Padé Approximation System for Power Functions

V. E. Kornilov

(Presented by the scientific seminar of the Department of Higher Mathematics)

In this article a system of Padé approximations ([1], pp. 20, 38–40) is proved for the functions \(y=(1+x)^a\) and \(y=x^a\).

In the article the abbreviated notation is used

\[ (a)_\kappa=a(a+1)\ldots(a+\kappa-1)=\frac{\Gamma(a+\kappa)}{\Gamma(a)}, \qquad (a)_0=1. \]

1. The system of rational approximations (Padé approximations) for the function \(y=(1+x)^a\) can be represented by the following formula:

\[ y=(1+x)^a \approx \frac{ \displaystyle \sum_{m=0}^{\,n-\kappa} C_{n-\kappa}^{m}\, \frac{(-a-\kappa)_m}{(-n)_m}\,x^m }{ \displaystyle \sum_{m=0}^{\,\kappa} C_{\kappa}^{m}\, \frac{(a-n+\kappa)_m}{(-n)_m}\,x^m } = \frac{P_{n-\kappa}(x)}{Q_{\kappa}(x)}, \tag{1} \]

where \(n=0,1,\ldots;\ \kappa=0,1,\ldots,n\).

Proof. Suppose that the system of rational approximations (1) is a system of Padé approximations; then, as is known ([1], p. 20), the expansion into a power series in \(x\) of the fraction (1) must differ from the series

\[ (1+x)^a=\sum_{m=0}^{\infty}\frac{(-a)_m}{(1)_m}(-x)^m \tag{2} \]

only by terms with \(x\) in powers greater than \(n\). Therefore, simplifying the difference of the series (1) and (2), we obtain:

\[ (1+x)^a-\frac{P_{n-\kappa}(x)}{Q_{\kappa}(x)} = \sum_{m=n+1}^{\infty} C_m x^m, \]

\[ (1+x)^a Q_{\kappa}(x)-P_{n-\kappa}(x) = \sum_{m=n+1}^{\infty} d_m x^m . \]

or

\[ \left[\sum_{m=0}^{\infty} \frac{(-a)_m}{(1)_m}(-x)^m\right] \left[\sum_{m=0}^{k} C_k^m \frac{(a-n+k)_m}{(-n)_m} x^m\right] - \sum_{m=0}^{n-k} C_{n-k}^m \frac{(-a-k)_m}{(-n)_m} x^m \equiv \sum_{m=n+1}^{\infty} d_m x^m, \tag{3} \]

where \(d_{n+1}\ne 0\).

If one takes into account only the exponents of the argument \(x\) not exceeding \(n\), then in this case the sum of the coefficients for an arbitrary exponent of \(x\) equal to \(l\) \((l=0,1,\ldots,n)\) is also identically equal to zero, namely:

\[ \sum_{m=0}^{l} C_k^m \frac{(a-n+k)_m(-a)_{l-m}}{(-n)_m(1)_{l-m}} (-1)^{l-m} \equiv C_{n-k}^l \frac{(-a-k)_l}{(-n)_l}, \tag{4} \]

where \(l=0,1,\ldots,n\).

After some transformations, identity (4) is brought to the following form:

\[ \sum_{m=0}^{l} \frac{(-l)_m(-n+m)_{l-m}(-k)_m(a-n+k)_m} {(1)_m(a-l+1)_m} \equiv \frac{(-n+k)_l(-a-k)_l}{(-a)_l}. \tag{5} \]

By direct calculation we verify that identity (5) is valid for \(l=0,1\). Suppose that identity (5) is valid; then, multiplying both sides of the equality by

\[ \frac{(-n+k+l)(a+k-l)}{a-l}, \]

we must obtain an analogous identity that contains one more term for the sum on the left-hand side of the equality. In this way we obtain the following identity:

\[ \sum_{m=0}^{l} \frac{(-l)_m(-n+m)_{l-m}(-k)_m(a-n+k)_m} {(1)_m(a-l+1)_m} \frac{(-n+k+l)(a+k-l)}{(a-l)} - \]

\[ - \sum_{m=0}^{l+1} \frac{(-l-1)_m(-n+m)_{l+1-m}(-k)_m(a-n+k)_m} {(1)_m(a-l)_m} \equiv 0, \]

which is transformed under one summation sign into

\[ \sum_{m=1}^{l+1} \frac{(-l)_{m-1}(-n+m)_{l-m}(-k)_{m-1}(a-n+k)_{m-1}} {(1)_m(a-l)_m} \{\,m(-n+m-1)\times \]

\[ \times(-n+k+l)(a+k-l) +(l+1)(-n+l)(-k+m+1)\times \]

\[ \times(a-n+k+m-1)\,\} -(-n)_{l+1}\equiv 0. \]

