Short-time-to-expiry expansion for a digital European put option under the CEV model November 1, 2017
Abstract In this paper I present a short-time-to-expiry asymptotic series expansion for a digital European put option in the CEV-model. In contrast to the digital European put in the geometric Brownian motion model presented in the lecture, this problem does not seem to have an analytic solution. I calculate all terms up to fourth order and use the resulting approximation to obtain the price of the option at earlier times numerically via a finite difference scheme. 1 Introduction and formulation of the problem In the CEV 1 -model the dynamics of an asset S is assumed to obey the stochastic differential equation ds t S t = rdt + σ St Ŝ ) α dw t 1) where r is the risk-free rate, σ 0 is a constant parameter that controls the volatility, Ŝ is a fixed reference price, W t is a Brownian motion and α 1 is the elasticity. For α 0 the volatility depends on the current value of the asset price, i. e. the CEV model is a local volatility model. For α = 0, Eq. 1) reduces to a simple geometric Brownian motion with constant variance σ. The CEV-model tries to capture the so-called leverage effect which is often observed in commodity and equity markets. In commodity markets, the volatility is often found to increase for large asset prices, whereas in equity markets the inverse behavior is observed [1]. The former case corresponds to α > 0 and the latter case to α < 0 in the model. The threshold that defines whether an asset price is small or large is controlled by the reference price Ŝ. There exist analytical solutions for European vanilla calls and puts in the CEV-model, c. f. Ref. [1], but there seem to exist no analytical solutions for the digital versions. The corresponding pricing equation for a derivative V S) in CEV-model can be obtained in the usual way via a hedging argument. We consider a portfolio Π t containing one option V t and t assets, Π t = V t S t. 2) According to Ito s lemma, the dynamics of Π t is given by V dπ t = t + rs V t S r ts t + 1 2 σ2 St 2 σ ) ) V 2α + σs t S St t dw t. Ŝ 1 Constant elasticity of variance St Ŝ ) ) 2α 2 V S 2 dt 3) 1
For t = V S, t)/ S, the portfolio becomes instantaneously risk-free and must therefore grow with the risk-free rate, i. e. dπ t = rπ t dt 4) must hold. Combining Eqs. 2), 3) and 4) for this choice of t we obtain V t t + 1 ) ) 2α 2 σ2 St 2 St 2 V V t + rs t rv t dt = 0. 5) Ŝ S t S 2 t As this must hold for all attainable values of S t and t, the pricing function V S, t) must obey the pricing equation V t + 1 ) S 2α 2 σ2 S 2 2 V V + rs Ŝ S2 S for all S > 0 and 0 t T. In this paper we will focus on a digital European put with payoff rv = 0 6) P S) = HS K) 7) where 1 x 0 Hx) = 0 x < 0 is the Heaviside function. This payoff implies the boundary condition 8) to the solution V S, t) of the pricing equation 6). V S, T ) = HS K) 9) In contrast to the α = 0 case discussed in the lectures, there seems to exist no analytic solution for this problem. The aim of this paper is to find an asymptotic series expansion for V S, t) in the limit t T which can be used calculate the value of the option at earlier times numerically via a finite difference scheme. The structure of this paper is as follows: The asymptotic expansion is derived in section 2. Section 3 deals with the numerical extension of the asymptotic solution to earlier times. Finally, this paper concludes with a short summary in section 4. 2 Asymptotic series expansion 2.1 Definition of an asymptotic series If fx) is a given function then the sum n S n x) = a j φ j x) 10) is called an asymptotic series expansion for fx) in the limit x x 0, iff 2 j=0
