On Convolution-Type Relations Between Legendre and Chebyshev Polynomials

V. M. Osipov

(Presented by the scientific-methodological seminar of the Department of Engineering and Computational Mathematics).

The Laplace transform of the exponential Legendre and Chebyshev polynomials (“\(e\)”-polynomials) [3] makes it possible to obtain a number of integral relations connecting both “\(e\)”-polynomials of a single index and the ordinary Legendre and Chebyshev polynomials, defined on the interval \((0,1)\) or \((-1,1)\).

Let us first consider the representation for the “\(e\)”-polynomial of Chebyshev of the first kind \(T_n^*(t)\).

Earlier [3] the correspondence was obtained

\[ \frac{e^{-\frac{at}{2}}}{\sqrt{1-e^{-at}}}\,T_n^*(t)\div W(p)\cdot \varphi_n(p), \tag{1} \]

where

\[ W(p)= \frac{\sqrt{\pi}\,\Gamma\!\left(\frac{p}{a}+\frac{1}{2}\right)} {a\Gamma\!\left(\frac{p}{a}\right)\cdot \frac{p}{a}} \div W(t)=\frac{e^{-\frac{at}{2}}}{\sqrt{1-e^{-a}}}, \tag{2} \]

\[ \varphi_n(p)= \frac{\frac{p}{a}\prod\limits_{k=1}^{n-1}\left(\frac{p}{a}-k\right)} {\prod\limits_{k=1}^{n}\left(\frac{p}{a}+k\right)} = (-1)^{n-1} \frac{p\prod\limits_{k=1}^{n-1}(ka-p)} {\prod\limits_{k=1}^{n}(ka+p)} \div \varphi_n(t). \tag{3} \]

According to the convolution theorem for transforms, we may write

\[ \frac{e^{-\frac{at}{2}}}{\sqrt{1-e^{-at}}}\,T_n^*(t) = \int_0^t W(t-\tau)\varphi_n(\tau)\,d\tau, \]

whence

\[ T_n^*(t)= e^{\frac{at}{2}}\sqrt{1-e^{-at}} \int_0^t W(t-\tau)\varphi_n(\tau)\,d\tau. \tag{4} \]

Let us determine the function \(\varphi_n(t)\). Its Laplace transform (3) can be represented as follows:

\[ \varphi_n(p)=(-1)^{n-1}p \frac{\displaystyle\prod_{k=1}^{n-1}(ka-p)} {\displaystyle\prod_{k=1}^{n}(ka+p)} = (-1)^{n-1}p[p+(n+1)a]\frac{1}{a}V_n(\,). \]

or, after expanding the brackets,

\[ \varphi_n(p)=(-1)^{n-1} \left[ \frac{1}{a}p^2V_n(p)+(n+1)pV_n(p) \right]. \tag{5} \]

If we now take into account that \(V_n(p)\) is the Laplace transform of the integral “\(e\)”-Legendre polynomial \(V_n^*(t)\) [4], i.e.

\[ V_n(p)=a \frac{\displaystyle\prod_{k=1}^{n-1}(ka-p)} {\displaystyle\prod_{k=1}^{n+1}(ka+p)} \div V_n^*(t), \tag{6} \]

then we easily find

\[ \begin{aligned} pV_n(p)&\div \frac{dV_n^*(t)}{dt},\\ p^2V_n(p)&\div \frac{d^2V_n^*(t)}{dt^2}+V_n^{*\,\prime}(0)\delta(t). \end{aligned} \tag{7} \]

Since

\[ V_n^{*\,\prime}(0)=-aP_n^*(0)=a(-1)^{n+1}, \]

the latter correspondence can be written in the form

\[ p^2V_n(p)\div \frac{d^2V_n^*(t)}{dt^2}+a(-1)^{n+1}\delta(t). \]

The delta function \(\delta(t)\) arises as a consequence of the discontinuity of the derivative \(V_n^{*\,\prime}(t)\) at \(t=0\), since we assume that \(V_n^*(t)\equiv 0\) for \(t<0\). Thus, the sought function \(\varphi_n(t)\) has the form

\[ \varphi_n(t)=(-1)^{n-1} \left[ \frac{1}{a}\frac{d^2V_n^*(t)}{dt^2} +(n+1)\frac{dV_n^*(t)}{dt} \right]+\delta(t). \tag{8} \]

The expression in brackets can be represented in terms of the “\(e\)”-Legendre polynomial \(P_n^*(t)\) and its derivative. We have [4]

\[ \frac{dV_n^*(t)}{dt}=-ae^{-at}P_n^*(t), \]

\[ \frac{d^2V_n^*(t)}{dt^2} =a^2e^{-at}P_n^*(t)-ae^{-at}\frac{dP_n^*(t)}{dt}. \]

Substitution gives

\[ \varphi_n(t)=(-1)^ne^{-at} \left[ naP_n^*(t)+\frac{dP_n^*(t)}{dt} \right]+\delta(t). \tag{9} \]

Thus, formula (4) takes the form

\[ T_n(t)=1+(-1)^ne^{\frac{at}{2}}\sqrt{1-e^{-at}} \int_0^t W(t-\tau)e^{-\alpha \tau} \left[ naP_n^*(\tau)+\frac{dP_n^*(t)}{dt} \right]d\tau. \tag{10} \]

This is the desired relation.

