType(a, 'real', @(value) piecewise(value < -tolerance | value > tolerance, value, 0))

a_new = 

Note that this does not give you exactly what you were hoping for. Your notation was unclear so I had to guess that 12X-e-6(X) represents 12*X-1e-6*(X) . But MATLAB would never hold those two coefficients of X separate: it would always combine them, and the only question is whether they are going to get combined as the rational 12 - 1/10^6 or as floating point 12.0 - 0.000001 .

The situation would be different if one of the coefficients had an additional variable; for example 12*X-1e-6*rho*(X) would not automatically combine.

If somehow you had an input of 12*X-1e-6*(X) and you wanted to do the coefficient reduction before combining the coefficients, then you would have to wrap each entry in a function or else you would need to do text manipulation on text expressions before converting the text to symbolic.