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 件のコメント

KALYAN ACHARJYA
KALYAN ACHARJYA 2018 年 10 月 10 日
What is your expectation (result)?
Anshuman S
Anshuman S 2018 年 10 月 10 日
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 2018 年 10 月 10 日
" for m1 =1:100000 aftermath due to this leads to more values "
Anshuman S
Anshuman S 2018 年 10 月 10 日
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

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 採用された回答

Anshuman S
Anshuman S 2018 年 10 月 10 日

0 投票

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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