Next, we replace the expression in braces by the identically following sum:

\[ m(-n+m-1)(-n+k+l)(a+k-l) +(l+1)(-n+l)\times \]

\[ \times(-k+m-1)(a-n+k+m-1) \equiv m(-n+m-1)\times \tag{6} \]

\[ \times(-n+l+m-1)(a-l+m-1) -(-k+m-1)\times \]

\[ \times(a-n+k+m-1)(-l+m-1)(-n+l+m), \]

then separate out, after the following equality sign, the first term of the first sum and increase the index \(m\) by one:

\[ \sum_{m=1}^{l+1} \frac{(-l)_{m-1}(-n+m-1)_{l+1-m}(-\kappa)_{m-1}(\alpha-n+\kappa)_{m-1}(-n+l+m-1)} {(1)_{m-1}(\alpha-l)_{m-1}} - \]

\[ -(-n)_{l+1} - \sum_{m=1}^{l+1} \frac{(-l)_m(-n+m)_{l-m}(-\kappa)_m(\alpha-n+\kappa)_m(-n+l+m)} {(1)_m(\alpha-l)_m} = \]

\[ = \sum_{m=1}^{l} \frac{(-l)_m(-n+m)_{l-m}(-\kappa)_m(\alpha-n+\kappa)_m(-n+l+m)} {(1)_m(\alpha-l)_m} -(-n)_{l+1} + \]

\[ +(-n)_{l+1} - \sum_{m=1}^{l} \frac{(-l)_m(-n+m)_{l-m}(-\kappa)_m(\alpha-n+\kappa)_m(-n+l+m)} {(1)_m(\alpha-l)_m} \equiv 0. \]

Thus, identity (5) is proved, and thereby it is confirmed that the system of rational approximations (1) is a system of Padé approximations of the function \((1+x)^\alpha\).

1. The denominator of the system of rational approximations (1) is identically transformed into the following function:

\[ \sum_{m=0}^{\kappa} (-1)^m \frac{(-\kappa)_m(\alpha-n+\kappa)_m} {(1)_m(-n)_m} x^m \equiv \tag{7} \]

\[ \equiv (-1)^\kappa \frac{(\alpha+1)_\kappa}{(-n)_\kappa} \sum_{m=0}^{\kappa} \frac{(-\kappa)_m(\alpha-n+\kappa)_m} {(1)_m(\alpha+1)_m} (1+x)^m . \]

Identity (7) is valid if the coefficients at equal powers of \(x\) are identically equal, namely:

\[ (-1)^m \frac{(-\kappa)_m(\alpha-n+\kappa)_m} {(1)_m(-n)_m} \equiv \tag{8} \]

\[ \equiv (-1)^\kappa \frac{(\alpha+1)_\kappa}{(-n)_\kappa} \sum_{l=m}^{\kappa} \frac{(m+1)_{l-m}(-\kappa)_l(\alpha-n+\kappa)_l} {(1)_{l-m}(1)_l(\alpha+1)_l}. \]

After some simplifications we obtain:

\[ \sum_{l=0}^{\kappa-m} \frac{(-\kappa+m)_l(\alpha-n+\kappa+m)_l} {(1)_l(\alpha+m+1)_l} \equiv \frac{(n-\kappa+1)_{\kappa-m}} {(\alpha+m+1)_{\kappa-m}}. \tag{9} \]

