How can i use a vector variable inside a function normally

I have this code i'm trying to get to run

% This is a driver program which calls different methods at same place

clc

format long

syms t y;

tspan = 2:0.25:3; % t-values = startpoint:h:endpoint

y2 = 2; % initial condition

functiona=@(t,y) -y+t*y^(1/2);%put f(t,y) here

df=@(t,y) y-(3/2)*t*y^(1/2)+y^(1/2)+(1/2)*t^2;%put f'(t,y) here

ddf=@(t,y) -y+(7/4)*t*y^(1/2)-2*y^(1/2)-(3/4)*t^2+(3/2)*t;%put f''(t,y) here

dddf=@(t,y) y+(11/4)*t*y^(1/2)+(13/4)*y^(1/2)+(7/8)*t^2-(5/2)*t+3/2;%put f'''(t,y) here

[t_values,y_values]=ode45(functiona,tspan,y2);

[t_values,w_values]=taylor_4th(functiona,df,ddf,dddf,tspan,y2);

syms y(t)

% This is the exact solution

f = @(tspan) (tspan-2+2^(1/2)*exp(1)*exp((-1/2)*tspan))^2;%put exact solution here;

error1 = abs(f(tspan)'-y_values);

error2 = abs(f(tspan)'-w_values);

varnames = {'t_values','ODE45', 'Taylor','exact_values', 'Errorode45','Error-Taylor'};

table(t_values, y_values,w_values, f(tspan)',error1,error2,'VariableNames',varnames)

the problem appears to be with the line 18

f = @(tspan) (tspan-2+2^(1/2)*exp(1)*exp((-1/2)*tspan))^2;%put exact solution here;

I think it has to do with needing to put periods in since 'tspan' is a vector and the operations need to be put in componentwise.

Answer 1

[Voss]

Looks like your hunch was correct: there was a missing period in that expression.

% This is a driver program which calls different methods at same place
clc
format long
syms t y;
tspan = 2:0.25:3; % t-values = startpoint:h:endpoint
y2 = 2; % initial condition
 
functiona=@(t,y) -y+t*y^(1/2);%put f(t,y) here
 
df=@(t,y) y-(3/2)*t*y^(1/2)+y^(1/2)+(1/2)*t^2;%put f'(t,y) here
ddf=@(t,y) -y+(7/4)*t*y^(1/2)-2*y^(1/2)-(3/4)*t^2+(3/2)*t;%put f''(t,y) here
dddf=@(t,y) y+(11/4)*t*y^(1/2)+(13/4)*y^(1/2)+(7/8)*t^2-(5/2)*t+3/2;%put f'''(t,y) here
 
[t_values,y_values]=ode45(functiona,tspan,y2);
% [t_values,w_values]=taylor_4th(functiona,df,ddf,dddf,tspan,y2);
syms y(t)
% This is the exact solution
% f = @(tspan) (tspan-2+2^(1/2)*exp(1)*exp((-1/2)*tspan))^2;%put exact solution here;
f = @(tspan) (tspan-2+2^(1/2)*exp(1)*exp((-1/2)*tspan)).^2;%put exact solution here;
error1 = abs(f(tspan)'-y_values);
% error2 = abs(f(tspan)'-w_values);

disp(error1);

   1.0e-09 *

   0.000000888178420
   0.175975678473606
   0.029717561744746
   0.082043261073750
   0.046412207410640

Answer 2

[Matthew Allred]

I was able to find the corrected code

f = @(tspan) (tspan-2+2.^(1/2).*exp(1).*exp(-tspan.*(1/2))).^2

It appears i just needed to put a period before every multiplication, division, exponential etc operator.

[Voss]

That's fine, but actually only the last period (.^) is necessary because all the other operations are between a vector and a scalar.