Muhammad Usman

235 Reputation

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10 years, 349 days
Beijing, China

MaplePrimes Activity


These are replies submitted by Muhammad Usman

@AmusingYeti 

First I am very thankful to you about your valuable comments. Now I am facing another problem in integration please see the attachment and fixed it as soon as possible. I am waiting your kind comments. Take crae.

Integration.mw

@tomleslie 

Dear how you get this solution, which command you used? Please correct the attached file and send me back. I am waiting your kind response.

@Markiyan Hirnyk 

I do not understand your question what you want to say.

@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

 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

@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

 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

@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

 

With my best regards and sincerely
 
Muhammad Usman

PhD (Scholar)
Department of Mathematics

HITEC University Taxila Cantt Pakistan

@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

 

With my best regards and sincerely
 
Muhammad Usman

PhD (Scholar)
Department of Mathematics

HITEC University Taxila Cantt Pakistan
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