Misplaced Pages

Legendre transform (integral transform)

Article snapshot taken from Wikipedia with creative commons attribution-sharealike license. Give it a read and then ask your questions in the chat. We can research this topic together.
This article is about an integral transform using Legendre polynomials. For the involution transform commonly used in classical mechanics and thermodynamics, see Legendre transformation.

In mathematics, Legendre transform is an integral transform named after the mathematician Adrien-Marie Legendre, which uses Legendre polynomials P n ( x ) {\displaystyle P_{n}(x)} as kernels of the transform. Legendre transform is a special case of Jacobi transform.

The Legendre transform of a function f ( x ) {\displaystyle f(x)} is

J n { f ( x ) } = f ~ ( n ) = ∫ − 1 1 P n ( x )   f ( x )   d x {\displaystyle {\mathcal {J}}_{n}\{f(x)\}={\tilde {f}}(n)=\int _{-1}^{1}P_{n}(x)\ f(x)\ dx}

The inverse Legendre transform is given by

J n − 1 { f ~ ( n ) } = f ( x ) = ∑ n = 0 ∞ 2 n + 1 2 f ~ ( n ) P n ( x ) {\displaystyle {\mathcal {J}}_{n}^{-1}\{{\tilde {f}}(n)\}=f(x)=\sum _{n=0}^{\infty }{\frac {2n+1}{2}}{\tilde {f}}(n)P_{n}(x)}

Associated Legendre transform

Associated Legendre transform is defined as

J n , m { f ( x ) } = f ~ ( n , m ) = ∫ − 1 1 ( 1 − x 2 ) − m / 2 P n m ( x )   f ( x )   d x {\displaystyle {\mathcal {J}}_{n,m}\{f(x)\}={\tilde {f}}(n,m)=\int _{-1}^{1}(1-x^{2})^{-m/2}P_{n}^{m}(x)\ f(x)\ dx}

The inverse Legendre transform is given by

J n , m − 1 { f ~ ( n , m ) } = f ( x ) = ∑ n = 0 ∞ 2 n + 1 2 ( n − m ) ! ( n + m ) ! f ~ ( n , m ) ( 1 − x 2 ) m / 2 P n m ( x ) {\displaystyle {\mathcal {J}}_{n,m}^{-1}\{{\tilde {f}}(n,m)\}=f(x)=\sum _{n=0}^{\infty }{\frac {2n+1}{2}}{\frac {(n-m)!}{(n+m)!}}{\tilde {f}}(n,m)(1-x^{2})^{m/2}P_{n}^{m}(x)}

