'입력 인수가 부족합니다' 이 오류를 해결하고 싶어요.

Question

gyul 2024-4-10

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2105046-

评论： gyul 2024-4-10

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);

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gyul 2024-4-10

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 2024-4-10

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

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Answer 1

Dyuman Joshi 2024-4-10

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2105046-#answer_1439036

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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