integral (numerical integration) base on r

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shahin hashemi
shahin hashemi on 4 Oct 2020
Answered: Walter Roberson on 4 Oct 2020
dear all
i want to use integral (numerical integration) base on r
for W for the code below :
clc
clear all
m=9;
a=1;
u=[3.8317 7.0156 10.1735 13.3237 16.4706 19.6159 22.7601 25.9037 29.0468 ];
syms r
YY1 = zeros(1,m,'sym');
for j=1:m
yy1 = zeros(1,100,'sym');
yy0 = zeros(1,100,'sym');
for i=1:100
yy0(i)=(((-1)^(i-1))*((((u(j))/a)*r)^(2*(i-1))))/((2^(2*(i-1)))*((factorial(i-1))^2)) ;
yy1(i)=(((-1)^(i-1))/((factorial(i-1))*(factorial(i))))*(((u(j))*r)/2)^(2*(i-1)+1) ;
end
YY1(j)=sum(yy1);
YY0(j)=sum(yy0);
W(j)=((YY1(j)/r)*YY0(j));
end
as it is obvious i have m=9 W that is base on r for diffrent u
i want to use numerical integration base on r in interval of 0 to 1
as i read matlab help for useing integral i should make function but i dont know to make function
i really appreciate if u could help me

Answers (1)

Walter Roberson
Walter Roberson on 4 Oct 2020
a = 1;
u = [3.8317 7.0156 10.1735 13.3237 16.4706 19.6159 22.7601 25.9037 29.0468 ];
m = length(u);
i = (1 : 100) .';
u = reshape(u, 1, 1, []);
Wfun = @(r) sum((((-1).^(i-1))./((factorial(i-1)).*(factorial(i)))).*(((u).*r)/2).^(2*(i-1)+1)) ./r .* ...
sum((((-1).^(i-1)).*((((u)/a).*r).^(2*(i-1))))./((2.^(2*(i-1))).*((factorial(i-1)).^2))) ;
W = squeeze( integral(Wfun, 0, 1, 'arrayvalued', true));
The output will be a vector the same length as u is.

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