Questions about the sequence of higher symbolic derivatives

[J�rgen Will]

Hallo,

I have some questions on the sequence of the consecutive higher symbolic
derivatives of a symbolically given function. Let f[n](x) be the n-th
symbolic derivative of the given function f(x) according to x. The
considered sequence is then: f[0](x), f[1](x), f[2](x), ..., f[n](x).

1.)
What is known in mathematics (or computer algebra or computer sciences)
about this sequence?

2.)
Is this sequence a topic of differential algebra?

3.)
Which kinds of functions give a periodic sequence, which a periodic sequence
from a certain degree of derivation, and which a non-periodic sequence? Is
this question known in mathematics? Has somebody an answer?

Thanks.

[Dave L. Renfro]

Jürgen Will wrote (in part):

http://groups.google.com/group/sci.math/msg/faf26201b9398274

> I have some questions on the sequence of the consecutive
> higher symbolic derivatives of a symbolically given function.
> Let f[n](x) be the n-th symbolic derivative of the given
> function f(x) according to x. The considered sequence is then:
> f[0](x), f[1](x), f[2](x), ..., f[n](x).
>
> 1.)
> What is known in mathematics (or computer algebra or
> computer sciences) about this sequence?

I'm not entirely sure what you're asking, but 1800s books
and journals are filled with computations leading to closed
form expressions for the n'th derivative of many different
function types, and most of this literature is now freely
available on the internet at google-books and other sites.
I can give you a lot of specific references if you're REALLY
interested, but it may take me a few days to do this. The
single best reference I know of off the top of my head is
not that old, but it's still probably dated and thus hard
to find (especially since it doesn't seem to be digitized
by google yet):

Isaac Joachim Schwatt, "An Introduction to the Operations
with Series", The Press of the University of Pennsylvania,
1924, x + 287 pages.

As for expressions similar to the Leibniz formula for the
n'th derivative of a product, here are some references you
may want to look at. The only one that might not be in
most university libraries happens to be freely available
on the internet, so I've supplied a URL for it.

AMM = Amer. Math. Monthly
MG = Math. Gazette

Closed form expressions for the n'th derivative of the
quotient of two functions: AMM 55 (1948), p. 491; MG 64
(1980), p. 52.

Closed form expressions for the n'th derivative of the
inverse of a function: AMM 69 (1962), pp. 904-907.

Closed form expressions for the n'th derivative of the
composition of two functions: AMM 50 (1943), pp. 9-12 & 356;
AMM 68 (1961), pp. 69-70; AMM 69 (1962), pp. 912-914;
MG 54 (1970), p. 389; MG 55 (1971), pp. 395-397; AMM 87
(1980), pp. 805-809; Mathematical Spectrum 30 #3 (1997/98),
pp. 54-57; AMM 109 (2002), pp. 217-234.

Closed form expressions for the n'th iterate of integration
by parts: Philosophical Magazine (3) 33 (1848), pp. 335-337
(see <http://books.google.com/books?id=PFYwAAAAIAAJ>); AMM 45
(1938), pp. 36-38; MG 35 (1951), pp. 122-123; AMM 67 (1960),
p. 372; Nieuw Archief voor Wiskunde (4) 7 (1989), pp. 95-105.

Dave L. Renfro

[I.N. Galidakis]

Addressing 3.) only, one can perform a very crude analysis, as follows:

Obviously polynomial functions P(x) have eventually periodic derivatives, since
Ek in N such that An>k:

P^{(n)}(x) = 0.

with f^{(n)}(x) denoting iterated differentiation.

As far as non-polynomial functions are concerned, let f(x) be such a function
with a periodic derivative. Then,

f^{(n+k)}(x) = f^{(n)}

Assuming wlog for the moment that the constants of integration are 0, and
denoting by Int^{(n)}(x) the iterated integration operator,

Int^{(n)}(f^{(n+k)}(x)dx^n)=
Int^{(n)}(f^{(n)}(x)dx^n) =>

This means that y = f(x) satisfies the differential equation:

y = y''...' (k primes), or

f(x) = d^{(k)}f(x)/dx^k

In other words, the k-th derivative of the function equals the function itself.

Try then the obvious:

f(x) = A*exp(B*x) and see what we get with Maple:

> f(x) = x-> A*exp(B*x);

> for k from 1 to 5 do
> solve(f(x)=diff(f(x),x$k),{A,B});
> od;

(k=1) {B = B, A = 0}, {B = 1, A = A}

(k=2) {B = B, A = 0}, {B = 1, A = A}, {A = A, B = -1}

(k=3) {B = B, A = 0}, {B = 1, A = A}, {A = A, B = -1/2+1/2*I*3^(1/2)}, {A = A, B
= -1/2-1/2*I*3^(1/2)}

(k=4) {B = B, A = 0}, {B = 1, A = A}, {A = A, B = -1}, {A = A, B = I}, {A = A, B
= -I}

(k=5) {B = B, A = 0}, {B = 1, A = A}, {A = A, B =
1/4*5^(1/2)-1/4+1/4*I*2^(1/2)*(5+5^(1/2))^(1/2)}, {A = A, B
= -1/4*5^(1/2)-1/4+1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)}, {A = A, B
= -1/4*5^(1/2)-1/4-1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)}, {A = A, B =
1/4*5^(1/2)-1/4-1/4*I*2^(1/2)*(5+5^(1/2))^(1/2)}

etc.

