Precise Implementation of Cad Primitives Using Rational Parametrizations of Stan
Precise Implementation of Cad Primitives Using Rational Parametrizations of Stan
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To quickly assess the difficulty of the text, read a short excerpt:
, ej_i respectively. Then EBF. Omits one row and one column of B^, and hence its rank is either r-1 or r-2.
-17- Suppose first that the rank of EBE is r-1, i. E. Identical to the rank of EB, so that these two matrices have the same range. Then there exists a u = Eu such that EB(e^^^^-u) = 0. Plainly ei^+^'u = = (Cei^)«u since Ce^ = e^+[ . Define a linear transformation T on X by putting Tx = X - (e|^+px)u + (Cu«x)e{^ - 1 /2(e^+^' x)(Cu« u)e^.
Then if Tx = 0, we have e^^^p Tx = e^^+px, so Tx = x... + (Cu'x)ei^. Hence = Cu« Tx = Cu« X + (Cu« x)(Cu« e^) = Cu* x, and therefore x = 0, showing that T is nonsingular. Plainly Te^ = e^^ for i = l... K. Moreover FBFej^ = 0, so that (Ty). FBF(Tx) = (y-(e^^^>. Y)u + ((Cu-y) - l/ZCe^^+l* y)(Gu. U) )e, J •FBF(x-(e^^-^px)u + ((Cu-x) - 1 /2(e, . +i. X) (Cu* u) )ei^) = ((e^. +l«y) (e^^+i-u) + Ey )• FBF( (e^+1' x) (e;^+^-u) + Ex) =yEBEx + (e^^py) (e^+i« x) (e^+i - u)« B(e|^+l-u) .
Thus in the basis Tej^, . . . Te^^, the form B has the representation A, ^o' where B^' has one less row and column than B^.
What to read after Precise Implementation of Cad Primitives Using Rational Parametrizations of Stan?
You can find similar books in the "Read Also" column, or choose other free books by S Ocken to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
, ej_i respectively. Then EBF. Omits one row and one column of B^, and hence its rank is either r-1 or r-2.
-17- Suppose first that the rank of EBE is r-1, i. E. Identical to the rank of EB, so that these two matrices have the same range. Then there exists a u = Eu such that EB(e^^^^-u) = 0. Plainly ei^+^'u = = (Cei^)«u since Ce^ = e^+[ . Define a linear transformation T on X by putting Tx = X - (e|^+px)u + (Cu«x)e{^ - 1 /2(e^+^' x)(Cu« u)e^.
Then if Tx = 0, we have e^^^p Tx = e^^+px, so Tx = x... + (Cu'x)ei^. Hence = Cu« Tx = Cu« X + (Cu« x)(Cu« e^) = Cu* x, and therefore x = 0, showing that T is nonsingular. Plainly Te^ = e^^ for i = l... K. Moreover FBFej^ = 0, so that (Ty). FBF(Tx) = (y-(e^^^>. Y)u + ((Cu-y) - l/ZCe^^+l* y)(Gu. U) )e, J •FBF(x-(e^^-^px)u + ((Cu-x) - 1 /2(e, . +i. X) (Cu* u) )ei^) = ((e^. +l«y) (e^^+i-u) + Ey )• FBF( (e^+1' x) (e;^+^-u) + Ex) =yEBEx + (e^^py) (e^+i« x) (e^+i - u)« B(e|^+l-u) .
Thus in the basis Tej^, . . . Te^^, the form B has the representation A, ^o' where B^' has one less row and column than B^.
What to read after Precise Implementation of Cad Primitives Using Rational Parametrizations of Stan?
You can find similar books in the "Read Also" column, or choose other free books by S Ocken to read onlineMoreLess
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