By direct calculation we verify that identity (9) is valid for \(\kappa-m=0,1\). Suppose that identity (9) holds; then, multiplying both sides of the equality by

\[ \frac{n-m+1}{\alpha+\kappa+1}, \]

we must obtain (to confirm our supposition) an analogous identity having on the left-hand side one term more than in equality (9), and this is equivalent to proving the identity:

\[ \sum_{l=0}^{\kappa-m} \frac{(-\kappa+m)_l(\alpha-n+\kappa+m)_l(n-m+1)} {(1)_l(\alpha+m+1)_l(\alpha+\kappa+1)} - \]

\[ - \sum_{l=0}^{\kappa-m+1} \frac{(-\kappa+m-1)_l(\alpha-n+\kappa+m)_l} {(1)_l(\alpha+m+1)_l} \equiv 0 \]

or

\[ \sum_{l=0}^{\kappa-m} \frac{(-\kappa+m)_l\,(a-n+\kappa+m)_l}{(1)_l\,(a+m+1)_l} \left[ \frac{n-m+1}{a+\kappa+1} - \frac{\kappa-m+1}{\kappa-m-l+1} \right] - \]

\[ - (-1)^{\kappa-m+1} \frac{(a-n+\kappa+m)_{\kappa-m+1}}{(a+m+1)_{\kappa-m+1}} \equiv 0. \]

Let us replace the expression in square brackets by the identically equal value:

\[ \frac{n-m+1}{a+\kappa+1} - \frac{\kappa-m+1}{\kappa-m-l+1} \equiv \]

\[ \equiv \frac{ -(\kappa-m-l+1)(a-n+\kappa+m+l)-l(a+m+l) }{ (a+\kappa+1)(\kappa-m-l+1) }, \tag{10} \]

then, after the equality sign, we separate off the last term of the first sum and decrease the index \(l\) by one. Thus we have:

\[ -\sum_{l=0}^{\kappa-m} \frac{(-\kappa+m)_l\,(a-n+\kappa+m)_{l+1}} {(1)_l\,(a+m+1)_l\,(a+\kappa+1)} + \]

\[ + \sum_{l=1}^{\kappa-m} \frac{(-\kappa+m)_{l-1}\,(a-n+\kappa+m)_l} {(1)_{l-1}\,(a+m+1)_{l-1}\,(a+\kappa+1)} + (-1)^{\kappa-m} \frac{(a-n+\kappa+m)_{\kappa-m+1}} {(a+m+1)_{\kappa-m+1}} = \]

\[ = -\sum_{l=1}^{\kappa-m} \frac{(-\kappa+m)_{l-1}\,(a-n+\kappa+m)_l} {(1)_{l-1}\,(a+m+1)_{l-1}\,(a+\kappa+1)} + \]

\[ + \sum_{l=1}^{\kappa-m} \frac{(-\kappa+m)_{l-1}\,(a-n+\kappa+m)_l} {(1)_{l-1}\,(a+m+1)_{l-1}\,(a+\kappa+1)} =0. \]

Identity (9), and hence identity (7), are proved.

Analogously, the identity

\[ \sum_{m=0}^{n-\kappa} C_{n-\kappa}^{m} \frac{(-a-\kappa)_m}{(-n)_m} x^m = \]

\[ = (-1)^{\,n-\kappa} \frac{(1-a)_{n-\kappa}}{(-n)_{n-\kappa}} \sum_{m=0}^{n-\kappa} \frac{(\kappa-n)_m\,(-a-\kappa)_m}{(1)_m\,(1-a)_m} (1+x)^m \tag{11} \]

is proved.

Making the change of variable in equalities (1), (7), and (11), namely replacing \(1+x\) by the variable \(x\), we obtain in this way:

\[ y=x^a \approx \frac{ \displaystyle \frac{(1-a)_{n-\kappa}}{(1)_{n-\kappa}} \sum_{m=0}^{n-\kappa} \frac{(-n+\kappa)_m\,(-a-\kappa)_m}{(1)_m\,(1-a)_m} x^m }{ \displaystyle \frac{(a+1)_\kappa}{(1)_\kappa} \sum_{m=0}^{\kappa} \frac{(-\kappa)_m\,(a-n+\kappa)_m}{(1)_m\,(a+1)_m} x^m }. \tag{12} \]

The system of rational approximations (12) is also a system of Padé approximations for the function \(y=x^a\), in view of the identical transformations of \(Q_\kappa(x)\) and \(P_{n-\kappa}(x)\) in equality (1).

LITERATURE

L. A. Lyusternik, O. A. Chervonenkis, A. R. Yanpolsky. Mathematical Analysis (Calculation of Elementary Functions), Moscow, Fizmatgiz, 1963.