In asymptotic analysis, one kind of asymptotic expansions is the asymptotic power series, which is closely related to the formal power series. Both Copson [1] and Erdélyi [2] gave definitions of asymptotic expansions and asymptotic power series. The asymptotic expansion of the form

$ f(x) = a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots \quad \text { as } x \rightarrow \infty $

is called asymptotic power series.

It is interesting to consider the asymptotic power series of the composition $ f(x)=h(g(x)) $ where $ g(x) = \sum\limits_{n=0}^{\infty} b_n x^{-n} $, $ b_n\in \mathbb{R} $, and $ h $ is some elementary function. In fact, there are some properties on the composition of such power series.

Lemma 1.1   [[2], p. 20] If the function $ h $ has expansion into power series

$ h(x)=\sum\limits_{n=0}^{\infty} c_n x^n, \quad \text { as } x \rightarrow 0 $

and $ g(x) = \sum_{n=0}^{\infty} b_n x^{-n} $ with leading coefficient $ b_0=0 $, then $ h(g(x)) $ has asymptotic expansion whose coefficients can be obtained by formal substitution and rearrangement of terms.

Lemma 1.2   [[2], p. 21] If the function $ f(x) = \sum_{n=0}^{\infty} a_n x^{-n} $ is differentiable and if $ f^{\prime} $ possesses an asymptotic power series expansion, then

$ f^{\prime}(x) = -\sum\limits_{n=1}^{\infty} na_n x^{-n-1}, \quad \text { as } x \rightarrow \infty \text {. } $

In 2013, Chen et al. [3] gave respectively these recursive formula for the asymptotic power series compounded with the power function, the logarithm function and exponential function.

Lemma 1.3   [[3], Lemma 3] Let $ g(x) = \sum_{n=1}^{\infty} b_n x^{-n} . $ Then $ f(x)=\exp (g(x)) $ has asymptotic power series

$ f(x)= \sum\limits_{n=0}^{\infty} a_n x^{-n}, $

where $ a_0=1 $ and

$ a_n=\frac{1}{n} \sum\limits_{k=1}^n k b_k a_{n-k}, \quad n \geq 1 . $

Lemma 1.4   [[3], Lemma 4] Let $ g(x) =\sum_{n=0}^{\infty} b_n x^{-n} $, $ b_0 \neq 0 $. Then $ f(x)=\ln (g(x)) $ has asymptotic power series

$ f(x) = \sum\limits_{n=1}^{\infty} a_n x^{-n}, $

where

$ a_n=\frac{b_n}{b_0}-\frac{1}{n b_0} \sum\limits_{k=1}^{n-1} k a_k b_{n-k}, \quad n \geq 1 . $

Lemma 1.5   [[3], Lemma 5] Let $ g(x) = \sum_{n=0}^{\infty} b_n x^{-n} $, $ b_0 \neq 0 $. Then $ f(x)=[g(x)]^r $, for all real $ r $, has asymptotic power series

$ f(x)=\sum\limits_{n=0}^{\infty} a_n x^{-n}, $

where

$ a_0 =b_0^r, \quad a_n =\frac{1}{n b_0} \sum\limits_{k=1}^n[k(1+r)-n] b_k a_{n-k}. $

These recursive formulas have a very wide range of applications to asymptotic expansions, for example, the new asymptotic expansions for Gamma function [4], the asymptotic expansion of the integral mean [5], the asymptotic expansion coefficients of Glaisher - Kinkelin type constants [6], Landau constants, Euler-Mascheroni constant, Stirling series [7] and explicit formulas for the Bell numbers and logarithmic polynomials [8].

Besides the above three elementary functions, it seems that no asymptotic expansion formula is given for the asymptotic power series compounded with trigonometric functions. The purpose of this paper is to present these recursive formula for the asymptotic power series compounded with seven trigonometric functions.

As an application, we use these results to give the asymptotic expansion of the roots of some equations, and our recursive method is more efficient than Lagrange's inverse theorem.

First, we gives the asymptotic power series of $ f(x)=\tan(g(x)) $.