1. φ j x) ) is an asymptotic sequence as x x 0, i. e. φ j x) = o φ j 1 x) ) for j = 1, 2,..., n 11) and 2. for each m = 0, 1, 2,..., n m fx) a j φ j x) = oφ m x)) as x x 0 12) j=0 It is important to note that S n x) will converge to fx) for fixed n as x x 0, whereas for fixed x, S n x) will no necessarily converge to fx) as n. This implies that adding more terms to the expansion does not necessarily improve the quality of the approximation. Our goal is to perform an expansion as in Eq. 10) for the solution to the pricing equation 6) with the time-to-expiry as expansion parameter. In order to do so, it is convenient to simplify the problem first. 2.2 Simplification of the problem We introduce the dimensionless parameters S = S K, t = σ 2 K Ŝ ) 2α T t), V S, t ) = V S, t). In terms of these variables, the pricing problem described by Eqs. 6) and 9) becomes where k is given by V t = 1 2 S 2+2α 2 V V + ks S 2 S kv, V S, 0) = H1 S ), 13) Eq. 13) can be simplified further by the the transformation which eliminates the discounting term and yields ) 2α k = r Ŝ σ 2 14) K V S, t ) = e kt US, t ), 15) U t = 1 2 S 2+2α 2 U U + ks S 2 S, US, 0) = H1 S ) 16) 3
2.3 Definition of the short-time-to-expiry We assume that the parameter t can be written as t = ɛ 2 where is assumed to be O1) and ɛ is the small parameter that will be used for the expansion. Hence, short-time-to-expiry means that t is Oɛ 2 ). The transformation changes Eq. 16) to t = ɛ 2, US, t ) = us, ) 17) 1 u ɛ 2 = 1 2 S 2+2α 2 u u + ks S 2 S, us, 0) = H1 S ) 18) It can be shown that a regular series expansion us, ; ɛ) = u 0 S, ) + ɛu 1 S, ) + ɛ 2 u 2 S, ) +... 19) fails for this problem as it eliminates the second derivative term from Eq. 16) for the leading order PDE for u 0 x, ). The problem is thus singular and has to be addressed with a different approach. 2.4 Definition of the inner region To proceed with the singular perturbation problem one can perform another variable transformation x = S 1, us, ) = vx, ) 20) ɛ and define an inner region as the region where x is O1). In terms of these variables Eq. 18) can be written as v = 1 2 1 + ɛx)2+2α 2 v v + ɛk1 + ɛx), vx, 0) = H x), 21) x2 x i. e. the second derivative is not eliminated in this case. 2.5 Derivation of PDEs for the terms of the asymptotic expansion We now expand the solution vx, ) as well as the term 1 + ɛx) 2+2α in powers of ɛ, vx, ; ɛ) = v 0 x, ) + ɛv 1 x, ) + ɛ 2 v 2 x, ) +..., 22) 1 + ɛx) 2+2α = c n ɛ n x n, 23) where c n is given by c 0 = 1, 24) c n = 1 n 2 + 2α k 1)), n! n 1 25) k=1 4
and plug Eq. 22) and Eq. 23) in Eq. 21). Equating equal powers of ɛ yields the set of coupled PDEs ɛ 0 : ɛ 1 : ɛ 2 : ɛ 3 :. ɛ n : v 0 1 v 0 2 x 2 = 0 26) v 1 1 v 1 2 x 2 = 1 2 c 1x v 0 x 2 + k v 0 x 27) v 2 1 v 2 2 x 2 = 1 2 c 2x 2 v 0 x 2 + 1 2 c 1x 2 v 1 x 2 + kx v 0 x + k v 1 x 28) v 3 1 v 3 2 x 2 = 1 2 c 3x 3 v 0 x 2 + 1 2 c 2x 2 2 v 1 x 2 + 1 2 c 1x 2 v 2 x 2 + kx v 1 x + k v 2 x 29) v n 1 v n 2 x 2 = 1 n 2 j=1 c j x j 2 v n j x 2 + kx v n 2 x + k v n 1 x 30) with the boundary conditions H x) n = 0 v n x, 0) =. 31) 0 n > 0 This problem is much more complicated than the α = 0 case discussed in the lecture. Here, all previous solutions v 0 x, ) to v n 1 x, ) appear as inhomogeneous terms on the right hand side of the PDE for v n x, ), c. f. Eq. 30), whereas only the two previous solutions v n 1 x, ) and v n 2 x, ) appear for α = 0. As the series in Eq. 25) terminates at n = 2 for α = 0, Eq. 30) reduces to the PDE given in the lecture notes for the case α = 0 [2]. 2.6 Solution for n = 0 The leading order PDE 26) is the same as for the α = 0 case and has the α-independent) solution v 0 x, ) = N x ) 32) which can be easily verified. To zeroth order, the short-time-to-expiry approximation is thus indifferent to the elasticity α. 2.7 Solution for n = 1 As c 1 = 21 + α), we expect the solution v 1 x, ) of Eq. 27) to depend on α as well. In order to solve Eq. 27) we introduce the Green s function Gx, ) = e x2 /2) 2π 33) 5
