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A recursive algorithm for an efficient and accurate computation of incomplete Bessel functions

Richard M. Slevinsky and Hassan Safouhi§,111The corresponding author (HS) acknowledges the financial support from the Natural Sciences and Engineering Research Council of Canada (NSERC) - Grant RGPIN-2016-04317

S2.29 Mathematical Institute
University of Oxford
Andrew Wiles Building, Radcliffe Observatory Quarter
Woodstock Road, Oxford UK OX2 6GG

§
Mathematical Section
Campus Saint-Jean, University of Alberta
8406, 91 Street, Edmonton (AB), Canada T6C 4G9

MSC Classification:  65B05; 65D30.

Abstract   In a previous work, we developed an algorithm for the computation of incomplete Bessel functions, which pose as a numerical challenge, based on the Gn(1)G_{n}^{(1)} transformation and Slevinsky-Safouhi formula for differentiation. In the present contribution, we improve this existing algorithm for incomplete Bessel functions by developing a recurrence relation for the numerator sequence and the denominator sequence whose ratio forms the sequence of approximations. By finding this recurrence relation, we reduce the complexity from 𝒪(n4){\cal O}(n^{4}) to 𝒪(n){\cal O}(n). We plot relative error showing that the algorithm is capable of extremely high accuracy for incomplete Bessel functions.

Keywords:  Incomplete Bessel functions. Extrapolation methods. The GG transformation. Numerical Integration. The Slevinsky-Safouhi formulae.

1 Introduction

Incomplete Bessel functions, which are a computational challenge, were a subject of significant research. These functions appear when describing a plethora of phenomena in hydrology, statistics, and quantum mechanics [1, 2, 3, 4, 5, 6, 7, 8, 9]. Incomplete Bessel functions of zero order are also involved in a numerous applications to electromagnetic waves [10, 11, 12, 13, 14]. By introducing a recursive algorithm for the GG transformation for tail integrals and applying it to incomplete Bessel functions, the article [15] has received some attention recently [16, 17, 18, 19, 20, 21]. Most of this attention has been due to the algorithm for incomplete Bessel functions.

Integral representations of the incomplete Bessel functions are given by [22]:

Kν(x,y)=1exty/ttν+1dt,K_{\nu}(x,y)=\int_{1}^{\infty}\frac{e^{-x\,t-y/t}}{t^{\nu+1}}{\rm\,d}t, (1)

The algorithm for the GG transformation [23, 24, 25, 26] applied to incomplete Bessel functions [15] appears as:

G~n(1)(x,y,ν)=xν𝒩~n(x,y,ν)𝒟n(x,y,ν),\tilde{G}^{(1)}_{n}(x,y,\nu)=x^{\nu}\frac{\tilde{\cal N}_{n}(x,y,\nu)}{{\cal D}_{n}(x,y,\nu)}, (2)

where:

𝒟n(x,y,ν)=(xy)nxν+1ex+yr=0n(nr)(y)ri=0rArixi,{\cal D}_{n}(x,y,\nu)=(-x\,y)^{n}\,x^{\nu+1}\,e^{x+y}\sum_{r=0}^{n}\binom{n}{r}(-y)^{-r}\sum_{i=0}^{r}A_{r}^{i}\,x^{i}, (3)

where the AriA_{r}^{i} are the coefficients in the Slevinsky-Safouhi formula I [27] and are given by:

Aki={1fori=k,(nν(k1)(μ+1))Ak10fori=0,k>0,(nν+i(m+1)(k1)(μ+1))Ak1i+Ak1i1for0<i<k,A^{i}_{k}=\left\{\begin{array}[]{lll}1&\quad\textrm{for}&i=k,\\ \displaystyle(n-\nu-(k-1)(\mu+1))A_{k-1}^{0}&\quad\textrm{for}&i=0,\leavevmode\nobreak\ k>0,\\ (n-\nu+i(m+1)-(k-1)(\mu+1))A^{i}_{k-1}+A^{i-1}_{k-1}&\quad\textrm{for}&0<i<k,\end{array}\right. (4)

with (μ,ν,m,n)=(2,ν1,0,0)(\mu,\nu,m,n)=(-2,-\nu-1,0,0).

