# What or where is the series expansion of the function $\ln\bigl(\frac{\tan x}{x}-1\bigr)$ or $\ln(\tan x-x)$ around $x=0$?

It is known that $$\begin{equation*} \tan x=\sum_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k}-1\bigr)}{(2k)!}|B_{2k}|x^{2k-1}, \quad |x|<\frac{\pi}{2} \end{equation*}$$ and $$\begin{equation*} \ln\tan x=\ln x+\sum_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k-1}-1\bigr)}{k(2k)!}|B_{2k}|x^{2k}, \quad 0 where $$B_{2k}$$ denotes the classical Bernoulli numbers. From the second series expansion, we acquire $$\begin{equation*} \ln\frac{\tan x}{x}=\sum_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k-1}-1\bigr)}{k(2k)!}|B_{2k}|x^{2k}, \quad |x|<\frac{\pi}{2}. \end{equation*}$$ My question is:

What and where is the series expansion of the function $$\ln\bigl(\frac{\tan x}{x}-1\bigr)$$ or $$\ln(\tan x-x)$$ around $$x=0$$?

References

1. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.
• Equivalently, what is the series expansion of the function $$\ln\frac{3(\tan x-x)}{x^3}=\ln\biggl[\frac{3}{x^2}\biggl(\frac{\tan x}{x}-1\biggr)\biggr]$$ around $x=0$ for $|x|<\frac{\pi}{2}$? Apr 7 at 1:31
• The first several terms are $3 \log(x) - \log(3) + \frac{2 x^2}{5}+\frac{43 x^4}{525}+\frac{524 x^6}{23625}+\frac{40897 x^8}{6063750}+\frac{19393844 x^{10}}{8868234375}+\cdots$. Any computer algebra system can compute lots more terms. (I typed Series[Log[Tan[x] - x], {x, 0, 10}] in Mathematica.) Do you have any reason to believe that there is a better answer than this? Apr 7 at 2:31
• @DavidESpeyer Yes, you are right: any computer algebra system can compute lots more terms. I can do that by Wolfram Mathematica too. I did that already. But I cannot write out the expression (formula) of the general term for the coefficients of the series expansion of the function $\ln(\tan x-x)$, or $\ln\bigl(\frac{\tan x}{x}-1\bigr)$, or $\ln\frac{3(\tan x-x)}{x^3}=\ln\bigl[\frac{3}{x^2}\bigl(\frac{\tan x}{x}-1\bigr)\bigr]$ around $x=0$. Can you please? Apr 7 at 2:47
• I can't, but I have no reason to expect that there is a closed formula. The Taylor series of a randomly chosen function usually doesn't have one. Apr 7 at 2:49
• @DavidESpeyer I think that the general term of the coefficients in the Maclaurin series expansion of any one of the functions mentioned here has a closed-form formula (expression) , because all the derivatives of the function $\ln\frac{3(\tan x-x)}{x^3}$, for example, $$\biggl[\ln\frac{3(\tan x-x)}{x^3}\biggr]'=\frac{3 \tan x-x \sec ^2x-2 x}{x (x-\tan x)},$$ are all elementary functions and are analytic at $x=0$. Apr 7 at 3:08

Another approach is obtained by introducing the Bessel functions to express \begin{align} f(x)&=\ln\left(\frac{\tan x-x}{x^3}\right)\\ &=\ln\left( \frac{\sin x-x\cos x}{x^3\cos x} \right)\\ &=\ln\left( \frac{J_{3/2}(x)}{x^2J_{-1/2}(x)} \right) \end{align} In a paper by Dickinson a uniformly convergent series expansion is given for the logarithm of a Bessel function: $$$$\ln J_{\nu}(x)=\ln \left(\frac{(x / 2)^{\nu}}{\Gamma(\nu+1)}\right)+\sum_{n=1}^{\infty} \ln \left(1-\frac{x^2}{j_{\nu, n}}\right)$$$$ when $$\nu>-1$$ and $$0 ($$j_{\nu,n}$$ is the $$n$$-th zero of the Bessel function of order $$\nu$$). Then, \begin{align} \ln J_{\nu}(x)&=\ln \left(\frac{(x / 2)^{\nu}}{\Gamma(\nu+1)}\right)-\sum_{n=1}^{\infty}\sum_{k=0}^{\infty} \frac{1}{k+1}\frac{x^{2k+2}}{j_{\nu,n}^{2k+2}}\\ &=\ln \left(\frac{(x / 2)^{\nu}}{\Gamma(\nu+1)}\right)-\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\sum_{n=1}^{\infty}\frac{1}{j_{\nu,n}^{2k+2}} \end{align} (all the summands in the convergent double series are positive and we may interchange the order of summation). Thus \begin{align} f(x)&=-2\ln x+\ln J_{3/2}(x)-\ln J_{-1/2}(x)\\ &=-\ln3 +\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\left(\sum_{n=1}^{\infty} \frac{1}{j_{-1/2,n}^{2k+2}}-\sum_{n=1}^{\infty}\frac{1}{j_{3/2,n}^{2k+2}} \right) \end{align}

The sum of inverse even powers of zeros of Bessel functions appears in problems involving the diffusion equation and a broad literature exists on the topic. The quantity $$$$\sigma(p,\nu)=\sum_{n=1}^\infty\frac{1}{j_{\nu,n}^{2p}}$$$$ is discussed here. A recent analysis can be found in a paper by Jorge L. deLyra where a general formula for certain linear combinations of these sums is given ; it can be used to derive the formulas for $$\sigma(p, \nu)$$ by purely linear-algebraic means. The first values for $$\nu=3/2$$ are \begin{align} &\sigma(1,3/2)=\frac{1}{10}\;;\;\sigma(2,3/2)=\frac{1}{350}\;;\;\sigma(3,3/2)=\frac{1}{7875}\\ &\sigma(4,3/2)=\frac{37}{6063750}\;;\;\sigma(5,3/2)=\frac{59}{197071875} \end{align} while $$j_{-1/2,n}=(2n-1)\pi/2$$. Then \begin{align} \sigma(k+1,-1/2)&=\sum_{n=1}^\infty\frac{1}{j_{-1/2,n}^{2k+2}}\\ &=\sum_{n=1}^\infty\left(\frac{2}{(2n-1)\pi} \right)^{2k+2}\\ &=\left( \frac{2}{\pi} \right)^{2k+2}(2^{2k+2}-1)\zeta(2k+2)\\ &=(2^{2k+2}-1)\frac{2^{2k+1}}{(2k+2)!}|B_{2k+2}| \end{align} Latter identity is obtained expressing the Zeta function at even values in terms of the Bernoulli numbers. Finally, \begin{align} f(x)&=-\ln3 +\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\left(\sigma(k+1,-1/2)-\sigma(k+1,3/2)\right)\\ &=-\ln3 +\sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}\left((2^{2k+2}-1)\frac{2^{2k+1}}{(2k+2)!}|B_{2k+2}|-\sigma(k+1,3/2)\right) \end{align} which leads to the same first terms as in the expansion given by other users.

