Laguerre polynomials

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Prove that the system of Laguerre polynomials

Ln(t)=etn!dndtn(tne−t),n=0,1,…, L_{n}(t) = \frac{e^{t}}{n!} \frac{d^{n}}{d t^{n}}\left(t^{n} e^{-t}\right), \quad n = 0,1, \ldots ,

is an orthonormal basis in the space L2((0,∞),e−tdt)L_{2}\left((0, \infty ), e^{-t} d t\right).

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If the sequence e−x/2,xe−x/2,x2e−x/2,…e^{-x/2}, xe^{-x/2}, x^2 e^{-x/2}, \ldots is orthogonalized in L2(0,∞)L^2(0, \infty ), the sequence en(x)=e−x/2Ln(x)/n!e_n(x) = e^{-x/2} L_n(x)/n! is obtained, where

Ln(x)=ex(ddx)n(xne−x). L_n(x) = e^x \left(\frac{d}{dx}\right)^n (x^n e^{-x}).

The functions LnL_n are called Laguerre polynomials.

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The Chebyshev--Laguerre polynomials are defined as follows:

Lm(x)=ex(xme−x)(m) (m=0,1,2,⋯ ). L_{m}(x)=e^{x}(x^{m}e^{-x})^{(m)} \ (m=0,1,2,\cdots ) .

Find an explicit expression for the polynomial Lm(x)L_{m}(x).

Prove that Lm(x)L_{m}(x) satisfies the equation

xLm′′(x)+(1−x)Lm′(x)+mLm(x)=0. xL''_{m}(x)+(1-x)L'_{m}(x)+mL_{m}(x)=0 .
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Prove that the Chebyshev--Laguerre polynomial

Ln(x)=exdndxn(xne−x) L_{n}(x)=e^{x}\frac{d^{n}}{dx^{n}}(x^{n}e^{-x})

has all its roots positive.

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Find the generating functions of the polynomials Pn(x)P_{n}(x), Pn(α,β)(x)P_{n}^{(\alpha ,\beta )}(x), Ln(α)(x)L_{n}^{(\alpha )}(x), defined by the formulas

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(1)

Pn(x)=12nn!dndxn(x2−1)nP_{n}(x) = \dfrac {1}{2^{n}n!}\dfrac {d^{n}}{dx^{n}}(x^{2}-1)^{n} (Legendre's polynomials);

(2)

(1−x)α(1+x)βPn(α,β)(x)=(−1)n2nn!dndxn(1−x)n+α(1+x)n+β(1-x)^{\alpha }(1+x)^{\beta }P_{n}^{(\alpha ,\beta )}(x) = \dfrac {(-1)^{n}}{2^{n}n!}\dfrac {d^{n}}{dx^{n}}(1-x)^{n+\alpha }(1+x)^{n+\beta }, α>−1\alpha >-1, β>−1\beta >-1 (Jacobi's polynomials);

(3)

e−xxαLn(α)(x)=1n!dndxne−xxn+αe^{-x}x^{\alpha }L_{n}^{(\alpha )}(x) = \dfrac {1}{n!}\dfrac {d^{n}}{dx^{n}}e^{-x}x^{n+\alpha }, α>−1\alpha >-1 (generalized Laguerre's polynomials).

Remark.
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Cf.Part VI, problems 84, 98, 99. The generating function of the Legendre polynomials is the series

P0(x)+P1(x)w+P2(x)w2+⋯+Pn(x)wn+⋯=1+xw+3x2−12w2+⋯ , P_{0}(x) + P_{1}(x)w + P_{2}(x)w^{2} + \cdots + P_{n}(x)w^{n} + \cdots = 1 + xw + \frac{3x^{2}-1}{2}w^{2} + \cdots ,

the sum of which has to be found as a function of xx and ww; similarly in the other two cases.