Theorem 2.1   Suppose that $ g(x) = \sum\limits_{n=0}^{\infty} b_n x^{-n} $. Then $ f(x)=\tan(g(x)) $ has asymptotic power series

$ f(x) =\sum\limits_{n=0}^{\infty} a_n x^{-n}, $

where $ a_0=\tan (b_0) $, and

$ \begin{eqnarray*} a_n&=&\frac{1}{n}\sum\limits_{k=1}^nk b_{k} c_{n-k}, \quad n \geq 1, \\ c_0&=&1+a_0^2, c_{n}=\sum\limits_{k=0}^na_{k} a_{n-k}, \quad n \geq 1; \end{eqnarray*} $

Proof   Taking the derivative of $ f (x) = \tan (g (x)) $ on both sides, and using the identity $ \sec^ 2 (g (x)) = 1 + \tan ^ 2(g (x)) $, we have

$ f^{\prime}(x)=\left(1+f^2(x)\right) g^{\prime}(x). $

Let $ 1+f^2(x) =\sum_{n=0}^{\infty} c_n x^{-n} $. Then

$ c_0=1+a_0^2, \quad c_n=\sum\limits_{k=0}^n a_k a_{n-k}, \quad n \geq 1. $

From $ f^{\prime}(x) = \sum_{n=0}^{\infty} (-n)a_n x^{-n-1} $, $ g^{\prime}(x) = \sum_{n=0}^{\infty}(-n) b_nx^{-n-1} $, we see that

$ \begin{eqnarray*} \left(\sum\limits_{n=1}^{\infty} (-n)b_{n} x^{-n-1}\right)\left(\sum\limits_{n=0}^{\infty} c_n x^{-n}\right)= \sum\limits_{n=1}^{\infty} \sum\limits_{k=1}^{n}(-k) b_{k}c_{n-k} x^{-n-1} =\sum\limits_{n=1}^{\infty} (-n) a_nx^{-n-1}. \end{eqnarray*} $

Consequently,

$ a_0=\tan b_0, \quad a_n=\frac{1}{n}\sum\limits_{k=1}^nk b_{k} c_{n-k}, \quad n\geq 1. $

This completes the proof of theorem 2.1.

Remark 2.1   With the similar argument, we can obtain the asymptotic power series of the composition with the cotangent function, arc-tangent function and arc-cotangent function.

Suppose that $ g(x) = \sum_{n=0}^{\infty} b_n x^{-n} $. Then the following results hold:

(ⅰ) $ f(x)=\cot(g(x)) $ has asymptotic power series

$ f(x) =\sum\limits_{n=0}^{\infty} a_n x^{-n}, $

where $ a_0=\cot (b_0) $, and

$ \begin{eqnarray*} a_n&=&-\frac{1}{n}\sum\limits_{k=1}^nk b_{k} c_{n-k}, \quad n \geq 1, \\ c_0&=&1+a_0^2, c_n=\sum\limits_{k=0}^na_{k} a_{n-k}, \quad n \geq 1; \end{eqnarray*} $

(ⅱ) $ f(x)=\arctan(g(x)) $ has asymptotic power series

$ f(x) =\sum\limits_{n=0}^{\infty} a_n x^{-n}, $

where $ a_0=\arctan (b_0) $, and

$ a_1=\frac{b_1}{1+b_0^2}, \quad a_n=\frac{b_n}{1+b_0^2}-\frac{1}{n(1+b_0^2)} \sum\limits_{k=1}^{n-1} k a_k c_{n-k}, \quad n \geq 2, $

$ c_0=1+b_0^2, \quad c_n=\sum\limits_{k=0}^n b_k b_{n-k}, \quad n \geq 1; $

(ⅲ) $ f(x)={\rm arccot}(g(x)) $ has asymptotic power series

$ f(x) =\sum\limits_{n=0}^{\infty} a_n x^{-n}, $

where $ a_0={\rm arccot} (b_0) $, and

$ a_1=-\frac{b_1}{1+b_0^2}, \quad a_n=-\frac{b_n}{1+b_0^2}-\frac{1}{n(1+b_0^2)} \sum\limits_{k=1}^{n-1} k a_k c_{n-k}, \quad n \geq 2, $