• This answer surely gives a Maclaurin power series expansion of the function $f(x)$. Regrettably, its coefficients aren't of closed-form, since $\sigma\bigl(k+1,\frac32\bigr)$ isn't of closed-form. What I emphasized is that the coefficients are of closed-form. The answer at mathoverflow.net/a/444485 is still the best one, although it was voted down by somebodies. Apr 12 at 16:51
• Comparing this answer with the one at mathoverflow.net/a/444485 results in $$\sigma\biggl(k,\frac32\biggr)=\frac{(2^{2k}-1)2^{2k-1}|B_{2k}|+3^{2k}k D_{2k}}{(2k)!}$$ for $k\ge1$. This means that the sequence $\sigma\bigl(k,\frac32\bigr)$ for $k\ge1$ can be expressed by a closed-form formula, not an infinite series. Therefore, if one can directly find a closed-form expression for $\sigma\bigl(k,\frac32\bigr)$, this answer will be perfect. Apr 12 at 18:42
• Let $j_{\nu,n}$ for $n\in\mathbb{N}$ denote the zeros of $\frac{J_\nu(z)}{z^{v}}$, where \begin{equation*} J_\nu(z)=\biggl(\frac{z}2\biggr)^\nu\sum_{n=0}^\infty\frac{(-1)^n(z/2)^{2n}}{n!\Gamma(\nu+n+1)}, \quad z\in\mathbb{C} \end{equation*} is the Bessel function of the first kind and $\nu\in\mathbb{C}\setminus\{-1,-2,\dotsc\}$ is called the order of $J_\nu(z)$. The Bessel zeta function is defined by \begin{equation*} \zeta_\nu(q)=\sum_{n=1}^{\infty}\frac{1}{j_{\nu,n}^q}, \quad q>1. \end{equation*} I think that the quantity $\sigma(p,\nu)=\zeta_\nu(2p)$. Apr 16 at 3:20
• If $\sigma(p,\nu)=\zeta_\nu(2p)$ is true, then the quantity $\sigma(p,\nu)$ has a closed-form expression $$\sigma(k,\nu+1)=\zeta_{\nu+1}(2k) =(-1)^{k+1}\frac{[\Gamma(\nu+2)]^{2k+1}}{(2k)!} \begin{vmatrix} P_{2k+1,1}(\nu) & Q_{2k+1,2k}(\nu) \end{vmatrix}_{(2k+1)\times(2k+1)}$$ for $k\in\mathbb{N}$ and $\nu\in\mathbb{C}\setminus\{-1,-2,\dotsc\}$. For more details, please refer to my paper at doi.org/10.32604/cmes.2021.016431. Then this answer is perfect: the coefficients in the series expansion are closed-form expressions. Apr 16 at 3:34
• In the above paper, we also have $$\sigma(k,\nu)=\zeta_{\nu}(2k) =\frac{(-1)^{k+1}\nu}{(2k)!} \begin{vmatrix} \frac{1}{(\nu)_0} & \frac{1}{(\nu+1)_0} & \dotsm & 0 & 0\\ 0 & 0 & \dotsm& 0 & 0\\ \frac{1}{2(\nu)_1} & \frac{1}{2(\nu+1)_1} & \dotsm & 0 &0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ \frac{(2k-3)!!}{2^{k-1}(\nu)_{k-1}} & \frac{(2k-3)!!}{2^{k-1}(\nu+1)_{k-1}} &\dotsm & \frac{1}{(\nu+1)_0} & 0\\ 0 & 0 & \dotsm & 0 & \frac{1}{(\nu+1)_0}\\ \frac{(2k-1)!!}{2^{k}(\nu)_k} & \frac{(2k-1)!!}{2^{k}(\nu+1)_k} & \dotsm & \binom{2k}{2k-2}\frac{1}{2(\nu+1)_1} & 0 \end{vmatrix}.$$ Apr 16 at 3:48

You can write the series coefficients of \begin{align} f(x)&=\log[3x^{-3}(\tan x-x)] \tag{1a}\\ &=\sum_{n=1}^{\infty} f^{(n)}(0) \frac{x^n}{n!} \tag{1b}\\ &=\frac{2 x^2}{5}+\frac{43 x^4}{525} +\frac{524 x^6}{23625}+\frac{40897 x^8}{6063750} +\frac{19393844 x^{10}}{8868234375} +O(x^{12}) \tag{1c} \end{align} according to \begin{align} f^{(n)}(0) = Y\left(\left[-(k-1)!,\quad-2^{k+2}\textstyle\binom{k+3}{k}^{-1}\cos(\pi k/2)E_{k+3}(0)\right]_{k=1}^n\right). \tag{2}\label{2} \end{align} Here, $$Y()$$ is the generalized Bell polynomial of the given $$n\times2$$ matrix, with Euler polynomials $$E_k(x)$$. The corresponding Mathematica code reads:

Series[Log[3/x^3 (Tan[x] - x)], {x,0,10}]
Sum[BellY[Table[{-(k-1)!,-2^(k+2) Cos[Pi k/2] EulerE[k+3,0]/
Binomial[k+3,k]},{k,n}]] x^n/n!, {n,1,10}]


Eq. \eqref{2} was obtained by writing $$f(x)=g(h(x))$$, with \begin{align} g(x) &= \log(1-x) = -\sum_{k=0}^\infty \frac{x^k}{k}, \tag{3a}\\ h(x) &= 1-\frac{3}{x^{3}}(\tan x-x) = -\sum_{k=1}^\infty 3 \cos(\pi k/2) \frac{2^{k+3} E_{k+3}(0)}{(k+3)!} x^k, \tag{3b} \end{align} and using the Arbogast–Faà di Bruno formula (https://en.wikipedia.org/wiki/Faà_di_Bruno's_formula, http://reference.wolfram.com/language/ref/BellY.html).

