However, that seems to be working only for integer exponents,

simplify(x+x^4+x^(7/2)+x^Pi,{x^3=0});

                            (7/2)    Pi
                           x      + x   + x

The following works for real exponents, too,

applyrule(x^n::satisfies(n->is(n>2))=0,
    x+x^4+x^(7/2)+x^Pi);

                                  x

Alec

In general, 2 normal matrices are unitary equivalent iff they have the same eigenvalues, which is easy to check in this case.

As usual, there is a bug related to that:

LinearAlgebra:-IsSimilar(A,B);
Error, (in LinearAlgebra:-LA_Main:-SmithForm) 
expecting a Matrix of rational polynomials in x

Alec

G:=combinat:-permute(18,4):

nops(G);

                                73440

G is not a set though - it is a list, but it can be converted to a set if necessary.

Alec

Without (1+g^(1/3))^5, that can be done using expand - however, everything is getting divided by 12, and that can be fixed by normal,

f := -(1/12)*(-g*vb-g+z*g-g^(2/3)*z+g^(2/3)*vb)*(g^(1/3)+1)^5/g^(2/3);

normal(expand(f/op(3,f))*op(3,f));

          (1/3)       (1/3)    (1/3)               (1/3)     5
        (g      vb + g      - g      z + z - vb) (g      + 1)
        ------------------------------------------------------
                                  12

By the way, factoring produced the following unexpected result,
factor(f,g^(1/3));

                              (2/3)    (1/3)
  (g - z g + g vb + z - vb + g      + g     )

                 (5/3)      (4/3)      (2/3)    2      (1/3)
        (-1 + 4 g      + 5 g      - 5 g      + g  - 4 g     )/(12

        (g - 1))

with (g-1) in the denominator, which seems like a bug, but it's interesting anyway.
Alec

member(7,Stuff[9]);

                                false

member(5,Stuff[9]);

                                 true

{1,2,3,4,5,6} minus {1,2,3};

                              {4, 5, 6}

To cut and paste the code, click Source in the editor, then type <pre>, paste the code and type </pre>. Then click Source again to return to the previous view.

Alec

For example,

L:=[seq(ithprime(n),n=1..20000)]:

L1:=select(n->isprime(n+4),L):

nops(L1);

                                 2355

L1[2009],L1[2009]+4;

                            186583, 186587

Alec

is(-(1/21)*cos(3*x)^6*sin(3*x)+(1/105)*cos(3*x)^4*sin(3*x)+
(4/315)*cos(3*x)^2*sin(3*x)+(8/315)*sin(3*x)=
(1/9)*sin(3*x)^3-(2/15)*sin(3*x)^5+(1/21)*sin(3*x)^7);

                                 true

It would be nice to have some consistency, but it would be also nice to have the correct answer at least, which Maple can't provide at the moment in many cases. In this case, the answers coincide (more or less), but that still doesn't mean that they are correct.

They may be, but without doing that by hand, I wouldn't be absolutely sure about that,

One way of discovering how Maple gets this or that answer is to increase printlevel - set it to 100 or 1000, and then follow all the procedures used with something like showstat.

Sometimes that helps and sometimes it doesn't, especially for integration, because most integration procedures are written in the style which is very far not only from so-called structured programming, but also from common sense.

Alec

That's, generally, more or less obvious - no need of using Maple.

First - the number of conjugacy classes is the same as the number of (unequivalent) irreducible unitary representations. Now, the sum of the squares of the dimensions of the (unequivalent)  irreducible unitary representations is equal to the group order which is 8 in this case - that means that the dimension of a representation can not be greater than 2 - it can be either 1 or 2. Also, the dimension of the trivial representation is 1, so there can not be two unequivalent irreducible 2-dimensional representations - the sum of their squares would give 8 and there wouldn't be a space for an additional (trivial) representation of dimension 1. So - for groups of order 8, there are only 2 possibilities - either all irreducible representations have dimension 1, in which case the group is commutative, or a group has one 2-dimensional representation and four 1-dimensional, up to the equivalency, total 5.

Since the quaternion group is not commutative, it has 5 conjugacy classes.

Now, what these classes could possibly be?

First, the center elements, 1 and -1 - that (i.e. the sets of them, {1} and {-1}) give 2 classes.

Now, we need 3 more classes, each containing at least 2 elements (because all the other elements are not in the center) out of 6 other elements, so each (other) class contains 2 elements. Now, ij=-ji tells that (by multiplications by i^-1 from the right hand side) j is conjugate with -j. Similarly, i is conjugate with -i and k is conjugate with -k.

That's it.

