Download
1 / 8
80 likes | 218 Views
Climate Modeling. In-Class Discussion: Legendre Polynomials. Legendre Polynomials 0 - 6 . P 0 (x) = 1 P 1 (x) = x P 2 (x) = (3x 2 - 1)/2 P 3 (x) = (5x 3 - 3x)/2 P 4 (x) = (35x 4 - 30x 2 + 3)/8 P 5 (x) = (63x 5 - 70x 3 + 15x)/8 P 6 (x) = (231x 6 - 315x 4 + 105x 2 - 5)/16.
E N D
Climate Modeling In-Class Discussion: Legendre Polynomials
Legendre Polynomials 0 - 6 P0(x) = 1 P1(x) = x P2(x) = (3x2 - 1)/2 P3(x) = (5x3 - 3x)/2 P4(x) = (35x4 - 30x2 + 3)/8 P5(x) = (63x5 - 70x3 + 15x)/8 P6(x) = (231x6 - 315x4 + 105x2 - 5)/16
Plots: Even Polynomials P0(x) = 1 P2(x) = (3x2 - 1)/2 P4(x) = (35x4 - 30x2 + 3)/8 P6(x) = (231x6 - 315x4 + 105x2 - 5)/16
Plots: Odd Polynomials P1(x) = x P3(x) = (5x3 - 3x)/2 P5(x) = (63x5 - 70x3 + 15x)/8
Basis Functions: Legendre Polynomials (1) Why? Convenient properties on the sphere when using x = sin(lat) Some examples: (a) Even Pn (e.g., above) satisfy boundary conditions 1 & 2 All = 0 at x = 0. All are finite at x = 1.
Basis Functions: Legendre Polynomials (2) Why? Convenient properties on the sphere when using x = sin(lat) (b) Eigenfunctions of this operator on the sphere. Simplifies evaluation of the derivatives (calculus becomes algebra).
Basis Functions: Legendre Polynomials (3) Why? Convenient properties on the sphere when using x = sin(lat) (c) Polynomials of different degrees are orthogonal. NOTE: The integral above is like taking the dot product with vectors: (A1,B1).(A2,B2) = A1A2 + B1B2 = 0 if the vectors are orthogonal The “components” of Pn are its values at each x.
