INTEGRAL LEGENDRE POLYNOMIALS AND APPROXIMATION OF FUNCTIONS

V. M. OSIPOV

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

I. Integral Legendre Polynomials

The ordinary Legendre polynomials can be defined by Rodrigues’ formula

\[ P_n(z)=\frac{1}{2^n n!}\,\frac{d^n}{dz^n}(z^2-1)^n \quad (-1<z<1). \tag{1-1} \]

By the change of variable \(z=1-2x\), the interval \((-1,1)\) is transformed into the interval \((0,1)\), and we obtain the so-called shifted Legendre polynomials [1]

\[ P_n^*(x)=\frac{1}{n!}\,\frac{d^n}{dx^n}x^n(1-x)^n . \tag{1-2} \]

An explicit expression for \(P_n^*(x)\) can be obtained from the representation by a hypergeometric function

\[ P_n^*(x)=F(-n,n+1;1;x)=\sum_{k=0}^{n}(-1)^k \frac{(n+k)!}{(k!)^2(n-k)!}x^k . \tag{1-3} \]

Let us introduce a new system of polynomials by the formula

\[ V_n^*(x)=\frac{1}{n!}\,\frac{d^{\,n-1}}{dx^{n-1}}x^n(1-x)^n \quad (n=1,2,3,\ldots), \tag{1-4} \]

i.e.,

\[ \frac{dV_n^*(x)}{dx}=P_n^*(x) \tag{1-5} \]

or

\[ V_n^*(x)=\int_0^x P_n^*(x)\,dx . \tag{1-6} \]

We shall call these polynomials the integral shifted Legendre polynomials. Let us consider their basic properties. First we find their representation by a hypergeometric function. The following identity is known:

\[ \frac{d^n}{dx^n}\left[x^{n+c-1}(1-x)^{b-1}\right] =(c)_n x_c^{-1}(1-x)^{b-c-n}F(-n,b;c;x). \tag{1-7} \]

Replacing \(n\) by \(n-1\) and putting \(c=2,\ b=n+2\), we obtain

\[ \frac{d^{\,n-1}}{dx^{n-1}} x^n(1-x)^n = n!\,x(1-x)F(-n+1,n+2;2;x) \]

and, consequently,

\[ V_n^*(x)=x(1-x)F(-n+1,n+2;2;x), \tag{1-8} \]

whence it is immediately seen that

\[ V_n^*(0)=V_n^*(1)=0. \tag{1-9} \]

If the hypergeometric series in (1-8) is expanded, we obtain an explicit expression for \(V_n^*(x)\)

\[ V_n^*(x)= \frac{x(1-x)}{n(n+1)} \sum_{k=0}^{n-1} (-1)^k \frac{(n+k+1)!\,x^k}{(n-k-1)!(k+1)!k!}. \tag{1-10} \]

For the first few values of “\(n\)” we shall have

\[ \begin{aligned} V_1^*(x)&=x(1-x),\\ V_2^*(x)&=x(1-x)(1-2x),\\ V_3^*(x)&=x(1-x)(1-5x+5x^2),\\ V_4^*(x)&=x(1-x)(1-9x+21x^2-14x^3),\\ V_5^*(x)&=x(1-x)(1-14x+56x^2-84x^3+42x^4). \end{aligned} \]

Differentiate (1-3):

\[ \frac{dP_n^*(x)}{dx} = \frac{d}{dx}F(-n,n+1;1;x) = \frac{(-n)_1(n+1)_1}{(1)_1} F(-n+1,n+2;2;x). \]

Since

\[ \frac{(-n)_1(n+1)_1}{(1)_1}=-n(n+1), \]

we may write

\[ -\frac{1}{n(n+1)}\,\frac{dP_n^*(x)}{dx} = F(-n+1,n+2;2;x). \]

