Two algorithms for computing the Randles-Sevcik function from electrochemistry Chelo Ferreira1 , Jos´e L. L´opez2 and Pedro J. Miana3
1 Departamento de Matem´atica Aplicada.
Universidad de Zaragoza.
e-mail:
cfer-
[email protected]
2 Departamento de Matem´atica e Inform´atica. Universidad P´ublica de Navarra. e-mail:
[email protected]
3 Departamento de Matem´aticas. Universidad de Zaragoza. e-mail:
[email protected] Abstract √ We derive two expansions of the Randles-Sevcik function πχ(x): an √ asymptotic expansion of πχ(x) for x → ∞ and its Taylor expansion at any x0 ∈ R. These expansions are accompanied by error bounds for the remainder at any order of the approximation.
2000 Mathematics Subject Classification: 41A60, 30E20 Keywords & Phrases: Randles-Sevcik function, Taylor expansions, asymptotic expansions, error bounds.
1
Introduction
√ The Randles-Sevcik function πχ(x) arises in electrochemistry. This function characterizes the dependence of electric current on cell voltage under certain electrolysis conditions [2],[10],[12]. Several authors have investigated its analytical properties and different methods for computation, see for example [4], [5], [6], [7], [8] and [11]. √ Reinmouth gave a rapidly convergent series for πχ(x) which permits its evaluation for negative values of x [11]. However, the chemically interesting region is contained in the positive real axis. The region x ' 1.1 is particularly interesting because this function has a relatively steep peak there [4], [8]. In order to approximate √ πχ(x) in this region, Oldham, by using Weyl fractional calculus, provided a series reformulation of the Randles-Sevcik function on the whole real axis [8], [9]: √ πχ(x) =
µ ¶1 X π 2 ∞
2
γn (x),
x ∈ R,
(1)
n=1 1
1
where γn (x) = βn−3 (βn − x) 2 (βn + 2x), βn = (x2 + b2n ) 2 and bn = (2n − 1)π. But the convergence of Oldham’s seris is extremely slow [7]. More recently, Lether has 1
obtained a integral representation of the Randles-Sevcik function from which he has √ derived a two-parameter expansion of πχ(x) valid ∀x ∈ R [7]. This expansion has the form √ πχ(x) = SN (x) + EN,K (x), (2) where SN is the N th partial sum of (1) and EN,K is an error correction term given by µ ¶ K X 1 π 2 2 2 −(k+ 41 ) EN,K (x) ≡ dk (x + 4N π ) sin (2k + )θN + , 2 4 k=0 where θN ≡ tan−1 (x/2N π) and the coefficients dk are defined in terms of Bernoulli numbers and the gamma function: dk ≡ 2(1 − 22k−1 )B2k π 2k−1
Γ(2k + 12 ) . (2k)!
Moreover, Lether presents a method for selecting appropriate values of N and K for √ computing πχ(x) to a specified number of decimal digits of accuracy. √ However, complete asymptotic expansions including error bounds of πχ(x) are not fully investigated. The purpose of this paper is to obtain an asymptotic √ expansions of πχ(x) for large x and its Taylor expansion at any point x0 ∈ R, in particular at x0 = 1.1. These expansions are derived in section 2. We obtain there easy algorithms to compute the coefficients of the expansions as well as error bounds at any order of the approximation. They are derived by using the ideas of Watson’s lemma [[13], chap 1]. We compare our expansion with Lether’s expansion and give numerical examples in section 3.