I guess there is no "closed-form" solution, as the involved polynomials become more and more complicated.

• I am sure that there is a closed-form expression for the general term of the coefficients in the Mackaurin power series expansion of the functions mentioned in the question and its first comment by me. I am sure! I am working on it now. Apr 7 at 13:37
• How did you get the relation $f^{(n)}(0)=Y(…)$? Can you give a reference to it? I repeat again that I have found a closed-form expression. @FredHucht Apr 7 at 14:31
• I have added a short explanation to my answer. I am looking forward to your closed-form expression. Apr 7 at 16:20
• @FredHucht The function $h(x)$ can be expanded into $$h(x)=1-3\sum_{k=0}^{\infty}\frac{2^{2k+4}\bigl(2^{2k+4}-1\bigr)} {(2k+4)!}|B_{2k+4}| x^{2k}$$ for $|x|<\frac{\pi}{2}$. By the Faa di Bruno formula, we have $$f^{(n)}(0)=-\sum_{k=0}^n\frac{(k-1)!}{[1-h(0)]^{k}} B_{n,k}\bigl(h'(0),h''(0),\dotsc,h^{(n-k+1)}(0)\bigr)$$ with $$h^{(2k+1)}(0)=0$$ and $$h^{(2k)}(0)=3\bigl(2^{2k+4}-1\bigr)2^{2k+4}\frac{(2k)!}{(2k+4)!}|B_{2k+4}|$$ for $k\ge0$, where $B_{n,k}=Y_{n,k}$ is the partial Bell polynomials. Is it easy to compute $$B_{n,k}\bigl(h'(0),h''(0),\dotsc,h^{(n-k+1)}(0)\bigr)$$ for all $n\ge k\ge0$? Apr 7 at 18:47
• @qifeng618 I don't think your determinant answer is more "closed-form" than my BellY answer, both are recursion relations. Btw, you can reduce the matrix size from $2k \times 2k$ to $k \times k$ by rearranging the rows and columns, first the odd and then the even rows/cols. The resulting $2\times 2$ block diagonal matrix has one trivial and one non-trivial determinant. Further, you should consider to vote up useful answers. Apr 11 at 17:30

