Abstract

It is known that power series expansion of certain functions such as diverges beyond a finite radius of convergence. We present here an iterative power series expansion (IPS) to obtain a power series representation of that is convergent for all . The convergent series is a sum of the Taylor series of and a complementary series that cancels the divergence of the Taylor series for . The method is general and can be applied to other functions known to have finite radius of convergence, such as . A straightforward application of this method is to solve analytically nonlinear differential equations, which we also illustrate here. The method provides also a robust and very efficient numerical algorithm for solving nonlinear differential equations numerically. A detailed comparison with the fourth-order Runge-Kutta method and extensive analysis of the behavior of the error and CPU time are performed.

1. Introduction

It is well-known that the Taylor series of some functions diverge beyond a finite radius of convergence [1]. For instance, by way of example not exhaustive enumeration, the Taylor series of and diverge for and , respectively. Increasing the number of terms in the power series does not increase the radius of convergence; it only makes the divergence sharper. The radius of convergence can be increased only slightly via some functional transforms [2]. Among the many different methods of solving nonlinear differential equations [3–9], the power series is the most straightforward and efficient [10]. It has been used as a powerful numerical scheme for many problems [11–19] including chaotic systems [20–23]. Many numerical algorithms and codes have been developed based on this method [10–12, 20–24]. However, the above-mentioned finiteness of radius of convergence is a serious problem that hinders the use of this method to wide class of differential equations, in particular the nonlinear ones. For instance, the nonlinear Schrödinger equation (NLSE) with cubic nonlinearity has the as a solution. Using the power series method to solve this equation produces the power series of a , which is valid only for .

A review of the literature reveals that the power series expansion was exploited by several researchers [10–12, 20–24] to develop powerful numerical methods for solving nonlinear differential equations. Therefore, this paper is motivated by a desire to extend these attempts to a develop a numerical scheme with systematic control on the accuracy and error. Specifically, two main advances are presented in this paper: a method of constructing a convergent power series representation of a given function with an arbitrarily large radius of convergence and a method of obtaining analytic power series solution of a given nonlinear differential equation that is free from the finite radius of convergence. Through this paper, we show robustness and efficiency of the method via a number of examples including the chaotic Lorenz system [25] and the NLSE. Therefore, solving the problem of finite radius of convergence will open the door wide for applying the power series method to much larger class of differential equations, particularly the nonlinear ones.

It is worth mentioning that the literature includes several semianalytical methods for solving nonlinear differential equations; such as homotopy analysis method (HAM), homotopy perturbation method (HPM), and Adomian decomposition method (ADM); for more details see [26–29] and the references therein. Essentially, these methods generate iteratively a series solution for the nonlinear systems where we have to solve a linear differential equation at each iteration. Although these methods prove to be effective in solving most of nonlinear differential equations and in obtaining a convergent series solution, they have few disadvantages such as the large number of terms in the solution as the number of iterations increases. One of the most important advantages of the present technique is the simplicity in transforming the nonlinear differential equation into a set of simple algebraic difference equations which can be easily solved.

The paper is thus divided into two, seemingly separated, but actually connected main parts. In the first (Section 2), we show, for a given function, how a convergent power series is constructed out of the nonconverging one. In the second part (Section 3.1), we essentially use this idea to solve nonlinear differential equations. In Section 3.2, we investigate the robustness and efficiency of the method by studying the behavior of its error and CPU time versus the parameters of the method. We summarise our results in Section 4.

2. Iterative Power Series Method

This section describes how to obtain a convergent power series for a given function that is otherwise not converging for all . In brief, the method is described as follows. We expand the function in a power series as usual, say around . Then we reexpress the coefficients, , in terms of . This establishes a recursion relation between the higher-order coefficients, , and the lowest order ones, and , and thus the power series is written in terms of only these two coefficients. Then the series and its derivative are calculated at , where is much less than the radius of convergence of the power series. A new power series expansion of is then performed at . Similarly, the higher-order coefficients are reexpressed in terms of the lowest order coefficients and . The value of the previous series and its derivative calculated at are then given to and , respectively. Then a new expansion around is performed with the lowest order coefficients being taken from the previous series, and so on. This iterative process is repeated times. The final series will correspond to a convergent series at .

