# Preface

This section discusses a special technique for power series representation of nonlinear equations, known as the modified decomposition method (MDM for short). Its first concept was proposed by Randolph Rach during his discussions with G. Adomian in 1989 that was crystallized later in a paper published with his colleagues G. Adomian and R.E. Meyers in 1992. That is why this technique is sometimes referred to as the Rach--Adomian--Meyers modified decomposition method. The main idea of the MDM is to decompose the nonlinear term into a power series with the aid of the Adomian polynomials (that are genuine polynomials in this case). Then substitution of the power series solutions along with Adomian decomposition leads to full-history recurrence for determination of its coefficients. This and next two sections illustrate applications of modified decomposition method for solving initial value problems for first and second order differential equations.

The MDM and its modifications and generalizations have been extensively applied in physics, chemistry, mechanics, hydrology, engineering, economics, biology, epidemiology, etc. Some references can be found in the following link.

# List of Publications using the Modified Decomposition Method

The crucial aspect of the Adomian decomposition method (ADM) is the assumption that both, the solution to the problem (for simplicity, we illustrate it on the initial value problem for the first order autonomous differential equation)

$$\frac{{\text d}y}{{\text d}x} = f(y), \qquad y(x_0 ) = y_0 , \label{EqMDM.1}$$
and the nonlinear term f(y) in the equation can be expanded via components determined recursively. More precisely, it is assumed that the solution to a (nonlinear) problem is represented by an infinite series
$$y = u_0 + u_1 + u_2 + \cdots = \sum_{n\ge 0} u_n , \label{EqMDM.2}$$
where components un are determined recursively by solving some auxiliary problems. Although the ADM can be applied for linear problems, it gives no advantage compared to well known other methods. Therefore, we consider only equations containing nonlinearities. If such nonlinearity is described by a holomorphic function f(y), the ADM decomposes it via so called Adomian polynomials
$f(y) = f \left( \sum_{n\ge 0} u_n \right) = \sum_{n\ge 0} A_n \left( u_0 , u_1 , \ldots u_n \right) .$
The employment of the "Adomian polynomials" An to represent the nonlinear portion of the equation as a convergent series actually does not depend on a particular form of its components u0, u1, … . So the Adomian decomposition for a nonlinear analytic function can be used for arbitrary sequence { un }n≥0. To determine Adomian polynomials, it is convenient to introduce the generating function corresponding to the given sequence { un }n≥0:
$Y(\lambda ) = \sum_{n\ge 0} u_n \lambda^n ,$
where λ is a grouping parameter that we later set equal to 1. Suppose that a holomorphic function f(y) is given, for which we assign the generating function
$F(\lambda ) = f \left( \sum_{n\ge 0} u_n \lambda^n \right) = \sum_{n\ge 0} A_n \lambda^n ,$
where coefficients, which we call the Adomian polynomials, are determined by the formula
$$A_n = \frac{1}{n!} \, \frac{\partial^n}{\partial \lambda^n} \left( f\left[ \sum_{n\ge 0} u_n \lambda^n \right] \right)_{\lambda =0} , \qquad n=0,1,2,\ldots . \label{EqAdomian.1}$$

Another algorithm was proposed by Tamer A. Abassy:

$$A_n = \frac{1}{(2n)!} \left[ \frac{{\text d}^{2n}}{{\text d} \lambda^{2n}} \,f \left( \sum_{i\ge 0} \lambda^i f_i x^i \right) \right]_{\lambda = 0} + \frac{1}{(2n+1)!} \left[ \frac{{\text d}^{2n+1}}{{\text d} \lambda^{2n+1}} \,f \left( \sum_{i\ge 0} \lambda^i f_i x^i \right) \right]_{\lambda = 0} , \qquad n=0,1,2, \ldots . \label{EqAbassy.1}$$

A natural question is for what sequence of functions { un }n≥0 the Adomian polynomials admit simplification? Looking at the structure of these polynomials, we see that if the elements of the sequence { un } satisfy the group property

$\phi_n (x)\,\phi_m (x) = \phi_{n+m} (x) , \qquad n,m =0,1,2,\ldots ,$
then all Adomian polynomials will be proportional to particular terms of the sequence. Two of well-known sequences are good candidates for such sequences:
• monomials ϕn(x) = xn;
• trigonometric monomials $$\displaystyle \phi_n (x) = e^{{\bf j} \omega x} = \cos (\omega x) + {\bf j}\,\sin (\omega x),$$ where j is a unit vector in positive vertical direction on the complex plane ℂ and ω is a real scalar.
There are exist other sequences of functions that are suitable for similar simplification, for instance, Chebyshev's polynomials of the first kind due to the formula $$\displaystyle 2\,T_n (x) T_m (x) = T_{n+m} (x) + T_{|n-m|} (x) .$$ However, they are adequate for solving first order differential equations because Appell's identify $$\displaystyle \frac{\text d}{{\text d}x}\,u_n (x) = n\,u_{n-1} (x)$$ is not valid for Chebyshev polynomials.

Each of the above two sequences (or their shifted versions) generates either Maclaurin power series (or Taylor) expansion or a Fourier series expansion. Since in this section we are looking for power series, we postpone applications of the latter to the second tutorial.

If we are given a sequence of functions { ϕn(x) }n≥0 that satisfies the group property $$\displaystyle \phi_n (x)\, \phi_m (x) = \phi_{n+m} (x) ,$$ we can expand the solution into (convergent) series

$y(x) = c_0 \phi_0 + c_1 \phi_1 (x) + \cdots = \sum_{k\ge 0} c_k \phi_k (x) ,$
where ck are coefficients (scalars) of the expansion above. Correspondingly, we expand the nonlinear term f(y) into the series over the same set of functions:
$f(y) = f \left( \sum_{k\ge 0} c_k \phi_k (x) \right) = \sum_{k\ge 0} A_k \left( c_0 \phi_0 , c_1 \phi_1 , \ldots , c_n \phi_k \right) \phi_k (x) = \sum_{k\ge 0} A_k \left( c_0 , c_1 , \ldots , c_k \right) \phi_k (x) ,$
where we use the same notation for Adomian's polynomials An in both cases:
$A_k \left( c_0 \phi_0 , c_1 \phi_1 , \ldots , c_k \phi_n \right) = A_k \left( c_0 , c_1 , \ldots , c_k \right) \phi_k (x) , \qquad k=0,1,2,\ldots .$
To make connection with the ADM, we denote uk = ckϕk(x). Then the formulas above becomes
$A_n \left( u_0 , u_1 , \ldots , u_n \right) = A_n \left( c_0 , c_1 , \ldots , c_n \right) \phi_n (x) , \qquad n=0,1,2,\ldots .$
Therefore ,
\begin{align*} A_0 \left( u_0 \right) &= A_0 \left( c_0 \phi_0 \right) = A_0 \left( c_0 \right) = f \left( c_0 \right) , \\ A_1 \left( u_0 , u_1 \right) &= A_1 \left( c_0 \phi_0 , c_1 \phi_1 \right) = u_1 f^{(1)} \left( u_0 \right) = c_1 \phi_1 (x)\,f^{(1)} \left( c_0 \right) = \phi_1 (x)\,A_1 \left( c_0 , c_1 \right) , \\ A_2 \left( u_0 , u_1 , u_2 \right) &= A_2 \left( c_0 \phi_0 , c_1 \phi_1 , c_2 \phi_2 \right) = u_2 f^{(1)} \left( u_0 \right) + \frac{1}{2}\, u_1^2 f^{(2)} \left( u_0 \right) \\ &= \phi_2 c_2 f^{(1)} \left( u_0 \right) + \phi_2 \,\frac{c_1^2}{2}\,f^{(2)} \left( u_0 \right) = \phi_2 A_2 \left( c_0 , c_1 , c_2 \right) , \end{align*}
and so on. Now we can state the theorem.
Theorem: If f(y) is a holomorphic function and { ϕn(x) }n≥0 is an arbitrary sequence of the functions that satisfies the group property $$\displaystyle \phi_n (x)\, \phi_m (x) = \phi_{n+m} (x) ,$$ then
$f \left( \sum_{n\ge 0} c_n \phi_n (x) \right) = \sum_{n\ge 0} \phi_n (x)\, A_n \left( c_0 , c_1 , \ldots , c_n \right) . \qquad \Box$
Actually, Mathematica can find any Adomian polynomial for polynomial nonlinearities with a dedicated command Coefficient or CoefficientList; so sometimes the above formula is not needed when there is an access to the computer algebra system. Recall that if N is not a polynomial, then Mathematica cannot provide you an answer, and you have to use a special script to determine coefficients.