Looks to me like there are infinitely many such functions.

> Thanks.
--
Ioannis

[Martin Eisenberg]

J�rgen Will wrote:

> I have some questions on the sequence of the consecutive higher
> symbolic derivatives of a symbolically given function. Let
> f[n](x) be the n-th symbolic derivative of the given function
> f(x) according to x. The considered sequence is then: f[0](x),
> f[1](x), f[2](x), ..., f[n](x).
>
> 1.)
> What is known in mathematics (or computer algebra or computer
> sciences) about this sequence?

I'm doubtful that much of interest can be said without *some* kind of
a priori assumption on the nature of f. Perhaps this old post
interests you:

http://groups.google.com/group/comp.programming/msg/6a626db1c157c6b0

Martin

--
Quidquid latine scriptum est, altum videtur.

[dave.rusin]

On Sep 24, 4:49 pm, "I.N. Galidakis" <morph...@olympus.mons> wrote:
> Jürgen Will wrote:

> > I have some questions on the sequence of the consecutive higher symbolic
> > derivatives of a symbolically given function. Let f[n](x) be the n-th
> > symbolic derivative of the given function f(x) according to x. The
> > considered sequence is then: f[0](x), f[1](x), f[2](x), ..., f[n](x).

> > 3.)

> > Which kinds of functions give a periodic sequence, which a periodic sequence
> > from a certain degree of derivation, and which a non-periodic sequence? Is
> > this question known in mathematics? Has somebody an answer?
>
> Addressing 3.) only, one can perform a very crude analysis, as follows:

> This means that y = f(x) satisfies the differential equation:

>
> y = y''...' (k primes), or

... which is a linear differential equation with constant
coefficients,
a topic covered in elementary differential-equation courses. The
solution
set (over any interval) is the complex vector space spanned by the
functions
e^{r x} where r is a primitive k-th root of unity. Adding and
subtracting
pairs of such functions with complex-conjugate values of r gives
real-valued
functions which also serve as a basis, which might be handy if you
want to
study only real-valued functions. You'll see only linear combinations
of
certain functions of the form exp(a x) cos(b x) or exp(a x) sin(b
x) .

Note that this vector space is closed under differentiation: if f[k]
= f,
then (f')[k] = f[k+1] = (f[k])' = f', so f' is in the vector space
too.

If the sequence is to be "eventually periodic", that means there exist
values of n and k for which f[n] is one of the functions F
with
F[k] = F, which I have just characterized. Choose s so that n+s
is
a multiple m k of k : then F = F[k] = ... = F[mk] = F[n+s] = (F[s])
[n],
that is, F is the n-th derivative of something (namely G = F[s] )
which is
in the vector space. So from f[n] = F and F = G[n] we deduce that
f-G has n-th derivative equal to 0, and hence is a polynomial of
degree < n.

We conclude that the general function whose sequence of derivatives is
eventually periodic, is exactly the span of the functions x^m
(m=0,1,2,...)
and exp(r x) (r = root of unity).

dave

[I.N. Galidakis]

dave.rusin wrote:
[snip]

>> This means that y = f(x) satisfies the differential equation:
>>
>> y = y''...' (k primes), or
>
> ... which is a linear differential equation with constant
> coefficients,
> a topic covered in elementary differential-equation courses.

Ah, yes! The priviledges of teaching: You remember stuff because you teach them.

The characteristic equation of:

y = y''...' (k primes), is of course:

r^k = 1, which admits the k-th roots of unity as solutions for r. Therefore the
general solution is spanned by exp(r*x).

I will now go to the corner of the classroom and wear my dunce hat as a
punishment for not remembering a class from 1982 :-)

[snip]

> dave
--
Ioannis

[J�rgen Will]

"Dave L. Renfro" <renf...@cmich.edu> schrieb im Newsbeitrag
news:3cbef8a3-cc49-4b1d...@d34g2000vbm.googlegroups.com...

>> 1.)
>> What is known in mathematics (or computer algebra or
>> computer sciences) about this sequence?
> I'm not entirely sure what you're asking

Actually I meant only the *sequence* here - the sequence for arbitrary
functions or for some classes of functions.

But your information and literature is very helpful for me.