Some Legendre transform pairs

f ( x ) {\displaystyle f(x)\,} f ~ ( n ) {\displaystyle {\tilde {f}}(n)\,}
x n {\displaystyle x^{n}\,} 2 n + 1 ( n ! ) 2 ( 2 n + 1 ) ! {\displaystyle {\frac {2^{n+1}(n!)^{2}}{(2n+1)!}}}
e a x {\displaystyle e^{ax}\,} 2 π a I n + 1 / 2 ( a ) {\displaystyle {\sqrt {\frac {2\pi }{a}}}I_{n+1/2}(a)}
e i a x {\displaystyle e^{iax}\,} 2 π a i n J n + 1 / 2 ( a ) {\displaystyle {\sqrt {\frac {2\pi }{a}}}i^{n}J_{n+1/2}(a)}
x f ( x ) {\displaystyle xf(x)\,} 1 2 n + 1 [ ( n + 1 ) f ~ ( n + 1 ) + n f ~ ( n − 1 ) ] {\displaystyle {\frac {1}{2n+1}}}
( 1 − x 2 ) − 1 / 2 {\displaystyle (1-x^{2})^{-1/2}\,} π P n 2 ( 0 ) {\displaystyle \pi P_{n}^{2}(0)}
[ 2 ( a − x ) ] − 1 {\displaystyle ^{-1}\,} Q n ( a ) {\displaystyle Q_{n}(a)}
( 1 − 2 a x + a 2 ) − 1 / 2 ,   | a | < 1 {\displaystyle (1-2ax+a^{2})^{-1/2},\ |a|<1\,} 2 a n ( 2 n + 1 ) − 1 {\displaystyle 2a^{n}(2n+1)^{-1}}
( 1 − 2 a x + a 2 ) − 3 / 2 ,   | a | < 1 {\displaystyle (1-2ax+a^{2})^{-3/2},\ |a|<1\,} 2 a n ( 1 − a 2 ) − 1 {\displaystyle 2a^{n}(1-a^{2})^{-1}}
∫ 0 a t b − 1 d t ( 1 − 2 x t + t 2 ) 1 / 2 ,   | a | < 1   b > 0 {\displaystyle \int _{0}^{a}{\frac {t^{b-1}\,dt}{(1-2xt+t^{2})^{1/2}}},\ |a|<1\ b>0\,} 2 a n + b ( 2 n + 1 ) ( n + b ) {\displaystyle {\frac {2a^{n+b}}{(2n+1)(n+b)}}}
d d x [ ( 1 − x 2 ) d d x ] f ( x ) {\displaystyle {\frac {d}{dx}}\leftf(x)\,} − n ( n + 1 ) f ~ ( n ) {\displaystyle -n(n+1){\tilde {f}}(n)}
{ d d x [ ( 1 − x 2 ) d d x ] } k f ( x ) {\displaystyle \left\{{\frac {d}{dx}}\left\right\}^{k}f(x)\,} ( − 1 ) k n k ( n + 1 ) k f ~ ( n ) {\displaystyle (-1)^{k}n^{k}(n+1)^{k}{\tilde {f}}(n)}
f ( x ) 4 − d d x [ ( 1 − x 2 ) d d x ] f ( x ) {\displaystyle {\frac {f(x)}{4}}-{\frac {d}{dx}}\leftf(x)\,} ( n + 1 2 ) 2 f ~ ( n ) {\displaystyle \left(n+{\frac {1}{2}}\right)^{2}{\tilde {f}}(n)}
ln ⁡ ( 1 − x ) {\displaystyle \ln(1-x)\,} { 2 ( ln ⁡ 2 − 1 ) , n = 0 − 2 n ( n + 1 ) , n > 0 {\displaystyle {\begin{cases}2(\ln 2-1),&n=0\\-{\frac {2}{n(n+1)}},&n>0\end{cases}}\,}
f ( x ) ∗ g ( x ) {\displaystyle f(x)*g(x)\,} f ~ ( n ) g ~ ( n ) {\displaystyle {\tilde {f}}(n){\tilde {g}}(n)}
∫ − 1 x f ( t ) d t {\displaystyle \int _{-1}^{x}f(t)\,dt\,} { f ~ ( 0 ) − f ~ ( 1 ) , n = 0 f ~ ( n − 1 ) − f ~ ( n + 1 ) 2 n + 1 , n > 1 {\displaystyle {\begin{cases}{\tilde {f}}(0)-{\tilde {f}}(1),&n=0\\{\frac {{\tilde {f}}(n-1)-{\tilde {f}}(n+1)}{2n+1}},&n>1\end{cases}}\,}
d d x g ( x ) ,   g ( x ) = ∫ − 1 x f ( t ) d t {\displaystyle {\frac {d}{dx}}g(x),\ g(x)=\int _{-1}^{x}f(t)\,dt} g ( 1 ) − ∫ − 1 1 g ( x ) d d x P n ( x ) d x {\displaystyle g(1)-\int _{-1}^{1}g(x){\frac {d}{dx}}P_{n}(x)\,dx}

References

  1. Debnath, Lokenath; Dambaru Bhatta (2007). Integral transforms and their applications (2nd ed.). Boca Raton: Chapman & Hall/CRC. ISBN 9781482223576.
  2. Churchill, R. V. (1954). "The Operational Calculus of Legendre Transforms". Journal of Mathematics and Physics. 33 (1–4): 165–178. doi:10.1002/sapm1954331165. hdl:2027.42/113680.
  3. Churchill, R. V., and C. L. Dolph. "Inverse transforms of products of Legendre transforms." Proceedings of the American Mathematical Society 5.1 (1954): 93–100.


Stub icon

This mathematical physics-related article is a stub. You can help Misplaced Pages by expanding it.

Categories: