Math 2.0

Special functions Definition

Shifted Legendre polynomials

Builds on Legendre polynomials

Definition

Let PnP_n denote the Legendre polynomial of degree n≥0n \ge 0. The shifted Legendre polynomial of degree nn is the polynomial P~n(x)=Pn(2x−1).\tilde{P}_n(x) = P_n(2x - 1). Equivalently, P~n(x)=(−1)n∑k=0n(nk)(n+kk)(−x)k=1n!dndxn(x2−x)n.\tilde{P}_n(x) = (-1)^n \sum_{k=0}^{n} \binom{n}{k} \binom{n+k}{k} (-x)^k = \frac{1}{n!} \frac{d^n}{dx^n} \left( x^2 - x \right)^n.

Remarks

The first ones are P~0=1\tilde{P}_0 = 1, P~1=2x−1\tilde{P}_1 = 2x - 1 and P~2=6x2−6x+1\tilde{P}_2 = 6x^2 - 6x + 1. Check for P~2\tilde{P}_2: with P2(t)=(3t2−1)/2P_2(t) = (3t^2 - 1)/2 and t=2x−1t = 2x - 1, one gets (12x2−12x+2)/2=6x2−6x+1(12x^2 - 12x + 2)/2 = 6x^2 - 6x + 1. They are orthogonal on [0,1][0, 1]: ∫01P~m(x)P~n(x) dx=δmn2n+1\int_0^1 \tilde{P}_m(x) \tilde{P}_n(x) \, dx = \frac{\delta_{mn}}{2n + 1}. For instance ∫01(2x−1) dx=0\int_0^1 (2x - 1) \, dx = 0. Non-example: the unshifted polynomials are not orthogonal on [0,1][0, 1], since ∫01P0(x)P1(x) dx=∫01x dx=12≠0\int_0^1 P_0(x) P_1(x) \, dx = \int_0^1 x \, dx = \frac12 \ne 0. Some authors, and some software libraries, use the reversed convention Pn(1−2x)=(−1)nP~n(x)P_n(1 - 2x) = (-1)^n \tilde{P}_n(x).

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Further reading

  1. Web https://en.wikipedia.org/wiki/Legendre_polynomials
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DefinesShifted Legendre polynomials
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