Items tagged with composition

how are iterated functions represented in maple? as in f(f(f(x))) is f^3(x)(x)  in conventional notation where by the reader knows it is refering to the iteration conducted 3 times on the argument x, but what does maple use to differentiate between iteration ,exponentiation and differentiation?

I have entered three functions into Maple and I would like to create a set of all possible two-function and three-function compositions involving the three functions. For instance, for my three functions f(x), g(x) and h(x), the set would contain f(g(x)), g(f(x)), f(g(h(x))), f(f(f(x))), etc. 

I'm also looking for a method that will be generalisable to larger numbers of initial functions than just three.

The expression (D[1]@D[2])(f) is equivalent to D[1,2](f). On grounds of that, I would naively have expected that (D[1]@D[1])(f) was equivalent to D[1,1](f). But it seems not to be, their lprint-versions being different:

expr1 := (D[1]@D[1])(f):
expr2 := D[1,1](f):
simplify(expr1 - expr2);
simplify(expr1 - expr2,`@`);
lprint(expr1);
lprint(expr2);

(D[1]@@2)(f)
D[1, 1](f)

I am curious to know why it has been implemented this way? Have I fundamentally misunderstood something?

This is not actually a question, but an interesting problem found in the recent book (2nd edition, 2015):
Mathematica®: A Problem-Centered Approach
by  Hazrat Roozbeh
Springer

I hope that you will enjoy the problem too.

Define the functions f, h : N --> N by
f(n) = the sum of the squares of the digits of n; e.g. f(25) = 2^2 + 5^2 = 29.
h(n) = min {f(n), f(f(n)), f(f(f(n))), ... };  e.g. h(7)  = min{49,97,130,10,1} = 1.

A natural number n is happy if h(n) = 1.
Find all the happy ages, i.e., happy numbers up to 100.
Conclude that happy ages are mostly before one gets a job or after retirement!

hi. i am tottaly new to the maple and i have a problem.

consider function f with variables x & y which are not independent. x & y are functions of t and the relation is unknown.

for example i wanna to the below job:

define f as f=x^2+y^2

differentiate it with respect to t : diff(f,t) which should give me 2(dx/dt)x+2(dy/dt)y

i've googled it alot and i couldn't find anything usefull.

(the problem is how to set x as a function of t with unknown relation and use it in another function and then differentiate it with respect to t)

thanks alot :)

 

Here the potential of maple 2015 to the quantitative study of the decomposition of a vector table is shown in two dimensions. Application for the exclusive use of engineering students, which was implemented with embedded components.

Atte.

Lenin Araujo Castillo

Archivo Corregido:  Decomposición_Vectorial_Corregido.mw

Hello All,

I have the PDE system shown below. It is a simple system for 2 unknown functions f1(x,t) and f2(x,t). Also, say we have x=x(t)=e^t for example. How does one solve such PDE system with Maple? I tried including the condition x=e^t in the PDE system itself, but got "System inconsistent" error message. x=x(t) can be looked at as an additional constraint and I am baffled how do I feed it into the PDE solver. 

 

Perhaps someone has experience with such systems?

 

 

f(x, g(x,y)) + f(g(x,y), y) >= f(g(x,y), g(x,y))
f(x, g(x,y)) + f(x, g(x,y)) >= f(g(x,y), y)

f(x, g(x,y)) + f(x, g(x,y)) >= g(f(x,y), y)

f(x, g(x,y))*f(x, g(x,y)) >= f(g(x,y), y)

how to create a combinations of function of another function in maple

Consider the following situation: I have a function f, say

f:=x -> 1+x^2/n;

I want to compute the composition of f with itself; e.g.

g:=f@@n;
eval(g(x),n=3);

So it appears eval does evaluate the threefold composition but not at n=3. Obviously I can wrap this example in a subs() or another eval to get the replacement done; but it is a bit curious and I wonder whether this behaviour is as designed or an "undocumented feature."

