Compact formula for the $n^\text{th}$ derivative of $\operatorname{sech}^2(x)$?

Question

In short, is there a clean formula for $\frac{d^n}{dx^n}\operatorname{sech}^2(x)?$

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For convenience, let

\begin{align} f(x)&=\operatorname{sech}^2(x),\\\\ g(x)&=\tanh(x). \end{align}

Then

\begin{align} f'(x)&=-2\operatorname{sech}^2(x)\tanh(x)& &=af(x)g(x),\\\\ g'(x)&=\operatorname{sech}^2(x)& &=f(x), \end{align}

where $a=-2.$ We also have $g^2(x)=1-f(x).$

Hence $f^{(n)}(x)$ can always be written as a sum where each term is of the form $c_{k,p} [f(x)]^k[g(x)]^p$ where $p=0$ or $1$. So the question becomes: is there a formula for the $c_{k,p}$'s?

I would expect some pattern to emerge from repeated uses of the power rule combined with the recursion of the above formulas, but I'm not sure how to proceed. Also, I know sometimes coefficients of special polynomials can encode information about $n^\text{th}$ derivatives (e.g., Hermite polynomials and $e^{-x^2}$), and I would find it acceptable to express the answer in terms of such coefficients if possible.

Answer 1

Using the General Leibniz rule, $$\mathrm D^n\operatorname{sech}^2=\sum_{k=0}^n\binom{n}{k}~\mathrm D^{n-k}\operatorname{sech}~\cdot~\mathrm D^{k}\operatorname{sech}$$ So our problem reduces to finding $\mathrm D^n \operatorname{sech}$.

To clean up the notation a bit, let $f_n(\cdot,\cdot)$ be a polynomial such that $$\mathrm D^n(\operatorname{sech})(t)=f_n(\operatorname{sech}(t),\tanh(t))$$ The first few of them are $$ \begin{array}{l} f_{0}( x,y) =x\\ f_{1}( x,y) =-xy\\ f_{2}( x,y) =-x^{3} +xy^{2}\\ f_{3}( x,y) =5x^{3} y-xy^{3}\\ f_{4}( x,y) =5x^{5} -18x^{3} y^{2} +xy^{4}\\ f_{5}( x,y) =-61x^{5} y+58x^{3} y^{3} -xy^{5}\\ f_{6}( x,y) =-61x^{7} +479x^{5} y^{2} -179x^{3} y^{4} +xy^{6}\\ f_{7}( x,y) =1385x^{7} y-3111x^{5} y^{3} +543x^{3} y^{5} -xy^{7}\\ f_{8}( x,y) =1385x^{9} -19\ 028x^{7} y^{2} +18\ 270x^{5} y^{4} -1636x^{3} y^{6} +xy^{8} \end{array}$$ Hmm. Some patterns are appearing in the first and last terms, but the middle terms are a bit murky. Still, it is clear that only $\operatorname{sech}$ and $\tanh$ will be present in the result. With that in mind it is pertinent to try to evaluate $$\mathrm D(\operatorname{sech}^l\tanh^m)$$ Where the superscript denotes exponentiation, not composition. One finds $$\mathrm D(\operatorname{sech}^l\tanh^m)=m\operatorname{sech}^{2+l}\tanh^{m-1}-l\operatorname{sech}^{l}\tanh^{m+1}$$ So in our polynomials, terms of the form $x^ly^m$ in $f_n(x,y)$ change into $mx^{2+l}y^{m-1}-lx^ly^{m+1}$ in $f_{n+1}(x,y)$. IMPORTANT: This also makes it clear that $f_n(x,y)$ will only contain terms of the form $x^{n+1-k}y^k$. This suggests the amazing recurrence relation $$f_{n}(x,y)=x^2~\partial_y f_{n-1}(x,y)-xy~\partial_xf_{n-1}(x,y)$$ So, letting $$f_n(x,y)=\sum_{k=0}^{n+1}a_{n,k}x^{n+1-k}y^k$$ We have $$\sum_{k=0}^{n+1}a_{n,k}x^{n+1-k}y^k=x^2\sum_{k=0}^{n}a_{n-1,k}~k~x^{n-k}y^{k-1}-xy\sum_{k=0}^{n}a_{n-1,k}(n-k)x^{n-k-1}y^k$$ In the first sum we now change index $k= j+1$ and in the second sum we change $k=l-1$ to obtain $$\sum _{k=0}^{n+1} a_{n,k} x^{n+1-k} y^{k} $$ $$=x^{2}\sum _{j=-1}^{n-1} a_{n-1,j+1}( j+1) x^{n-j-1} y^{j} -xy\sum _{l=1}^{n+1} a_{n-1,l-1}( n-l+1) x^{n-l} y^{l-1}$$ Removing the first summand from the first sum (since it's zero), renaming our indices, and distributing the outer factors we get

