why the fourier series coefficient answer in matlab different compare to my manually calculation

i was use matlab to slove the fourier coefficient, why the answer come out different compare to my manually caculation(1/pi)*((-cos(0)/n)+(cos(n*-pi)/n^2)+(cos(n*pi)/n^2)-(cos(0)/n^2)).

sin pi*n should be get zero right?

syms n t pi

f1=(-t)

f2=(t)

a0=(1/(2*pi))*(int(f1,t,-pi,0)+int(f2,t,0,pi))

an1=int(f1*cos(n*t),t,-pi,0);

a1=subs(an1,sin(n*pi),0);

an2=int(f2*cos(n*t),t,0,pi);

a2=subs(an2,sin(n*pi),0);

an=(1/pi)*((an1+an2))

-(2*(2*sin((pi*n)/2)^2 - n*pi*sin(pi*n)))/(n^2*pi) (matlab answer)

[John D'Errico]

Is n an integer? Given your statement that sin(n*pi) should be an integer, can we infer that n is an integer?

Should a computer know that, without being told? I see only that n is some symbolic parameter, in terms of what you tell MATLAB.

[jason tan]

alright tq Mr john

Answer 1

[Ameer Hamza]

As pointed by John in his comment. You need to specify that the symbolic variable n is integer explicitly. You assume() to inform MATLAB that n is an integer and then use simplify() to apply the assumption. Also, don't directly define pi as a symbolic variable. pi is built-in constant.

syms n t
PI = sym(pi);
 
f1=(-t);
f2=(t);
 
a0=(1/(2*PI))*(int(f1,t,-PI,0)+int(f2,t,0,PI));
an1=int(f1*cos(n*t),t,-PI,0);
a1=subs(an1,sin(n*PI),0);
an2=int(f2*cos(n*t),t,0,PI);
a2=subs(an2,sin(n*PI),0);
an = (1/PI)*((an1+an2));
assume(n/2, 'integer');
simplify(an)
assume((n+1)/2, 'integer');
simplify(an)

Result

ans =
0
 
ans =
-4/(n^2*pi)

[jason tan]

why an need to minus (2*(2*sin((PI*n)/2)^2 - n*PI*sin(PI*n)))/(n^2*PI)?

[Ameer Hamza]

Sorry! You don't need to minus that. I think that I accidentally pasted it with your code. Check the updated answer.

[jason tan]

but the answer in matlab still different compare to my manually calculation.It is something wrong in my coding?

[Ameer Hamza]

In the original question, you didn't specify the assumption about even and odd. Check the updated code.

[jason tan]

ya,mistake on that.

Tq Ameer Hamza

[Ameer Hamza]

Glad to be of help.