CoefficientList[(c0 + c1 x + c2 *x^2 + c3 x^3 + c4 x^4 + c5 x^5)^2 , x]
{c0^2, 2 c0 c1, c1^2 + 2 c0 c2, 2 c1 c2 + 2 c0 c3, c2^2 + 2 c1 c3 + 2 c0 c4, 2 c2 c3 + 2 c1 c4 + 2 c0 c5, c3^2 + 2 c2 c4 + 2 c1 c5, 2 c3 c4 + 2 c2 c5, c4^2 + 2 c3 c5, 2 c4 c5, c5^2}
Mathematica code for evaluating Adomian polynomials (Jun-Sheng Duan, An efficient algorithm for the multivariable Adomian polynomials, Applied Mathematics and Computation, 2010, 217, pp. 2456--2467):
Subscript[A, 0] = f[Subscript[u, 0]];
Z = Table[0, {i, 1, n}, {j, 1, i}];
Do[Z[[m, 1]] = Subscript[u, m], {m, 1, n}];
For[m = 2, m <= n, m++,
For[k = 2, k <= m, k++,
Z[[m, k]] = Expand[Subscript[u, 1]*Z[[m - 1, k - 1]]];
If[Head[Z[[m, k]]] === Plus, Z[[m, k]] = Map[#/Exponent[#, Subscript[u, 1]]&, Z[[m, k]]], Z[[m, k]] = Map[#/Exponent[#, Subscript[u, 1]]&, Z[[m, k]],{0}]]];
For[k = 2, k <= Floor[m/2], k++, Z[[m, k]] = Z[[m, k]] + (Z[[m - k, k]]/. Subscript[u, s_] -> Subscript[u, s + 1])]];
dir = Table[D[f[Subscript[u, 0]], {Subscript[u, 0], k}], {k, 1, n}];
Do[Subscript[A, m] = Take[dir, m].Z[[m]], {m, 1, n}];
Table[Subscript[A, m], {m, 0, n}]]

Example: We start with a simple power nonlinearity: N[y] = y². Using Mathematica, we can find first few terms without a problem.

CoefficientList[(c0 + c1 x + c2 *x^2 + c3 x^3 + c4 x^4 + c5 x^5)^2 , x]
{c0^2, 2 c0 c1, c1^2 + 2 c0 c2, 2 c1 c2 + 2 c0 c3, c2^2 + 2 c1 c3 + 2 c0 c4, 2 c2 c3 + 2 c1 c4 + 2 c0 c5, c3^2 + 2 c2 c4 + 2 c1 c5, 2 c3 c4 + 2 c2 c5, c4^2 + 2 c3 c5, 2 c4 c5, c5^2}

In this case, we have the general expression $$A_n = \sum_{k=0}^n u_k u_{n-k} .$$ Therefore,

A0(u0) = u0² A0(c0) = c0²
A1(u0, u1) =2 u0 u1 A1(c0, c1) =2 c0 c1
A2(u0, u1, u2) = u1² + 2 u0 u2 A2(c0, c1, c2) = c1² + 2 c0 c2
A3(u0, u1, u2, u3) = 2 u1u2 + 2 u0 u3 A3(c0, c1, c2, c3) = 2 c1c2 + 2 c0 c3
This allows us to expand a square of Maclaurin series as another Maclaurin series:
$\left( c_0 + c_1 x + c_2 x^2 + \cdots \right)^2 = c_0^2 + 2\,c_0 c_1 x + \left( c_1^2 + 2\,c_0 c_2 \right) x^2 + \cdots .$
So we don't need to multiply two series and use the convolution rule:
\begin{align*} {\bf c}\ast {\bf c} &= \left( c_0 + c_1 x + c_2 x^2 + \cdots \right)^2 = \left( c_0 + c_1 x + c_2 x^2 + \cdots \right) \left( c_0 + c_1 x + c_2 x^2 + \cdots \right) \\ &= \sum_{n\ge 0} \left( \sum_{k=0}^n c_k c_{n-k} \right) x^n . \end{align*}
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Example: Our next nonlinearity is f(y) = y³. The general formula is also available that represents the general term as a double convolution:

\begin{align*} {\bf c}\ast {\bf c} \ast {\bf c} &= \left( c_0 + c_1 x + c_2 x^2 + \cdots \right)^3 = \left( c_0 + c_1 x + c_2 x^2 + \cdots \right) \left( c_0 + c_1 x + c_2 x^2 + \cdots \right) \left( c_0 + c_1 x + c_2 x^2 + \cdots \right) \\ &= \sum_{n\ge 0} \left( \sum_{k=0}^n \sum_{j=0}^k c_{k-j} c_j c_{n-k} \right) x^n . \end{align*}
Coefficient[(u0 + s*u1 + s^2 *u2 + u3*s^3 + s^4 *u4)^3, s, 3]
u1^3 + 6 u0 u1 u2 + 3 u0^2 u3
We can obtain the same values of coefficients c0, c1, c2, … using the Adomian polynomials. Application of the Adomian algorithm to the function f(y) = y³ yields
A0(u0) = u0³ A0(c0) = c0³
A1(u0, u1) = 3 u0² u1 A1(c0, c1) =2 c0² c1
A2(u0, u1, u2) = 3u0u1² + 3 u0² u2 A2(c0, c1, c2) = 3c0c1² + 3 c0² c2
A3(u0, u1, u2, u3) = 3 u0² u3 + 6 u0 u1 u2 + u1³ A3(c0, c1, c2, c3) = 3 c0² c3 + 6 c0 c1 c2 + c1³
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Example: Given the artangent Maclaurin series

$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots = \sum_{k\ge 0} \frac{(-1)^k}{2k+1}\, x^{2k+1} ,$
we find the corresponding power series for its square
$\left( \arctan x \right)^2 = \left( \sum_{k\ge 0} \frac{(-1)^k}{2k+1}\, x^{2k+1} \right) \left( \sum_{k\ge 0} \frac{(-1)^k}{2k+1}\, x^{2k+1} \right) .$
Although we can multiply these two series and use the convolution rule, we show how coefficients of arctan²x can be determined with Adomian polynomials. Using the condition that all even powers are not present in arctangent series:
$c_{2k} = 0 , \qquad c_{2k+1} = \frac{(-1)^k}{2k+1} , \quad k=0,1,2,\ldots ,$
we calculate
A0(c0) = A0( 0 ) = 0² = 0; A0(c0) = c0²
A1(0, 1) =2 * 0 * 1 = 0; A1(c0, c1) =2 c0 c1
A2(0, 1, 0) = 1² + 2 * 0 = 1, A2(c0, c1, c2) = c1² + 2 c0 c2
A3(0, 1, 0, 1/3) = 2*1*0 + 2*0*1/3 = 0 A3(c0, c1, c2, c3) = 2 c1c2 + 2 c0 c3
Therefore,
Series[(ArcTan[x])^2 , {x, 0, 14}]
$\left( \arctan x \right)^2 = x^2 - \frac{2}{3}\, x^4 + \frac{23}{45}\, x^6 - \frac{44}{105}\,x^8 + \frac{563}{1575}\,x^{10} - \frac{3254}{10395}\, x^{12} + \frac{88069}{315315}\, x^{14} - \cdots .$
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Example: Let us consider the exponential function $$\displaystyle u = e^x = \sum_{n\ge 0} \frac{x^n}{n!} = \sum_{n\ge 0} c_n x^n ,$$ where $$\displaystyle c_n = \frac{1}{n!} , \quad n=0,1,2,\ldots .$$ Raising it to the third power, we get