$ c_0=1+b_0^2, \quad c_n=\sum\limits_{k=0}^n b_k b_{n-k}, \quad n \geq 1; $

Next, we give the asymptotic power series of $ f(x)=\sin(g(x)) $. In the following Theorem, without loss of generality, we assume that $ \cos(g(x))>0 $. In fact, the case $ \cos(g(x))\leq 0 $ is similar to the case $ \cos(g(x))>0 $.

Theorem 2.2   Suppose that $ g(x) = \sum_{n=0}^{\infty} b_n x^{-n} $ and $ \cos(g(x))>0 $. Then $ f(x)=\sin(g(x)) $ has asymptotic power series

$ f(x) = \sum\limits_{n=0}^{\infty} a_n x^{-n}, $

where $ a_0=\sin b_0 $, and

$ \begin{eqnarray*} a_n&=&\frac{1}{n}\sum\limits_{k=1}^{n} kb_kd_{n-k}, \quad n\geq 1, \\ d_0&=&\sqrt{1-a_0^2}, d_n=\frac{1}{nc_0}\sum\limits_{k=1}^n (\frac{3k}{2}-n)c_k d_{n-k}, \quad n\geq 1, \\ c_0&=&1-a_0^2, c_n=-\sum\limits_{k=0}^n a_k a_{n-k}, \quad n\geq 1; \end{eqnarray*} $

Proof   From $ \cos^2 (g(x))+\sin^2 (g(x))=1 $ and $ \cos(g(x))>0 $, differentiating both sides of $ f(x)=\sin (g(x)) $ yields that

$ \begin{equation} g^{\prime}(x) \sqrt{1-f^2(x)} =f^{\prime}(x). \end{equation} $

(2.1)

Put $ 1-(f(x))^2= \sum_{n=0}^{\infty} c_n x^{-n} $. Then we have

$ c_0=1-a_0^2, \quad c_n=-\sum\limits_{k=0}^n a_k a_{n-k} , \quad n\geq 1, $

since

$ (f(x))^2 =\left(\sum\limits_{n=0}^{\infty} a_n x^{-n}\right)\left(\sum\limits_{n=0}^{\infty} a_n x^{-n}\right) = \sum\limits_{n=0}^{\infty} \sum\limits_{k=0}^n a_k a_{n-k} x^{-n}. $

For the sake of convenience, put $ \sqrt{1-f^2(x)} = \sum_{n=0}^{\infty} d_n x^{-n} $. By Lemma 1.5, one see that

$ d_0=\sqrt{1-a_0^2}, \quad d_n=\frac{1}{nc_0}\sum\limits_{k=1}^n (\frac{3k}{2}-n)c_k d_{n-k} , \quad n\geq 1. $

Together with

$ g^{\prime}(x)=\sum\limits_{n=1}^{\infty}(-n) b_n x^{-n-1}, \quad f^{\prime}(x) =\sum\limits_{n=1}^{\infty} a_n(-n) x^{-n-1}, $

it follows from Eq. (2.1) that

$ \sum\limits_{n=1}^{\infty} (-n)b_n x^{-n-1}\sum\limits_{n=0}^{\infty} d_n x^{-n}= \sum\limits_{n=1}^{\infty} (-n)a_n x^{-n-1}. $

Comparing the coefficients of $ x^{-n-1} $, one can see that

$ \sum\limits_{k=1}^{n} kb_kd_{n-k}=na_n. $

Therefore,

$ a_0=\sin b_0, \quad a_n=\frac{1}{n}\sum\limits_{k=1}^{n} kb_kd_{n-k}, \quad n\geq 1. $

This completes the proof of Theorem 2.2.

Remark 2.2   With the similar argument, we can obtain the asymptotic power series of the composition with the arc-sine function and arc-cosine function.