The numerator 𝒩~n(x,y,ν)\tilde{\cal N}_{n}(x,y,\nu) in (2) is given by:

𝒩~n(x,y,ν)=exyxνyr=1n(nr)𝒟nr(x,y,ν)(xy)rs=0r1(r1s)ysi=0sAsi(x)i,\tilde{\cal N}_{n}(x,y,\nu)=\frac{e^{-x-y}}{x^{\nu}\,y}\sum_{r=1}^{n}\binom{n}{r}\,{\cal D}_{n-r}(x,y,\nu)\,(x\,y)^{r}\sum_{s=0}^{r-1}\binom{r-1}{s}\,y^{-s}\sum_{i=0}^{s}A_{s}^{i}(-x)^{i}, (5)

where the AsiA_{s}^{i} are the coefficients in the Slevinsky-Safouhi formula I with (μ,ν,m,n)=(2,ν1,0,0)(\mu,\nu,m,n)=(-2,\nu-1,0,0).

Therefore, the original algorithm involves sums over sums over sums, effectively making the complexity of the computation of the numerator an 𝒪(n3){\cal O}(n^{3}) process and the complexity of the computation of the denominator an 𝒪(n2){\cal O}(n^{2}) process. Therefore, calculating a sequence of approximations {Gn}n0\left\{G_{n}\right\}_{n\geq 0} results in an 𝒪(n4){\cal O}(n^{4}) algorithm.

In the following, we introduce an algorithm for the GG transformation that reduce the calculation to an easily programmable and parallel four-term recurrence relation: with one initialization, it computes the numerator; with another initialization, it computes the denominator. As there are only a finite number of arithmetic operations required in the recurrence relation, the resulting algorithm for calculating a sequence of approximations {Gn}n0\left\{G_{n}\right\}_{n\geq 0} is an 𝒪(n){\cal O}(n) algorithm.

The inductive proof of this recurrence relation follows and the notation is heavy because of the two variables xx and yy and the parameter ν\nu already included in the incomplete Bessel function. As well, with this new algorithm comes the ability to program the GG transformation to unprecedentedly high order. We show some numerical results in the form of figures of the relative error for six different values. The reduction in complexity of the original algorithm comes with the benefit of a stable algorithm for high order.

2 The algorithm

Theorem 2.1.

[28] Let f(x)f(x) be integrable on [0,)[0,\infty) (i.e. 0f(t)dt<\int_{0}^{\infty}f(t)\,\mathrm{d}t\,<\,\infty) and satisfy a linear differential equation of order mm of the form:

f(x)=k=1mpk(x)f(k)(x),f(x)=\displaystyle\sum_{k=1}^{m}p_{k}(x)\,f^{(k)}(x), (6)

where pkp_{k} for k=1,2,,mk=1,2,\ldots,m have asymptotic expansions as xx\to\infty, of the form:

p(x)\displaystyle p(x) \displaystyle\sim xiki=0aixiwithikk.\displaystyle x^{i_{k}}\sum_{i=0}^{\infty}\frac{a_{i}}{x^{i}}\quad\textrm{with}\quad i_{k}\,\leq\,k. (7)

If for 1im1\leq i\leq m and ikmi\leq k\leq m, we have:

limxpk(i1)(x)f(ki)(x)=0,\displaystyle\lim_{x\to\infty}p_{k}^{(i-1)}(x)\,f^{(k-i)}(x)=0, (8)

and for every integer l1l\geq-1, we have:

k=1ml(l1)(lk+1)pk,01,\displaystyle\sum_{k=1}^{m}l(l-1)\cdots(l-k+1)p_{k,0}\neq 1, (9)

where pk,0=limxxkpk(x)p_{k,0}=\displaystyle\lim_{x\to\infty}x^{-k}\,p_{k}(x) for 1km1\leq k\leq m, then as xx\to\infty, we have:

xf(t)dtk=0m1xjkf(k)(x)(β0,k+β1,kx+β2,kx2++),\int_{x}^{\infty}f(t)\,\mathrm{d}t\sim\sum_{k=0}^{m-1}x^{j_{k}}\,f^{(k)}(x)\left(\beta_{0,k}+\frac{\beta_{1,k}}{x}+\frac{\beta_{2,k}}{x^{2}}+\cdots+\right), (10)

where jkmax(ik+1,ik+21,,imm+k+1)fork=0,1,,m1j_{k}\leq\mathrm{max}(i_{k+1},i_{k+2}-1,\ldots,i_{m}-m+k+1)\quad for\quad k=0,1,\ldots,m-1.

