Math Problems Vault

Problem 5578

Prove the identity:

t\frac{e^{tz}-1}{e^{t}-1} = \sum _{n=1}^{\infty } \frac{\varphi _{n}(z)}{n!}t^{n}.

Here $\varphi_{n}(z)$ are the Bernoulli polynomials, for which

\varphi _{n}(z+1)-\varphi _{n}(z) = nz^{n-1}.

For integer $z > 0$ we have

\frac{1}{n}\varphi _{n}(z) = 1^{n-1}+2^{n-1}+\ldots +(z-1)^{n-1}.

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Remark.

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A function $\Phi (z,t)$ is called generating for the functions $\varphi_{n}(z)$ if

\Phi (z,t) = \sum _{n=0}^{\infty } \varphi _{n}(z)t^{n}.

If $\Phi (z,t)$ is comparatively simple, then from its properties one can derive a range of properties of the functions $\varphi_{n}(z)$ .

Below are given examples of generating functions for several important classes of polynomials, together with the principal properties of these polynomials that are to be proved using generating functions, as well as by other means.

Problem 5579

Prove that the generating function

e^{-tz-\frac{t^{2}}{2}} = \sum _{n=0}^{\infty } H_{n}(z)t^{n}

determines the Chebyshev--Hermite polynomials $H_{n}(z)$ , whose principal properties are expressed by the identities:

H_{n}(z) = \frac{1}{n!}e^{\frac{z^{2}}{2}}\frac{d^{n}e^{-\frac{z^{2}}{2}}}{dz^{n}}; \quad H_{n}'(z) = -H_{n-1}(z);

nH_{n}(z)+zH_{n-1}(z)+H_{n-2}(z) = 0;

H_{n}''(z)-zH_{n}'(z)+nH_{n}(z) = 0.

\int _{-\infty }^{\infty } e^{-\frac{z^{2}}{2}}H_{n}(z)H_{m}(z)\, dz = \begin{cases} 0, & \text{if } m \neq n; \\ \dfrac {\sqrt{2\pi }}{m!}, & \text{if } m = n. \end{cases}

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Problem 5580

Prove that the generating function

\frac{e^{-\frac{zt}{1-t}}}{(1-t)^{a+1}} = \sum _{n=0}^{\infty } L_{n}^{(a)}(z)t^{n}

determines the generalized Chebyshev--Laguerre polynomials $L_{n}^{(a)}(z)$ with the properties:

L_{n}^{(a)}(z) = \frac{1}{n!}e^{z}z^{-a}\frac{d^{n}z^{n+a}e^{-z}}{dz^{n}},

nL_{n}^{(a)}(z)+(z-2n-a+1)L_{n-1}^{(a)}(z)+(n+a-1)L_{n-2}^{(a)}(z) = 0,

z\left(L_{n}^{(a)}(z)\right)''+(a+1-z)\left(L_{n}^{(a)}(z)\right)'+nL_{n}^{(a)}(z) = 0,

\int _{0}^{\infty } z^{a}e^{-z}L_{n}^{(a)}(z)L_{m}^{(a)}(z)\, dz = \begin{cases} 0 & \text{for } m \neq n; \\ \dfrac {\Gamma (a+n+1)}{n!} & \text{for } m = n. \end{cases}

For $a = 0$ these become the ordinary Laguerre polynomials.

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Problem 5581

Prove that the generating function

\frac{2^{\alpha +\beta }}{r}(1-t+r)^{-\alpha }(1+t+r)^{-\beta } = \sum _{n=0}^{\infty } P_{n}^{\alpha ,\beta }(x)t^{n}; \qquad r = \sqrt{1-2xt+t^{2}}

determines the Jacobi polynomials $P_{n}^{\alpha ,\beta }$ with the properties:

(1-x)^{\alpha }(1+x)^{\beta }P_{n}^{\alpha ,\beta }(x) = \frac{(-1)^{n}}{2^{n}\cdot n!}\frac{d^{n}(1-x)^{n+\alpha }(1+x)^{n+\beta }}{dx^{n}}.

\int _{-1}^{1} (1-x)^{\alpha }(1+x)^{\beta }P_{n}^{\alpha ,\beta }(x)P_{m}^{\alpha ,\beta }(x)\, dx = 0, \text{ if } m \neq n.

