Different answers for the result of integration and summation

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Yuriy Yerin 2018-7-5

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评论： Yuriy Yerin 2018-7-7

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Hello. I'd like to get the result of this expression for k=1 and m=1

I prepared the simple code

function z=self_energy1
t=-0.000689609;
T_c=0.242731;
mu=0.365908;
k=1;
N=-1000:1000;
nn=10^4;
fun1=@(a,q) a.*tanh((a.^2-mu)/(2*T_c)).*log((2*a.^2+2*a*q+q^2-2*mu-1i*2*pi*N*T_c)./(2*a.^2-2*a*q+q^2-2*mu-1i*2*pi*N*T_c))/q-2;
fun2=@(q) t+integral(@(a)fun1(a,q),0,nn,'AbsTol',1e-6,'RelTol',1e-3,'ArrayValued',true);
fun3=@(q,m) T_c*q/k*log((-k^2+2*k*q-q^2+mu+1i*2*pi*N*T_c-1i*(2*m+1)*pi*T_c)./(-k^2-2*k*q-q^2+mu+1i*2*pi*N*T_c-1i*(2*m+1)*pi*T_c));
fun4=@(q,m) fun3(q,m)/fun2(q);
S=@(q,m) sum(fun4(q,m));
y1=integral(@(q)S(q,1),0.0001,nn,'AbsTol',1e-6,'RelTol',1e-3,'ArrayValued',true)
end

but the final result depends the point in the function4, i.e. if I set point fun4=@(q,m) fun3(q,m)./fun2(q) then I have one result and if no point there fun4=@(q,m) fun3(q,m)/fun2(q) I get another answer.

Could you tell me what is the correct syntax for the evaluating this expression? Thank you in a advance for your help.

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Aquatris 2018-7-6

The dot division is element-wise division, meaning i th element of the first matrix is divided by i th element of the second matrix. If there is no dot, it is a matrix division. In your case, I think you need the dot division since you need to evaluate the expression seperatly for each q and N values.

Yuriy Yerin 2018-7-7

Thank you for you response. I know about the meaning of the dot division but I was confused that I got wrong result of this expression obtained earlier by another approach (Fortran). So that's why I decided to ask community in hope that somebody pays attention on my possible mistake. Anyway thanks again.

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