Substituting the expression obtained into (1-8), we shall have

\[ n(n+1)V_n^*(x) = -x(1-x)\frac{dP_n^*(x)}{dx}. \tag{1-11} \]

From (1-5), taking (1-11) into account, it follows that

\[ \frac{d^2V_n^*(x)}{dx^2} = \frac{dP_n^*(x)}{dx} = -\frac{n(n+1)}{x(1-x)}V_n^*(x). \]

or, in another form,

\[ x(1-x)\frac{d^2V_n^*(x)}{dx^2} + n(n+1)V_n^*(x)=0, \tag{1-12} \]

i.e. the polynomials \(V_n^*(x)\) satisfy the differential equation (1-12). They can be expressed directly in terms of shifted Legendre polynomials.

The following recurrence relation is known for the ordinary Legendre polynomials [2]:

\[ \frac{dP_{n+1}(z)}{dz} - \frac{dP_{n-1}(z)}{dz} = (2n+1)P_n(z). \]

Making the substitution \(z=1-2x\), we obtain an analogous identity for the shifted polynomials

\[ -\frac{dP^*_{n-1}(x)}{dx}+\frac{dP^*_{n+1}(x)}{dx} =2(2n+1)P^*_n(x). \tag{1-13} \]

If we now integrate (1-13) over the interval from zero to \(x\) and take (1-6) into account, then we shall have

\[ 2(2n+1)V^*_n(x)=P^*_{n-1}(x)-P^*_{n+1}(x)\quad (n=1,2,3,\ldots), \tag{1-14} \]

i.e., the polynomials \(V^*_n(x)\) are represented as the difference of two neighboring Legendre polynomials. The latter form a complete system in the space \(L^2(0,1)\); consequently, the system of polynomials \(V^*_n(x)\) \((n=0,1,2,\ldots)\) is complete in the same space. Moreover, we shall show that it is also orthogonal with weight \(\dfrac{1}{x(1-x)}\).

We shall proceed from the orthogonality of the Legendre polynomials

\[ \int_0^1 P^*_n(x)P^*_m(x)\,dx = \begin{cases} 0, & n\ne m,\\[4pt] \dfrac{1}{2n+1}, & n=m. \end{cases} \tag{1-15} \]

Taking (1-5) and (1-11) into account, we have

\[ \int_0^1 P^*_n(x)P^*_m(x)\,dx = \int_0^1 P^*_n(x)\frac{dV^*_m(x)}{dx}\,dx = -\int_0^1 V^*_m(x)\frac{dP^*_n(x)}{dx}\,dx = n(n+1)\int_0^1 \frac{V^*_n(x)V^*_m(x)}{x(1-x)}\,dx. \]

Thus,

\[ \int_0^1 \frac{V^*_n(x)V^*_m(x)}{x(1-x)}\,dx = \begin{cases} 0, & m\ne n,\\[4pt] \dfrac{1}{n(n+1)(2n+1)}, & m=n, \end{cases} \tag{1-16} \]

which proves our assertion.

It follows from (1-5) that the extrema of the polynomial \(V^*_n(x)\) are located at the zeros of the polynomial \(P^*_n(x)\). In particular, the polynomials \(V^*_n(x)\) with odd indices attain an extremum in the middle of the interval \(\left(x=\dfrac12\right)\), which is directly evident from representation (1-4), if one takes into account that the binomial \(x^n(1-x)^n\) has a maximum at \(x=\dfrac12\). This maximum is the largest in absolute value. As for the polynomials \(V^*_n(x)\) with even indices, at \(x=\dfrac12\) they vanish, and their extrema are located symmetrically with respect to the line \(x=\dfrac12\).

