Hi Rit,

I read your question about Maple software.  I have had some success with difiningt procedures, but I have not made any modules.  I stripped down your code to remove the warnings.

m := Matrix([[1, 2], [3, 4]]);

L := MatrixMatrixMultiply(m, m)

Maple coding is hard and your hint looks valid.

Regards,
Matt

Hi Kurt,

I found a treaste on groups that may be helpful to you.  It enumerates all possible groups for small number orders.

http://www.math.niu.edu/~beachy/courses/algebra/pedersen/small_groups.html

Regards,

Matt

@Glowing 

Hi Glowing,

Thank you for your email.  Your problem is to expand the expression (x+1)(y+z).

I would use the FOIL method.  This stands for First Outside Inside Last.  These 4 are the things that get multiplied together.  For example to expand -

(x+3)(y+7) you do the 4 steps and write the 4 expressions.

xy + 3y + 7x + 21.

Hopefully this sets you on the path to solve this with pencil and paper.

Regards,

Matt

@Carl Love 

I am not that guy.

Hi Kurt,

Thank you for asking about groups.  There is a difference between a dihedral group and a cyclic group. 

Any group with a prime number of elements is cyclic.  This is a theorem.

I imagine a 6 sided piece of cardboard with the corners numbered 1 to 6.   The group operation is to partially rotate the cardboard a number of times.  The choices for amount of clockwise rotation is 1/6, 2/6, … until all the way around.

A simpler cyclic group is what I will call C3, a group with 3 elements.  Imagine the triangle that can be rotated a third, two thirds or left alone.

Now for a dihedral group.  Now our cardboard not only has the 3 edges numbered 0, 1 and 2 but it has a black side and a white side and you are allowed to turn the cardboard over.

Later on in your book you will learn about exotic groups like the quaternion group.  Simpler than that is the symmetric group.  For example S3, the symmetric group of order 3 has 3 factorial elements.  S3 has order 6.  This last sentence is a theorem, or at least true.

Since I don't have my textbook in front of me, I cannot give you a more concise answer.

Best of Luck with your studies.

Regards, Matt

@Carl Love 

That is a clever observation Man.

Hello Man,

I'm glad you have asked this question about a logarithmic function. 

The way I learned it is, concave down is like an umbrella.  The second derivative is negative.

Also, concave up is like a coffe cup.  And its second derivative is positive.

I hope this helps clear things up for you.

Regards,

Matt

Hi Sarra :),

Here is my little reply to your question.

# a way to do reduced eschelon form of a matrix by Matt C. Anderson 4 17 2014

# rref on some calculators

https://docs.google.com/file/d/0B6s70Xwo_mBXZUNpY1VxYzZJM2s/edit

Regards,

Matt

@Alejandro Jakubi 

I can see that this is not the first answer, but I have some short code.

>
> f := proc (x, y) description "evaluates 2x+3y"; 2*x+3*y end proc;
proc(x, y) ... end;
> f(4, 5);
23
>

> f := proc (xid, yid) options operator, arrow; (sum(x[i], i = 1 .. xid))*(sum(y[i], i = 1 .. yid)) end proc;
/ xid \ / yid \
|----- | |----- |
| \ | | \ |
| ) | | ) |
(xid, yid) -> | / x[i]| | / y[i]|
|----- | |----- |
\i = 1 / \i = 1 /
> firstprocedure := proc (xid, yid) (sum(x[i], i = 1 .. xid))*(sum(y[i], i = 1 .. yid)) end proc;
proc(xid, yid) ... end;
> firstprocedure(4, 5);
(x[1] + x[2] + x[3] + x[4]) (y[1] + y[2] + y[3] + y[4] + y[5])

Although this does not evaluate to an integer, at least it does something.

HOpe this helps.

Regards,

Matt
>

Thanks, Carl, for taking the time to look at my code.  When Maple crashes, the application stops appearing in the applications tab of my task manager ant the task bar at the bottom of my Windows 7 system.  However mserver.exe appears in the Processes tab of the task manager.  I can't see the memory and time numbers after it crashes, but I just started another run, and the Memory says 60.61M, and the time is just incrementing.  The kernelopts(cpulimit) and the kernelopts(datalimit) commands both return the infinity symbol.  

I check for small factors with the igcd() command before I run the isprime() command because it make the code run faster.  I am looking for clusters of 12 prime numbers.  It is faster to make a quick check that all 12 do not have any small factors before running the slower isprime() command.  Also, I have two levels of pretest and this seems to be the fastest way that I have found.  I know about showstat(isprime) and that gave me the idea for a quick check.  

In fact on line 12 of showstat(isprime) it says - 

r := gmp_isprime(n);

Do you know how I can see the code for the gmp_isprime(n); function?

Again thank you for your help.  Hopefully I can find a way to do long calcualtions without having to spend too much money.

Regards,
Matt 

Thanks, Carl, for taking the time to look at my code.  When Maple crashes, the application stops appearing in the applications tab of my task manager ant the task bar at the bottom of my Windows 7 system.  However mserver.exe appears in the Processes tab of the task manager.  I can't see the memory and time numbers after it crashes, but I just started another run, and the Memory says 60.61M, and the time is just incrementing.  The kernelopts(cpulimit) and the kernelopts(datalimit) commands both return the infinity symbol.  

I check for small factors with the igcd() command before I run the isprime() command because it make the code run faster.  I am looking for clusters of 12 prime numbers.  It is faster to make a quick check that all 12 do not have any small factors before running the slower isprime() command.  Also, I have two levels of pretest and this seems to be the fastest way that I have found.  I know about showstat(isprime) and that gave me the idea for a quick check.  

In fact on line 12 of showstat(isprime) it says - 

r := gmp_isprime(n);

Do you know how I can see the code for the gmp_isprime(n); function?

Again thank you for your help.  Hopefully I can find a way to do long calcualtions without having to spend too much money.

Regards,
Matt 

Fm = 2^(2^m)+1

Every Fermat number Fm, when represented in binary has the form
10000...00001 with 2^m-1 zeros inside.

So it would be possible to just count zeros, and not have to store them all.

There are some downloadable software that check for divisors of fermat numbers at fermatsearch.

Regards,
Matt

Fm = 2^(2^m)+1

Every Fermat number Fm, when represented in binary has the form
10000...00001 with 2^m-1 zeros inside.

So it would be possible to just count zeros, and not have to store them all.

There are some downloadable software that check for divisors of fermat numbers at fermatsearch.

Regards,
Matt