with the properties G = 1 2 G 2 x 2, G x ) x = Gx, ). 34) In terms of this function we can express the derivatives on the right hand side of Eq. 27) as v 0 x = Gx, ), 2 v 0 x 2 = x ) Gx, ) 35) and hence Eq.27) becomes v 1 1 ) v 1 1 + α 2 x 2 = x 2 k Gx, ). 36) The properties of Gx, ) motivate an ansatz of the form v 1 x, ) = A 1,0 ) + A 1,2 )x 2) Gx, ). 37) Plugging this ansatz into Eq. 27) yields [ ) A1,0 ) A A1,2 1,2) + k + ) + 2A 1,2) which finally gives us the two coupled ODEs A 1,0 ) ] 1 + α) x 2 Gx, ) = 0 38) ) A 1,2) + k = 0, 39) 2 A 1,2 ) ) 1 + α) = 0 40) with boundary condition A 1,0 0) = 0 and A 1,2 0) determined by a minimum singularity condition. The solutions are given by 1 + α A 1,0 ) = 2 A 1,2 ) = 1 + α 2 and thus v 1 x, ) is given by 1 + α v 1 x, ) = 2 ) k ) k, 41) 42) + 1 + α ) x 2 Gx, ). 43) 2 Note that for α = 0 the solution reduces to the result given in the lecture [2]. 2.8 Solution for n = 2 The derivatives on the right hand side of the PDE 28) can again be expressed in terms of Gx, ), v 2 1 [ v 2 2k 2 2 x 2 = + α 1)k + 1 + α)3 + α) x 2 1 + α)5 + k + 2α) x 3 1 + α)3 + 2 2 2 x 5 ] Gx, ), 44) 6
which motivates an ansatz of the form v 2 x, ) = A 2,1 )x + A 2,3 )x 3 + A 2,5 )x 5) Gx, ). 45) Plugging this in Eq.28) to yields the following set of coupled ODEs: A 2,1 ) + A 2,1) A 2,3 ) + 3A 2,3) 3A 2,3 ) = 1 2 2k 2 + kα 1) + 1 + α)3 + α) ), 46) 10A 2,5 ) = 1 1 + α)5 + k + 2α), 47) 2 A 2,5 ) + 5A 2,5) = 1 + α) 2. 48) The solutions can be obtained e. g. with a symbolic algebra software and are given by The complete solution v 2 x, ) is thus given by A 2,1 ) = 1 4k + 4k2 + 4α + 3α 2, 8 49) α + 2k 5)1 + α) A 2,3 ) =, 12 50) 1 + α)2 A 2,5 ) =. 8 51) [ 1 4k + 4k 2 + 4α + 3α 2 v 2 x, ) = x + 8 + ] 1 + α)2 x 5 Gx, ) 8 α + 2k 5)1 + α) x 3 12 52) 2.9 Solution for n 3 In order to solve the PDE 30) for n 3 it is reasonable to proceed systematically. The forms of v 1 x, ) and v 2 x, ) imply that the derivatives occurring on the right hand side of the PDE for n = 3, c. f. Eq. 29), can again be expressed in terms of Gx, ) by using Eq. 34). A detailed analysis shows that the right hand side is of the functional form px, )Gx, ) 53) where px, ) is a polynomial of finite degree in x with coefficients depending on. This motivates the ansatz v 3 x, ) = qx, )Gx, ) 54) where qx, ) is a again a polynomial in x of the same degree as px, ) with coefficients depending on and vanishing coefficients in px, ) implying vanishing coefficients in qx, ). 7
In particular, we find for n = 3: px, ) = 1 4k 3 t 4k 2 t + 3α 2 kt + 4αkt + kt ) 8 + 1 α 3 15α 2 35α 4k 3 4αk 2 + 8k 2 α 2 k 8αk 7k 21 ) x 2 8 23α 3 + 127α 2 + 217α + 4αk 2 + 4k 2 + 13α 2 k + 26αk + 13k + 113 ) + 24t 19α 3 79α 2 101α 3α 2 k 6αk 3k 41 ) + α 3 + 3α 2 + 3α + 1 ) + 24t 2 x 6 8t 3 x 8 55) qx, ) = A 3,0 ) + A 3,2 )x 2 + A 3,4 )x 4 + A 3,6 )x 6 + A 3,8 )x 8 56) Noting that a term A n,k )x k Gx, ) transforms under the operator L := 1 2 ) 2 x 2 as for k = 0, as for k = 1 and as [ ] L A n,0 )Gx, ) x 4 57) = A n,0 )Gx, ) 58) [ ] An,1 L A n,1 )xgx, ) = ) + x ) Gx, ) 59) [ ] L A n,k )x k Gx, ) = [ An,k ) + ka ) n,k) x k 12 ] kk 1)A n,k)x k 2 Gx, ) 60) for k 2, we find that the left hand side of Eq. 29) is of the same functional form as the right hand side, c. f. Eq. 53). By matching the coefficients of x in the two polynomials one can derive a set of coupled ODEs for the coefficients A n,k ). It is tedious, but straightforward, to solve these ODEs using a symbolic algebra package to obtain the solution v 3 x, ), which is given in the appendix. The necessary steps to calculate v n x, ) can be summarized as follows: 1. Insert the already calculated solutions {v 0 x, ),..., v n 1 x, )} in the right hand side of Eq. 30) to obtain px, ). 2. Make an appropriate ansatz for qx, ) and calculate the left hand side of Eq. 30) using Eqs. 58), 59) and 60). 8
