why the 0.1 is a high require in f2 line

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JICHAO ZHANG
JICHAO ZHANG 2023-6-20
评论： JICHAO ZHANG 2023-6-20
f = @(r,theta,phi,xi) r.^3 .* sin(theta).^2 .* sin(phi);
f1=@(theta,phi,xi)integral(@(r)f(r,theta,phi,xi),0,2,'ArrayValued',1);
df2=@(theta,phi,xi)cell2mat(arrayfun(@(theta,phi,xi)f1(theta,phi,xi),theta,phi,xi,'UniformOutput',0));
f2=integral3(@(theta,phi,xi)f(theta,phi,xi),0,pi,0,pi,0,2*pi,'AbsTol', 0,'RelTol',0.1);
ret=f2
  1 个评论
Walter Roberson
Walter Roberson 2023-6-20
f = @(r,theta,phi,xi) r.^3 .* sin(theta).^2 .* sin(phi);
So f expects 4 input parameters
f2=integral3(@(theta,phi,xi)f(theta,phi,xi),0,pi,0,pi,0,2*pi,'AbsTol', 0,'RelTol',0.1);
but here f is invoked with three input parameters

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回答（1 个）

Naman
Naman 2023-6-20
Hi Jichao Zhang,
The 'RelTol' parameter in the integral3 function sets the relative error tolerance for the numerical integration.
In the given code, 'RelTol' is set to 0.1, which allows for a maximum relative error of 10% in the numerical integration. This value is relatively high and may not be appropriate for some applications that require high accuracy. However, depending on the specific requirements of the application, a 'RelTol' of 0.1 may be sufficient to achieve reasonable accuracy while still maintaining computational efficiency.
Hope it helps.

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