Make the substitutions \(x=e^{-at}\) and \(y=e^{-a\tau}\); then the \(e\)-polynomial \(T_n^*(t)\) is transformed into the shifted Chebyshev polynomial of the first kind \(T_n^*(x)\), and the \(e\)-polynomial \(P_n(\tau)\) into the shifted Legendre polynomial \(P_n^*(y)\) [1, 3]. Further, taking into account the equalities

\[ W(t-\tau)= \frac{e^{-\frac{a}{2}(t-\tau)}}{\sqrt{1-e^{-a(t-\tau)}}} = \frac{e^{-\frac{a}{2}t}}{\sqrt{e^{-a\tau}-e^{-at}}} = \frac{\sqrt{x}}{\sqrt{y-x}}, \]

\[ \frac{dP_n^*(\tau)}{d\tau} = -\frac{dP_n^*(y)}{dy}\,y\cdot a; \qquad d\tau=-\frac{dy}{y\cdot a}, \]

we immediately obtain

\[ T_n^*(x) = 1-(-1)^n\sqrt{1-x} \int_{1}^{x} \frac{1}{\sqrt{y-x}} \left[ nP_n^*(y)-y\frac{dP_n^*(y)}{dy} \right]dy \]

or, interchanging the limits, we may write

\[ T_n^*(x) = 1+(-1)^n\sqrt{1-x} \int_{x}^{1} \frac{1}{\sqrt{y-x}} \left[ nP_n^*(y)-y\frac{dP_n^*(y)}{dy} \right]dy \qquad (0\le x\le 1). \tag{11} \]

Finally, making the substitution [1, 3]

\[ x\to \frac{1+x}{2} \quad\text{and}\quad y\to \frac{1+y}{2}, \]

we obtain the relation for the ordinary polynomials

\[ T_n(x) = 1+\frac{(-1)^n}{2}\sqrt{1-x} \int_{x}^{1} \frac{ nP_n(y)-(1-y)\dfrac{dP_n(y)}{dy} }{ \sqrt{y-x} }\,dy \qquad (-1\le x\le 1). \tag{12} \]

Let us establish an analogous connection between a Chebyshev polynomial of the second kind and a Legendre polynomial.

For the \(e\)-polynomial of Chebyshev of the second kind \(U_{n-1}^*(t)\), taking into account (10), and also the equality [3]

\[ U_{n-1}^*(t) = -\frac{1}{2na}e^{at}\frac{dT_n^*(t)}{dt}, \tag{13} \]

we immediately obtain

\[ U_{n-1}^*(t) = \frac{(-1)^{n+1}}{2na} e^{at}\frac{d}{dt} e^{\frac{at}{2}}\sqrt{1-e^{-at}} \int_{0}^{t} W(t-\tau)e^{-a\tau} \left[ naP_n^*(\tau)+ \frac{dP_n^*(\tau)}{d\tau} \right]d\tau . \tag{14} \]

For shifted polynomials, instead of (13) we shall have

\[ U_{n-1}^*(x) = \frac{1}{2n}\, \frac{dT_n^*(x)}{dx} \qquad (0\le x\le 1). \tag{15} \]

Thus, differentiating (11) with respect to \(x\) and substituting into (15), we find

\[ U_{n-1}^*(x) = \frac{(-1)^n}{2n}\, \frac{d}{dx}\sqrt{1-x} \int_{x}^{1} \frac{ nP_n^*(y)-y\dfrac{dP_n^*(y)}{dy} }{ \sqrt{y-x} }\,dy \qquad (0\le x\le 1). \tag{16} \]

For ordinary polynomials the relation has the form

\[ U_{n-1}(x)=\frac{(-1)^n}{n}\frac{d}{dx}\sqrt{1-x}\int_x^1 \frac{nP_n(y)-(1+y)\dfrac{dP_n(y)}{dy}}{\sqrt{y-x}}\,dy \qquad (-1\le x\le 1). \tag{17} \]