n=0 n=1 n=2 n=3 n=4 ɛ = 0.1 ɛ = 0.01 error 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0-0.002 0.6 0.8 1 1.2 1.4 S 7e-05 6e-05 5e-05 4e-05 3e-05 2e-05 1e-05 0-1e-05 0.9 0.95 1 1.05 1.1 S Figure 1: Error of the short-time-to-expiry approximation including the first n terms for K = S 0 = 1, σ = 0.3, r = 0.05, T t = ɛ/σ) 2 this implies = 1) and α = 0 for ɛ = 0.1 left figure) and ɛ = 0.01 right figure). The maximum error for the n = 0 approximation for ɛ = 0.01 is smaller than 0.0014 not shown in the figure). 3. Match the coefficients of x in Eq. 30) to obtain a set of coupled ODEs for the coefficients {A n,k )}. 4. Solve the ODEs and obtain the solution v n x, ). This algorithm can be iterated to calculate the solutions up to arbitrary order n. solutions up to n = 4 are given in the appendix. The 2.10 Discussion of the series-expansion We first focus on the case α = 0 for which there exists an analytic solution that can be used as a reference. Figure 1 shows the error of short-time-to-expiry approximation for ɛ = 0.1 and ɛ = 0.01. One can observe that while the error decreases drastically if the first term is included in the expansion, it does decrease significantly further if higher order terms are included. As expected the error is smaller for ɛ = 0.01 than for ɛ = 0.1 Figure 2 shows the influence of a non-zero elasticity α on the value of the option. One can clearly observe that, compared to the α = 0 case, the option price increases for α > 0, whereas it decreases for α < 0. This behavior can be explained as follows: For α > 0 the volatility increases as the asset price increases above Ŝ, which we set equal to K. Hence the probability for an out-of-the-money put to run in-the-money increases. At the same time, the probability for an in-the-money put to remain in-the-money increases as well since the 9
V S, t) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 α = 1 α = 0.5 anal. α = 0 α = +1 α = +2 α = +3 0 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 S Figure 2: The short-time-to-expiry approximation for K = S 0 = 1, σ = 0.3, r = 0.05, T t = ɛ/σ) 2 this implies = 1), ɛ = 0.1 and different values of α. volatility becomes smaller for S < Ŝ. Consequently, the value of the option must be larger than for α = 0 and it must increase with α. For α < 0, on the other hand, the opposite is true and thus the value is expected to decrease with decreasing α. 3 Numerical extension of the short-time-to-expiry approximation to earlier times The aim of this section is to show that the short-time-to-expiry approximation derived in section 2 can be used to improve the accuracy of numerical methods. The underlying idea is to replace the discontinuous payoff of the option with the smooth short-time-to-expiry approximation and to work out the solution at earlier times numerically. In this paper we focus on finite differences which is the method-of-choice for the pricing problem considered here due to its low dimension. It should however be noted that the approach presented in this section would also work with Monte-Carlo. In particular, it would allow the calculation of sensitivities with the pathwise sensitivities method. 3.1 Crank-Nicolson scheme for the CEV-model The Crank-Nicolson finite-difference scheme for the standard Black-Scholes model has to be altered to account for the elasticity in the CEV-model. Starting with the pricing equation 6) one can e. g. follow the procedure shown in Ref. [3] to discretize the derivatives on a lattice 10