$\left( e^x \right)^3 = \sum_{n\ge 0} A_n x^n = e^{3x} = \sum_{k\ge 0} \frac{(3x)^k}{k!} = \sum_{k\ge 0} \left( \frac{3^k}{k!} \right) x^k \qquad \Longrightarrow \qquad A_n = \frac{3^n}{n!} .$
Now we evaluate the Adomian polynomials using  CoefficientList
CoefficientList[(1 + 1 x + 1/2 *x^2 + 1/3! x^3 + 1/4! x^4 + 1/5! x^5)^3 , x]
{1, 3, 9/2, 9/2, 27/8, 81/40, 121/120, 17/40, 49/320, 409/8640, \ 361/28800, 9/3200, 181/345600, 1/12800, 1/115200, 1/1728000}
CoefficientList[(c0 + c1 x + c2 *x^2 + c3 x^3 + c4 x^4 + c5 x^5)^3 , x]
\begin{align*} A_0 \left( u_0 \right) &= A_0 \left( c_0 \right) = c_0^3 = 1, \\ A_1 \left( u_0 , u_1 \right) &= 3\,c_0^2 c_1 = 3 , \\ A_2 \left( u_0 , u_1 , u_2 \right) &= 3\,c_0 c_1^2 + 3\,c_0^2 c_2 = 9/2 , \\ A_3 \left( u_0 , u_1 , u_2 , u_3 \right) &= c_1^3 + 6\. c_0 c_1 c_2 + 3\, c_0^2 c_3 = 9/2 , \\ A_4 \left( u_0 , u_1 , u_2 , u_3 , u_4 \right) &= 3\, c_1^2 c_2 + 3\, c_0 c_2^2 + 6\, c_0 c_1 c_3 + 3\, c_0^2 c_4 = 27/8 , \end{align*}
and so on.    ■

Example: Let us consider the reciprocal function f(y) = 1/y. In this case, we actually need to find the inverse of power series. It was done previously using multiplication of two series and solving recurrence relation. Here we show that the required series is easily determined with Adomian's decomposition.

A0(u0) = 1/u0 A0(c0) = 1/c0
$$\displaystyle A_1 (u_0 , u_1 ) = -\frac{u_1}{u_0^2} .$$ $$\displaystyle A_1 (c_0 , c_1 ) = -\frac{c_1}{c_0^2} .$$
$$\displaystyle A_2 \left( u_0 , u_1 , u_2 \right) = \frac{u_1^2}{u_0^3} - \frac{u_2}{u_0^2} .$$ $$\displaystyle A_2 \left( c_0 , c_1 , c_2 \right) = \frac{c_1^2}{c_0^3} - \frac{c_2}{c_0^2} .$$
$$\displaystyle A_3 \left( u_0 , u_1 , u_2 , u_3 \right) = -\frac{u_1^3}{u_0^4} + \frac{2\,u_1 u_2}{u_0^3} - \frac{u_3}{u_0^2} .$$ $$\displaystyle A_3 \left( c_0 , c_1 , c_2 , c_3 \right) = -\frac{c_1^3}{c_0^4} + \frac{2\,c_1 u_2}{c_0^3} - \frac{c_3}{c_0^2} .$$
$$\displaystyle A_4 \left( u_0 , u_1 , u_2 , u_3 , u_4 \right) = \frac{u_1^4}{u_0^5} - \frac{3\,u_1^2 u_2}{u_0^4} + \frac{u_2^2 + 2\,u_1 u_3}{u_0^3} - \frac{u_4}{u_0^2} .$$ $$\displaystyle A_4 \left( c_0 , c_1 , c_2 , c_3 , c_4 \right) = -\frac{c_1^4}{c_0^5} - \frac{3\,c_1^2 c_2}{c_0^4} + \frac{c_2^2 + 2\,c_1 u_3}{c_0^3} - \frac{c_4}{c_0^2} .$$
$$\displaystyle A_5 \left( u_0 , u_1 , u_2 , u_3 , u_4 , u_5 \right) = -\frac{u_1^5}{u_0^6} + \frac{4\,u_1^3 u_2}{u_0^5} - \frac{3\,u_1 u_2^2 + 3\,u_1^2 u_3}{u_0^4} + \frac{2}{u_0^3} \left( u_2 u_3 + u_1 u_4 \right) - \frac{u_5}{u_0^2} .$$ $$\displaystyle A_5 \left( c_0 , c_1 , c_2 , c_3 , c_4 , c_5 \right) = -\frac{c_1^5}{c_0^6} + \frac{4\,c_1^3 c_2}{c_0^5} - \frac{3\,c_1 c_2^2 + 3\,c_1^2 c_3}{c_0^4} + \frac{2}{c_0^3} \left( c_2 c_3 + c_1 c_4 \right) - \frac{c_5}{c_0^2} .$$

