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matlab 求二元函数最小值

update:I found that if make the result meet the condition"(y^2-x^2)^0.5≥4", there is no complex number left.code is as follows:[x y]=meshgrid(15:0.01:20);a=2*(y.^2-x.^2).^0.5+(pi/2-2*acos(x./y)).*y;b=((y.^2-x.^2).^0.5>=4);[t i]=min...

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