Error using inline/subsref (line 12) Not enough inputs to inline function.

Hi, I've been tested so many times and tried all the answers from here and still not working.
if the expression is
with
m=4.541*TN*S0/T
T=1:0.1:100
TN=70
S0=0.99
integration with simpson 1/3
here my code:
clear all; clc;
f1=inline('3/2*(cos(x)^2)*exp(m*(3/2*(cos(x)^2))-1/2)*sin(x)','m','x');
f2=inline('exp(m*(3/2*(cos(x)^2)-1/2))','m','x');
T0=1;
TM=100;
dT=0.1;
TN=70;
S0=0.99;
for T=T0:dT:TM;
m=4.541*TN*S0/T;
a=0;b=pi;
c=0;d=pi;
n=100;
h=(b-a)/n;
f1a=f1(a);f1b=f1(b);
f2a=f2(a);f2b=f2(b);
jumeven=0;jumodd=0;
for i=1:n-1
p=mod(1,2);
if p==0
xi=x+i*h;
jumeven=jumeven+f(xi);
else
xi=x+i*h;
jumodd=jumodd+f(xi);
end
end
I1=(h/3)*(f1a+4*jumodd+2*jumeven+f1b);
I2=(h/3)*(f2c+4*jumodd+2*jumeven+f2d);
S=I1/I2;
S0=S;
end
disp([S' T']);
disp([S]);
plot(S,T,'ob-','linewidth',3);
grid on;
xlabel('S','fontsize',18);
ylabel('T','fontsize',18);

댓글 수: 4

What advantage do you see in using inline ? Have you benchmarked it for your situation and found it gives better performance? Have you found something using inline that you have not been able to express in other ways?
I ask because Mathworks has recommended against using inline() for roughly the last 18 years.
Humm sorry, I don't know any other way, and why is it inline? because, it's already in my lecturer's guide. And btw, any suggestion from you? I'm happy to try it
Stephen23
Stephen23 2022년 3월 12일
편집: Stephen23 2022년 3월 12일
"it's already in my lecturer's guide."
Your lecturer wrote their (mis)guide back in the dark ages. You should ask for your money back.
"And btw, any suggestion from you? I'm happy to try it"
You do what the MATLAB documentation recommends:
f1 = @(m,x) 3/2*(cos(x).^2).*exp(m.*(3/2*(cos(x).^2))-1/2).*sin(x);
f2 = @(m,x) exp(m.*(3/2*(cos(x).^2)-1/2));

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 채택된 답변

These lines:
f1a=f1(a);f1b=f1(b);
f2a=f2(a);f2b=f2(b);
Should be:
f1a=f1(m,a);f1b=f1(m,b);
f2a=f2(m,a);f2b=f2(m,b);
As the functions f1, f2 need two inputs.

댓글 수: 5

another error appears: Unrecognized function or variable 'f'.
jumodd=jumodd+f(xi);
thanks before
It is because, there is no function with the name f. Either you have to define it or it could be a typo error for f1 or f2.
and 1 more, how can i define f as f1/f2?
okayy got it, thankyou so much @KSSV and @Walter Roberson 🙏

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2022년 3월 11일

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