Reciprocal of exponential function

It is well-known that
$\frac{1}{e^x} = e^{-x} = \sum_{n\ge 0} \frac{(-1)^n}{n!}\,x^n = \sum_{n\ge 0} A_n x^n ,$
where An are corresponding Adomian polynomials. We find its coefficients An by calculating the reciprocal to $$\displaystyle e^x = \sum_{n\ge 0} \frac{1}{n!}\,x^n$$ when coefficients cn = 1/n! are known. Using the previous formulas, we find the corresponding coefficients:
Reciprocal coefficients for x General formula
A0(c0) = 1/c0 = 1 A0(c0) = 1/c0
$$\displaystyle A_1 (u_0 , u_1 ) = -\frac{c_1}{c_0^2} = - \frac{1}{1} = -1 .$$ $$\displaystyle A_1 (c_0 , c_1 ) = -\frac{c_1}{c_0^2} .$$
$$\displaystyle A_2 \left( c_0 , c_1 , c_2 \right) = \frac{c_1^2}{c_0^3} - \frac{c_2}{c_0^2} = 1 - \frac{1}{2} = \frac{1}{2} .$$ $$\displaystyle A_2 \left( c_0 , c_1 , c_2 \right) = \frac{c_1^2}{c_0^3} - \frac{c_2}{c_0^2} .$$
$$\displaystyle A_3 \left( 1 , 1 , 1/2 , 1/3! \right) = -1 + \frac{2}{2} - \frac{1}{3!} = - \frac{1}{3!} .$$ $$\displaystyle A_3 \left( c_0 , c_1 , c_2 , c_3 \right) = -\frac{c_1^3}{c_0^4} + \frac{2\,c_1 u_2}{c_0^3} - \frac{c_3}{c_0^2} .$$
$$\displaystyle A_4 \left( 1 , 1 , 1/2 , 1/3! , 1/4! \right) = \frac{1}{1} - \frac{3}{2} + \frac{1}{2^2} + \frac{2}{3!} - \frac{1}{4!} = \frac{1}{4!} .$$ $$\displaystyle A_4 \left( c_0 , c_1 , c_2 , c_3 , c_4 \right) = -\frac{c_1^4}{c_0^5} - \frac{3\,c_1^2 c_2}{c_0^4} + \frac{c_2^2 + 2\,c_1 u_3}{c_0^3} - \frac{c_4}{c_0^2} .$$
$$\displaystyle A_5 \left( 1 , 1 , \frac{1}{2} , \frac{1}{3!} , \frac{1}{4!} , \frac{1}{5!} \right) = -\frac{1}{1} + \frac{4}{2} - \frac{3}{4} - \frac{3}{3!} + 2 \left( \frac{1}{2}\,\frac{1}{3!} + \frac{1}{4!} \right) - \frac{1}{5!} = - \frac{1}{5!} .$$ $$\displaystyle A_5 \left( c_0 , c_1 , c_2 , c_3 , c_4 , c_5 \right) = -\frac{c_1^5}{c_0^6} + \frac{4\,c_1^3 c_2}{c_0^5} - \frac{3\,c_1 c_2^2 + 3\,c_1^2 c_3}{c_0^4} + \frac{2}{c_0^3} \left( c_2 c_3 + c_1 c_4 \right) - \frac{c_5}{c_0^2} .$$
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Example: Let us consider the arccotangent function on the interval |x| < 1