$$\sum _{k=0}^{n+1} a_{n,k} x^{n+1-k} y^{k} =\begin{matrix} \sum _{k=0}^{n-1} a_{n-1,k+1}( k+1) x^{n+1-k} y^{k}\\ -\sum _{k=1}^{n+1} a_{n-1,k-1}( n-k+1) x^{n+1-k} y^{k} \end{matrix}$$ Adding and subtracting some terms we get $$\sum _{k=0}^{n+1} a_{n,k} x^{n+1-k} y^{k} =\begin{matrix} -a_{n-1,n+1}( n+1) xy^{n} -a_{n-1,n+2}( n+2) y^{n+1}\\ +\sum _{k=0}^{n+1} a_{n-1,k+1}( k+1) x^{n+1-k} y^{k}\\ +a_{n-1,-1}( n+1) x^{n+1}\\ -\sum _{k=0}^{n+1} a_{n-1,k-1}( n-k+1) x^{n+1-k} y^{k} \end{matrix}$$ But for obvious reasons, we must have $$a_{n-1,n+1}=a_{n-1,n+2}=a_{n-1,-1}=0$$ So in all we have $$\sum _{k=0}^{n+1} a_{n,k} x^{n+1-k} y^{k} =\sum _{k=0}^{n+1}( a_{n-1,k+1}( k+1) -a_{n-1,k-1}( n-k+1)) x^{n+1-k} y^{k}$$ And so by a linear algebra argument we have a recurrence relation $$\boxed{a_{n,k}=(k+1)a_{n-1,k+ 1}-(n-k+1)a_{n-1,k-1}}$$ With an initial conditions $a_{0,0}=1,a_{0,1}=0$ and boundary conditions $$a_{n,k}=0\text{ if }k<0\text{ or }k>n+1$$

A quick test: we know that for example we should have $$a_{1,0}=0,a_{1,1}=-1,a_{1,2}=0$$ We have $$a_{1,0}=a_{0,1}-(1-0+1)a_{0,-1}=0$$ $$a_{1,1}=2a_{0,2}-(1-1+1)a_{0,0}=-1$$ $$a_{1,2}=3a_{0,3}-(1-2+1)a_{0,1}=0$$ Our formula is working as expected. So using this definition of $a_{n,k}$ we write $$\mathrm{D}^n(\operatorname{sech})=\sum_{k=0}^{n+1}a_{n,k}\operatorname{sech}^{n+1-k}\tanh^k$$ Therefore finally,

$$\mathrm{D}^n(\operatorname{sech}^2)=\sum_{k=0}^n\left[\binom{n}{k}\left(\sum_{j=0}^{n-k+1}a_{n-k,j}\operatorname{sech}^{n-k+1-j}\tanh^{j}\right)\left(\sum_{l=0}^{k+1}a_{k,l}\operatorname{sech}^{k+1-l}\tanh^l\right)\right]$$ I don't think we can simplify this any further!