To solve for the unknowns βk,i\beta_{k,i}, we must set up and solve a system of linear equations.

The Gn(m)G_{n}^{(m)} transformation [23] truncates the asymptotic expansions (10) after nn terms each and the system is formed by differentiation. The approximation Gn(m)G_{n}^{(m)} to 0f(t)dt\int_{0}^{\infty}f(t)\mathrm{d}t is given as the solution of the system of mn+1mn+1 linear equations [23]:

dldxl{Gn(m)0xf(t)dtk=0m1xσkf(k)(x)i=0n1β¯k,ixi}=0,l=0,1,,mn,\frac{\mathrm{d}^{l}}{\mathrm{d}x^{l}}\left\{G^{(m)}_{n}-\int_{0}^{x}f(t)\,\mathrm{d}t-\sum_{k=0}^{m-1}x^{\sigma_{k}}f^{(k)}(x)\sum_{i=0}^{n-1}\frac{\bar{\beta}_{k,i}}{x^{i}}\right\}=0,\quad l=0,1,\ldots,mn, (11)

where it is assumed that dldxlGn(m)0,l>0\displaystyle\frac{\mathrm{d}^{l}}{\mathrm{d}x^{l}}G^{(m)}_{n}\equiv 0,\forall l>0. In the above system (11), σk=min(sk,k+1)\sigma_{k}=\mathrm{min}(s_{k},k+1) where sks_{k} is the largest of the integers ss such that limxxsf(k)(x)=0\displaystyle\lim_{x\to\infty}x^{s}f^{(k)}(x)=0 holds, k=0,1,,m1k=0,1,\ldots,m-1. Also, Gn(m)G^{(m)}_{n} and β¯k,i\bar{\beta}_{k,i} are the respective set of mn+1mn+1 unknowns.

The Gn(1)G^{(1)}_{n} transformation can be written as the solution to the linear system (11) with m=1m=1. Instead of solving the system of linear equations each time for each order nn, it would be ideal to resolve each approximation Gn(1)G^{(1)}_{n} in a recursive manner [15].

By considering the equation (11) for l=0l=0:

Gn(1)F(x)=xσ0f(x)i=0n1β¯0,ixiwithF(x)=0xf(t)dt,G^{(1)}_{n}-F(x)=x^{\sigma_{0}}f(x)\sum_{i=0}^{n-1}\frac{\bar{\beta}_{0,i}}{x^{i}}\quad\textrm{with}\quad F(x)\,=\,\int_{0}^{x}f(t)\mathrm{d}t, (12)

and by isolating the summation on the right hand side, we obtain:

Gn(1)F(x)xσ0f(x)=i=0n1β¯0,ixi.\displaystyle\frac{\displaystyle G^{(1)}_{n}-F(x)}{x^{\sigma_{0}}f(x)}=\sum_{i=0}^{n-1}\frac{\bar{\beta}_{0,i}}{x^{i}}. (13)

To eliminate the summation, and consequently all of the unknowns β¯0,i\bar{\beta}_{0,i}, we apply the (x2ddx)\displaystyle\left(x^{2}\frac{\mathrm{d}}{\mathrm{d}x}\right) operator nn times, obtaining:

(x2ddx)n[Gn(1)F(x)xσ0f(x)]=0Gn(1)=(x2ddx)n(F(x)xσ0f(x))(x2ddx)n(1xσ0f(x)),\left(x^{2}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{n}\left[\frac{\displaystyle G^{(1)}_{n}-F(x)}{x^{\sigma_{0}}f(x)}\right]=0\quad\Longrightarrow\quad G^{(1)}_{n}=\displaystyle\frac{\left(x^{2}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{n}\left(\frac{\displaystyle F(x)}{x^{\sigma_{0}}f(x)}\right)}{\left(x^{2}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{n}\left(\frac{1}{x^{\sigma_{0}}f(x)}\right)}, (14)

which leads to a recursive algorithm for the Gn(1)G^{(1)}_{n} transformation.

