MATLAB only display numeric values up to 4 decimal places. (MATLAB r2020a)

I created two functions called 'newton' and 'fjacob' to solve nonlinear equations using Newton's method.

When i type [xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(@fjacob,[8;5],1e-8,15) in command window,

The results show only 4 decimal places (shown in attached picture), even when I set to format 'long'.

%--newton--%
function [xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(FunctionName,x0,epsilon,IterationMax)
x = sym('x',[1 2]).';
IFLAG = "Fail to converge";
[f,J] = FunctionName(x);
Xk = x0.';
Jk = [];
IterationUsed = 0;
xsol = x0;
iter = [0];
for i=1:IterationMax
    fsub = subs(f,x,xsol);
    Jsub = subs(J,x,xsol);
    Fk(i,:) = fsub.';
    Jk = [Jk;Jsub];
    s = -inv(Jsub)*fsub;
    x_new = xsol + s;
    Xk = [Xk;x_new'];
    IterationUsed = IterationUsed + 1;
    iter(i+1) = i;
    if norm(s,inf) < epsilon
        xsolution = x_new;
        IFLAG = "Converges to xsolution";
        break
    end
    if i == IterationMax; break , end
    xsol = vpa(x_new);
end
end

%--fjacob--%
function [f,J] = fjacob(x)

f = [x(1)^2 + x(2)^2 - 1 ; 5*x(1)^2 - x(2) - 2];
J = jacobian(f,x);
end

Answer 1

[Star Strider]

I cannot run thst.

It appears that you are using the Symbolic Math Toolbox and the vpa funciton. Use the digits function to see what the digits (displayed precision) setting is, and change it if necessary.

.

[Natthaphat G.Mahtani]

It is set to 32 digits.

I update the code a bit and got this result (see picture).

Why does Xk only have 4 decimal place?

[Star Strider]

As I mentioned, I cannot run the posted code (since I do not have any of the arguments), so I cannot determine what the problem is. I was simply suggesting a possible solution.

Also:

format 'long'

(the command form with the quotes) will not work, however

format long

(without the quotes)

or the function form

format('long')

(with the quotes) will.

I am not certain if that is actually the problm, however.

.

[Natthaphat G.Mahtani]

This is my updated 'newton' code:

function [xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(FunctionName,x0,epsilon,IterationMax)
x = sym('x',[1 2]).';
IFLAG = "Fail to converge";
[f,J] = FunctionName(x);
Xk = x0.';
IterationUsed = 0;
xsol = x0;
iter = [0];
for i=1:IterationMax
    fsub = subs(f,x,xsol);
    Jsub = subs(J,x,xsol);
    Fk(i,:) = fsub.';
    Jk(2*i-1:2*i,:) = Jsub;
    s = -Jsub\fsub;
    x_new = xsol + s;
    Xk(i+1,:) = x_new.';
    IterationUsed = IterationUsed + 1;
    iter(i+1) = i;
    if norm(s,inf) < epsilon
        xsolution = x_new;
        IFLAG = "Converges to xsolution";
        break
    end
    if i == IterationMax; break , end
    xsol = vpa(x_new);
end
end

[Natthaphat G.Mahtani]

and here is the 'fjacob' code

function [f,J] = fjacob(x)

f = [x(1)^2 + x(2)^2 - 1 ; 5*x(1)^2 - x(2) - 2];
J = jacobian(f,x);
end

[Star Strider]

I still cannot run it.

[Natthaphat G.Mahtani]

You don't need to run it. Just type [xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(@fjacob,[8;5],1e-8,15) into the command window.

[Star Strider]

I just did, and I get the expected result, with full precision —

[xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(@fjacob,[8;5],1e-8,15) 

xsolution = 

Xk = 

Fk = 

Jk = 

IFLAG = "Converges to xsolution"

IterationUsed = 8

%--newton--%
function [xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(FunctionName,x0,epsilon,IterationMax)
x = sym('x',[1 2]).';
IFLAG = "Fail to converge";
[f,J] = FunctionName(x);
Xk = x0.';
Jk = [];
IterationUsed = 0;
xsol = x0;
iter = [0];
for i=1:IterationMax
    fsub = subs(f,x,xsol);
    Jsub = subs(J,x,xsol);
    Fk(i,:) = fsub.';
    Jk = [Jk;Jsub];
    s = -inv(Jsub)*fsub;
    x_new = xsol + s;
    Xk = [Xk;x_new'];
    IterationUsed = IterationUsed + 1;
    iter(i+1) = i;
    if norm(s,inf) < epsilon
        xsolution = x_new;
        IFLAG = "Converges to xsolution";
        break
    end
    if i == IterationMax; break , end
    xsol = vpa(x_new);
    Xk = vpa(Xk);
end
end

%--fjacob--%
function [f,J] = fjacob(x)

f = [x(1)^2 + x(2)^2 - 1 ; 5*x(1)^2 - x(2) - 2];
J = jacobian(f,x);
end

Since I do not have access to your computer to see what the problem is, this is likely something you will need to solve. My suggestions with respect to digits and format are the best I can do from here.

I added:

Xk = vpa(Xk);

however made no other changes. That only affected the format of the output, nothing else.

.

[Natthaphat G.Mahtani]

As I have mentioned earlier,

I type digits into command window and it returns 32

format is set to long.