Let $$\begin{equation*} f(x)=\begin{cases} \ln\dfrac{\tan x-x}{x^3}, & 0<|x|<\dfrac{\pi}{2};\\ -\ln3, & x=0. \end{cases} \end{equation*}$$ Then the even function $$f(x)$$ has the Maclaurin power series expansion \label{ln-tan-x-cubic-ser-expans} \begin{aligned} f(x)&=-\ln3-\sum_{k=1}^{\infty}\frac{3^{2k}D_{2k}}{(2k)!}x^{2k}\\ &=-\ln3+\frac{2 x^2}{5}+\frac{43 x^4}{525}+\frac{524 x^6}{23625}+\frac{40897 x^8}{6063750} +\frac{19393844 x^{10}}{8868234375}+\dotsm \end{aligned} for $$|x|<\frac{\pi}{2}$$, where \begin{align*} D_{2k}&=\begin{vmatrix} 0 & Q_0 & 0 & \dotsm&0& 0& 0\\ Q_1 & 0 & Q_0 & \dotsm& 0&0& 0\\ 0 & Q_1 & 0 & \dotsm& 0& 0&0\\ \dotsm & \dotsm & \dotsm & \ddots&\dotsm& \dotsm& \dotsm\\ Q_{k-1} & 0 & \binom{2k-3}{1}Q_{k-2} & \dotsm& 0 & Q_0 & 0\\ 0 & Q_{k-1} & 0 & \dotsm& \binom{2k-2}{2k-4}Q_1 & 0 & Q_0\\ Q_{k} & 0 & \binom{2k-1}{1}Q_{k-1} & \dotsm& 0 & \binom{2k-1}{2k-3}Q_1 & 0 \end{vmatrix}\\ &=\begin{vmatrix} e_{i,j} \end{vmatrix}_{2k\times2k}, \quad k\ge1,\\ e_{i,j}&=\begin{cases} 0, & (i,j)=(2\ell-1,1),\quad 1\le\ell\le k;\\ Q_\ell, & (i,j)=(2\ell,1),\quad 1\le\ell\le k;\\ 0, & 1\le i\le j-2\le2k-2;\\ 0, & i-j=2\ell, \quad 0\le\ell\le k-1;\\ \binom{i-1}{j-2}Q_{\ell}, & i-j=2\ell-1, \quad 0\le\ell\le k-1,\quad j\ge2, \end{cases}\\ Q_m&=\frac{2^{2m+2}\bigl(2^{2m+4}-1\bigr)}{(m+1)(m+2)(2m+1)(2m+3)}|B_{2m+4}|, \quad m\ge0, \end{align*} and $$B_{2m+4}$$ denotes the Bernoulli numbers generated by $$\begin{equation*} \frac{z}{\operatorname{e}^z-1}=\sum_{k=0}^\infty B_k\frac{z^k}{k!}=1-\frac{z}2+\sum_{k=1}^\infty B_{2k}\frac{z^{2k}}{(2k)!}, \quad |z|<2\pi. \end{equation*}$$ As corollaries of the above Maclaurin power series expansion, we deduce the series expansions $$\begin{equation*} \ln(\tan x-x)=-\ln3+3\ln x-\sum_{k=1}^{\infty}\frac{3^{2k}D_{2k}}{(2k)!}x^{2k}, \quad 0 and $$\begin{equation*} \ln\biggl(\frac{\tan x}{x}-1\biggr) =-\ln3+2\ln|x|-\sum_{k=1}^{\infty}\frac{3^{2k}D_{2k}}{(2k)!}x^{2k}, \quad 0<|x|<\frac{\pi}{2}. \end{equation*}$$ Since the Bernoulli numbers can be expressed by closed-form formulas, so the sequence $$D_{2k}$$ is surely a closed-form formula.

About closed-form formulas for the Bernoulli numbers and about some conclusions on the tangent function, I would like to recommend the following papers.

References

1. Xue-Yan Chen, Lan Wu, Dongkyu Lim, and Feng Qi, Two identities and closed-form formulas for the Bernoulli numbers in terms of central factorial numbers of the second kind, Demonstratio Mathematica 55 (2022), no. 1, 822--830; available online at https://doi.org/10.1515/dema-2022-0166.
2. Feng Qi and Jacques Gelinas, Revisiting Bouvier's paper on tangent numbers, Advances and Applications in Mathematical Sciences 16 (2017), no. 8, 275--281.
3. Feng Qi and Bai-Ni Guo, An explicit formula for derivative polynomials of the tangent function, Acta Universitatis Sapientiae Mathematica 9 (2017), no. 2, 348--359; available online at https://doi.org/10.1515/ausm-2017-0026.
4. Feng Qi, Derivatives of tangent function and tangent numbers, Applied Mathematics and Computation 268 (2015), 844--858; available online at https://doi.org/10.1016/j.amc.2015.06.123.
5. Jiao-Lian Zhao, Qiu-Ming Luo, Bai-Ni Guo, and Feng Qi, Remarks on inequalities for the tangent function, Hacettepe Journal of Mathematics and Statistics 41 (2012), no. 4, 499--506.
6. Chao-Ping Chen and Feng Qi, A double inequality for remainder of power series of tangent function, Tamkang Journal of Mathematics 34 (2003), no. 4, 351--355; available online at https://doi.org/10.5556/j.tkjm.34.2003.236.
7. https://math.stackexchange.com/a/4248328
8. https://math.stackexchange.com/a/4248341
9. https://math.stackexchange.com/a/4248488
10. https://math.stackexchange.com/a/4254493
11. https://math.stackexchange.com/a/4254500
12. https://math.stackexchange.com/a/4256893
13. https://math.stackexchange.com/a/4256913
14. https://math.stackexchange.com/a/4256915
15. https://math.stackexchange.com/a/4656534
• What is the use of this expansion? Apr 10 at 5:45
• At the beginning, I wish to find the series expansion of the function $f(x)$ for proving the function $$\frac{\ln\frac{3(\tan x-x)}{x^3}}{\ln\frac{\tan x}{x}} =\frac{\ln\bigl[\frac{3}{x^2}\bigl(\frac{\tan x}{x}-1\bigr)\bigr]}{\ln\frac{\tan x}{x}}$$ to be decreasing from $\bigl(0,\frac{\pi}{2}\bigr)$ onto $\bigl(1,\frac{6}{5}\bigr)$. Till now I am not successful to prove the monotonicity yet. The question at mathoverflow.net/q/444490 is an alternative try to prove the monotonicity which was guessed by a colleague last week. Apr 10 at 13:22
• This answer is too complicated for me to prove that the function $$\frac{\ln\frac{3(\tan x-x)}{x^3}}{\ln\frac{\tan x}{x}} =\frac{\ln\bigl[\frac{3}{x^2}\bigl(\frac{\tan x}{x}-1\bigr)\bigr]}{\ln\frac{\tan x}{x}}$$ is decreasing from $\bigl(0,\frac{\pi}{2}\bigr)$ onto $\bigl(1,\frac{6}{5}\bigr)$. It would be more possible for sufficiently presenting a proof of the monotonicity if the determinant $D_{2k}$ is simplified in some form. Apr 10 at 13:48
• After getting the Maclaurin power series expansion of the function $f(x)$ in this answer, which, in my opinion, is much beautiful in mathematics, the next problem is: how to compute or expand the determinant $D_{2k}$ of a Hessenberg matrix (en.wikipedia.org/wiki/Hessenberg_matrix) in a simple form? Apr 11 at 14:18
• You can reduce the matrix size from $2k \times 2k$ to $k \times k$ by rearranging the rows and columns, first the odd and then the even rows/cols. The resulting $2 \times 2$ block diagonal matrix has one trivial and one non-trivial determinant. Apr 11 at 17:36

