Using a while loop to solve a Taylor Series with approximate values.

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stephen 2022-10-14

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https://ww2.mathworks.cn/matlabcentral/answers/1825688-using-a-while-loop-to-solve-a-taylor-series-with-approximate-values

评论： VBBV 2022-10-14

f = atan(1);
n = 0;
i = -1;
err_approx = 100;
while (err_approx<=.5)
    i = i + 2;
    y_prev = y_approx
    y_approx = (-1)^(i)(1/i+2)*1^(i+2)
    err_approx = abs((y_approx-y_prev)/y_prev)*100
    err_true = abs((y_approx-f)/f)*100
    n = n + 1
[y_approx, err_approx, err_true, n]
end

This is my current code. I am using it to solve the taylor series expansion with approximate values to solve for arctan(1). I am having 2 errors. It is stating y_approx is undefined, and there is no output.

the expansion is in the form -x+x^(3)/3-x^(5)/5+x^(7)/7

Thanks in advance

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回答（1 个）

VBBV 2022-10-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1825688-using-a-while-loop-to-solve-a-taylor-series-with-approximate-values#answer_1074263

编辑：VBBV 2022-10-14

在 MATLAB Online 中打开

f = atan(1);
n = 0;
i = -1;
err_approx = 100;
while (err_approx >= .5)   % check the condition here
    i = i + 2;
    
    y_approx = ((-1)^(i))*(1/(i+2))*1^(i+2)
    y_prev = y_approx;
    err_approx = abs((y_approx-y_prev)/y_prev)*100
    err_true = abs((y_approx-f)/f)*100;
    n = n + 1
[y_approx, err_approx, err_true, n];
end
y_approx = -0.3333
err_approx = 0
n = 1
y_approx
y_approx = -0.3333

Use a inequality opertor which satisfies the condition for while loop.

In your code, while loop is not satisfied, hence you get an error

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stephen 2022-10-14

I am trying to get it to output a matrix of all values that were used

VBBV 2022-10-14

在 MATLAB Online 中打开

f = atan(1);
n = 0;
i = -1;
err_approx = 100;
y_approx = 0; % give an initial value for y_approx
while (err_approx >= .5)   % check the condition here
    i = i + 2;
    y_prev = y_approx;
    y_approx = ((-1)^(i))*(1/(i+2))*1^(i+2);
    
    err_approx = abs((y_approx-y_prev)/y_prev)*100;
    err_true = abs((y_approx-f)/f)*100;
    n = n + 1;
    Y_p(n) = y_approx;
    E_p(n) = err_approx;
    E_t(n) = err_true; % error
    N(n) = n; % iter
end
 
 [Y_p;E_p;E_t;N].'
ans = 200×4
   -0.3333       Inf  142.4413    1.0000
   -0.2000   40.0000  125.4648    2.0000
   -0.1429   28.5714  118.1891    3.0000
   -0.1111   22.2222  114.1471    4.0000
   -0.0909   18.1818  111.5749    5.0000
   -0.0769   15.3846  109.7942    6.0000
   -0.0667   13.3333  108.4883    7.0000
   -0.0588   11.7647  107.4896    8.0000
   -0.0526   10.5263  106.7013    9.0000
   -0.0476    9.5238  106.0630   10.0000