i understand quotient group is

G/(normal subgroup)

= G composite with inverse permutation group of normal subrgoup

is this understanding correct?

If do not have subgroup or normal subgroup, how to do quotient group?

if i shift second row many times in order to find Subgroup satisfy G*Subgroup = Subgroup*G

after find Subgroup then test

inverse(g)*Subgroup*g = Subgroup

how to test whether inverse(g)*Subgroup*g belong to Subgroup?

or just use equal in inverse(g)*Subgroup*g = Subgroup?

a*b - b*a    where a , b are permutation group

how to minus this?

if a + b , then how to plus permutation group

Hello, 

I have to show cubicspline(...) and some kind of known function on the same axis, but it seems I can only show cubicspline(..) using Draw, but it cannot plot the function, so I can do only one at a time. Does anyone know how to do it?

Hello everyone,

I need help to type in the following type of initial condition.

diff(1/x*diff(F(x),x),x)=0 at x =0.

Thanks

im trying to input a number between 0-100 and have the operation return the grade a,b,c,d,f. etc though long i though this might work.

Grades:=proc(x)
local a,b,c,d,f;
a:=(100..89.5);
b:=(89.4..79.5);
c:=(79.4..69.5);
d:=(69.4..59.5);
f:=(59.4..0);
if x=(100..89.5) then
display(a);
else
if x=(89.4..79.5)then
display(b);
else
if x=(79.4..69.5) then
display(c);
else
if x=(69.4..59.5) then
display(d);
else
if x=(59.4..0) then
display(f)
end;
end;
end;
end;
end;
end;

count the number of primes less than using an if-then statement.  Implement your code where j goes from 2 to 15. 

im at a loss i need a little nudge in the right direction.

Dear All,

I have a simple question I try to find the Fourrier Series of:

 f(x)=x*e^{I*x}

with maple or without maple.

Thanks

Salut,

Comment factoriser l'expression ci-dessous :

P(x,y):=-20736*x^7*y+72576*x^6*y^2-72576*x^5*y^3+72576*x^3*y^5-72576*x^2*y^6+20736*x*y^7+(-24192*3^(1/2)*x^6*y^2+72576*3^(1/2)*x^5*y^3-96768*3^(1/2)*x^4*y^4+72576*3^(1/2)*x^3*y^5-24192*3^(1/2)*x^2*y^6)*I;

Merci d'avance,

Gérard.

Dear all,

here, I propose two methods for Adams Moulton Methos, but which one can I used.

The n-step Adams Moulton method to solve y'(x)=F(x,y(x)) is defined by the stencil

y(x+h)=y(x)+h *sum_{j=-1}^{n-1} beta_j F( x-j*h, y(x-j*h) ) + O(h^{n+2})

I want a procedure with single argument ''n'' that calculates and return the ''beta_i'' coefficients

I get two Methods. Which one correspond to my question please, and I don't understand the procedure proposed.

For me; the first give the iterative schemae used, but don't return the vector coefficients ( beta_i) and this methode method an interpolation of the function.

The second method, there is a function f, how this function is maded, and the same for the matrux A and the vector b...

the First Method:

> Adamsmoulton := proc (k::posint)

local P, t, f, y, n;

P := interp([t[n]+h, seq(t[n]-j*h, j = 0 .. k-2)], [f[n+1], seq(f[n-j], j = 0 .. k-2)], x);

y[n+1] = y[n]+simplify(int(P, x = t[n] .. t[n]+h))

end proc;

Second Method:

f:=proc(x,y) if x =0 and y=0 then 1 else x**y fi end;

n:=3; A:=matrix(n,n,(i,j)->f(1-j,1-i));

b:=vector(n, i->1/i);

linsolve(A,b);

I'm trying to run two statement sequences, one after the other, numerous times. I have the statement sequences:

>for j from 1 to N do

>S[j]:=V[j]+t;

>S[j]:=S[j]+3;

>end do:

>for j from 1 to N do

>if S[j]>99 then S[j]:=0

>end if:

>end do:

I can manage to run one of them multiple times, but when I try to encompass both of them within my 

>for counter from initial to final do statementsequence

end do:

it doesn't seem to work.

Thanks in advance

assume the word equation is

a_i *a_j - a_j *a_i = 0

how to find which permutation group is a_i and a_j

my understanding is to try all rotations

a book use underscript i and j

can i see them as upper script for i rotations which is shift i times to left for second row

and try all combination and composite them in two for loop? 

Whassup homies?

http://www.mathsisfun.com/puzzles/who-lives-in-the-city--solution.html

tried to solve this using C.Loves program, but didn't quite get their solution...

Who_Lives_in_the_Cit.mw

Vars:= [PN,Name, TV, Dest,Ages,Hair,Lives]:
PN:=[$1..5]:
Name:= [Bob, Keeley, Rachael, Eilish, Amy]:
TV:=[Simpsons, Coronation, Eastenders, Desperate, Neighbours]:
Dest:= [Fra, Aus, Eng, Afr,Ita]:
Ages:= [14, 21, 46, 52, 81]:
Hair:=[afro, long, straight, curly , bald]:
Lives:= [town, city, village, farm, youth]:
Con1:= Desperate=3: Con2:= Bob=1: Con3:= NextTo(Simpsons,youth,PN): Con4:= Succ(Afr,Rachael,PN): Con5:= village=52: Con6:= Aus=straight: Con7:= Afr=Desperate: Con8:= 14=5: Con9:= Amy=Eastenders: Con10:= Ita=long: Con11:= Keeley=village: Con12:= bald=46: Con13:= Eng=4: Con14:= NextTo(Desperate,Neighbours,PN): Con15:= NextTo(Coronation,afro,PN): Con16:= NextTo(Rachael,afro,PN): Con17:= 21=youth: Con18:= Coronation=long: Con19:= 81=farm: Con20:= Fra=town: Con21:= Eilish<>straight:

read "LogicProblem.mpl"; City:= LogicProblem(Vars): with(City);

we use modern computer algebra books

i) computer the GSO of (22,11,5),(13,6,3),(-5,-2,-1) belong to R^3.

ii)trace algorithm 16.10 on computer a reduced basis of the lattice in Z^3 spanned by the vectors form(i).

trace also the values of the d_i and of D, and compare the number of exchange steps to the theoretical upper bound from section 16.3

we use Modern Computer Algebra

let f=x^15-1 belong to Z[x]. take a nontrivial factorization f≡gh mod 2 with g,h belong to Z[x] monic and of degree at least 2. computer g*,h* belong to Z[x] such that   f≡g*h* mod 16 ,deg g*=deg g, g*≡g mod 2.

show your  intermediate. can  you guess some factors of f in Z[x]?

we use Modern Computer Algebra book  

trace algorithm 15.2 on factoring f=30x^5+39x^4+35x^3+25x^2+9x+2 belong to Z[x].choose the prime p=5003 in step.