L04_Solving_Cubic_Polynomials

L04_Solving_Cubic_Polynomials - CE507 Lecture 4 I The Cubic...

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CE507 Lecture 4 Numerical Solution of the Cubic Polynomial I. The Cubic Polynomial From [ ] A vv λ = ±± Using 0 AI −= to solve for 11 12 13 21 22 23 31 32 33 0 aa a a a −− = () 22 23 21 23 21 22 11 12 13 32 33 31 33 31 32 0 a a a a a λλ −+ = ( )( ) 11 22 33 23 32 12 21 33 13 32 21 13 31 22 0 a a a a a a a a a a a a ⎡⎤ = ⎣⎦ Simplifying 32 12 3 11 1 2 2 3 3 2 21 3 0 1 2 det ii ij ji ij II I I aaa a a a Ia +− = ==+ + =− = I.1 Example 211 234 112 A ⎛⎞ ⎜⎟ = ⎝⎠ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 2323 1 9 2 13 34 1 1 1 1 4 2 2 1 2 24 23 I I I =+−= = −×+×+× −+×+×+× ×+ × − = + = so 33 0 + = with factors ( )( )( ) 113 0 λλλ + =
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II. Cardan’s formula for the Roots of Cubic Polynomials To solve 32 12 3 0 xI x I −+ = Let 1 3 I x ξ =− or 1 3 I x =+ 3 2 11 1 1 2 2 22 1 3 33 2 2 1 1 1 23 2 1 2 3 9 2 9 39 3 9 2 7 7 I II I x I x I III I x ξξ ⎛⎞ + + ⎜⎟ ⎝⎠ + + + + + + + The cubic polynomial is transformed to; 2 2 1 2 0 2 7 I I +− −− + = Define 2 3 2 3 3 9 29 2 7 54 p rp I I q = = = In fact 3 0 pq = With 1 3/2 cos q p θ = , the three roots are
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1 2 3 2c o s 3 2 o s 3 4 o s 3 r r r θ ξ π ⎛⎞ = ⎜⎟ ⎝⎠ + = + =
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This note was uploaded on 02/28/2008 for the course CE 507 taught by Professor Lee during the Fall '07 term at USC.

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L04_Solving_Cubic_Polynomials - CE507 Lecture 4 I The Cubic...

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