  1. 1.

    Set:

    𝒩0(x)=F(x)xσ0f(x)and𝒟0(x)=1xσ0f(x).{\cal N}_{0}(x)=\frac{F(x)}{x^{\sigma_{0}}f(x)}\qquad\textrm{and}\qquad{\cal D}_{0}(x)=\frac{1}{x^{\sigma_{0}}f(x)}. (15)
  2. 2.

    For n=1,2,,n=1,2,\ldots, compute 𝒩n(x){\cal N}_{n}(x) and 𝒟n(x){\cal D}_{n}(x) recursively from:

    𝒩n(x)=(x2ddx)𝒩n1(x)and𝒟n(x)=(x2ddx)𝒟n1(x).{\cal N}_{n}(x)=\left(x^{2}\frac{\mathrm{d}}{\mathrm{d}x}\right){\cal N}_{n-1}(x)\qquad\textrm{and}\qquad{\cal D}_{n}(x)=\left(x^{2}\frac{\mathrm{d}}{\mathrm{d}x}\right){\cal D}_{n-1}(x). (16)
  3. 3.

    For all nn, the approximations Gn(1)(x)G^{(1)}_{n}(x) to (0x+x)f(t)dt\displaystyle\left(\int_{0}^{x}+\int_{x}^{\infty}\right)f(t)\,\mathrm{d}t are given by:

    Gn(1)(x)=𝒩n(x)𝒟n(x).G^{(1)}_{n}(x)=\frac{{\cal N}_{n}(x)}{{\cal D}_{n}(x)}. (17)

For the incomplete Bessel functions, σ0=0\sigma_{0}=0, and the algorithm for the Gn(1)G_{n}^{(1)} transformation takes the form below. Let:

f(t)=exty/ttν+1andF(t)=0tf(s)ds.f(t)=\dfrac{e^{-xt-y/t}}{t^{\nu+1}}\quad\textrm{and}\quad\displaystyle F(t)=\int_{0}^{t}f(s){\rm\,d}s. (18)
  1. 1.

    Set:

    𝒩0(x,y,ν,t)=F(t)f(t)and𝒟0(x,y,ν,t)=1f(t).{\cal N}_{0}(x,y,\nu,t)=\frac{F(t)}{f(t)}\qquad\textrm{and}\qquad{\cal D}_{0}(x,y,\nu,t)=\frac{1}{f(t)}. (19)
  2. 2.

    For n=1,2,,n=1,2,\ldots, compute 𝒩n(x,y,ν,t){\cal N}_{n}(x,y,\nu,t) and 𝒟n(x,y,ν,t){\cal D}_{n}(x,y,\nu,t) recursively from:

    𝒩n(x,y,ν,t)\displaystyle{\cal N}_{n}(x,y,\nu,t) =\displaystyle= (t2ddt)𝒩n1(x,y,ν,t)\displaystyle\left(t^{2}\frac{\mathrm{d}}{\mathrm{d}t}\right){\cal N}_{n-1}(x,y,\nu,t)
    𝒟n(x,y,ν,t)\displaystyle{\cal D}_{n}(x,y,\nu,t) =\displaystyle= (t2ddt)𝒟n1(x,y,ν,t).\displaystyle\left(t^{2}\frac{\mathrm{d}}{\mathrm{d}t}\right){\cal D}_{n-1}(x,y,\nu,t). (20)
  3. 3.

