'입력 인수가 부족합니다' 이 오류를 해결하고 싶어요.

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gyul on 10 Apr 2024
Commented: gyul on 10 Apr 2024
syms ('x')
xo=2000;
tolerance=0.03;
Re=1.0;
fid=fopen('solution1.txt', 'w');
while Re>tolerance
assignment2v=assignment2_2f(xo);
assignment22v=assignment2_2ff(xo);
xn=xo-assignment2v/assignment22v;
Re=abs((xn-xo)/xo);
fprintf(fid, 'approx. solution is : x=5.2%f\n', xn);
xo=xn; fo=fn; ffo=ffn;
end
fprintf(fid, 'The coverged solution is : x=5.2%f\n', xo);
type solution1.txt
----------------------------
function y=assignment2_2f(x)
y=x/400+10.245*sin(4.4934*x/4000)-2;
end
---------
function g=assignment2_2ff(x)
g=diff(assignment2_2f,x);
end
-----------------------------
assignment2_2
입력 인수가 부족합니다.
오류 발생: assignment2_2f (2번 라인)
y=x/400+10.245*sin(4.4934*x/4000)-2;
오류 발생: assignment2_2 (6번 라인)
fo=int(assignment2_2f,xo); ffo=int(assignment2_2ff,xo);
3 CommentsShow 1 older commentHide 1 older comment
gyul on 10 Apr 2024
I wanted to define the substitution of 'xo' for function that 'assinment2_2f'. Now that I look at it, I don't think it's necessary. However, deleting it does not change the error.
Is the 'Re>tolerance' part correct that the one limitation you mentioned?
Dyuman Joshi on 10 Apr 2024
If you want to define substitution use subs instead of int.

Dyuman Joshi on 10 Apr 2024
syms('x')
%% Define the functions as symbolic functions
assignment2_2f(x) = x/400+10.245*sin(4.4934*x/4000)-2;
assignment2_2ff(x) = diff(assignment2_2f, x);
xo=2000;
%% Calculate the values with xo as inputs
%Note that you can use this syntax with symbolic functions as well
fo = assignment2_2f(xo);
ffo = assignment2_2ff(xo);
tolerance=0.03;
Re=1.0;
fid=fopen('solution1.txt', 'w');
while Re>tolerance
assignment2v=assignment2_2f(xo);
assignment22v=assignment2_2ff(xo);
xn=xo-assignment2v/assignment22v;
Re=abs((xn-xo)/xo);
%% I assume that fn and ffn are calculated based on xn
fn = assignment2_2f(xn);
ffn = assignment2_2ff(xn);
fprintf(fid, 'approx. solution is : x=5.2%f\n', xn);
xo=xn; fo=fn; ffo=ffn;
end
fprintf(fid, 'The coverged solution is : x=5.2%f\n', xo);
%% Also, you should close the file after data writing has been completed.
fclose(fid);
type solution1.txt
approx. solution is : x=5.24338.898407 approx. solution is : x=5.24629.292712 approx. solution is : x=5.24562.954523 The coverged solution is : x=5.24562.954523
gyul on 10 Apr 2024
ohh,,,thank you for saving me,,

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