Here is a detailed description of the method. The function is expanded in a Taylor series, , around . The infinite Taylor series is an exact representation of for where is the radius of convergence. For the series diverges. We assume that is divided into small intervals such that . Expanding around the beginning of each interval we obtain convergent Taylor series representing in each intervalwhere denotes the Taylor series expansion of around and is the th derivative of calculated at . It is noted that we use as the independent variable for the th Taylor series expansion to distinguish it from . However, these series can not be combined in a single series since their ranges of applicability are different and do not overlap. To obtain a single convergent power series out of the set of series , we put forward two new ideas, which constitute the basis of our method; namely:

Reexpress in terms of as , where the functional is determined by direct differentiation of for times and then reexpressing the result in terms of only. We conjecture that this is possible for a wide class of functions if not all. At least for the two specific functions considered here, this turned out to be possible. Equation (1) then takes the formwhere we have renamed by and by for a reason to be obvious in the next section. Thus, the coefficients for all are determined only by .

Calculate from at which amounts to assigning the value of the Taylor series at the end of an interval to of the consecutive one. Equation (3) captures the essence of the recursive feature of our method; is calculated recursively from by repeated action of the right-hand-side on . While represents the function within an interval of width , the sequence corresponds to the values of the function at the end of the intervals. In the limit , or equivalently , the discrete set of values and render to the continuous function and its independent variable , respectively. Formally, the convergent power series expansion of around will thus be given bywhere denotes the iteration of

As an illustrative example, we apply the method to . The infinite Taylor series expansion of this function diverges sharply to infinity at . The first step is to determine the coefficients , which are the coefficients of the seriesThe next step is to reexpress the higher-order coefficients, , in terms of the zeroth-order coefficient . The property is used to that end. It is noticed, however, that while it is possible to express the even- coefficients in terms of only, the odd- coefficients can only be expressed terms of both and . In the context of solving differential equations using the power series method, this reflects the fact that the solution is expressed in terms of two independent parameters (initial conditions). The sech function is indeed a solution of a second-order differential equation, which is solved using this method in the next section. Equation (6) then takes the formCalculating series at the end of its interval of applicability, , we getwhere the “recursion” coefficients are given byFinally, we assign to The second independent coefficient is determined by the derivative of calculated at While, in the limit , corresponds to the function , the sequence corresponds to . Therefore, the power series expansion of and its first derivative are given bywhere the superscript of the matrix on the right-hand-side, , denotes the th iteration of the matrix. The superscript has been removed since the functional form of the recursion coefficients does not depend on . The procedure of calculating the power series recursively is described as follows. First, and are substituted in the right-hand-side of the last equation. Then the result of the upper element is taken as the updated value of , and, similarly, the lower element updates . The two updated values are then resubstituted back in the right-hand-side. The process is repeated times. To obtain an explicit form of the series we truncate the Taylor series at and use iterations. The resulting expansion takes the formIt is noted that the higher-order coefficients, which correspond to ratios of large integers, are represented in real numbers for convenience. Already with such a small number of iterations, , the number of terms equals . By inspection, we find that the number of terms equals . Here, is even due to the fact that is an even function.

It is also noted that the series (13) is composed of the Taylor expansion of around zero, represented by the first three terms, and a series of higher-order terms generated from the nonlinearity in the recursion relations of . In fact, we prove in the next section that this property holds for any , , and function , provided that the Taylor series of the later exists. Therefore, the power series expansion of , given by (12), can be put in the suggestive formwhere is the infinite Taylor series of about and is a complementary series. It turns out that the complementary series increases the radius of convergence of for . For finite , the effect of is to shift the radius of convergence, , to a larger value such that for . In Figure 1 we plot the convergent power series obtained by the present method as given by (12) using and . The curve is indistinguishable from the plot of . Both the Taylor series expansion, , and the complementary series, , diverge sharply at . Since is essentially zero for , it will not affect the sum . However, its major role is to cancel the divergency for . In the limit , will be an exact representative of for and will equal zero in the same interval. For , the divergences in and cancel each other with a remainder that is an exact representative of . In this manner, will represent for all .