2
A Taylor Expansion and an Asymptotic Expansion
The starting point is the integral representation [7]: π 1/2 χ(x) =
Z ∞ 0
µ
t1/2 cschπt sin xt +
¶
π dt. 4
(3)
Using the exponential representation of the csch and sin functions, we can write this integral as ( ) Z ∞ t1/2 1/2 iπ/4 itx π χ(x) = 2= e e dt . (4) eπt − e−πt 0 Then, expansions of π 1/2 χ(x) follow immediately from expansions of the Fourier transform of the function t1/2 /(eπt − e−πt ): Z ∞ 0
t1/2 eitx dt. eπt − e−πt
2
(5)
Theorem 1 For n = 1, 2, 3, ..., the Taylor expansion of the integral (5) at x0 , convergent for |x − x0 | < π, is given by ¶
µ
Z ∞
n−1 X ik Γ (k + 3/2) t1/2 eitx 3 1 i dt = ζ k+ , − x0 (x − x0 )k + Rn (x0 , x), πt −πt k+3/2 e −e 2 2 2π k!(2π) 0 k=0 (6) where ζ(z, a) is the Hurwitz zeta function. The remainder term is bounded by
|Rn (x0 , x)| ≤
Γ(n + 3/2)ζ(n + 3/2) |x − x0 |n , n!π n+3/2
(7)
where ζ(z) is the Riemann zeta function. Proof. We write (5) in the form: Z ∞ 1/2 itx0 t e 0
eπt − e−πt
eit(x−x0 ) dt,
(8)
and consider the Taylor expansion of the exponential function at the origin: e
it(x−x0 )
=
n−1 X
[it(x − x0 )]k + rn (t, x − x0 ). k! k=0
(9)
Introducing this expansion in (8), interchanging sum and integral and using the integral representation of the Hurwitz zeta function [[14], 2.2.1, p.46], we obtain (6) with Z ∞ 1/2 itx0 t e Rn (x0 , x) ≡ rn (t, x − x0 )dt. eπt − e−πt 0 To derive (7), consider the explicit expression given by the Lagrange form for the remainder rn (t, x − x0 ) of the Taylor expansion (9): rn (t, x − x0 ) =
eiξ [it(x − x0 )]n , n!
ξ ∈ (0, t(x − x0 )),
n = 1, 2, 3, ....
Therefore |rn (t, x − x0 )| ≤ tn |x − x0 |n /n!, and 1 |Rn (x0 , x)| ≤ n!
Z ∞ 0
tn+1/2 dt|x − x0 |n . eπt − e−πt
Using that eπt − e−πt ≥ eπt − 1 ∀ t ≥ 0 and [[1], eq. 23.2.7], we obtain (7).
]
Corollary 1 For |x| < π, µ
¶
∞ 1 X (−1)[k/2] k+3/2 3 χ(x) = 2 (2 − 1)Γ (k + 3/2) ζ k + xk . k 2π k=0 k!(2π) 2
Proof. Set x0 = 0 in (6) and use (4) and [[1], eq. 23.2.20].
3
(10) ]
Theorem 2 For n = 2, 3, ..., an asymptotic expansion of the integral (5) for x → ∞ is given by √ Z ∞ t1/2 1 π itx e dt = √ + πt −πt 1/2 e −e 2 π(π − ix) 4(π − ix)3/2 0 (11) n−1 X (2π)2k−1 B2k Γ(2k + 1/2) + + Rn (x). (2k)!(π − ix)2k+1/2 k=1 The remainder term verifies n ³
´
´h ³
³
´i
2 4 1 + 3(22n−1 Γ 2n + 12 − Γ 2n + 21 , π|π − ix| −1) ´¾ ³ √ ´2n ³ 1 2 1 2 2 +C π π + x Γ 2n + 2 , π|π − ix| , |π−ix|2n+1/2
|Rn (x)| ≤
(12)
w where √ C is a bound of |w/(e − 1)| in the region W ≡ {w ∈ C, |w − t(π − ix)| < 2 2 2 π / π + x , 0 ≤ t < ∞} (see fig. 1 (a)). A possible value for C is given in [[3], eq. (16)].
Proof. From (5), Z ∞
Z
t1/2 1 ∞ −1/2 (−2πt) −(π−ix)t itx e dt. e dt = t eπt − e−πt 2π 0 e−2πt − 1 0 Consider the Taylor expansion [[1], eq. 23.1.1], n−1 X B2k −2πt = 1 + πt + (2πt)2k + rn (t), e−2πt − 1 (2k)! k=1
n = 1, 2, 4, 6, 8, ...
(13)
(14)
where rn (t) = O(t2n ) when t → 0+ and n = 2, 4, 6, 8, .... Introducing this expansion into the second integral in (13) and interchanging sum and integral, we obtain (11) with Z 1 ∞ −1/2 Rn (x) ≡ t rn (t)e−(π−ix)t dt. 2π 0 After the change of variable t → u/(π − ix), we obtain Rn (x) =
1 2π(π − ix)1/2
Z ∞(π−ix) 0
µ
u−1/2 rn
¶
u e−u du. π − ix
The integrand is an analytic function of u in the sector |Arg(u)−Arg(π − ix)| < π/2 and is exponentially small when u → ∞ with Arg(π − ix)
