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133467 p2a3
1.
1.- Lee el
documentoPOLINOMIODENEWTON.pdf ydeterminalasoluciónde lossiguientes sistemasde ecuacionesutilizandoel métodoantesmencionadoa) Buscarel polinomiode Newton cuya funcióntabule: X0 X1 X2 X3 xi 0 1 2 3 yi -1 6 31 18 f(x0) f(x1) F(x2) F(x3) ٨ f(X) f(x1,x0) = f(x1) – f(x0) /x1–x0 = 6 – (-1) /1-0 = 7 f(x2,x1) = f(x2)-f(x1) /x2-x1=31 – 6/2-1= 25 f(x3,x2)=f(x3)-f(x2) /x3-x2= 18-31 / 3-2 = -13 xi yi 0 -1 1 6 f(x1,x0)=7 2 31 f(x2,x1)=25 3 18 f(x3,x2)=-13 ٨ ˆ2f(X) f(x2,x1,x0)=f(x2,x1) –f(x1,x0) /x2 – x0 = 25 – 7 / 2-0 = 9 f(x3,x2,x1) =f(x3,x2) – f(x2,x1) /x3-x1=-13-25 /3-1= -19 xi yi 0 -1 1 6 f(x1,x0)=7 2 31 f(x2,x1)=25 f(x2,x1,x0)=9
2.
3 18 f(x3,x2)=-13
f(x3,x2,x1) =-19 ٨ ˆ3f(X) f(x3,x2,x1,x0) =f(x3,x2,x1) –f(x2,x1,x0) /x3–x0 = -19 – 9 /3-0 = -28/ 3 xi yi 0 -1 1 6 f(x1,x0)=7 2 31 f(x2,x1)=25 f(x2,x1,x0)=9 3 18 f(x3,x2)=-13 f(x3,x2,x1) =-19 f(x3,x2,x1,x0)=-28/3 A0=-1 A1=7 A2=9 A3=-28/3 Sustituyendoen A0+A1(x – x0) + A2 (x-x0)(x-x1) +A3 (x-x0)(x-x1)(x-x2) = -1 + 7(x-0) + 9(x-0)(x-1) –28/3 (x-0)(x-1)(x-2) = -1 + 7x + 9xˆ2 – 9x- 28/3 xˆ3 + 28 xˆ2 – 56/3 x = -1 – 62/3 x + 37𝒙𝟐 – 28/3 𝒙𝟑
3.
(0,-1) (1,6)
4.
(2,31)
5.
(3,18)
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