Not able to solve this expression

I am not able to code this expression , I have tried but it's not giving correct answer ! My try is given below:
if true
sol= zeros(4,4);
s=0;
for i = 1:5
for j = 1:5
r = (j-1)*0.25 ;
fi = 2*pi*(i-1)/4 ;
disp(r)
disp(fi)
disp('r,fi')
for m1 =1:100000
s = s + (2*((r/0.03)^(2*m1))*cos(2*m1*fi)) / ((4*m1*m1 - 1)*(1 + (2*m1*40)/(20*0.03)));
end
sol(i,j)= 1/pi - (1/pi)*s + ((r/0.03)*sin(fi))/(2*(1+ (45/(20*0.03))));
end
end
disp(sol)
end
The r & phi are taken for 5 and 5 points respectively . r = 0, 0,25, 0,50, 0,75 1.0 and simillarly phi = 0, pi/2 , pi , 3pi/2 , 2pi . But my solution is coming very large than the expected value.

4 comentarios

KALYAN ACHARJYA
KALYAN ACHARJYA el 10 de Oct. de 2018
What is your expectation (result)?
Anshuman S
Anshuman S el 10 de Oct. de 2018
It must come in some real number , as Q(Theta) is non dimmensionalised it won't be to much in magnitude, may be less than 100 only . Thus each value coming in range of 1 to 100. ( or something close to that only )
madhan ravi
madhan ravi el 10 de Oct. de 2018
" for m1 =1:100000 aftermath due to this leads to more values "
Anshuman S
Anshuman S el 10 de Oct. de 2018
it's given in the problem to take many number of terms, m1 shows the number of terms in the series, in the problem it's taken to be infinite

Iniciar sesión para comentar.

 Respuesta aceptada

Anshuman S
Anshuman S el 10 de Oct. de 2018

0 votos

I have solved the problem now , The r term was not within the bounds. It should be less than r0 ( 0.03 cm ) . thus r = 0.03*(j-1)*0.25 ;
Thanks for your time and help !

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