Also, using the group package in Maple, that can be done as

group:-transgroup(8,names);

  [{"C(8)"}, {"4[x]2"}, {"E(8)", "2[x]2[x]2"}, {"D_8(8)", "[4]2"},

        {"Q_8(8)"}, {"D(8)"}, {"1/2[2^3]4"}, {"2D_8(8)", "[D(4)]2"},

        {"E(8):2", "D(4)[x]2"}, {"[2^2]4"}, {"Q_8:2", "1/2[2^3]E(4)"},

        {"2A_4(8)", "SL(2,3)"}, {"E(8):3", "A(4)[x]2"},

        {"S(4)[1/2]2", "1/2(S_4[x]2)"}, {"[1/4cD(4)^2]2"},

        {"1/2[2^4]4"}, {"[4^2]2"}, {"E(8):E_4", "[2^2]D(4)"},

        {"E(8):4", "[1/4eD(4)^2]2"}, {"[2^3]4"},

        {"1/2[2^4]E(4)", "[1/4dD(4)^2]2"}, {"E(8):D_4", "[2^3]2^2"},

        {"2S_4(8)", "GL(2,3)"}, {"E(8):D_6", "S(4)[x]2"},

        {"E(8):7", "F_56(8)"}, {"1/2[2^4]eD(4)"}, {"[2^4]4"},

        {"1/2[2^4]dD(4)"}, {"E(8):D_8", "[2^3]D(4)"},

        {"1/2[2^4]cD(4)"}, {"[2^4]E(4)"}, {"[2^3]A(4)"},

        {"E(4):6", "E(8):A_4", "[1/3A(4)^2]2"},

        {"E(4)^2:D_6", "1/2[E(4)^2:S_3]2"}, {"[2^4]D(4)"},

        {"E(8):F_21"}, {"L(8)", "PSL(2,7)"}, {"[2^4]A(4)"},

        {"[2^3]S(4)"}, {"1/2[2^4]S(4)"},

        {"E(4)^2:D_12", "E(8):S_4", "[E(4)^2:S_3]2"}, {"[A(4)^2]2"},

        {"L(8):2", "PGL(2,7)"}, {"[2^4]S(4)"}, {"[1/2S(4)^2]2"},

        {"1/2[S(4)^2]2"}, {"[S(4)^2]2"}, {"AL(8)", "E(8):L_7"},

        {"A(8)"}, {"S(8)"}]

The quaternion group Q_8 is number 5 in the list, so

G:=permgroup(8,group:-transgroup([8,5],generators));

  G := permgroup(8,

        {[[1, 2, 3, 8], [4, 5, 6, 7]], [[4, 2, 6, 8], [1, 7, 3, 5]]})

ListTools:-Categorize(curry(group:-areconjugate,G),[group:-elements(G)[]]);

  [[]], [[[1, 2, 3, 8], [4, 5, 6, 7]], [[1, 8, 3, 2], [4, 7, 6, 5]]],

        [[[1, 4, 3, 6], [2, 7, 8, 5]], [[1, 6, 3, 4], [2, 5, 8, 7]]],

        [[[1, 5, 3, 7], [2, 4, 8, 6]], [[1, 7, 3, 5], [2, 6, 8, 4]]],

        [[[1, 3], [2, 8], [4, 6], [5, 7]]]

It's not that easy in this notation to realize that i is conjugate with -i etc. The group package (dated back to 1980s, or, maybe, to 1990s) is really in a serious need to be updated - who would want to use it instead of much more powerful GAP (included in Sage)

Alec

The main thing that Maple is missing in Windows in external calling is that Maple is using only stdcall convention, while many (if not most) dlls are using a different convention - either cdecl or fastcall usually.

Neither Python nor Mathematica have problems with that - only Maple.

Alec

Do you realize that the number you wrote as the answer is greater than 1?

To answer the question, one needs to know also the probability of a successful throw.

If it is 5/6, the answer is

with(Statistics):
X := Binomial(7,5/6):
ProbabilityFunction(X, 6);

                                109375
                                ------
                                279936

evalf(%);

                             0.3907143061

and if it is 1/6, the 5/6 above should be replaced with 1/6.

Alec

series(A,t,2) and taylor(A,t,2) seem to be working.

On the other hand,

mtaylor(A,[t],2);
Error, (in PDEtools/useD/diff_to_D) invalid input: 
diff received t*t, which is not valid for its 2nd argument

which may be a bug.

Alec

The standard way is to go to a graduate school and to get a PhD.

But be careful what you wish for.

The job market is overloaded and the salaries are low.

It is much better to be a Medicinæ Doctor.

Alec

Numerically,

sum(((n+1)/n-1)*(12*n^2+12*n+1)/
((1+((n+1)/n)^(1/3))*(12*n^3+18*n^2+6*n)), n = 1 .. infinity);

evalf(%,60);

    0.577664051207838413728862917951293462227701145226200958019115

Alec

How to make it beautiful? Use LaTeX.

Alec

Try exp(x) instead of e^x.

Alec