Let us find, for the polynomials \(V^*_n(x)\) with odd indices, the greatest extremum in absolute value, corresponding to \(x=\dfrac12\). Replacing \(n\) by \(2n-1\) in (1-14), we obtain as a result

\[ 2(4n-1)V^*_{2n-1}(x)=P^*_{2(n-1)}(x)-P^*_{2n}(x). \]

If we now take into account that [2]

\[ P^{*}_{2(n-1)}\left(\frac{1}{2}\right) = (-1)^{\,n-1}\frac{1\cdot 3\cdot \ldots \cdot (2n-3)}{2^{\,n-1}(n-1)!} \]

\[ P^{*}_{2n}\left(\frac{1}{2}\right) = (-1)^n \frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2^n n!}, \]

then we may write

\[ 2(4n-1)V^{*}_{2n-1}\left(\frac{1}{2}\right) = (-1)^{\,n-1} \frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-3)} {2^{\,n-1}(n-1)!} \left(1+\frac{2n-1}{2n}\right), \]

whence it follows that

\[ \left|V^{*}_{2n-1}\left(\frac{1}{2}\right)\right| = |V^{*}_{2n-1}(x)|_{\max} = \frac{1}{4n} \frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-3)} {2^{\,n-1}(n-1)!} = \]

\[ = \frac{(2n-3)!!}{2^{\,n+1}n!}. \]

The last expression can be written differently, namely:

\[ \frac{(2n-3)!!}{2^{\,n+1}n!} = \frac{\Gamma(2n-1)}{2^{\,n-1}(n-1)!2^{\,n+1}n!} = \frac{\Gamma(2n-1)}{2^{\,2n}n[\Gamma(n)]^2}, \]

where \(\Gamma(n)\) is the gamma function. Thus, we have the estimate

\[ |V^{*}_{2n-1}(x)| \leq \frac{\Gamma(2n-1)}{2^{\,2n}n[\Gamma(n)]^2}. \]

For any \(n>1\) the right-hand side of the inequality is a continuous function. Replacing \(n\) by \(\dfrac{n+1}{2}\), we shall have, for any \(n\),

\[ |V^{*}_{n}(x)| \leq \frac{\Gamma(n)} {2^n(n+1)\left[\Gamma\left(\dfrac{n+1}{2}\right)\right]^2} = \]

\[ = \frac{(n-1)!} {2^n(n+1)\left[\Gamma\left(\dfrac{n+1}{2}\right)\right]^2}. \tag{1-17} \]

We can now introduce a scale factor and consider the polynomials

\[ \overline{V}^{*}_{n}(x) = \frac{ 2^n(n+1)\left[\Gamma\left(\dfrac{n+1}{2}\right)\right]^2 }{(n-1)!} V^{*}_{n}(x), \tag{1-18} \]

for which, obviously, the estimate has the form

\[ |\overline{V}^{*}_{n}(x)|\leq 1. \]

For the first few values of \(n\), (1-18) gives

\[ \overline{V}^{*}_{1}(x)=4V^{*}_{1}(x); \quad \overline{V}^{*}_{2}(x)=3\pi V^{*}_{2}(x); \quad \overline{V}^{*}_{3}(x)=16V^{*}_{3}(x); \]

\[ \overline{V}^{*}_{4}(x)=\frac{15}{2}\pi V^{*}_{4}(x); \quad \overline{V}^{*}_{5}(x)=32V^{*}_{5}(x). \]

Let us consider the question of expanding an arbitrary function in a series in the polynomials \(V^{*}_{n}(x)\).

2. Expansion of functions in series in the polynomials \(V_n^*(x)\)

Let \(\varphi(x)\) be a continuous function defined on the interval \((0,1)\) and vanishing at the endpoints of the interval. Suppose that the expansion

\[ \varphi(x)=\sum_{n=1}^{\infty} C_n V_n^*(x). \tag{2-1} \]

holds. Proceeding formally, multiply both sides of (2-1) by

\[ \frac{V_m^*(x)}{x(1-x)} \]

and integrate from 0 to 1. Taking (1-16) into account, we shall have

\[ C_n=n(n+1)(2n+1)\int_0^1 \frac{\varphi(x)V_n^*(x)}{x(1-x)}\,dx \tag{2-2} \]

or, taking (1-14) into account and integrating by parts,

\[ C_n=-(2n+1)\int_0^1 \varphi(x)\frac{dP_n^*(x)}{dx}\,dx =(2n+1)\int_0^1 \varphi'(x)P_n^*(x)\,dx. \tag{2-3} \]