    For all n0n\geq 0, the approximations G~n(1)(x,y,ν)\tilde{G}^{(1)}_{n}(x,y,\nu) to Kν(x,y)K_{\nu}(x,y) are given by:

    G~n(1)(x,y,ν)=𝒩~n(x,y,ν,1)𝒟n(x,y,ν,1),\tilde{G}^{(1)}_{n}(x,y,\nu)=\frac{\tilde{\cal N}_{n}(x,y,\nu,1)}{{\cal D}_{n}(x,y,\nu,1)}, (21)

    where:

    𝒩~n(x,y,ν,t)=𝒩n(x,y,ν,t)F(t)𝒟n(x,y,ν,t).\tilde{\cal N}_{n}(x,y,\nu,t)={\cal N}_{n}(x,y,\nu,t)-F(t){\cal D}_{n}(x,y,\nu,t). (22)
Proposition 2.2.

Let:

N~0(x,y,ν)=0\displaystyle\tilde{N}_{0}(x,y,\nu)=0 andN~1(x,y,ν)=1,\displaystyle\qquad\textrm{and}\qquad\tilde{N}_{1}(x,y,\nu)=1, (23)
D0(x,y,ν)=ex+y\displaystyle D_{0}(x,y,\nu)=e^{x+y} andD1(x,y,ν)=(x+ν+1y)D0(x,y,ν),\displaystyle\qquad\textrm{and}\qquad D_{1}(x,y,\nu)=(x+\nu+1-y)D_{0}(x,y,\nu), (24)

and:

N~1(x,y,ν)=D1(x,y,ν)=0.\tilde{N}_{-1}(x,y,\nu)=D_{-1}(x,y,\nu)=0. (25)

Let also:

(n+1)Qn+1(x,y,ν)\displaystyle(n+1)Q_{n+1}(x,y,\nu) =\displaystyle= (x+ν+1+2ny)Qn(x,y,ν)\displaystyle(x+\nu+1+2n-y)\,Q_{n}(x,y,\nu) (26)
+\displaystyle+ (2yνn)Qn1(x,y,ν)yQn2(x,y,ν),\displaystyle(2y-\nu-n)\,Q_{n-1}(x,y,\nu)-y\,Q_{n-2}(x,y,\nu),

where the Qn(x,y,ν)Q_{n}(x,y,\nu) stand for either the N~n(x,y,ν)\tilde{N}_{n}(x,y,\nu) or the Dn(x,y,ν)D_{n}(x,y,\nu).

Then:

G~n(1)(x,y,ν)=N~n(x,y,ν)Dn(x,y,ν).\tilde{G}_{n}^{(1)}(x,y,\nu)=\dfrac{\tilde{N}_{n}(x,y,\nu)}{D_{n}(x,y,\nu)}. (27)
Proof..

We begin the proof with the denominators.

Let 𝒟n(t)=𝒟n(x,y,ν,t){\cal D}_{n}(t)={\cal D}_{n}(x,y,\nu,t) for short and let 𝒟2(t)=𝒟1(t)=0{\cal D}_{-2}(t)={\cal D}_{-1}(t)=0.

The sequence {𝒟n(t)}n0\{{\cal D}_{n}(t)\}_{n\geq 0} is generated by the one-term recurrence:

𝒟n(t)=(t2ddt)𝒟n1(t).{\cal D}_{n}(t)=\left(t^{2}\dfrac{\rm d}{{\rm d}t}\right){\cal D}_{n-1}(t). (28)

Then we show that 𝒟n(t){\cal D}_{n}(t) satisfies, for n0n\geq 0:

𝒟n+1(t)\displaystyle{\cal D}_{n+1}(t) =\displaystyle= (xt2+(ν+1+2n)ty)𝒟n(t)\displaystyle(xt^{2}+(\nu+1+2n)t-y){\cal D}_{n}(t) (29)
+\displaystyle+ (2ntyn(n1)t2n(ν+1)t2)Dn1(t)n(n1)t2yDn2(t).\displaystyle(2nty-n(n-1)t^{2}-n(\nu+1)t^{2})D_{n-1}(t)-n(n-1)t^{2}yD_{n-2}(t).

