Newton-interpolation

Vad?

Diskret mängd datapunkter

(x_0, y_0), (x_1, y_1), ..., (x_n, y_n)
(x0,y0),(x1,y1),...,(xn,yn)(x_0, y_0), (x_1, y_1), ..., (x_n, y_n)
  • Kvadratiska polynom
  • Spline-interpolation
  • Newton-interpolation
  1. Linjär interpolation
  2. Polynominterpolation
f_{n-1}(x)=b_1+b_2(x-x_1)+b_3(x-x_1)(x-x_2)+ ...+b_n(x-x_1)(x-x_2)...(x-x_{n-1})
fn−1(x)=b1+b2(x−x1)+b3(x−x1)(x−x2)+...+bn(x−x1)(x−x2)...(x−xn−1)f_{n-1}(x)=b_1+b_2(x-x_1)+b_3(x-x_1)(x-x_2)+ ...+b_n(x-x_1)(x-x_2)...(x-x_{n-1})
b_1=f(x_1)
b1=f(x1)b_1=f(x_1)
b_3=f[x_3,x_2,x_1]
b3=f[x3,x2,x1]b_3=f[x_3,x_2,x_1]
b_2=f[x_2,x_1]
b2=f[x2,x1]b_2=f[x_2,x_1]
b_n=f[x_n,x_{n-1},...,x_2,x_1]
bn=f[xn,xn−1,...,x2,x1]b_n=f[x_n,x_{n-1},...,x_2,x_1]

Vad?

f[x_i,x_j]=\frac{f(x_i)-f(x_j)}{x_i-x_j}
f[xi,xj]=f(xi)−f(xj)xi−xjf[x_i,x_j]=\frac{f(x_i)-f(x_j)}{x_i-x_j}
f[x_i,x_j,x_k]=\frac{f[x_i,x_j]-f[x_j,x_k]}{x_i-x_k}
f[xi,xj,xk]=f[xi,xj]−f[xj,xk]xi−xkf[x_i,x_j,x_k]=\frac{f[x_i,x_j]-f[x_j,x_k]}{x_i-x_k}
f[x_n,x_{n-1},...,x_2,x_1]=\frac{f[x_n,x_{n-1},...,x_2]-f[x_{n-1},x_{n-2},...,x_1]}{x_n-x_1}
f[xn,xn−1,...,x2,x1]=f[xn,xn−1,...,x2]−f[xn−1,xn−2,...,x1]xn−x1f[x_n,x_{n-1},...,x_2,x_1]=\frac{f[x_n,x_{n-1},...,x_2]-f[x_{n-1},x_{n-2},...,x_1]}{x_n-x_1}
f_{n-1}(x)=f(x_1)+(x-x_1)f[x_2,x_1]+(x-x_1)(x-x_2)f[x_3,x_2,x_1]
fn−1(x)=f(x1)+(x−x1)f[x2,x1]+(x−x1)(x−x2)f[x3,x2,x1]f_{n-1}(x)=f(x_1)+(x-x_1)f[x_2,x_1]+(x-x_1)(x-x_2)f[x_3,x_2,x_1]
+...+(x-x_1)(x-x_2)...(x-x_{n-1})f[x_n,x_{n-1},...,x_2,x_1]
+...+(x−x1)(x−x2)...(x−xn−1)f[xn,xn−1,...,x2,x1]+...+(x-x_1)(x-x_2)...(x-x_{n-1})f[x_n,x_{n-1},...,x_2,x_1]

Image source: https://en.wikipedia.org/wiki/Newton_polynomial

 
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Hur?

Hur?

f(x) = 2x^2
f(x)=2x2f(x) = 2x^2
f(x) = sin(x)
f(x)=sin(x)f(x) = sin(x)
f(x) = \frac{1}{1 + 25x^2}
f(x)=11+25x2f(x) = \frac{1}{1 + 25x^2}

Tack!

Newton-interpolering

By Jakob von Essen

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Newton-interpolering

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