The obtained integral formulas can be written more compactly. Introduce the system of “\(e\)-polynomials”

\[ D_n^*(t)=\frac{1}{2}\left[P_n^*(t)-P_{n-1}^*(t)\right]\qquad (n=1,2,\ldots), \tag{18} \]

where \(P_n^*(t)\) is an “\(e\)-polynomial” of Legendre. This system is, obviously, complete in the same Hilbert space in which the system of Legendre “\(e\)-polynomials” \(P_n^*(t)\) \((n=0,1,2,\ldots)\) is complete [3]. We also note the equalities

\[ D_n^*(0)=1 \quad \text{and} \quad D_n^*(\infty)=0. \]

We now show that

\[ e^{-at}\left[naP_n^*(t)+\frac{dP_n^*(t)}{dt}\right] = \frac{dD_n^*(t)}{dt}. \tag{19} \]

The following recurrence relation holds for the Legendre “\(e\)-polynomials”:

\[ (1-2e^{-at})\frac{dP_n^*(t)}{dt} - \frac{dP_{n-1}^*(t)}{dt} = 2nae^{-at}P_n^*(t). \tag{20} \]

It is obtained from the relation for ordinary Legendre polynomials, defined on the interval \((-1,1)\) [2], if one makes the substitution \(x\to 1-2e^{-at}\). From (20) we find

\[ e^{-at}\frac{dP_n^*(t)}{dt} = -\frac{1}{2}\left[ \frac{dP_n^*(t)}{dt} - \frac{dP_{n-1}^*(t)}{dt} \right] - nae^{-at}P_n^*(t) \]

or, in another form,

\[ e^{-at}\frac{dP_n^*(t)}{dt} = -\frac{dD^*(t)}{dt} - nae^{-at}P_n^*(t), \]

whence identity (19) follows immediately. For shifted polynomials we shall have \((x=e^{-at})\)

\[ nP_n^*(x)-x\frac{dP_n^*(x)}{dx} = \frac{dD_n^*(x)}{dx}, \qquad 0\le x\le 1. \tag{21} \]

Making the substitution \(x\to \dfrac{1+x}{2}\), we obtain for ordinary polynomials

\[ nP_n(x)-(1+x)\frac{dP_n(x)}{dx} = 2\frac{dD_n(x)}{dx}, \qquad -1\le x\le 1. \tag{22} \]

Instead of relations (10), (11), and (12) we can now write:

\[ \left. \begin{aligned} T_n^*(t) &= 1+(-1)^n e^{\frac{at}{2}}\sqrt{1-e^{-at}} \int_0^t W(t-\tau)\frac{dD_n^*(\tau)}{d\tau}\,d\tau \qquad (0<t<\infty),\\[6pt] T_n^*(x) &= 1+(-1)^n\sqrt{1-x} \int_x^1 \frac{1}{\sqrt{y-x}}\cdot \frac{dD_n^*(y)}{dy}\,dy \qquad (0<x<1),\\[6pt] T_n(x) &= 1+(-1)^n\sqrt{1-x} \int_x^1 \frac{1}{\sqrt{y-x}}\frac{dD_n(y)}{dy}\,dy \qquad (-1<x<1). \end{aligned} \right\} \tag{23} \]

Similarly, formulas (14), (16), and (17) can be written down. Let us now solve the inverse problem, i.e., express the Legendre polynomial in terms of the Chebyshev polynomial.

First let us find the Laplace transform of the “shifted” polynomial \(D_n^*(t)\). We have [3]

\[ P_n^*(t) \div \prod_{k=1}^{n}\frac{ka-p}{ka+p} \quad \text{and} \quad P_{n-1}^*(t) \div \prod_{k=1}^{n-1}\frac{ka-1}{ka+p}. \]

Thus,

\[ D_n^*(t)=-\frac{1}{2}\bigl[P_n^*(t)-P_{n-1}^*(t)\bigr]\div -\frac{\displaystyle\prod_{k=1}^{n-1}(ka-p)} {\displaystyle\prod_{k=1}^{n}(ka+p)} =D_n(p). \tag{24} \]

Introduce the notation

\[ \frac{e^{-\frac{at}{2}}}{\sqrt{1-e^{-at}}}\,T_n^*(t)\div \psi_n(p)=W(p)\cdot \varphi_n(p). \tag{25} \]

