Error with equation using sym/subsindex

Question

emma 2021-4-6

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/794097-error-with-equation-using-sym-subsindex

评论： emma 2021-4-7

When I try to run my code, I get "Error using sym/subsindex...Invalid indexing or function definition..." and "Error in line with f(A) = x.*tan(A)-(g./(2.*V.^2.*((cos(A)).^2))).* x.^2-y". Can someone please show me how I can fix my code below?

% Variables
V = 1500;
g = 32.2;
x = 500; 
y = 50; 
tol = 0.000001; % Error tolerance
 
% Function
syms A
f(A) =  x.*tan(A)-(g./(2.*V.^2.*((cos(A)).^2))).* x.^2-y;
 
x_coor = linspace(0, 0.2, 1000);
y_coor = f(x_coor);
 
figure
plot(x_coor, y_val)
xline(0.103247833251953)
yline(0)
root = bisection(f,0.1,0.2,tol);    % Root from Bisection Method
function [c] = bisection(f,a,b,tol) % Function file
% Inputs: f = function; a,b = range or interval for roots to be found where a>b; tol = error tolerance
err = 1/tol;
% Bisection method
while err>tol 
c = (a+b)/2; 
    if(abs(f(c))== 0) %if f(c) is equal to 0, then stop
    break
    end
    if f(a)*f(c)<0
    b=c;
    else
    a=c; 
    end
    err=abs(a-b); 
end
end

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Answer 1

Walter Roberson 2021-4-6

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/794097-error-with-equation-using-sym-subsindex#answer_668777

编辑：Walter Roberson 2021-4-6

在 MATLAB Online 中打开

% Variables

V = 1500;

g = 32.2;

x = 500;

y = 50;

tol = 0.000001; % Error tolerance

% Function

syms A

f(A) = x.*tan(A)-(g./(2.*V.^2.*((cos(A)).^2))).* x.^2-y;

x_coor = linspace(0, 0.2, 1000);

y_coor = f(x_coor);

figure

plot(x_coor, y_coor) %you used undefined variable y_val

xline(0.103247833251953)

yline(0)

[root, iters] = bisection(f,0.1,0.2,tol); % Root from Bisection Method

root

root = 0.1032

iters

iters = 17

function [c, iters] = bisection(f,a,b,tol) % Function file

% Inputs: f = function; a,b = range or interval for roots to be found where a>b; tol = error tolerance

err = 1/tol;

% Bisection method

iters = 0;

while err>tol

iters = iters + 1;

c = (a+b)/2;

if(abs(f(c))== 0) %if f(c) is equal to 0, then stop

break

end

if f(a)*f(c)<0

b=c;

else

a=c;

end

err=abs(a-b);

end

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Walter Roberson 2021-4-7

在 MATLAB Online 中打开

% Variables

V = 1500;

g = 32.2;

x = 500;

y = 50;

tol = 0.000001; % Error tolerance

% Function

syms A

f(A) = x.*tand(A)-(g./(2.*V.^2.*((cosd(A)).^2))).* x.^2-y; %DEGREES

x_coor = linspace(0, 0.2, 1000);

y_coor = f(x_coor);

figure

plot(x_coor, y_coor) %you used undefined variable y_val

xline(0.103247833251953)

yline(0)

[root, iters] = bisection(f,0.1,10,tol); % Root from Bisection Method %CHANGE UPPER BOUND

root

root = 5.9157

iters

iters = 24

function [c, iters] = bisection(f,a,b,tol) % Function file

% Inputs: f = function; a,b = range or interval for roots to be found where a>b; tol = error tolerance

err = 1/tol;

% Bisection method

iters = 0;

while err>tol

iters = iters + 1;

c = (a+b)/2;

if(abs(f(c))== 0) %if f(c) is equal to 0, then stop

break

end

if f(a)*f(c)<0

b=c;

else

a=c;

end

err=abs(a-b);

end

emma 2021-4-7

Thank you so much! You are a godsend.

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Answer 2

Sulaymon Eshkabilov 2021-4-6

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/794097-error-with-equation-using-sym-subsindex#answer_668337

在 MATLAB Online 中打开

In your code, there are a few confusions and flaws. You've define: f(A) = x.*tan(A)-(g./(2.*V.^2.*((cos(A)).^2))).* x.^2-y; f as a function of A. At the same time, y_coor = f(x_coor). The syntax of y_coor = f(x_coor) is not correct as defined above.

Some confusions need to be fixed, y_coor vs. y_val, x vs. x_coor.

Moreover, without the value of A, y_coor can't be computed.

Here is a corrected code:

% Variables
V = 1500;
g = 32.2;
x = 500; 
y = 50; 
tol = 0.000001; % Error tolerance 
% Function
syms A
f(A)=  x.*tan(A)-(g./(2.*V.^2.*((cos(A)).^2))).* x.^2-y;
root = bisection(f,0.1,0.2,tol);    % Root from Bisection Method
A = root;
fprintf('Computed root is: %f \n ', root);
% Plot:
x = linspace(0, 0.2, 1000);
Y = @(x)(x.*tan(A)-(g./(2.*V.^2.*((cos(A)).^2))).* x.^2-y);
y_val=Y(x); 
figure
plot(x, y_val)
xline(0.103247833251953)
yline(0)
function [c] = bisection(f,a,b,tol) % Function file
% Inputs: f = function; a,b = range or interval for roots to be found where a>b; tol = error tolerance
err = 1/tol;
% Bisection method
while err>tol 
c = (a+b)/2; 
    if(abs(f(c))== 0) %if f(c) is equal to 0, then stop
    break
    end
    if f(a)*f(c)<0
    b=c;
    else
    a=c; 
    end
    err=abs(a-b); 
end
end

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emma 2021-4-6

Thank you! This was a big help. Is there any way I can count the number of iterations for the plot?

Walter Roberson 2021-4-6

The syntax of y_coor = f(x_coor) is not correct as defined above.

That is not correct. The poster defined a symbolic function. Just like regular functions and anonymous functions, the named parameters will be substituted with whatever is passed to the function when it is invoked, so it is fine for numeric values or a variable of a different name to be passed to f.

Perhaps the user edited the original code after you posted, but for the code currently posted at least, there are almost no changes needed to get the plot -- only one misnamed variable.

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