\int _{-1}^{1} (1-x)^{\alpha }(1+x)^{\beta }\left(P_{n}^{\alpha ,\beta }(x)\right)^{2}\, dx = \frac{2^{\alpha +\beta +1}}{\alpha +\beta +2n+1}\cdot \frac{\Gamma (\alpha +n+1)\Gamma (\beta +n+1)}{\Gamma (n+1)\Gamma (\alpha +\beta +n+1)};

(n+1)(n+\alpha +\beta +1)(2n+\alpha +\beta )P_{n+1}^{\alpha ,\beta }(x)-

-\frac{1}{2}(2n+\alpha +\beta +1)\left((2n+\alpha +\beta +2)(2n+\alpha +\beta )x+(\alpha ^{2}-\beta ^{2})\right)P_{n}^{\alpha ,\beta }(x)+

+(n+\alpha )(n+\beta )(2n+\alpha +\beta +2)\times P_{n-1}^{\alpha ,\beta }(x) = 0.

(1-x^{2})y''+\left(\beta -\alpha -(\alpha +\beta +2)x\right)y'+n(n+\alpha +\beta +1)y = 0; \qquad y = P_{n}^{\alpha ,\beta }(x).

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Problem 5582

Prove that the generating function

\frac{1}{\sqrt{1-2tx+t^{2}}} = \sum _{n=0}^{\infty } P_{n}(x)t^{n}

determines the Legendre polynomials $P_{n}(x)$ , which are obtained from the Jacobi polynomials (problem 5581) at $\alpha = \beta = 0$ .

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Problem 5583

Prove that the generating function

\frac{4-t^{2}}{4-4tx+t^{2}} = \sum _{n=0}^{\infty } T_{n}(x)t^{n}

determines the Chebyshev polynomials $T_{n}(x)$ :

T_{0}(x) = 1, \quad T_{n}(x) = \frac{1}{2^{n-1}}\cos (n\arccos x) \quad (n = 1,2,\ldots ).

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Problem 5584

Prove the identity

\frac{1}{1-2tx-t^{2}} = \sum _{n=0}^{\infty } u_{n}(x)t^{n},

where

u_{n}(x) = \frac{(x+\sqrt{x^{2}+1})^{n+1}-(x-\sqrt{x^{2}+1})^{n+1}}{2\sqrt{1+x^{2}}}.

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Problem 5585

Using the expansion $(1-2xt+t^{2})^{-1/2} = \sum_{\nu =0}^{\infty } P_{\nu }(x)t^{-\nu -1}$ , prove for the Legendre polynomials (problem 5582) the identity:

P_{n}(\cos \theta ) = \frac{1}{2\pi i}\int _{C} \frac{t^{n}}{\sqrt{1-2t\cos \theta +t^{2}}}\, dt,

where $C$ is a closed contour inside which lie both singular points of the integrand.

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Problem 5586

From the preceding (problem 5585), obtain the Dirichlet--Mehler formula:

P_{n}(\cos \theta ) = \frac{2}{\pi }\int _{0}^{\theta } \frac{\cos \left(n+\frac{1}{2}\right)t}{\sqrt{2(\cos t-\cos \theta )}}\, dt.

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Problem 5587

Using the expansion

(1-2xt+t^{2})^{-\frac{1}{2}} = \sum _{\nu =0}^{\infty } P_{\nu }(x)t^{-\nu -1},

obtain Laplace's formula (for the Legendre polynomials, problem 5582):

P_{n}(z) = \frac{1}{\pi }\int _{-1}^{1} \left(z+\sigma \sqrt{z^{2}-1}\right)^{n}\frac{d\sigma }{\sqrt{1-\sigma ^{2}}}.

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Problem 5588

Applying the generating function (problem 5582) and the identity

1-2t\cos \theta +t^{2} = (1-te^{\theta i})(1-te^{-\theta i}),

prove the identity:

P_{n}(\cos \theta ) = \frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots 2n}\times

\times \left(2\cos n\theta +\frac{1}{2}\cdot \frac{2n}{2n-1}\cdot 2\cos (n-2)\theta +\frac{1\cdot 3}{2\cdot 4}\cdot \frac{2n(2n-2)}{(2n-1)(2n-3)}\cdot 2\cos (n-4)\theta +\ldots \right).

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