One may define the Legendre, the Laguerre and the Hermite polynomials by the formulae

Pn(x)=12nn!dndxn(x2−1)n,e−xLn(x)=1n!dndxne−xxn,e−x22Hn(x)=1n!dndxne−x22 P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n, \qquad e^{-x} L_n(x) = \frac{1}{n!} \frac{d^n}{dx^n} e^{-x} x^n, \qquad e^{-\frac{x^2}{2}} H_n(x) = \frac{1}{n!} \frac{d^n}{dx^n} e^{-\frac{x^2}{2}}

respectively (problems 84, 99, 100 of Part Six). Deduce from this definition that these polynomials have only real and simple zeros, which lie in the interior of the intervals

(−1,+1),(0,+∞),(−∞,+∞) (-1,+1), \qquad (0,+\infty ), \qquad (-\infty ,+\infty )

respectively (problems 97, 99, 100 of Part Six).

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Prove the analogues of problems 84, 85, 87--91, 97 for the generalized Laguerre polynomials

L0(α)(x),L1(α)(x),L2(α)(x),…,Ln(α)(x),…,α>−1, L_0^{(\alpha )}(x), \quad L_1^{(\alpha )}(x), \quad L_2^{(\alpha )}(x), \quad \ldots , \quad L_n^{(\alpha )}(x), \ldots , \qquad \alpha > -1,

defined by the following conditions: (1), (2) Ln(α)(x)L_n^{(\alpha )}(x) is of degree nn, has real coefficients and orthogonality property

\int _0^{\infty } e^{-x}x^\alpha L_m^{(\alpha )}(x)L_n^{(\alpha )}(x)\, dx = \begin{cases} 0, & \text{for } m \neq n, \\ ]4pt] \Gamma (\alpha +1)\dbinom {n+\alpha }{n}, & \text{for } m=n; \quad m,n=0,1,2,\ldots ; \end{cases} \[ (3) the coefficient of $x^n$ in $L_n^{(\alpha )}(x)$ has the sign $(-1)^n$. \]
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If Pn(α,β)(x)P_n^{(\alpha ,\beta )}(x) and Ln(α)(x)L_n^{(\alpha )}(x) are the functions defined in problems 98 and 99, respectively, then we have

lim⁡β→+∞Pn(α,β)(1−ε)=Ln(α)(x), \lim _{\beta \to +\infty } P_n^{(\alpha ,\beta )}(1-\varepsilon ) = L_n^{(\alpha )}(x),

if ε\varepsilon converges to 00 as β\beta increases in such a manner that

lim⁡β→+∞εβ=2x. \lim _{\beta \to +\infty } \varepsilon \beta = 2x.
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If Ln(α)(x)L_n^{(\alpha )}(x) and Hn(x)H_n(x) are the functions defined in problems 99 and 100, respectively, then we have

H2q(x)=(−1)q1⋅3⋅5⋯(2q−1)Lq(−1/2)(x22), H_{2q}(x) = \frac{(-1)^q}{1\cdot 3\cdot 5\cdots (2q-1)}L_q^{(-1/2)}\left(\frac{x^2}{2}\right), H2q+1(x)=(−1)q+11⋅3⋅5⋯(2q+1)xLq(1/2)(x22),q=0,1,2,…. H_{2q+1}(x) = \frac{(-1)^{q+1}}{1\cdot 3\cdot 5\cdots (2q+1)}xL_q^{(1/2)}\left(\frac{x^2}{2}\right), \qquad q=0,1,2,\ldots .
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Let α\alpha be a constant, α>−1\alpha >-1. Determine the maximum of ∣P(0)∣\left|P(0)\right| for all polynomials P(x)P(x) of nnth degree which satisfy the condition

∫0∞e−xxα[P(x)]2 dx=1. \int _0^{\infty } e^{-x}x^\alpha [P(x)]^2\, dx = 1.

How does this maximum behave for large values of nn?

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Compute

min⁡a,b,c∫0∞∣x3−a−bx−cx2∣2e−xdx \min _{a, b, c} \int _{0}^{\infty }\left|x^{3}-a-b x-c x^{2}\right|^{2} e^{-x} d x

State and solve the corresponding maximum problem, as in Exercise 14.

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References

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[Con07] Conway J.B. A Course in Functional Analysis. 2nd ed., Springer, 2007.
[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.