For n=0n=0:

𝒟1(t)\displaystyle{\cal D}_{1}(t) =\displaystyle= (t2ddt)𝒟0(t)\displaystyle\left(t^{2}\dfrac{\rm d}{{\rm d}t}\right){\cal D}_{0}(t) (30)
=\displaystyle= (t2ddt)tν+1ext+y/t\displaystyle\left(t^{2}\dfrac{\rm d}{{\rm d}t}\right)t^{\nu+1}e^{xt+y/t}
=\displaystyle= (xt2+(ν+1)ty)𝒟0(t).\displaystyle(xt^{2}+(\nu+1)t-y){\cal D}_{0}(t).

The induction argument assumes:

𝒟k+1(t)\displaystyle{\cal D}_{k+1}(t) =\displaystyle= (xt2+(ν+1+2k)ty)𝒟k(t)\displaystyle(xt^{2}+(\nu+1+2k)t-y){\cal D}_{k}(t) (31)
+\displaystyle+ (2ktyk(k1)t2k(ν+1)t2)Dk1(t)k(k1)t2yDk2(t).\displaystyle(2kty-k(k-1)t^{2}-k(\nu+1)t^{2})D_{k-1}(t)-k(k-1)t^{2}yD_{k-2}(t).

Differentiating and multiplying by t2t^{2}, we obtain:

𝒟k+2(t)\displaystyle{\cal D}_{k+2}(t) =(xt2+(ν+1+2k)ty)𝒟k+1(t)\displaystyle=(xt^{2}+(\nu+1+2k)t-y){\cal D}_{k+1}(t)
+(2xt3+(ν+1+2k)t2+2ktyk(k1)t2k(ν+1)t2)Dk(t)\displaystyle+(2xt^{3}+(\nu+1+2k)t^{2}+2kty-k(k-1)t^{2}-k(\nu+1)t^{2})D_{k}(t)
+(2kt2y2k(k1)t32k(ν+1)t3k(k1)t2y)Dk1(t)\displaystyle+(2kt^{2}y-2k(k-1)t^{3}-2k(\nu+1)t^{3}-k(k-1)t^{2}y)D_{k-1}(t)
2k(k1)t3yDk2(t).\displaystyle-2k(k-1)t^{3}yD_{k-2}(t). (32)

Multiplying (31) by 2t2t and subtracting it from (32), we obtain, after simplification:

𝒟k+2(t)\displaystyle{\cal D}_{k+2}(t) =(xt2+(ν+1+2(k+1))ty)𝒟k+1(t)\displaystyle=(xt^{2}+(\nu+1+2(k+1))t-y){\cal D}_{k+1}(t)
+(2(k+1)tyk(k+1)t2(k+1)(ν+1)t2)Dk(t)k(k+1)t2yDk1(t),\displaystyle+(2(k+1)ty-k(k+1)t^{2}-(k+1)(\nu+1)t^{2})D_{k}(t)-k(k+1)t^{2}yD_{k-1}(t), (33)

which proves (29) by induction.

As with the denominator, let 𝒩n(t)=𝒩n(x,y,ν,t){\cal N}_{n}(t)={\cal N}_{n}(x,y,\nu,t) and 𝒩~n(t)=𝒩~n(x,y,ν,t)\tilde{\cal N}_{n}(t)=\tilde{\cal N}_{n}(x,y,\nu,t) for short.

Since 𝒩0(t)=F(t)𝒟0(t){\cal N}_{0}(t)=F(t){\cal D}_{0}(t):

𝒩1(t)\displaystyle{\cal N}_{1}(t) =\displaystyle= F(t)𝒟1(t)+t2f(t)𝒟0(t)\displaystyle F(t){\cal D}_{1}(t)+t^{2}f(t){\cal D}_{0}(t) (34)
=\displaystyle= F(t)𝒟1(t)+t2.\displaystyle F(t){\cal D}_{1}(t)+t^{2}.