Answer 2

We may express $ \frac{d^n}{dx^n}\operatorname{sech}^2(x) = g^{(n+1)}(x)$ and we have

$g^{(1)}=1-g^2$,

$g^{(2)}=-2g(1-g^2)$,

$g^{(3)}=-2(1-g^2)^2+4g^2(1-g^2)=2(1-g^2)(3g^2 -1)$

$g^{(4)}=-4g(1-g^2)(3g^2-1)+2(1-g^2)6g(1-g^2)=8g(1-g^2)(2-3g^2)$ ...

suppose that $g^{(n)}=(1-g^2)P_{n-1}(g)$, where $P_{n-1}$ is a polynomial of order $n-1$ and $P_0=1$, then $g^{(n+1)}=-2g(1-g^2)P_{n-1}(g)+(1-g^2)^2P_{n-1}^{'}(g)$, then

$$P_{n}(g)= -2gP_{n-1}(g)+(1-g^2)P_{n-1}^{'}(g)$$

If $P_n(g)=\sum_k \alpha_{n,k}g^k$ , then $\alpha_{n,k}=-2\alpha_{n-1,k-1}+ (k+1)\alpha_{n-1,k+1}- (k-1)\alpha_{n-1,k-1}$

$$ \begin {align}\alpha_{n,k}&= (k+1)(\alpha_{n-1,k+1}- \alpha_{n-1,k-1})\\ \text{ with }a_{0,k}&= 0 \text{ for all non-zero } k \text{ and }a_{0,0}= 1 .\end{align}$$

Finally

$$ \frac{d^n}{dx^n}\mathbb{sech}^2x =\mathbb{sech}^2x \cdot P_{n}\big(\mathbb{tanh}x \big).$$

The polynomial $P_n$ can be computed recursively and the begining of the triangular table for $a_{n,k}$ is

n\k	0	1	2	3	4	5	6
0	1
1	0	-2
2	-2	0	6
3	0	16	0	-24
4	16	0	-120	0	120
5	0	-272	0	960	0	-720
6	-272	0	3696	0	-8400	0	5040

Up to the sign, we have the tangent numbers in the first two columns, and factorials in the main diagonal.

Actually, up to the sign, the table is $(k+1)!A059419(n+1,k+1)$ where A059419 are the coefficients of of $x^n/n!$ in expansion of $(\mathbb{tan} x)^k/k!$.

I don't know whether the polynomials $P_n$ have a name.

$P_n$ is even when $n$ is even.

$P_{2n}(g)= \sum_k a_{2n,2k}(g^2)^k=\sum_k a_{2n,2k}(1-f)^k$

$P_{2n}(g)=\sum_k a_{2n,2k}\sum_j(-1)^j {k\choose j}f^j$

$P_{2n}(g)=\sum_j f^j\sum_k (-1)^j{k\choose j}a_{2n,2k}$ $$ \frac{d^{2n}}{dx^{2n}}\mathbb{sech}^2x =\sum_{j=0}^n \Big(\sum_{k=j}^n (-1)^j{k\choose j}a_{2n,2k}\Big)\mathbb{sech}^{2j+2}x$$

and when $n$ is odd

$P_{2n+1}(g)=g \sum_{k=0}^n a_{2n+1,2k+1}(g^2)^k=g\sum_{k=0}^n a_{2n+1,2k+1}(1-f)^k$

$P_{2n+1}(g)=g\sum_{j=0}^n\sum_{k=j}^n (-1)^j{k \choose j}a_{2n+1,2k+1}f^j$ and

$$ \frac{d^{2n+1}}{dx^{2n+1}}\mathbb{sech}^2x =\mathbb{tanh}x\sum_{j=0}^n \Big(\sum_{k=j}^n (-1)^j{k\choose j}a_{2n+1,2k+1}\Big)\mathbb{sech}^{2j+2}x$$

Now putting everything together the requested formula is

$$ \frac{d^{n}f}{dx^{n}} =\sum_{j=0}^{\lfloor \frac {n}{2} \rfloor} c_{n,j}f^{j+1}g^p, $$

with $$p= \frac{1-(-1)^n}{2}$$ and $$c_{n,j}= \sum_{k=2j}^{n} (-1)^j{\lfloor \frac {k}{2} \rfloor\choose j}a_{n,k}.$$

The beginning of the table for $c_{n,j}$ is

n\j	0	1	2	3	4	5	6
0	1
1	-2
2	4	-6
3	-8	24
4	16	-120	120
5	-32	480	-720
6	64	-2016	6720	-5040

whereby, for instance :

$$ \frac{d^{5}}{dx^{5}}\mathbb{sech}^2x =-32\mathbb{sech}^{2}x\cdot\mathbb{tanh}x +480\mathbb{sech}^{4}x\cdot\mathbb{tanh}x-720\mathbb{sech}^{6}x\cdot\mathbb{tanh}x .$$