(ⅰ) $ f(x)=\arcsin(g(x)) $ has asymptotic power series $ f(x) = \sum_{n=0}^{\infty} a_n x^{-n}, $ where

$ \begin{eqnarray*} a_0=\arcsin b_0, && a_1=\frac{b_1}{\sqrt{1-b_0^2}}, \quad a_n=\frac{b_n}{d_0}-\frac{1}{nd_0}\sum\limits_{k=1}^{n-1} ka_kd_{n-k}, \quad n\geq2, \\ d_0=\sqrt{1-b_0^2}, && d_n=\frac{1}{nc_0}\sum\limits_{k=1}^n (\frac{3k}{2}-n)c_k d_{n-k} , \quad n\geq 1, \\ c_0=1-b_0^2, && c_n=-\sum\limits_{k=0}^n b_k b_{n-k} , \quad n\geq 1; \end{eqnarray*} $

(ⅱ) $ f(x)=\arccos(g(x)) $ has asymptotic power series $ f(x) = \sum_{n=0}^{\infty} a_n x^{-n}, $ where

$ \begin{eqnarray*} a_0=\arccos b_0, && a_1=\frac{b_1}{\sqrt{1-b_0^2}}, \quad a_n=-\frac{b_n}{d_0}-\frac{1}{nd_0}\sum\limits_{k=1}^{n-1} ka_kd_{n-k}, \quad n\geq2, \\ d_0=\sqrt{1-b_0^2}, && d_n=\frac{1}{nc_0}\sum\limits_{k=1}^n (\frac{3k}{2}-n)c_k d_{n-k} , \quad n\geq 1, \\ c_0=1-b_0^2, && c_n=-\sum\limits_{k=0}^n b_k b_{n-k} , \quad n\geq 1. \end{eqnarray*} $

In this section, we will use Theorem 2.1 to give the asymptotic behavior of roots of the equation $ x=\cot x $.

Example 3.1   We can see that there exists a unique root $ x_n $ of the equation $ x=\cot x $ such that $ x_n \in(n \pi, (n+1) \pi) $ for every fixed $ n\in \mathbb{Z} $. Then $ x_n $ has asymptotic power series

$ x_n=n \pi+\frac{1}{n \pi}+\frac{b_2}{(n \pi)^2}+\frac{b_3}{(n \pi)^3}+\cdots , \quad n \rightarrow \infty, $

where

$ b_1=1; b_{2n}=0; $

$ b_3=-\frac{4}{3};b_5=\frac{53}{15}; b_7=-\frac{1226}{105}; b_9=\frac{13597}{315}; b_{11}=-\frac{65782}{385}; $

$ b_{13}=\frac{478551932}{675675}; b_{15}=-\frac{6152345618}{2027025}; b_{17}=\frac{3217326704057}{241215975}. $

Proof   Let $ \omega=n\pi $ and $ z=\sum_{n=1}^{\infty}b_n \omega^{-n} $. It follows from Theorem 2.1 that

$ \tan(z)=\sum\limits_{n=1}^{\infty}a_n \omega^{-n}, $

where

$ b_0=0, a_0=0, c_0=1, b_1=1, $

$ a_n=\frac{1}{n}\sum\limits_{k=1}^{n}kb_kc_{n-k}, \quad c_n= \sum\limits_{k=0}^{n}a_ka_{n-k}, \quad n\geq 1. $

From $ (\omega +z)\tan(\omega +z)=(\omega +z)\tan(z)=1 $, we see that

$ \sum\limits_{n = 1}^\infty {a_n}{\omega^{ - n + 1}} + \left( {\sum\limits_{n = 1}^\infty {{b_n}{\omega^{ - n}}} } \right)\left( {\sum\limits_{n = 1}^\infty {{a_n}{\omega^{ - n}}} } \right) = \sum\limits_{n = 0}^\infty {a_{n + 1}}{\omega^{ - n}} + \sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^\infty {{a_k}{b_{n - k}}{\omega^{ - n}}} } = 1, $

where $ b_0=1, a_1=1 $. Then

$ \sum\limits_{n = 1}^\infty {{a_{n + 1}}{\omega^{ - n}} + \sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^n {{a_k}{b_{n - k}}{\omega^{ - n}}} } = 0}. $