Taking (2), (3), and (24) into account, we can write

\[ \psi_n(p)=(-1)^{n-1} \frac{\sqrt{\pi}\,\Gamma\!\left(\frac{p}{a}+\frac{1}{2}\right)} {\Gamma\!\left(\frac{p}{a}\right)} \cdot \frac{\displaystyle\prod_{k=1}^{n-1}(ka-p)} {\displaystyle\prod_{k=1}^{n}(ka+p)} = \]

\[ =(-1)^n \frac{\sqrt{\pi}\,\Gamma\!\left(\frac{p}{a}+\frac{1}{2}\right)} {\Gamma\!\left(\frac{p}{a}\right)} D_n(p). \]

Whence

\[ D_n(p)=(-1)^n \frac{\Gamma\!\left(\frac{p}{a}\right)} {\sqrt{\pi}\,\Gamma\!\left(\frac{p}{a}+\frac{1}{2}\right)} \psi_n(p) \tag{26} \]

Let

\[ \frac{\Gamma\!\left(\frac{p}{a}\right)} {\sqrt{\pi}\,\Gamma\!\left(\frac{p}{a}+\frac{1}{2}\right)} \div f(t). \]

By the shift theorem we shall have

\[ \frac{\pi}{a}e^{-\frac{at}{2}}f(t)\div \frac{\sqrt{\pi}\,\Gamma\!\left(\frac{p}{a}+\frac{1}{2}\right)} {a\Gamma\!\left(\frac{p}{a}\right)\cdot \frac{p}{a}}. \]

Comparing with (2), we arrive at the conclusion that

\[ f(t)=\frac{a}{\pi\sqrt{1-e^{-at}}}. \tag{27} \]

Thus, passing to the originals in the operator equality (26), we obtain, according to the convolution theorem,

\[ D_n^*(t)=(-1)^n\int_0^t f(t-\tau)\, \frac{e^{-\frac{a\tau}{2}}}{\sqrt{1-e^{-a\tau}}}\, T_n^*(\tau)\,d\tau . \tag{28} \]

or, in expanded form,

\[ D_n^*(t)=\frac{(-1)^n a}{\pi}\int_0^t \frac{1}{\sqrt{1-e^{-a(t-\tau)}}}\cdot \frac{e^{-\frac{a\tau}{2}}}{\sqrt{1-e^{-a\tau}}}\, T_n^*(\tau)\,d\tau \tag{29} \]

If we take into account that [4]

\[ D_n^*(t)+D_{n+1}^*(t)=\frac{1}{2}\,[P_{n+1}^*(t)-P_{n-1}^*(t)] =-(2n+1)V_n^*(t), \tag{30} \]

we easily find

\[ V_n^*(t)=\frac{(-1)^{n+1}a}{\pi(2n+1)} \int_0^t \frac{e^{-\frac{a\tau}{2}}\,[T_{n+1}^*(\tau)-T_n^*(\tau)]} {\sqrt{1-e^{-a(t-\tau)}}\sqrt{1-e^{-a\tau}}}\,d\tau \tag{31} \]

Finally, from the identity

\[ \frac{dV_n^*(t)}{dt}=-ae^{-at}P_n^*(t). \]

there follows the relation for the “\(e\)” polynomials

\[ P_n^*(t)=\frac{(-1)^n e^{at}}{\pi(2n+1)}\, \frac{d}{dt}\int_0^t \frac{e^{-\frac{a\tau}{2}}\,[T_{n+1}^*(\tau)-T_n^*(\tau)]} {\sqrt{1-e^{-a(t-\tau)}}\sqrt{1-e^{-a\tau}}}\,d\tau . \tag{32} \]

For the shifted polynomials, making the substitution \(x=e^{-at}\) and \(y=e^{-a\tau}\), we shall have

\[ P_n^*(x)=\frac{(-1)^n}{\pi(2n+1)}\, \frac{d}{dx}\int_1^x \frac{T_{n+1}^*(y)-T_n^*(y)} {\sqrt{(y-x)(1-y)}}\,dy \quad (0<x<1). \tag{33} \]

For ordinary polynomials the relation takes the form

\[ P_n(x)=\frac{(-1)^n 2}{\pi(2n+1)}\, \frac{d}{dx}\int_1^x \frac{T_{n+1}(y)-T_n(y)} {\sqrt{(y-x)(1-y)}}\,dy \quad (-1<x<y) \tag{34} \]

In conclusion, we note that all the relations obtained can easily be checked by direct integration.

REFERENCES

1. K. Lanczos. Practical Methods of Applied Analysis. Fizmatgiz, Moscow, 1961.

2. N. N. Lebedev. Special Functions and Their Applications. Fizmatgiz, Moscow–Leningrad, 1963.

3. V. M. Osipov. “Exponential polynomials and the expansion of certain typical signals,” Izv. TPI, vol. 180, Tomsk, 1969.

4. V. M. Osipov. “On the question of approximate inversion of the Laplace transform.” (This volume).