Answer 3

The $n$-th derivative of $\tanh(x)$ are obtained in this article by Boyadzhiev. Denoting $z=\tanh x$ and $C_{n}\left( \tanh x \right)=\frac{d^{n}}{dx^{n}}\left( \tanh x \right) $, we have \begin{equation} C_{n}\left(z \right)=2^{n}\left( 1+ z\right)\sum_{k=0}^{n}\frac{k!}{2^k}S\left( n,k \right)\left( z-1 \right)^k \end{equation} where $S(n,k)$ are the Stirling numbers of the second kind. Then, \begin{align} \frac{d^n}{dx^n}\operatorname{sech}^2(x)&=\frac{d^{n+1}}{dx^{n+1}}\tanh(x)\\ &=C_{n+1}\left( \tanh(x) \right)\\ &=2^{n+1}\left( 1+ \tanh(x)\right)\sum_{k=0}^{n+1}\frac{k!}{2^k}S\left( n+1,k \right)\left( \tanh(x)-1 \right)^k \end{align}

Remarking that $S(n+1,0)=0$ , the polynomials can also be expressed as \begin{equation} C_{n+1}\left(z \right)=\left( 1-z^2 \right)P_{n}(z) \end{equation} where $P_{n}(z) $ is a polynomial of degree $n$: \begin{align} \frac{d^n}{dx^n}\operatorname{sech}^2(x)&=\operatorname{sech}^2(x)P_n(\tanh x)\\ P_n(\tanh x)&=-2^{n+1}\sum_{k=1}^{n+1}\frac{k!}{2^k}S\left( n+1,k \right)\left( \tanh(x)-1 \right)^{k-1} \end{align} These polynomials are identical to the $P_n$ of @RenéGy 's answer.

Answer 4

The hyperbolic secant function is defined by \begin{equation*} \textrm{sech}\,z=\frac{2}{\textrm{e}^z+\textrm{e}^{-z}}=\frac{2\textrm{e}^z}{1+\textrm{e}^{2z}}. \end{equation*} Then, when setting $u=u(z)=1+\textrm{e}^{2z}$, by virtue of the Leibnitz rule and the Faa di Bruno formula, we obtain \begin{align} \frac{\textrm{d}^n}{\textrm{d} z^n}(\textrm{sech}\,z)^\alpha&=2^\alpha \frac{\textrm{d}^n}{\textrm{d} z^n}\frac{\textrm{e}^{\alpha z}}{(1+\textrm{e}^{2z})^\alpha}\\ &=2^\alpha \sum_{k=0}^{n}\binom{n}{k}(\textrm{e}^{\alpha z})^{(n-k)}\bigl[\bigl(1+\textrm{e}^{2z}\bigr)^{-\alpha}\bigr]^{(k)}\\ &=2^\alpha \sum_{k=0}^{n}\binom{n}{k}\alpha^{n-k}\textrm{e}^{\alpha z} \sum_{\ell=0}^{k}(u^{-\alpha})^{(\ell)} B_{k,\ell}(u',u'',\dotsc, u^{(k-\ell+1)})\\ &=2^\alpha \sum_{k=0}^{n}\binom{n}{k}\alpha^{n-k}\textrm{e}^{\alpha z} \sum_{\ell=0}^{k}\langle-\alpha\rangle_\ell u^{-\alpha-\ell} B_{k,\ell}\bigl(2\textrm{e}^{2z}, 2^2\textrm{e}^{2z},\dotsc, 2^{(k-\ell+1)}\textrm{e}^{2z}\bigr)\\ &=2^\alpha \sum_{k=0}^{n}\binom{n}{k}\alpha^{n-k}\textrm{e}^{\alpha z} \sum_{\ell=0}^{k}\langle-\alpha\rangle_\ell\bigl(1+\textrm{e}^{2z}\bigr)^{-\alpha-\ell} 2^ke^{2\ell z} B_{k,\ell}(1,1,\dotsc,1)\\ &=\sum_{k=0}^{n}\binom{n}{k}\alpha^{n-k}\sum_{\ell=0}^{k}S(k,\ell) \langle-\alpha\rangle_\ell \frac{2^{\alpha+k}\textrm{e}^{(\alpha+2\ell)z}}{(1+\textrm{e}^{2z})^{\alpha+\ell}}\\ &=\sum_{k=0}^{n}\binom{n}{k}\alpha^{n-k}\sum_{\ell=0}^{k}S(k,\ell) \langle-\alpha\rangle_\ell2^{k-\ell}(\sinh z+\cosh z)^{\ell} \textrm{sech}^{\alpha+\ell}z\tag{Q}\label{QFSQF} \end{align} for $\alpha\in\mathbb{R}$, where $S(k,\ell)$ denotes the Stirling numbers of the second kind, $\langle-\alpha\rangle_\ell$ represents the falling factorial, and $B_{k,\ell}$ stands for the Bell polynomials of the second kind.

