'입력 인수가 부족합니다' 이 오류를 해결하고 싶어요.

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);

[Dyuman Joshi]

Definite integration requires two limits, where as you have provided only a single limit -

fo=int(assignment2_2f,xo); ffo=int(assignment2_2ff,xo);

What are you trying to do? Please specify.

[gyul]

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]

If you want to define substitution use subs instead of int.

Answer 1

[Dyuman Joshi]

I have made some changes to your code, please check the code below and refer to the comments for information -

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]

ohh,,,thank you for saving me,,