Error with equation using sym/subsindex

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

Answer 1

[Walter Roberson]

% 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
end

[emma]

If I am trying to find the root of the equation to find the angle A (in degrees), how can I edit my code to make the root equal to about 5.9156599 (The correct answer according to my professor). Do I change the intervals?

[Walter Roberson]

% 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
end

[emma]

Thank you so much! You are a godsend.

Answer 2

[Sulaymon Eshkabilov]

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

[emma]

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

[Walter Roberson]

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.