@tomleslie 

eta=1

@Carl Love 

Sorry! I was busy last three days, didnt check your mail, first I will checked then let you know. Thanks

@vv 

Dear Sir please help me to complete this program of square free factrization whose explanation below

Let

to be factored over the field with three elements.

The algorithm computes first

Since the derivative is non-zero we have w = f/c = x² + 2 and we enter the while loop. After one loop we have y = x + 2, z = x + 1 and R = x + 1 with updates i = 2, w = x + 2 andc = x⁸ + x⁷ + x⁶ + x²+x+1. The second time through the loop gives y = x + 2, z = 1, R = x + 1, with updates i = 3, w = x + 2 and c = x⁷ + 2x⁶ + x + 2. The third time through the loop also does not change R. For the fourth time through the loop we get y = 1, z = x + 2, R = (x + 1)(x + 2)⁴, with updates i = 5, w = 1 and c = x⁶ + 1. Since w = 1, we exit the while loop. Since c ≠ 1, it must be a perfect cube. The cube root of c, obtained by replacing x³ by x is x² + 1, and calling the square-free procedure recursively determines that it is square-free. Therefore, cubing it and combining it with the value of R to that point gives the square-free decomposition

New.mw

@Carl Love 

Dear Sir! I am waiting your kind response

@Carl Love 

Dear sir first of all very thansful for your nice cooperation. The above program give roungh result like

Let

to be factored over the field with three elements.

The algorithm computes first

Since the derivative is non-zero we have w = f/c = x² + 2 and we enter the while loop. After one loop we have y = x + 2, z = x + 1 and R = x + 1 with updates i = 2, w = x + 2 andc = x⁸ + x⁷ + x⁶ + x²+x+1. The second time through the loop gives y = x + 2, z = 1, R = x + 1, with updates i = 3, w = x + 2 and c = x⁷ + 2x⁶ + x + 2. The third time through the loop also does not change R. For the fourth time through the loop we get y = 1, z = x + 2, R = (x + 1)(x + 2)⁴, with updates i = 5, w = 1 and c = x⁶ + 1. Since w = 1, we exit the while loop. Since c ≠ 1, it must be a perfect cube. The cube root of c, obtained by replacing x³ by x is x² + 1, and calling the square-free procedure recursively determines that it is square-free. Therefore, cubing it and combining it with the value of R to that point gives the square-free decomposition

but the program given

(x+1)*(x+2)^4*(x^6+1)

Recheck.mw

@Carl Love 

Dear sir please try by taking p = 3. I am waiting your response as I need urgent.

@Carl Love 

p=3

@ThU 

Dear, I already doing same procdure but this particular solution is rongh, the correct one has attached

@Carl Love 

This algorithm is not run. Please send its maple file

@Kitonum 

Sir I am facing problem when i used this concept in my required file. I attached this file please check it and solve the problem. I will be veriy thankful to you for this kindness

PhD (Scholar)
Department of Mathematics

@Carl Love 

Sir you are so coprative and nice person. Whenever i post any question you will be the first who answered. I request to you please share if you have any lecture or book on maple which help us for programing in Maple

PhD (Scholar)
Department of Mathematics

@Carl Love 

I want to find in one step of the following expression

u := psi(x)^T*(P^2)^T*U*P^2*psi(t)-t*psi(x)^T*(P^2)^T*U*P^2*psi(1)-x*psi(1)^T*(P^2)^T*U*P^2*psi(t)+t*x*psi(1)^T*(P^2)^T*U*P^2*psi(1)

for given psi(x),P and U as mention above

@Kitonum Help.mw

@Carl Love 

AOA.. Dear this algorithm not working

AOA... Sir I am Muhammad Usman PhD Schloar al HITEC University Pakistan your help in Maple very useful in my research work... Thanks alot one again...

With my best regard and sincerely

M. Usman

@acer