Comparing the coefficients of $ \omega^{ - n} $, we obtain

$ {a_{n + 1}} = - \sum\limits_{k = 1}^n {{a_k}{b_{n - k}}}. $

Thus,

$ \frac{1}{n+1}\sum\limits_{k=1}^{n+1}kb_kc_{n+1-k}= b_{n+1}c_{0} +\frac{1}{n+1}\sum\limits_{k=1}^{n}kb_kc_{n+1-k}= - \sum\limits_{k = 1}^n {{a_k}{b_{n - k}}} , $

Therefore,

$ b_{n+1}=-\frac{1}{n+1}\sum\limits_{k=1}^{n}kb_kc_{n+1-k} - \sum\limits_{k = 1}^n {{a_k}{b_{n - k}}} , $

where

$ b_0=0, a_0=0, c_0=1, b_1=1, c_1=0, a_1=1, $

$ a_n=\frac{1}{n}\sum\limits_{k=1}^{n}kb_kc_{n-k}, \quad c_n= \sum\limits_{k=0}^{n}a_ka_{n-k}, \quad n\geq 1. $

Using the software MATLAB, it is easy to get the values of the previous terms:

$ b_1=1; b_{2n}=0; $

$ b_3=-\frac{4}{3};b_5=\frac{53}{15}; b_7=-\frac{1226}{105}; b_9=\frac{13597}{315}; b_{11}=-\frac{65782}{385}; $

$ b_{13}=\frac{478551932}{675675}; b_{15}=-\frac{6152345618}{2027025}; b_{17}=\frac{3217326704057}{241215975}. $

There is another method in [9, p. 24]. From the intersections of $ y=x $ and $ y=\cot x $, one can find that $ x=\cot x $ has infinitely many roots. Further, we have

$ x_n \in(n \pi, (n+1) \pi), \quad n= \pm 1, \pm 2, \pm 3, \cdots . $

Without loss of generality, assume $ x_n>0 $ and $ n \pi<x_n<(n+1) \pi $. Then

$ x_n=\cot x_n=\cot \left(x_n-n \pi\right) . $

Since $ x_n \rightarrow \infty $, one has $ x_n-n \pi \rightarrow 0 $. Let $ z=x_n-n \pi $ and $ \omega=n\pi $. Thus $ \omega+z=\cot z $. Consequently,

$ \omega^{-1}=\frac{\sin z}{\cos z-z \sin z} =\frac{z}{\frac{z(\cos z-z \sin z)}{\sin z}}. $

Using the Lagrange's inverse theorem [9, p. 22], one has

$ \begin{eqnarray*} z & =& \sum\limits_{n=1}^{\infty} \frac{1}{n !} \frac{d^{n-1}}{d \zeta^{n-1}}\left\{\left(\frac{\zeta(\cos \zeta-\zeta \sin \zeta)}{\sin \zeta}\right)^n\right\}_{\zeta=0} \omega^{-n} \\ & =&\omega^{-1}+b_2 \omega^{-2}+b_3 \omega^{-3}+\cdots. \end{eqnarray*} $

Namely,

$ x_n=n \pi+z=n \pi+\frac{1}{n \pi}+\frac{b_2}{(n \pi)^2}+\cdots=n \pi+\frac{1}{n \pi}+O\left(\frac{1}{n^2}\right) . $

Using the symbolic calculation software MAPLE, it is easy to get the values of the previous terms:

$ b_1=1; b_{2}=b_4=b_6=b_8=b_{10}=0; $

$ b_3=-\frac{4}{3};b_5=\frac{53}{15}; b_7=-\frac{1226}{105}; b_9=\frac{13597}{315}; b_{11}=-\frac{65782}{385}. $

Comparing these two methods, we can see that the recursive method is more efficient than Lagrange's inverse theorem. At the same time, we are sure that some higher derivatives can be solved by the recursive method.