taylor expanssion calculation result is wrong
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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?
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Ameer Hamza
2020년 11월 3일
편집: Ameer Hamza
2020년 11월 3일
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
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추가 답변 (1개)
Steven Lord
2020년 11월 3일
편집: Steven Lord
2020년 11월 3일
I want to walk through your code and comment on a few places.
function [SN] = sintaylorfunction(A, tol)
k = 0;
error = 0 ;
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.
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