But we now write F(t)=𝒩0(t)𝒟0(t)F(t)=\dfrac{{\cal N}_{0}(t)}{{\cal D}_{0}(t)} to conclude that:

𝒩1(t)=(t2x+(ν+1)ty)𝒩0(t)+t2.{\cal N}_{1}(t)=(t^{2}x+(\nu+1)t-y){\cal N}_{0}(t)+t^{2}. (35)

Differentiating and multiplying by t2t^{2}, we obtain:

𝒩2(t)=(t2x+(ν+1)ty)𝒩1(t)+(2xt3+(ν+1)t2)𝒩0(t)+2t3.{\cal N}_{2}(t)=(t^{2}x+(\nu+1)t-y){\cal N}_{1}(t)+(2xt^{3}+(\nu+1)t^{2}){\cal N}_{0}(t)+2t^{3}. (36)

Multiplying (35) by 2t2t and subtracting it from (36), we obtain, after simplification:

𝒩2(t)=(t2x+(ν+3)ty)𝒩1(t)+(2yt(ν+1)t2)𝒩0(t).{\cal N}_{2}(t)=(t^{2}x+(\nu+3)t-y){\cal N}_{1}(t)+(2yt-(\nu+1)t^{2}){\cal N}_{0}(t). (37)

But this is just (29) for n=1n=1 with the labels 𝒩n(t){\cal N}_{n}(t) interchanged for 𝒟n(t){\cal D}_{n}(t). Any further differentiation and multiplication by t2t^{2} will, therefore, ultimately lead to the same four-term recurrence relation, the difference being different initial conditions.

As a sequences:

{𝒩~n(t)}n0={𝒩n(t)F(t)𝒟n(t)}n0,\{\tilde{\cal N}_{n}(t)\}_{n\geq 0}=\{{\cal N}_{n}(t)-F(t){\cal D}_{n}(t)\}_{n\geq 0}, (38)

is a linear combination of both solutions 𝒩n(t){\cal N}_{n}(t) and 𝒟n(t){\cal D}_{n}(t).

Therefore, 𝒩~n(t)\tilde{\cal N}_{n}(t) satisfies (29) as well with the labels appropriately interchanged.

To complete the proof, we must return to the original sequences:

{N~n(x,y,ν)}n0and{Dn(x,y,ν)}n0.\{\tilde{N}_{n}(x,y,\nu)\}_{n\geq 0}\qquad\textrm{and}\qquad\{D_{n}(x,y,\nu)\}_{n\geq 0}. (39)

Indeed, the relationship is that:

Qn(x,y,ν)=𝒬n(x,y,ν,1)n!,Q_{n}(x,y,\nu)=\dfrac{{\cal Q}_{n}(x,y,\nu,1)}{n!}, (40)

where the Qn(x,y,ν)Q_{n}(x,y,\nu) stand for either the N~n(x,y,ν)\tilde{N}_{n}(x,y,\nu) or the Dn(x,y,ν)D_{n}(x,y,\nu) and the 𝒬n(x,y,ν,t){\cal Q}_{n}(x,y,\nu,t) stand for either the 𝒩~n(x,y,ν,t)\tilde{\cal N}_{n}(x,y,\nu,t) or the 𝒟n(x,y,ν,t){\cal D}_{n}(x,y,\nu,t). ∎

3 Figures

In Figures 1 and 2, the relative error of the GG transformation is shown for several different values of xx, yy, and ν\nu. This figure shows the excellent stability and convergence properties of the improved algorithm that come with the use of the stable four-term recurrence relation.

Refer to caption
Figure 1: Plots of the relative error of the GG transformation for K3(4,y)K_{3}(4,y), y=0,2,4y=0,2,4.
Refer to caption
Figure 2: Plots of the relative error of the GG transformation for K0(10,y)K_{0}(10,y), y=0,5,10y=0,5,10.

4 Conclusion

We improve an existing algorithm for incomplete Bessel functions based on the GG transformation [15] by developing a recurrence relation the numerator sequence and the denominator sequence whose ratio form the sequence of approximations. By finding this recurrence relation, we reduce the complexity from 𝒪(n4){\cal O}(n^{4}) to 𝒪(n){\cal O}(n). We plot relative error showing that the algorithm is capable of extremely high accuracy for incomplete Bessel functions. The stability and convergence appear to be remarkable.

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