taylor expanssion calculation result is wrong

2020 11 3
2 个回答
回答已采纳
更新时间：2020 11 4
2 次查看（30 天）

0 个投票

function [SN] = sintaylorfunction(A, tol)
k = 0;
error = 0 ;
sin = 0;
SN = sin;
sinplus1 = 0;
SM = sinplus1;
while E <= tol
sin = (-1)^k * A^(2*k+1)/factorial(2*k+1);
SN = SN + sin;
sinplus1 = (-1)^(k+1) * A^(2*(k+1)+1)/factorial(2*(k+1)+1);
SM = SM + sinplus1;
error= abs( max(SM - SN));
end
end
I wrote the sin function in this way. When I calculate sin(pi/2), I get sin(pi/2) = 1.5708. However, we know that sin(pi/2) = 1. Where is my mistake in my function?

请先登录，再回答此问题。

 采纳的回答

Ameer Hamza
Ameer Hamza 2020-11-3
编辑：Ameer Hamza 2020-11-3

0 个投票

kou are not incrementing 'k' inside the while-loop. Also, the while-condition is incorrect. Try the following code
sintaylorfunction(pi/2, 0.01)
function [SN] = sintaylorfunction(A, tol)
k = 0;
error = inf;
sin = 0;
SN = sin;
sinplus1 = 0;
SM = sinplus1;
while error >= tol
sin = (-1)^k * A^(2*k+1)/factorial(2*k+1);
SN = SN + sin;
sinplus1 = (-1)^(k+1) * A^(2*(k+1)+1)/factorial(2*(k+1)+1);
SM = SM + sinplus1;
error= abs( max(SM - SN));
k = k+1;
end
end

6 个评论
显示 4更早的评论 隐藏 4更早的评论

Zeynep Toprak
Zeynep Toprak 2020-11-3
thank very much!! I didnt know the increment term. It helps very much. Well, my another question is that will I use the same things for cos expansion in the same while loop except for its fomula ?
Ameer Hamza
Ameer Hamza 2020-11-3
Yes, the formula for taylor series of cos is a bit different as shown in your question too. The method will be same.
Zeynep Toprak
Zeynep Toprak 2020-11-3
sintaylorfunction(pi/2, 0.01)
function [SN CN] = sintaylorfunction(A, tol)
k = 0;
error = inf;
sin = 0;
SN = sin;
sinplus1 = 0;
SM = sinplus1;
while error >= tol
sin = (-1)^k * A^(2*k+1)/factorial(2*k+1);
SN = SN + sin;
sinplus1 = (-1)^(k+1) * A^(2*(k+1)+1)/factorial(2*(k+1)+1);
SM = SM + sinplus1;
error= abs( max(SM - SN));
k = k+1;
end
n=0;
error_cos=inf;
cos=1;
CN=cos;
cosplus1 = 1;
CM = cosplus1;
while error_cos >=tol
cos = (-1)^n * A^(2*n)/factorial(2*n);
CN = CN + cos;
cosplus1 = (-1)^(n+1) * A^(2*(n+1))/factorial(2*(n+1));
CM = CM + cosplus1;
error= abs( max(CM - CN));
n
= n+1;
end
end
again I get wrong result for cos. I cannot find my mistake.
Ameer Hamza
Ameer Hamza 2020-11-4
I have simplified your code a bit. Following will work
[q, w] = sintaylorfunction(pi/2, 0.01)
function [SN CN] = sintaylorfunction(A, tol)
k = 0;
error = inf;
SN = 0;
while error >= tol
sin = (-1)^k * A^(2*k+1)/factorial(2*k+1);
SN = SN + sin;
sinplus1 = (-1)^(k+1) * A^(2*(k+1)+1)/factorial(2*(k+1)+1);
error= abs(sinplus1-sin);
k = k+1;
end
n=0;
error_cos=inf;
CN=0;
while error_cos >=tol
cos = (-1)^n * A^(2*n)/factorial(2*n);
CN = CN + cos;
cosplus1 = (-1)^(n+1) * A^(2*(n+1))/factorial(2*(n+1));
error_cos= abs(cosplus1-cos);
n = n+1;
end
end
Zeynep Toprak
Zeynep Toprak 2020-11-4
many thanks dear Hamza. I see my missings.

请先登录，再进行评论。

更多回答（1 个）

Steven Lord
Steven Lord 2020-11-3
编辑：Steven Lord 2020-11-3

0 个投票

I want to walk through your code and comment on a few places.
function [SN] = sintaylorfunction(A, tol)
k = 0;
error = 0 ;
Don't create a variable named error. That identifier already has a meaning in MATLAB.
sin = 0;
The identifier sin also already has a meaning in MATLAB, so I recommend you choose a different variable name.
SN = sin;
sinplus1 = 0;
SM = sinplus1;
while E <= tol
The variable E doesn't exist.
sin = (-1)^k * A^(2*k+1)/factorial(2*k+1);
From the role this plays, a better name might be term.
SN = SN + sin;
sinplus1 = (-1)^(k+1) * A^(2*(k+1)+1)/factorial(2*(k+1)+1);
SM = SM + sinplus1;
error= abs( max(SM - SN));
In this loop you change neither the (nonexistent) variable E nor the variable tol. So if this were to enter the loop, you'd never exit.
FYI, you might want to check your answer using the funm function to compute the matrix cosine and matrix sine.

类别

在 帮助中心 和 File Exchange 中查找有关 Operators and Elementary Operations 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by