## 230 Reputation

10 years, 75 days
Beijing, China

eta=1

## @Carl Love  Sorry! I was busy last ...

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

## Need Help...

Let

$f = x^{11} + 2 x^9 + 2x^8 + x^6 + x^5 + 2x^3 + 2x^2 +1 \in \mathbf{F}_3[x],$

to be factored over the field with three elements.

The algorithm computes first

$c = \gcd(f, f') = x^9 + 2x^6 + x^3 + 2.$

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

$f= (x+1)(x^2+1)^3(x+2)^4.$
New.mw

## Need Help...

Dear Sir! I am waiting your kind response

## Request to recheck...

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

Let

$f = x^{11} + 2 x^9 + 2x^8 + x^6 + x^5 + 2x^3 + 2x^2 +1 \in \mathbf{F}_3[x],$

to be factored over the field with three elements.

The algorithm computes first

$c = \gcd(f, f') = x^9 + 2x^6 + x^3 + 2.$

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 + 2z = x + 1 and R = x + 1 with updates i = 2w = x + 2 andc = x8 + x7 + x6 + x2+x+1. The second time through the loop gives y = x + 2z = 1R = x + 1, with updates i = 3w = 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 = 1z = x + 2R = (x + 1)(x + 2)4, with updates i = 5w = 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

$f= (x+1)(x^2+1)^3(x+2)^4.$
but the program given
(x+1)*(x+2)^4*(x^6+1)
Recheck.mw

## Value of p...

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

p=3

## Again Problem...

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

## Problem...

This algorithm is not run. Please send its maple file

## Problem...

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

With my best regards and sincerely

PhD (Scholar)
Department of Mathematics

HITEC University Taxila Cantt Pakistan

## Request...

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

With my best regards and sincerely

PhD (Scholar)
Department of Mathematics

HITEC University Taxila Cantt Pakistan

## Solution of expression involve matrix...

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

## Not working...

AOA.. Dear this algorithm not working

## Thanks Alot...

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

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