This is not a complete answer for your question but just a key idea, I have used the first terms evaluated above by @ David E Speyer i have got the following discrete plot:

Using Mathematica the inverse series of the first 10 terms given by the following formula

$$O\left(\left(\sqrt{\frac{5}{2}} \sqrt{x-\log \left(\frac{x^3}{3}\right)}\right)^{10}\right)+\frac{47503588337 \left(\sqrt{\frac{5}{2}} \sqrt{x-\log \left(\frac{x^3}{3}\right)}\right)^9}{1601901100800000}-\frac{107033 \left(\sqrt{\frac{5}{2}} \sqrt{x-\log \left(\frac{x^3}{3}\right)}\right)^7}{181104000}+\frac{271 \left(\sqrt{\frac{5}{2}} \sqrt{x-\log \left(\frac{x^3}{3}\right)}\right)^5}{30240}-\frac{43}{420} \left(\sqrt{\frac{5}{2}} \sqrt{x-\log \left(\frac{x^3}{3}\right)}\right)^3+\sqrt{\frac{5}{2}} \sqrt{x-\log \left(\frac{x^3}{3}\right)}.$$

I'm not sure if there is such closed formula for your series expansion, but one can use polynomial approximation of the function $$\tan(x)$$ (second order) around $$x=0$$ and
using Inverse Fourier Transform one can get closed form by means of trigonometric functions and gamma function as: $$-3 i^n \Gamma (n) \left| x\right| ^{-n} \left(\cos \left(\frac{\pi n}{2}\right)+i \sin \left(\frac{\pi n}{2}\right) \text{sgn}(x)\right).$$

• i claimed that no closed form ,just i apprximated the tan function arround 0 then i got the last formula using inverse fourier transform .you should know that is wasting of time to search for such closed form regarding your given function Apr 7 at 20:37
• I have obtained a closed-form expression for the general term of all coefficients in the Maclaurin power series expansion of the function $\ln\bigl(\frac{\tan x}{x}-1\bigr)$, or $\ln(\tan x-x)$, or $\ln\frac{3(\tan x-x)}{x^3}=\ln\bigl[\frac{3}{x^2}\bigl(\frac{\tan x}{x}-1\bigr)\bigr]$. So I think that your claim is wrong. @zeraoulia-rafik Apr 8 at 1:52
• Are you sure that your picture is correct? Usually when Mathematica produces an image with a red background like that, it is indicating that there is an error in the code that produced it. Apr 10 at 12:57