Closed form chains of derivatives

[mathbalarka]

A recent post of chisigma rings me the bell of an old problem I thought of posting in a forum (either here or MMF).

Is there any particular approach to computing a closed form for derivatives of certain smooth and continuous functions of $\mathbb{R}$?

For example, it is easy to find the $n$-th derivative of $e^x$, which is $e^x$ in turn. But this gets considerably hard for some Weird functions. For example, I recall of having to fight with the derivatives of some loggamma function in order to get an asymptotic approximation for the inverse gamma function.

Leibniz's rule in general reduces considerable amount of work, although rational function, even after Leibniz rule applied can give one a hard time, especially the ones with denominators made of function composition. Speaking of the devil, is there any general $n$-derivative closed form for composition of functions?

Nevertheless, I think an approach would use a considerable amount of tools from fractional calculus. I give my apologies for not being able to show research effort on this question, I reckon I am getting lazy lately. I hope I would be able to work out something tomorrow. Meanwhile, please feel free to post any ideas / works you are familiar with. I would be happy to see any (well-accepted) literature on this.

 

[chisigma]

A recent post of chisigma rings me the bell of an old problem I thought of posting in a forum (either here or MMF).

Is there any particular approach to computing a closed form for derivatives of certain smooth and continuous functions of $\mathbb{R}$?

For example, it is easy to find the $n$-th derivative of $e^x$, which is $e^x$ in turn. But this gets considerably hard for some Weird functions. For example, I recall of having to fight with the derivatives of some loggamma function in order to get an asymptotic approximation for the inverse gamma function.

Leibniz's rule in general reduces considerable amount of work, although rational function, even after Leibniz rule applied can give one a hard time, especially the ones with denominators made of function composition. Speaking of the devil, is there any general $n$-derivative closed form for composition of functions?

Nevertheless, I think an approach would use a considerable amount of tools from fractional calculus. I give my apologies for not being able to show research effort on this question, I reckon I am getting lazy lately. I hope I would be able to work out something tomorrow. Meanwhile, please feel free to post any ideas / works you are familiar with. I would be happy to see any (well-accepted) literature on this.

I think that You have in mind the 'efforts' to compute the successive derivatives in x=0 of the function...

$\displaystyle f(x) = - \frac{1}{(x - \frac{3}{2}\ \pi)^{2}\ \cos x}\ (1)$

I'm afraid that a really 'comfortable' way to do that doesn't exist but there is a good idea to make the job a little less tedious. The idea is based of the relation... $\displaystyle \frac{d}{d x}\ \ln f(x) = \frac{f^{\ '}(x)}{f(x)} \implies f^{\ '} (x) = f(x)\ \frac{d}{d x} \ln f(x)\ (2)$

Applying (2) to (1) you obtain... $\displaystyle \frac{d}{dx} \ln f(x) = - \frac{d}{d x} \{2\ \ln (x - \frac{3}{2}\ \pi) + \ln \cos x \} = - \frac{2}{x - \frac{3}{2}\ \pi} + \tan x \implies f^{\ '} (x) = \frac{1}{(x - \frac{3}{2}\ \pi)^{2}\ \cos x}\ (\frac{2}{x - \frac{3}{2}\ \pi} + \tan x)\ (3)$

What matters is that You can use (3) to evaluate $\displaystyle f^{\ ''} (x)$ only evaluating $\displaystyle \frac{d} {d x} \ln f^{\ '}(x)$... Kind regards $\chi$ $\sigma$

 

[mathbalarka]

I appreciate your answer but I am afraid I wasn't asking for what you intended to answer. My question was whether it is possible to determine $n$-th derivative of a function without known the $n-1$-th.

 

[alyafey22]

For most cases, en equivalent form of a function whether (integral or series) can make differentiation a way much easier. I have two examples in mind

The digamma function is defined as the logarithmic derivative of the gamma function

$$\psi_0(x) = \frac{\Gamma'(x)}{\Gamma(x)}$$

This form doesn't seem handy when considering higher derivative but if we use

$$\psi_0(x)=-\gamma+\sum_{n\geq 0}\frac{1}{n+1}-\frac{1}{n+x} $$

The first derivative (so called Trigamma function)

$$\psi_1(x)=\sum_{n\geq 0}\frac{1}{(n+x)^2}=\zeta(2,x) $$

where we define

$$\zeta(s,a)=\sum_{n\geq 0}\frac{1}{(n+a)^s}$$

$$\zeta(s,1) \equiv \zeta(s) $$

By multiple differentiation we get

$$\psi_n(x)=(-1)^{n+1} n! \, \zeta(n+1,x) $$

Another example is my favorite (polylogarithms)

Define the following

$$\mathrm{Li}_s(x) = \sum_{n\geq 1}\frac{x^n}{n^s}$$

We can attain the following recrusive integral representation

$$\mathrm{Li}_s(x) = \int^x_0 \frac{\mathrm{Li}_{s-1}(x)}{x}\, dx$$

$$\frac{d}{dx}\mathrm{Li}_s(x) = \frac{\mathrm{Li}_{s-1}(x)}{x}$$

Now multiple differentiation becomes extremely easy.

By the way nice topic. I'll come back once I get other examples (Wait)

 

[chisigma]

I appreciate your answer but I am afraid I wasn't asking for what you intended to answer. My question was whether it is possible to determine $n$-th derivative of a function without known the $n-1$-th.

A good approach to Your request exists if f(x) is L-transformable. In that case if $\displaystyle F(s) = \mathcal{L} \{f(x)\}$ You have...

$\displaystyle \mathcal{L}\ \{\frac{d^{n}}{d x^{n}} f(x) \} = s^{n}\ F(s) - s^{n - 1}\ f(0) - s^{n - 2}\ f^{\ '}(0) - ... - s\ f^{(n-2)} (0) - f^{(n-1)} (0)\ (1)$

Kind regards

$\chi$ $\sigma$

 

[Prometheus1]

I believe you're looking for Faà di Bruno's formula.

 

[mathbalarka]

Good find. But my more general question remains still open.

 

[Prometheus1]

Good find. But my more general question remains still open.

I'm not sure what your question is then. Could you reiterate?

 

[mathbalarka]

Sure. I want a closed form for $n$-th derivative of a smooth, real and analytic function

 

[chisigma]

Sure. I want a closed form for $n$-th derivative of a smooth, real and analytic function

Analytic function means complex function inside a closed line $\gamma$ and in this case, for a point a inside $\gamma$ is...

$\displaystyle f^{(n)} (a) = \frac{n!}{2\ \pi\ i}\ \int_{\gamma} \frac{f(z)}{(z - a)^{n + 1}}\ dz\ (1)$

Kind regards

$\chi$ $\sigma$

 

[mathbalarka]

Not really. What you give there is the definition of holomorphic functions. What I am referring to is real analytic functions