@Carl Love 

Really thanks alot for your kind efforts. But still there is problem when I compare the coefficient having "x power 0" (used the command remove(has,f, x)). Please solve the problem the problem is attached.

Equate.mw

@tomleslie 

Dear friend!

I am very thankful you kind effort. Your file which you sent me I used it to found the solution for H=[0,1000,2000,3000,4000] using your program (attached). Then I calculate the solution using my program (attached) for the same values of H but they are different. My results are correct but I want to calculate the same result using your program. Please let me know where is the problem because solution dosent match. Your program more better than me, and convient to use. I am waiting your positive response. 

MHD_Flow_H=4000_R=100_alpha=5.mw

MHD_Flow_H_varies_R=100_alpha=5.mw

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

@tomleslie 

Thanks dear your reply really apperciate. Actually the orginal file are attached please fix it I will be really thankful to you.

Help.mw

@Carl Love 

Dear Sir I want to plot the above graph from -1..1 where the change will happen? I am waiting your positive response. 

@John Fredsted I required a sepcial lower triangular matrix as shown in attached fine. I know the built-in command which you sent me. If you fix my problem, i will be grateful to you. Thanks

@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 

@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.