Figure 1 

The solid curve corresponds to the convergent power series obtained by the present method as given by (12) using and . It is indistinguishable from the solid curve of . Dashed curve corresponds to the Taylor series expansion, , and the dotted curve corresponds to the complementary series, .

For finite values of and , the series is an approximate representative of . Truncating the Taylor series at introduces an error of order . This error will be magnified times due the recursive substitutions. The total error is then estimated byFor the parameters used in Figure 1, this error is of order at . This can be reduced to extremely small values such as with . However, the number of terms in the series will be of order which is extremely large and hinders any analytical manipulations.

As another example, we consider with Taylor series diverging at . Much of the formulation we followed for the previous case holds here and the specifics of the function alter only the recursion relations, (9):The convergent power series is obtained by using these recursion relations in (12). Plots similar to those of Figure 1 are obtained.

We present now a proof that the convergent power series produced by the recursive procedure always regenerates the Taylor series in addition to a complementary one.

Proposition 1. If we expand in a Taylor series, , around truncated at and use the recursive procedure, as described above, the resulting convergent power series always takes the form where is a power series of orders larger than . This is true for any number of iterations, , maximum power of the Taylor series, , and for all functions that satisfy the general differential equationwhere is an analytic real functional that does not contain derivatives.

Proof. It is trivial to prove this for a specific case, such as . For the general case, we prove this only for and . The Taylor series expansion of around isIn our recursive procedure, this is put in the equivalent formwhere the approximation stems from using finite and , and . For , , and , (20) is regenerated. However, in our recursive procedure and are kept as variables since they will be substituted for at each recursive step. Only at the last step are their numerical values inserted. For , we resubstitute in the last equation for and by their updated expressions, as follows:Substituting the updated expressions for and in (21), we getClearly for , the last equation gives the Taylor expansion, that is, (21) with . The complimentary series, , is absent here since we have terminated the expansions at . For , another step of resubstituting updated expressions is needed, and so on.
Now, we present the proof for the more general case, namely, when is unspecified but is a solution to (19). We start with the following Taylor series expansion of and its derivativeSubstituting on the right-hand-side for and by and , respectively, we getThe partial derivatives can not be calculated unless the functional forms of the recursion coefficients are known. One possibility is to specify the function being expanded, , as we did at the start of this proof. The other possibility is to exploit the differential equation that is a solution for, namely, (19). Substituting the Taylor expansion of , from (24) in (19) and expanding up to the fourth power in , we obtain the following relations:which lead toUsing these equations in (28), we obtainFor , the last equation regenerates the Taylor series of , namely, (24) with , and this completes the proof.

3. Application to Nonlinear Differential Equations

The method described in the previous section can be used as a powerful solver and integrator of nonlinear differential equations both analytically and numerically. In Section 3.1, we apply the method on a number of well-known problems. In Section 3.2, we show the power of the method in terms of detailed analysis of the error and CPU time.

3.1. Examples

For the sake of demonstration, we consider the following nonlinear differential equation:The reason for selecting this equation is that is one of its exact solutions. Substituting the power series expansion , we obtain, as usual, the recursion relationswhere and turn out to be independent parameters which in the present case correspond to the initial conditions on the solution and its first derivative. It is not surprising that these recursion relations are identical with those we found for the in the previous section, (9). Therefore, substituting the above recursion relations in we obtain the Taylor series expansion, , of . Removing the divergency in follows exactly the same steps as in the previous section, and thus an exact solution in terms of a convergent power series is obtained, as also plotted in Figure 1.

For , the relevant differential equation isSubstituting the power series expansion in this equation, the recursion relations will be given by (16)–(18). Similarly, the convergent series solution will be obtained, as in the previous section.