Difference between revisions of "Laguerre L"
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$$x\dfrac{y^2x}{dx^2}+(1-x)\dfrac{dy}{dx}+ny=0.$$ | $$x\dfrac{y^2x}{dx^2}+(1-x)\dfrac{dy}{dx}+ny=0.$$ | ||
One of the solutions of this differential equations are the Laguerre polynomials | One of the solutions of this differential equations are the Laguerre polynomials | ||
| − | $$L_n(x) = \displaystyle\sum_{k=0}^n (-1)^k \dfrac{n!}{(n-r)!(r!)^2}x^ | + | $$L_n(x) = \displaystyle\sum_{k=0}^n (-1)^k \dfrac{n!}{(n-r)!(r!)^2}x^k.$$ |
The first few Laguerre polynomials are given by | The first few Laguerre polynomials are given by | ||
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
Revision as of 19:16, 4 October 2014
Laguerre's equation is $$x\dfrac{y^2x}{dx^2}+(1-x)\dfrac{dy}{dx}+ny=0.$$ One of the solutions of this differential equations are the Laguerre polynomials $$L_n(x) = \displaystyle\sum_{k=0}^n (-1)^k \dfrac{n!}{(n-r)!(r!)^2}x^k.$$ The first few Laguerre polynomials are given by $$\begin{array}{ll} L_0(x) &= 1 \\ L_1(x) &= -x+1 \\ L_2(x) &= \dfrac{1}{2}(x^2-4x+2) \\ L_3(x) &= \dfrac{1}{6}(-x^3+9x^2-18x+6) \\ L_4(x) &= \dfrac{1}{24}(x^4-16x^3+72x^2-96x+24)\\ \vdots \end{array}$$
Properties
Theorem: The following formula holds: $$\dfrac{e^{\frac{-xt}{1-t}}}{1-t} = \displaystyle\sum_{k=0}^{\infty} L_k(x)t^k.$$
Proof: █
Theorem: The following formula holds: $$L_n(x) = \dfrac{e^x}{n!} \dfrac{d^n}{dx^n} (x^n e^{-x}).$$
Proof: █