$\mbox{arccot} (x) = \frac{\pi}{2} - \arctan (x) = \frac{\pi}{2} -x + \frac{x^3}{3} - \frac{x^5}{5} + \frac{x^7}{7} - \cdots = \frac{\pi}{2} + \sum_{k\ge 1} (-1)^k \frac{x^{2k+1}}{2k+1} , \qquad |x| < 1.$
This means that we have a Maclaurin expansion of this function that converges on the in the open interval |x| < 1:
$\mbox{arccot} (x) = \sum_{k\ge 0} c_k x^k \qquad\mbox{with} \quad c_0 = \frac{\pi}{2} , \quad c_{2k} =0 \quad\mbox{for} \quad k=1,2,\ldots , \quad c_{2k+1} = \frac{(-1)^k}{2k+1}\quad\mbox{for} \quad k=1,2,\ldots .$
Assuming[x > 0, Series[ArcCot[x], {x, 0, 8}]]
$$\frac{\pi}{2} -x + \frac{x^3}{3} - \frac{x^5}{5} + \frac{x^7}{7} + O\left[ x \right]^9$$
Suppose we want to find the Maclaurin series for 1/(arccot(x))³. We cannot use the build-in Mathematica command Coefficient because the given function 1/(arccot(x))³ is not a polynomial. Therefore, we use the standard command to expand this function into the Maclaurin series.
f[y_] = 1/(ArcCot[y])^3;
Assuming[x > 0, Series[f[x], {x, 0, 9}]]
which yields
$\frac{1}{\left( \mbox{arccot} (x) \right)^3} = \frac{8}{\pi^3} + \frac{48}{\pi^4}\,x + \frac{192}{\pi^5}\, x^2 - \frac{16 \left( \pi^2 - 40 \right)}{\pi^6}\, x^3 - \frac{128 \left( \pi^2 -15 \right)}{\pi^7}\, x^4 + \frac{16 \left( 9\pi^6 -784 \pi^4 + 11760 \pi^2 -48384 \right)}{5\, \pi^8}\, x^5 + \cdots$
Choosing f(y) = 1/y³, we expand the composition of two functions f and arccot into power series using Adomian's polynomials:
$\frac{1}{\left( \mbox{arccot} (x) \right)^3} = \left( \frac{2}{\pi} \right)^3 + \sum_{k\ge 1} A_k \left( c_0 , c_1 , \ldots , c_k \right) x^k ,$
where the first term is obviously
$A_0 (c_0 ) = f(c_0 ) = \frac{1}{c_0^3} , \qquad \mbox{with} \quad c_0 = \frac{\pi}{2} .$
Using the derivative of the arccotangent function, $${\text d}\,\mbox{arccot}(x)/{\text d}x = -\left( 1 + x^2 \right)^{-1} ,$$ we calculate the next Adomian's polynomial:
$A_1 = u_1 f'(u_0 ) = -\frac{3\,u_1}{u_0^4} = x\,c_1 f' (c_0 ) = -\frac{3\,c_1}{c_0^4} = \frac{48}{\pi^4} ,$
where c1 = -1, which follows from arccotangent Maclaurin power series expansion. The second Adomian polynomial is
$A_2 = u_2 f' (u_0 ) + \frac{1}{2}\, u_1^2 f'' (u_0 ) = - \frac{3\,u_2}{u_0^4} + \frac{6\, u_1^2}{u_0^5}$
Therefore,
$A_2 \left( c_0 , c_1 , c_2 \right) = \frac{6\, c_1^2}{c_0^5} - \frac{3\,c_2}{c_0^4} = \frac{6}{c_0^5} = \frac{6*2^5}{\pi^5} = \frac{192}{\pi^5}$
because c2 = 0 and c1 = -1. Next term is
$A_3 = u_3 f'(u_0 ) + \frac{1}{6}\,u_1^3 f''' (u_0 ) + u_1 u_2 f'' (u_0 ) = -u_3 \frac{3}{u_0^4} - \frac{60}{6}\,u_1^3 \frac{1}{u_0^6} + u_1 u_2 \,\frac{12}{u_0^5} .$
Using values c0 = π/2, c1 = -1, c2 = 0, and c3 = 1/3, we obtain
$A_2 \left( c_0 , c_1 , c_2 , c_3 \right) = A_2 \left( \frac{\pi}{2} , -1 , 0 , \frac{1}{3} \right) = -c_3 \frac{3}{c_0^4} + \frac{10}{c_0^6} = \frac{10}{c_0^6} - \frac{1}{c_0^4} = \frac{10-c_0^2}{c_0^6} = 2^4 \frac{40- \pi^2}{\pi^6} .$
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