I have uninstalled and install r2021a but the problem is still there.

How should I fix it?

[Star Strider]

‘How should I fix it?’

I already suggested everything I can think of. The only other option would be to check the sympref settings, since this apparently occurs entirely within the Symbolic Math Toolbox.

Another option might be to use double instead of vpa, since none of the results contain symbolic variables:

format long

[xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(@fjacob,[8;5],1e-8,15) 

xsolution = 

Xk = 

Fk = 

Jk = 

IFLAG = "Converges to xsolution"

IterationUsed = 
     8

xsolution = double(xsolution)

xsolution = 2×1

0.732260195229901 0.681024967590665

Xk = double(Xk)

Xk = 9×2

8.000000000000000 5.000000000000000 4.056372549019608 2.509803921568627 2.110076410041546 1.321768978064554 1.188938852687405 0.825405979306453 0.820914046228528 0.692288065630997 0.737056992686961 0.681105025200372 0.732275804664860 0.681024971693328 0.732260195396269 0.681024967590665 0.732260195229901 0.681024967590665

Fk = double(Fk)

Fk = 8×2

1.0e+02 * 0.880000000000000 3.130000000000000 0.217532739811611 0.777609873606305 0.051994956875876 0.189403433030045 0.010948706261045 0.042424719978418 0.001531626371104 0.006772112908455 0.000071570658219 0.000351600271434 0.000000228661675 0.000001142987945 0.000000000002436 0.000000000012182

Jk = double(Jk)

Jk = 16×2

16.000000000000000 10.000000000000000 80.000000000000000 -1.000000000000000 8.112745098039216 5.019607843137255 40.563725490196077 -1.000000000000000 4.220152820083093 2.643537956129108 21.100764100415464 -1.000000000000000 2.377877705374811 1.650811958612906 11.889388526874054 -1.000000000000000 1.641828092457056 1.384576131261993 8.209140462285278 -1.000000000000000

%--newton--%
function [xsolution,Xk,Fk,Jk,IFLAG,IterationUsed] = newton(FunctionName,x0,epsilon,IterationMax)
x = sym('x',[1 2]).';
IFLAG = "Fail to converge";
[f,J] = FunctionName(x);
Xk = x0.';
Jk = [];
IterationUsed = 0;
xsol = x0;
iter = [0];
for i=1:IterationMax
    fsub = subs(f,x,xsol);
    Jsub = subs(J,x,xsol);
    Fk(i,:) = fsub.';
    Jk = [Jk;Jsub];
    s = -inv(Jsub)*fsub;
    x_new = xsol + s;
    Xk = [Xk;x_new'];
    IterationUsed = IterationUsed + 1;
    iter(i+1) = i;
    if norm(s,inf) < epsilon
        xsolution = x_new;
        IFLAG = "Converges to xsolution";
        break
    end
    if i == IterationMax; break , end
    xsol = vpa(x_new);
    Xk = vpa(Xk);
    Jk = vpa(Jk);
end
end

%--fjacob--%
function [f,J] = fjacob(x)

f = [x(1)^2 + x(2)^2 - 1 ; 5*x(1)^2 - x(2) - 2];
J = jacobian(f,x);
end

.

[Natthaphat G.Mahtani]

I created another code to solve the same problem. However, I do not use function this time (and no symbolic substitution). I ran the code and the results are what i was looking for. What does this tell me about the problem?

clear; clc;
x0 = [8;5]; epsilon = 1e-8; IterationMax = 100;
x = sym('x',[1 2]).';
xsol = x0;
Xk = x0.';
iter = [0];
IterationUsed = 0;
for i=1:IterationMax
    f = [xsol(1)^2 + xsol(2)^2 - 1 ; 5*xsol(1)^2 - xsol(2) - 2];
    J = [2*xsol(1) 2*xsol(2);10*xsol(1) -1];
    fsub = f;
    Jsub = J;
    Fk(i,:) = fsub.';
    Jk(2*i-1:2*i,:) = Jsub;
    %Jk = [Jk;Jsub];
    s = -Jsub\fsub;
    x_new = xsol + s;
    Xk(i+1,:) = x_new.';
    %Xk = [Xk;x_new'];
    IterationUsed = IterationUsed + 1;
    iter(i+1) = i;
    if norm(s,inf) < epsilon
        xsolution = vpa(x_new);
        IFLAG = 'Converges to xsolution'
        break
    end
    if i == IterationMax
        IFLAG = 'Fail to converge'
        break , end
    xsol = double(x_new);
    Xk = double(Xk);
end
Fk = double([Fk;f.']);
Iteration = iter.'; x1 = Xk(:,1); x2 = Xk(:,2); f1 = Fk(:,1); f2 = Fk(:,2);
T = table(Iteration,x1,x2,f1,f2)

[Star Strider]

‘What does this tell me about the problem?’

The subs calls so not appear in this version, so they could be part of the problem. Beyond that, I honestly have no idea, since I was unable to replicate the error here or on my own computer. It will be necessary to go through the previous code (with the truncation error) and see if it is possible to isolate the error. I can’t do that.

.

[Natthaphat G.Mahtani]

I think i have fixed it.

I type sympref('default') and call out the function again, the results are same as yours.

Thanks very much for your help.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.