If we now take into account that

\[ \frac{dP_n^*(x)}{dx} =-n(n+1)F(-n+1,n+2;2;x)= \]

\[ =\sum_{k=0}^{n-1}(-1)^k \frac{(n+k+1)!x^k}{(n-k-1)!(k+1)!k!}, \tag{2-4} \]

then we may write

\[ C_n=(2n+1)\sum_{k=0}^{n-1}(-1)^k \frac{(n+k+1)!}{(n-k-1)!(k+1)!k!} \int_0^1 x^k\varphi(x)\,dx. \tag{2-5} \]

The quantity

\[ M_k=\int_0^1 x^k\varphi(x)\,dx. \tag{2-6} \]

is, evidently, the moment of order \(k\) of the function \(\varphi(x)\). Formula (2-5) takes the following form:

\[ C_n=(2n+1)\sum_{k=0}^{n-1}(-1)^k \frac{(n+k+1)!M_k}{(n-k-1)!(k+1)!k!}. \tag{2-7} \]

Whence, in particular, we easily find:

\[ \begin{aligned} C_1&=6M_0,\\ C_2&=30(M_0-2M_1),\\ C_3&=84(M_0-5M_1+5M_2),\\ C_4&=180(M_0-9M_1+21M_2-14M_3),\\ C_5&=330(M_0-14M_1+56M_2-84M_3+42M_4). \end{aligned} \]

Let us clarify the convergence conditions for the expansion (2-1). If we differentiate the series (2-1) and take (1-5) into account, we obtain an expansion of the derivative \(\varphi'(x)\) in a series in shifted Legendre polynomials

\[ \varphi'(x)=\sum_{n=1}^{\infty} C_n P_n^*(x). \tag{2-9} \]

The coefficients of this series are determined by formula (2-3). It will converge to \(\varphi'(x)\) at every interior point of the interval \((0,1)\) that is a point of continuity, if the following conditions [2] are satisfied:

1. \(\varphi'(x)\) is a piecewise-continuous function.

1. The integral \(\displaystyle \int_0^1 [\varphi'(x)]^2\,dx\) has a finite value.

From the convergence of the series (2-9) there follows the unconditional convergence of the expansion (2-1).

Instead of the expansion (2-1), it is practically more convenient to use the expansion in the polynomials \(\overline{V}_n^*(x)\)

\[ \varphi(x)=\sum_{n=1}^{\infty} A_n\overline{V}_n^*(x), \tag{2-10} \]

where

\[ A_n=\frac{(n-1)!}{2^n(n+1)\left[\Gamma\left(\frac{n+1}{2}\right)\right]^2}C_n. \tag{2-11} \]

If in (2-10) one restricts oneself to a finite number of terms, then the maximum absolute error will be approximately equal to the first of the discarded coefficients. We note that if the function \(f(x)\) does not vanish at the ends of the interval \((0,1)\), but takes the finite values \(f(0)\) and \(f(1)\), then the expansion takes the form

\[ f(x)=f(0)(1-x)+f(1)x+\sum_{n=1}^{\infty} A_n\overline{V}_n^*(x). \tag{2-12} \]

where \(A_n\) are the expansion coefficients for the function

\[ \varphi(x)=f(x)-f(0)(1-x)-f(1)x. \]

As an example, let us consider the approximation of the function \(\varphi(y)=\sin y\) on the interval \((0,2\pi)\) by a segment of the series (2-10). By the substitution \(y=2\pi x\), the indicated interval is transformed into \((0,1)\). We have

\[ M_k=\int_0^1 x^k\sin 2\pi x\,dx \]