with N points in S-direction and M points in t-direction. The resulting scheme can be written as a n V m 1 n 1 + b nv m 1 n with coefficients given by a n = 1 ) S 2α 4 tn2+2α σ 2 + 1 S 0 4 nr t, A n = a n, b n = 1 + 1 ) S 2α 2 t n 2+2α σ 2 + r) S 0 + c n V m 1 n+1 = A nv m n + B n V m n + C n V m n+1 61), B n = 1 1 2 t a n = 1 ) S 2α 4 tn2+2α σ 2 1 S 0 4 nr t, C n = c n, ) S 2α n 2+2α σ 2 + r), S 0 where V m n = V S n, t m ) denotes the value of the pricing function on the lattice point S n, t m ) and S and t are the step sizes in S and t direction. The scheme in Eq. 61) can be used to work out the value of the digital put option at all times t < T step-by-step backwards in time starting from the discretized payoff c. f. Eq. 9). V M n = HS n K), 62) Due to the discontinuity of the payoff at S = K the derivatives with respect to S cannot be approximated reasonably by finite differences at this point which limits the accuracy of the solution. By differentiating Eq. 6) and discretizing the derivatives as above, one can also derive the Crank-Nicolson scheme for the Delta of the option, a n m 1 n 1 + b n m 1 n + c n m 1 n+1 = A n m n + B n m n + C n m n+1 63) with coefficients given by a n = 1 ) S 2α 4 tn2+2α σ 2 + 1 ) S 2α S 0 4 n t 1 + α)n 2α σ 2 + r), S 0 b n = 1 + 1 ) S 2α 2 t n 2+2α σ 2 + r), S 0 c n = 1 ) S 2α 4 tn2+2α σ 2 1 S 0 4 n t A n = a n, B n = 1 1 ) S 2α 2 t n 2+2α σ 2 + r), S 0 C n = c n. 1 + α)n 2α σ 2 S S 0 ) 2α + r), 11
The terminal condition for M n = S n, t M ) can be obtained by differentiating the payoff Eq. 62) with respect to S n and is given by S n, t M ) = δs n K). 64) In principle, the calculation of the Delta in its own finite-difference scheme is more accurate than numerical differentiation of the option price obtained by the scheme in Eq. 61) as numerical differentiation amplifies the error of the solution. The δ-distribution in Eq. 64), however, implies that the finite difference scheme shown in Eq. 63) cannot be used to obtain the Delta of this particular option. 3.2 Solution with short-time-to-expiry approximation as boundary condition To overcome the problem with the boundary condition shown in Eq. 64) one can treat the smooth short-time-to-expiry approximation derived in section 2 as payoff of the option close to expiry and work out the solution at earlier times from there using the finite difference schemes given above. I implemented both finite difference schemes in Python and calculated the price and Delta of the option using the short-time-to-expiry approximation with n = 4 terms and ɛ = 0.1 as terminal condition. Figure 3 shows the error of this approach for α = 0 relative to the analytic solution. For comparison the the error of a finite difference scheme with the original discontinuous payoff is shown as well. One can clearly observe that using the short-time-to-expiry approximation with n 1 terms as terminal condition decreases the error in the price of the option significantly compared to using the original discontinuous payoff. Furthermore, this procedure allows us to calculate the Delta of the option by finite differences at all. Figures 4 and 5 show the value of the option and its Delta using the short-time-expiry approximation with n = 4 terms and ɛ = 0.1 as terminal condition for different values of α. One can again observe that the price of the option increases with α as it was already explained in section 2.10. The α-dependence of the price is also reflected in a slightly modified Delta compared to the α = 0 case, c. f. Fig. 5. It should again be emphasized that the Delta of the option cannot be obtained by finite differences if the original discontinuous payoff is used as terminal condition. 12
