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taylor expanssion calculation result is wrong

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Zeynep Toprak
Zeynep Toprak am 3 Nov. 2020
Kommentiert: Ameer Hamza am 4 Nov. 2020
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?

Akzeptierte Antwort

Ameer Hamza
Ameer Hamza am 3 Nov. 2020
Bearbeitet: Ameer Hamza am 3 Nov. 2020
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 Kommentare
Zeynep Toprak
Zeynep Toprak am 4 Nov. 2020
many thanks dear Hamza. I see my missings.
Ameer Hamza
Ameer Hamza am 4 Nov. 2020
I am glad to be of help!

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Weitere Antworten (1)

Steven Lord
Steven Lord am 3 Nov. 2020
Bearbeitet: Steven Lord am 3 Nov. 2020
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.

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