I ran across this when investigating whether or how to do the composition for an arbitrary n (so I can e.g. find the limit for n=infinity) and really wanted to ask about that, but I see that such a question was dealt with before by Joe Riel using rsolve so I'll see first how far that approach gets me. In my case the function will be a polynomial vector function with vectorial arguments, providing for some additional challenge.

Mac Dude.

Hello,

I would like to ask for help with factorization, collection or decomposition of matricies. If I have the symbolic product of matrices:

A := Matrix(2, 2, {(1, 1) = a[11], (1, 2) = a[12], (2, 1) = a[21], (2, 2) = a[22]})

B := Matrix(2, 2, {(1, 1) = b[11], (1, 2) = b[12], (2, 1) = b[21], (2, 2) = b[22]})

then C:= A*B :

Matrix(2, 2, {(1, 1) = a[11]*b[11]+a[12]*b[21], (1, 2) = a[11]*b[12]+a[12]*b[22], (2, 1) = a[21]*b[11]+a[22]*b[21], (2, 2) = a[21]*b[12]+a[22]*b[22]})

and my question follows:

Can I factor this result C and get the imput matrices A and B ? Is any function for this operation ? I would like to use it for matrices 3 time 3 not only for 2 times 2.

Thank you for your help,

vidocq

 

RandomCompositions:= proc(n::posint, k::posint)
local
C,
Compositions:= [seq(C-~1, C= combinat:-composition(n+k, k))],
Rand:= rand(1..nops(Compositions))
;
()-> Compositions[Rand()]
end proc:

R:= RandomCompositions(9,6):
n:= 2^13:
S:= 'R()' $ n:

 

I want to compile statistics each number in a sequence cannot  occur  over twice.

The sequences that do not fit the rule above must be ommitted.

The statistic is  Fermi-Dirac statistics.

confused the Bose-Einstein condensation and Fermi-Dirac statistics.

But the theory is right.

RandomCompositions:= proc(n::posint, k::posint)
local
C,
Compositions:= [seq(C-~1, C= combinat:-composition(n+k, k))],
Rand:= rand(1..nops(Compositions))
;
()-> Compositions[Rand()]
end proc:

R:= RandomCompositions(9,6):
n:= 10:
S:= 'R()' $ n;

S := [4, 1, 1, 1, 2, 0], [3, 2, 1, 1, 0, 2], [0, 1, 1, 0, 0, 7], [0, 1, 1, 5, 0, 2], [1, 0, 3, 1, 3, 1],

        [1, 3, 1, 1, 0, 3], [1, 4, 2, 0, 2, 0], [5, 0, 0, 3, 1, 0], [1, 1, 1, 4, 0, 2], [0, 1, 2, 1, 0, 5]

 

[4, 1, 1, 1, 2, 0] , [1, 1, 1, 4, 0, 2]  and [0, 1, 1, 5, 0, 2] , [0, 1, 2, 1, 0, 5]  are same number 

  but different order.

There are two same sequence. I want to  count  as one, and compile statistics the summation, and 

divide by 8.

the result

0=14/8

1=17/8

2=6/8

...

4=2/8

5=2/8

...

 

How to do composition for finite group?

is the composition like permutation group ?

if express finite group like permutation group,

if so, can elements in first row duplicate? can the second row duplicate?

however, do not know how to map when there are more than or equal two choices

my guess is that

if finite group can be expressed into permutation group

for example 3*3 matrx, each column is a permutation group

then there will be 3 permutation groups, when do composition , first column's permutation group composite with first column's permutation group , second composite with second etc

is it right?

(A o C) o Colimit = (B o C) o Colimit

if known A, B, C and framework of Colimit

can it be said colimit?

 

Bonus, what are A,B,C? <- this can be not answered

i have a sense that substitution represent associative x*y and x+y

is my understanding correct?

if i change associative property to gcd and lcd,

how to do composition?

if associative is a pair of function, x+y, x*y

does it mean that non associative should also a pair

 

is it possible replace substitution with two operation x+y and x*y?

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