I am getting wrong answer

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SANDIPKUMAR ROYADHIKARI
SANDIPKUMAR ROYADHIKARI 2021 年 8 月 28 日
I am trying to calculate the error of a function within some limit, but I am getting wrong answer. I have attached my code. Please help me to find out the problem. Thanx
clear all
clc
x=1;
err=10000;
while (err>=20)
err=131-x^2;
x=x+1;
end
After running the code, err is coming out less than 20. Please help me to fix it.
  6 件のコメント
SANDIPKUMAR ROYADHIKARI
SANDIPKUMAR ROYADHIKARI 2021 年 8 月 28 日
Are you saying this ?
x=1;
err=1000;
while err >20
err=131-x^2;
x=x-1;
end
This also doesn't give the correct answer

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回答 (2 件)

Awais Saeed
Awais Saeed 2021 年 8 月 28 日
編集済み: Awais Saeed 2021 年 8 月 28 日
You wrote a loop to run when err >= 20, its value is decreasing in the loop and when err<20 (in your case 10), why would it still run? It has to stop. See the output to know the reason
x=1;
err=10000;
while (err>=20)
err=131-x^2;
fprintf('x = %d, err = %d\n', x, err)
x=x+1;
end
x = 1, err = 130
x = 2, err = 127
x = 3, err = 122
x = 4, err = 115
x = 5, err = 106
x = 6, err = 95
x = 7, err = 82
x = 8, err = 67
x = 9, err = 50
x = 10, err = 31
x = 11, err = 10

Walter Roberson
Walter Roberson 2021 年 8 月 28 日
You are looking for an integer, x, such that
syms x
xsol = solve(131-x^2 == 20)
xsol = 
You can see from xsol that the x that solve that equation are not integers: they are numbers that are between 10 and 11 and the negative of that.
Changing to a smaller increment such as 0.1 will only get you closer . No matter what rational increment you use instead of 1, the actual solutions are irrational and so cannot be reached by the method you are using (though possibly you could find something that came out okay to within roundoff error

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