error 0.0005 0-0.0005-0.001-0.0015-0.002-0.0025 STA n = 0) STA n = 1) VS, t) -0.003 0 2 4 6 8 10 S STA n = 2) STA n = 3) 0.0025 0.002 0.0015 0.001 0.0005 0-0.0005-0.001-0.0015-0.002-0.0025-0.003 STA n = 4) disc. payoff S, t) 0 0.5 1 1.5 2 S Figure 3: Error of the option price V S, t) and its Delta S, t) for a Crank-Nicolson finite difference scheme with N = 1200 steps in S-direction and M = 500 steps in t-direction for the original discontinuous payoff and M = 490 steps for the short-time-to-expiry approximation as payoff, such that the step-size t is kept constant for both approaches. The other parameters are α = 0, r = 0.05, σ = 0.3, ɛ = 0.1, T t = 1/σ 2, K = Ŝ = 1. The Delta is not shown for the discontinuous payoff on the right figure as it cannot be calculated using Eq. 63) in this case. 13
0.6 0.5 0.4 α = 0.3 α = 0.2 α = 0.1 anal. α = 0 α = +0.1 α = +0.2 α = +0.3 V S, t) 0.3 0.2 0.1 0 0 2 4 6 8 10 S Figure 4: The value of the option obtained by using the short-time-to-expiry approximation with n = 4 terms and ɛ = 0.1 as terminal condition for the finite-difference scheme Eq. 61). Parameters are r = 0.05, σ = 0.3, T t = 1/σ 2, K = Ŝ = 1. 0-0.05-0.1-0.15 S, t) -0.2-0.25-0.3-0.35-0.4-0.45 α = 0.3 α = 0.2 α = 0.1 anal. α = 0 α = +0.1 α = +0.2 α = +0.3 0 2 4 6 8 10 S Figure 5: The Delta of the option obtained by using the short-time-to-expiry approximation with n = 4 terms and ɛ = 0.1 as terminal condition for the finite-difference scheme Eq. 63). Parameters are r = 0.05, σ = 0.3, T t = 1/σ 2, K = Ŝ = 1. 14
4 Summary In this paper I calculated the first four terms of a short-time-to-expiry series expansion of a digital European put in the CEV-model. For the special case of vanishing elasticity the series expansion has been compared to the analytic solution and shown very good agreement. The effect of a non-vanishing elasticity on the price and Delta of the option has been discussed and explained in detail. Finally, the short-time-to-expiry approximation was used as terminal condition in a finite difference scheme to obtain the price and the Delta of the option at earlier times with high accuracy. The results are of great importance as finite difference methods usually run into problems when applied to discontinuous payoffs, in particular if sensitivities have to be calculated. The work presented here could be extended straightforward to obtain e. g. the Gamma of the option. Additionally, one could analyze the optimal ratio between the expansion parameter ɛ and the number of discretization points for the finite difference scheme. This is, however, beyond the scope of this work which aims at a proof-of-concept of the presented method. References [1] J. Hull, Options, Futures and Other Derivatives, 7th edition, Pearson/Prentice Hall 2009) [2] J. Dewynne, Asymptotic methods in finance, Lecture Notes 2017). [3] C. Reisinger, Finite Differences, Lecture Notes 2017). 15
Appendix Solution for n = 3 The solution for n = 3 is given by v 3 x, ) = A 3,0 x, ) + A 3,2 x, )x 2 + A 3,4 x, )x 4 + A 3,6 x, )x 6 + A 3,8 x, )x 8) Gx, ) with coefficients given by A 3,0 x, ) = 1 48 2 3α 3 7α 2 15α + 8k 3 2α + 1)k 2 + 2 5α 2 + 4α 1 ) k 5 ), A 3,2 x, ) = 1 24 3α 3 + 7α 2 + 15α + 4k 3 + 2α 5)k 2 + α 2 + 4α + 5 ) k + 5 ), A 3,4 x, ) = 1 24 α + 1) 2 α 2 + α + 4 ) + k 2 + α + 1)k ), A 3,6 x, ) = α + 1)2 5α + 3k + 27), 120 α + 1)3 A 3,8 x, ) = 48 2 Solution for n = 4 The solution for n = 4 is given by with coefficients given by v 4 x, ) = A 4,1 x, )x + A 4,3 x, )x 3 + +A 4,5 x, )x 5 + A 4,7 x, )x 7 + A 4,9 x, )x 9 + A 4,11 x, )x 11) Gx, ) A 4,1 x, ) = 1 384 2 αα152 45α)α + 370) + 216) + 48k 4 96k 3 + 83αα + 3) + 8)k 2 8α + 1)6α 2 2α 3)k + 43), A 4,3 x, ) = 1 384 α + 1)αα21α 125) + 27) 3) + 16k4 + 16α 5)k 3 8α 2 + α 6)k 2 + 8αα + 1)3α + 2)k), A 4,5 x, ) = 1 960 α + 1)α173 7α)α + 91) + 8k3 + 45α 3)k 2 4α 11)α 4)k + 183), A 4,7 x, ) = α + 1)2 5α119α + 632) + 84k 2 + 485α + 134)k + 5785 ), 20160 A 4,9 x, ) = α + 1)3 49α + 8k + 153) 2688 2, α + 1)4 A 4,11 x, ) = 384 3 16