When taking $\alpha=2$ in the above formula \eqref{QFSQF} yields the required result: \begin{equation*} \frac{\textrm{d}^n}{\textrm{d} z^n}(\textrm{sech}\,z)^2 =2^{n}\sum_{k=0}^{n}\binom{n}{k} \sum_{\ell=0}^{k}(-1)^\ell\frac{(\ell+1)!S(k,\ell)}{2^{\ell}}(\sinh z+\cosh z)^{\ell} \textrm{sech}^{\ell+2}z. \end{equation*}

For some basic and accurate knowledge used above, please refer to the following references.

References

1. Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Series expansions of powers of arcsine, closed forms for special values of Bell polynomials, and series representations of generalized logsine functions, AIMS Mathematics 6 (2021), no. 7, 7494--7517; available online at https://doi.org/10.3934/math.2021438.
2. Feng Qi and Bai-Ni Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterranean Journal of Mathematics 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
3. Feng Qi, Dongkyu Lim, and Bai-Ni Guo, Explicit formulas and identities for the Bell polynomials and a sequence of polynomials applied to differential equations, Revista de la Real Academia de Ciencias Exactas Fisicas y Naturales Serie A Matematicas 113 (2019), no. 1, 1--9; available online at https://doi.org/10.1007/s13398-017-0427-2.
4. Feng Qi, Dongkyu Lim, and Yong-Hong Yao, Notes on two kinds of special values for the Bell polynomials of the second kind, Miskolc Mathematical Notes 20 (2019), no. 1, 465--474; available online at https://doi.org/10.18514/MMN.2019.2635.
5. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Bai-Ni Guo, Closed formulas and identities for the Bell polynomials and falling factorials, Contributions to Discrete Mathematics 15 (2020), no. 1, 163--174; available online at https://doi.org/10.11575/cdm.v15i1.68111.
6. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.
7. Feng Qi, Xiao-Ting Shi, Fang-Fang Liu, and Dmitry V. Kruchinin, Several formulas for special values of the Bell polynomials of the second kind and applications, Journal of Applied Analysis and Computation 7 (2017), no. 3, 857--871; available online at https://doi.org/10.11948/2017054.

Answer 5

$\textbf{Fa'a di Bruno's formula}:$ if $g$ and $f$ are functions with a sufficient number of derivatives, then \begin{multline} \frac{d^m}{dt^m}g(f(t)) =\sum \frac{m!}{b_1!b_2!\cdots b_m!}g^{(k)}(f(t))\left(\frac{f'(t)}{1!}\right)^{b_1}\left(\frac{f''(t)}{2!}\right)^{b_2}\cdots \left(\frac{f^{(m)}(t)}{m!}\right)^{b_m} \end{multline} where the sum is over all different solutions in nonnegative integers $b_1, b_2,\cdots, b_m$ of $b_1+2b_2+\cdots +mb_m=m$, and $k:=b_1+\cdots+b_m$. Let $g(x)=x^2$ and $f(x)=\text{sech}(x)$ then $h(x)= g\circ f=\text{sech}^2(x)$. Now, use the above formula to obtain the $nth$ derivative.