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Question: RE: eliminating terms caused by precision limit

Question:RE: eliminating terms caused by precision limit

pik1432 245 Maple
simplification floating-point
July 27 2020
2

Hello there, 

I wonder if there is any method to eliminate the terms ((6.*10^(-11))*I) in this worksheet. Apparently, they came from the limitation of the numerical precision. 


 

> 

restart;
> 

M2[transform] := <<2/(sqrt(3)*a),-1/(sqrt(3)*a),-1/(sqrt(3)*a)> | <0, -1/a, 1/a>>;

Matrix(3, 2, {(1, 1) = (2/3)*sqrt(3)/a, (1, 2) = 0, (2, 1) = -(1/3)*sqrt(3)/a, (2, 2) = -1/a, (3, 1) = -(1/3)*sqrt(3)/a, (3, 2) = 1/a})

 (1)
> 

Iload := <i[T]*exp(-I*Phi[T])|I*i[M]*exp(-I*Phi[M])>;

Vector[row](2, {(1) = i[T]*exp(-I*Phi[T]), (2) = I*i[M]*exp(-I*Phi[M])})

 (2)
> 

alpha := exp(I*convert(120*degrees, radians));

-1/2+((1/2)*I)*3^(1/2)

 (3)
> 

ABC2SEQ :=  1/3 * <<1,1,1> | <alpha, alpha^2, 1> | <alpha^2, alpha, 1>>;

Matrix(3, 3, {(1, 1) = 1/3, (1, 2) = -1/6+((1/6)*I)*sqrt(3), (1, 3) = (1/3)*(-1/2+((1/2)*I)*sqrt(3))^2, (2, 1) = 1/3, (2, 2) = (1/3)*(-1/2+((1/2)*I)*sqrt(3))^2, (2, 3) = -1/6+((1/6)*I)*sqrt(3), (3, 1) = 1/3, (3, 2) = 1/3, (3, 3) = 1/3})

 (4)
> 

Iabc := M2[transform] . (Iload^(%T));

Vector(3, {(1) = (2/3)*sqrt(3)*i[T]*exp(-I*Phi[T])/a, (2) = -(1/3)*sqrt(3)*i[T]*exp(-I*Phi[T])/a-I*i[M]*exp(-I*Phi[M])/a, (3) = -(1/3)*sqrt(3)*i[T]*exp(-I*Phi[T])/a+I*i[M]*exp(-I*Phi[M])/a})

 (5)
> 

Iabcx := subs[eval]({a=1.0} , Iabc);

Vector(3, {(1) = .6666666667*sqrt(3)*i[T]*exp(-I*Phi[T]), (2) = -.3333333333*sqrt(3)*i[T]*exp(-I*Phi[T])-(1.000000000*I)*i[M]*exp(-I*Phi[M]), (3) = -.3333333333*sqrt(3)*i[T]*exp(-I*Phi[T])+(1.000000000*I)*i[M]*exp(-I*Phi[M])})

 (6)
> 

Ipnz := ABC2SEQ . Iabcx:
> 

simplify( Ipnz );

Vector(3, {(1) = (.5773502693-0.6000000000e-10*I)*i[T]*exp(-I*Phi[T])+.5773502694*i[M]*exp(-I*Phi[M]), (2) = (.5773502693-0.6000000000e-10*I)*i[T]*exp(-I*Phi[T])-.5773502694*i[M]*exp(-I*Phi[M]), (3) = 0.})

 (7)
> 

 


 

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