Calculations give

\[ M_0=0;\quad M_1=-\frac{1}{2\pi};\quad M_2=-\frac{1}{2\pi};\quad M_3=-\frac{1}{2\pi}+\frac{6}{(2\pi)^3}; \]

\[ M_4=-\frac{1}{2\pi}+\frac{12}{(2\pi)^3};\quad M_5=-\frac{1}{2\pi}+\frac{20}{(2\pi)^3}-\frac{120}{(2\pi)^5}. \]

By formulas (2-7) and (2-11) we find

\[ A_1=0;\quad A_2=\frac{10}{\pi^2}\simeq 1.01321;\quad A_3=0;\quad A_4=\frac{12}{\pi^4}(2\pi^2-21)\simeq -0.15532 \]

\[ A_5=0;\quad A_6=\frac{208}{5\pi^6}(\pi^4-60\pi^2+4\cdot 75)\simeq 0.01007. \]

If one restricts oneself to only two terms of the expansion, i.e., assumes that

\[ \sin 2\pi x \simeq \frac{10}{\pi^2}\overline{V}_2^*(x)+\frac{12}{\pi^4}(2\pi^2-21)\overline{V}_3^*(x), \]

then the absolute maximum error will be of order \(0.01\), and at the ends of the interval and in its middle \(\left(x=\frac{1}{2}\right)\) the error is equal to zero. Adding one more term will reduce the maximum absolute error by an order of magnitude.

In conclusion, let us compare the expansion in the polynomials \(V_n^*(x)\) of a certain function \(\varphi(x)\), defined on the interval \((0,1)\), with the expansion

of this same function in Legendre polynomials \(P_n^*(x)\). Suppose that the function \(\varphi(x)\) is specified by its values at the endpoints of the interval and by several of its first moments \(M_k\) \((k=0, 1, 2, \ldots, v)\). It is required to reconstruct the function \(\varphi(x)\) from these data with the greatest accuracy. A segment of the series in the polynomials \(P_n^*(x)\) will contain \(v+1\) terms (starting with the zeroth), and the information on the values of the function \(\varphi(x)\) at the endpoints of the interval will not be used. The expansion in the polynomials \(\overline{V}_n^*(x)\) will contain the same number of terms (starting with the first), and the additional information on the boundary conditions will be used completely. This latter circumstance ensures a considerably higher accuracy of approximation of the function \(\varphi(x)\) by a segment of the series in \(\overline{V}_n^*(x)\). Thus, in our example, if the first 6 moments are used, we obtain

\[ \sin 2\pi x \approx \overline{A}_2 \overline{V}_2^*(x) + \overline{A}_4 \overline{V}_4^*(x) + \overline{A}_6 \overline{V}_6^*(x). \]

The use of the polynomials \(P_n^*(x)\) gives the expression

\[ \sin 2\pi x \approx C_1' P_1^*(x) + C_3' P_3^*(x) + C_5' P_5^*(x), \]

where

\[ C_1'=\frac{3}{\pi}\approx 0.95493;\qquad C_3'=\frac{7}{\pi^3}(\pi^2-15)\approx -1.15824; \]

\[ C_5'=\frac{11}{\pi}\left[1-\frac{105(\pi^2-9)}{\pi^4}\right]\approx 0.21929. \]

The absolute maximum error of the first of these approximations is an order of magnitude smaller than the error of the second.

Thus, the integral Legendre polynomials \(V_n^*(x)\), in the indicated sense, are a considerably more effective means of approximating functions than the ordinary polynomials \(P_n^*(x)\).

REFERENCES

1. K. Landosh. Practical Methods of Applied Analysis, Fizmatgiz, Moscow, 1961.
2. N. N. Lebedev. Special Functions and Their Applications. Fizmatgiz, Moscow–Leningrad, 1963.