# Nth derivative of $\tan^m x$

$m$ is positive integer,

$n$ is non-negative integer.

$$f_n(x)=\frac {d^n}{dx^n} (\tan ^m(x))$$

$P_n(x)=f_n(\arctan(x))$

I would like to find the polynomials that are defined as above

$P_0(x)=x^m$

$P_1(x)=mx^{m+1}+mx^{m-1}$

$P_2(x)=m(m+1)x^{m+2}+2m^2x^{m}+m(m-1)x^{m-2}$

$P_3(x)=(m^3+3m^2+2m)x^{m+3}+(3m^3+3m^2+2m)x^{m+1}+(3m^3-3m^2+2m)x^{m-1}+(m^3-3m^2+2m)x^{m-3}$

I wonder how to find general formula of $P_n(x)$?

I also wish to know if any orthogonal relation can be found for that polynomials or not?

EDIT:

I proved Robert Isreal's generating function. I would like to share it.

$$g(x,z) = \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) = \tan^m(x+z)$$

$$\frac {d}{dz} (\tan^m(x+z))=m \tan^{m-1}(x+z)+m \tan^{m+1}(x+z)=m \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^{m-1}(x)+m \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^{m+1}(x)= \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} (m\tan^{m-1}(x)+m\tan^{m+1}(x))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} (\dfrac{d}{dx}(\tan^{m}(x)))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^{n+1}}{dx^{n+1}} (\tan^{m}(x))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^{n+1}}{dx^{n+1}} (\tan^{m}(x))$$

$$\frac {d}{dz} ( \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) )= \sum_{n=1}^\infty \dfrac{z^{n-1}}{(n-1)!} \dfrac{d^n}{dx^n} \tan^m(x) = \sum_{n=1}^\infty \dfrac{z^{n-1}}{(n-1)!} \dfrac{d^n}{dx^n} \tan^m(x)=\sum_{k=0}^\infty \dfrac{z^{k}}{k!} \dfrac{d^{k+1}}{dx^{k+1}} \tan^m(x)$$

I also understood that it can be written for any function as shown below .(Thanks a lot to Robert Isreal)

$$\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} h^m(x) = h^m(x+z)$$

I also wrote $P_n(x)$ as the closed form shown below by using Robert Israel's answer.

$$P_n(x)=\frac{n!}{2 \pi i}\int_0^{2 \pi i} e^{nz}\left(\dfrac{x+\tan(e^{-z})}{1-x \tan(e^{-z})}\right)^m dz$$

I do not know next step how to find if any orthogonal relation exist between the polynomials or not. Maybe second order differential equation can be found by using the relations above. Thanks for advice.

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Note that the formula for the derivatives of the tangent are already sufficiently complicated..‌​. you might be able to combine that result with Faà di Bruno's formula. – J. M. Jul 13 '12 at 7:47
I added an answer explaining how to generally view the generating function derivation. – Bill Dubuque Jul 13 '12 at 14:00
@J.M.: I checked the other question but I am looking for $tan^m(x)$ and related polinomials. Really it is not duplicate. Thanks for the link. – Mathlover Jul 15 '12 at 18:19

I don't know if this will help:

The exponential generating function of $f_n(x)$ is $$g(x,z) = \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) = \tan^m(x+z) = \left(\dfrac{\tan(x)+\tan(z)}{1-\tan(x)\tan(z)}\right)^m$$ So the exponential generating function of $P_n(x)$ is $$G(x,z) = g(\arctan(x),z) = \left(\dfrac{x+\tan(z)}{1-x \tan(z)}\right)^m$$

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Fantastic generating function you offered. Thanks. Can we find any orthogonal relation for polynoms via using that generating function? Maybe second order differential equation can be found by using it? Thanks for advice. – Mathlover Jul 13 '12 at 6:39
Could you please add or send a link how we can prove that generating function works? Thanks a lot for helps – Mathlover Jul 13 '12 at 6:45

The formula used to obtain the exponential generating function in Robert's answer is most easily seen with a little operator calculus. Let $\rm\:D = \frac{d}{dx}.\,$ Then the operator $\rm\,{\it e}^{\ zD} = \sum\, (zD)^k/k!\:$ acts as a linear shift operator $\rm\:x\to x+z\,\:$ on polynomials $\rm\:f(x)\:$ since

$$\rm {\it e}^{\ zD} x^n =\, \sum\, \dfrac{(zD)^k}{k!} x^n =\, \sum\, \dfrac{z^k}{k!} \dfrac{n!}{(n-k)!}\ x^{n-k} =\, \sum\, {n\choose k} z^k x^{n-k} =\, (x+z)^n$$

so by linearity $\rm {\it e}^{\ zD} f(x) = f(x\!+\!z)\:$ for all polynomials $\rm\:f(x),\:$ and also for all formal power series $\rm\,f(x)\,$ such that $\rm\:f(x\!+\!z)\,$ converges, i.e. where $\rm\:ord_x(x\!+\!z)\ge 1,\:$ e.g. for $\rm\: z = tan^{-1} x = x -x^3/3 +\, \ldots$

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Very good operator. Great (+1). $\rm {\it e}^{\ zD} f(x) = f(x\!+\!z)\:$ : Does not it come from taylor series expansion? If you put the proof in your answer to inform reader, it could be perfect. It is just idea. Thanks a lot for answer and advice. – Mathlover Jul 13 '12 at 14:20
@Mathlover Yes, you can view it as an formal operational form of Taylor's formula. What further proof do you seek? – Bill Dubuque Jul 13 '12 at 14:29
Your answer is ok. I understood what you mean in your answer. The generator is an application of Taylor's formula. Thanks – Mathlover Jul 13 '12 at 14:35
@Mathlover To be clear, it's not intended as a complete answer to your question but, rather, to complement Robert's answer. – Bill Dubuque Jul 13 '12 at 14:53
:It is very clear. Thank you for your kindness – Mathlover Jul 13 '12 at 14:56

I have been working on the problem of finding the nth derivative and the nth anti derivative of elementary and special functions for years. You are asking a question regarding a class of functions I have called "the class of meromorphic functions with infinite number of poles. I refer you to the chapter in my Ph.D. thesis (UWO, 2004) where you can find some answers.

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Your thesis on Fractional Derivatives and Integrals are very useful , really good reference. Everything is very clear. There is example for nth derivative of $tan x$ in page 109. What about nth derivative of $tan^m x$. how can it be found? Thanks a lot for sharing and answer. – Mathlover Jul 14 '12 at 15:34
read from page 111-114, there I gave identities for some powers of tan(x) in terms of the Psi function, then it is a matter of differentiating the Psi function n times. Try to work it out for tan(x)^m. – Mhenni Benghorbal Jul 15 '12 at 6:50
@MhenniBenghorbal: I read that part. you gave example till $m=5$ I could not reach general formula for $\tan^m(x)$. How to find the patern for general formula. It also looks problem to find coeffiency of Psi function for general formula. Thanks for advice. – Mathlover Jul 15 '12 at 16:12
I am just wondering how you reached this problem? – Mhenni Benghorbal Jul 18 '12 at 1:30

For $m \ge 1$, $P_n(x) = m x^{m-n} (1+x^2) R_n(x^2)$ where $R_n(t)$ is a polynomial of degree $n-1$ such that $R_1(t) = 1$ and $$R_{n+1}(t) = 2 t (t+1) R'_n(t) + (m-n + (m-n+2) t) R_n(t)$$

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looks Very nice result. Could you please give me clue how you got that result. thanks a lot for answer and advice. – Mathlover Jul 14 '12 at 15:22