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Non-default Set Orderings...

Asked by: Adam Ledger 360
set list
June 23 2018
0  4

I want to learn about why sometimes different algebraic expressions are ordered differently, and what would be the best way to instruct maple to chose a different "rule" for which a set of arithmetic functions are ordered, while still maintaining the axiom of uniqueness. 

 

I figured this would be accomplished on a case by case basis, and it may involve just using lists instead,defining a procedure to produce the desired ordering for any given set of functions, and using the "remove" command to replace the set "difference" operator and the Join command for the lists in place of the union operator. Then of course using the "Remove Duplicates" option from ListTools to impose the axiom of unique elements. 

 

Is this the best way to go about this, or is there a much simpler way, that includes an abstract algebra package that I'm unaware of thus far?

How to get phase portrait sketch for homogeneous ...

Asked by: ANANDMUNAGALA 175
differential-equations phaseportrait
June 23 2018
0  2

Dear sir, I hereby request you to suggest an appropriate method to plot phase portrait sketches for the above cited subject

in 2D and 3D for the problem  

 

 

With thanks and regards.

 

Mr M ANAND

Associate Profesoor in Mathematics.

evalf/xxx, printlevel, typesetting...

Asked by: vv 13852
evalf printlevel frac
June 23 2018
4  12

This question is related to an answer I gave here:
So, please look at a simple worksheet containing only a few lines; the resuts are in the # comments.

restart;
evalf(frac(Pi^20));
#                              23.
restart;
printlevel:=40:
evalf(frac(Pi^20));
  ###  prinlevel stuff
#                              0.


And now the questions.
1. Why the first evalf(frac(Pi^20))  does not  call  `evalf/frac`?
     (the second does, trace(`evalf/frac`)  shows this  if inserted).
     Note that  `evalf/frac`(Pi^20)    returns  0.
2. Why evalf(frac(Pi^20))    depends on printlevel?
    Note that  if  printlevel is changed to 20 (say)  the result is again 23.
3. Why if we set interface(typesetting=standard)  in a fresh session
     the results are both 23?

 

Problem with a while do loop...

Asked by: Ronan 1341
loop while
June 22 2018
0  3

I can't get a While do loop to work as expected.

For i from 2 while M[i,1]<>M[1,1] and i<25 do..   It doesn't catch row 15 where M[15,1] =M[1,1] but it does stop at i = 24 ok.
 

restart

``

M := Matrix(60, 3, {(1, 1) = Vector(2, {(1) = -3, (2) = 5}), (1, 2) = Vector(2, {(1) = -2, (2) = -2}), (1, 3) = Vector(2, {(1) = 0, (2) = 0}), (2, 1) = Vector(2, {(1) = -5, (2) = 3}), (2, 2) = Vector(2, {(1) = -1, (2) = -3}), (2, 3) = Vector(2, {(1) = 1, (2) = -1}), (3, 1) = Vector(2, {(1) = -6, (2) = 0}), (3, 2) = Vector(2, {(1) = 0, (2) = -3}), (3, 3) = Vector(2, {(1) = 1, (2) = 0}), (4, 1) = Vector(2, {(1) = -6, (2) = -3}), (4, 2) = Vector(2, {(1) = 1, (2) = -2}), (4, 3) = Vector(2, {(1) = 1, (2) = 1}), (5, 1) = Vector(2, {(1) = -5, (2) = -5}), (5, 2) = Vector(2, {(1) = 2, (2) = -1}), (5, 3) = Vector(2, {(1) = 1, (2) = 1}), (6, 1) = Vector(2, {(1) = -3, (2) = -6}), (6, 2) = Vector(2, {(1) = 3, (2) = 0}), (6, 3) = Vector(2, {(1) = 1, (2) = 1}), (7, 1) = Vector(2, {(1) = 0, (2) = -6}), (7, 2) = Vector(2, {(1) = 3, (2) = 1}), (7, 3) = Vector(2, {(1) = 0, (2) = 1}), (8, 1) = Vector(2, {(1) = 3, (2) = -5}), (8, 2) = Vector(2, {(1) = 2, (2) = 2}), (8, 3) = Vector(2, {(1) = -1, (2) = 1}), (9, 1) = Vector(2, {(1) = 5, (2) = -3}), (9, 2) = Vector(2, {(1) = 1, (2) = 3}), (9, 3) = Vector(2, {(1) = -1, (2) = 1}), (10, 1) = Vector(2, {(1) = 6, (2) = 0}), (10, 2) = Vector(2, {(1) = 0, (2) = 3}), (10, 3) = Vector(2, {(1) = -1, (2) = 0}), (11, 1) = Vector(2, {(1) = 6, (2) = 3}), (11, 2) = Vector(2, {(1) = -1, (2) = 2}), (11, 3) = Vector(2, {(1) = -1, (2) = -1}), (12, 1) = Vector(2, {(1) = 5, (2) = 5}), (12, 2) = Vector(2, {(1) = -2, (2) = 1}), (12, 3) = Vector(2, {(1) = -1, (2) = -1}), (13, 1) = Vector(2, {(1) = 3, (2) = 6}), (13, 2) = Vector(2, {(1) = -3, (2) = 0}), (13, 3) = Vector(2, {(1) = -1, (2) = -1}), (14, 1) = Vector(2, {(1) = 0, (2) = 6}), (14, 2) = Vector(2, {(1) = -3, (2) = -1}), (14, 3) = Vector(2, {(1) = 0, (2) = -1}), (15, 1) = Vector(2, {(1) = -3, (2) = 5}), (15, 2) = Vector(2, {(1) = -2, (2) = -2}), (15, 3) = Vector(2, {(1) = 1, (2) = -1}), (16, 1) = Vector(2, {(1) = -5, (2) = 3}), (16, 2) = Vector(2, {(1) = -1, (2) = -3}), (16, 3) = Vector(2, {(1) = 1, (2) = -1}), (17, 1) = Vector(2, {(1) = -6, (2) = 0}), (17, 2) = Vector(2, {(1) = 0, (2) = -3}), (17, 3) = Vector(2, {(1) = 1, (2) = 0}), (18, 1) = Vector(2, {(1) = -6, (2) = -3}), (18, 2) = Vector(2, {(1) = 1, (2) = -2}), (18, 3) = Vector(2, {(1) = 1, (2) = 1}), (19, 1) = Vector(2, {(1) = -5, (2) = -5}), (19, 2) = Vector(2, {(1) = 2, (2) = -1}), (19, 3) = Vector(2, {(1) = 1, (2) = 1}), (20, 1) = Vector(2, {(1) = -3, (2) = -6}), (20, 2) = Vector(2, {(1) = 3, (2) = 0}), (20, 3) = Vector(2, {(1) = 1, (2) = 1}), (21, 1) = Vector(2, {(1) = 0, (2) = -6}), (21, 2) = Vector(2, {(1) = 3, (2) = 1}), (21, 3) = Vector(2, {(1) = 0, (2) = 1}), (22, 1) = Vector(2, {(1) = 3, (2) = -5}), (22, 2) = Vector(2, {(1) = 2, (2) = 2}), (22, 3) = Vector(2, {(1) = -1, (2) = 1}), (23, 1) = Vector(2, {(1) = 5, (2) = -3}), (23, 2) = Vector(2, {(1) = 1, (2) = 3}), (23, 3) = Vector(2, {(1) = -1, (2) = 1}), (24, 1) = Vector(2, {(1) = 6, (2) = 0}), (24, 2) = Vector(2, {(1) = 0, (2) = 3}), (24, 3) = Vector(2, {(1) = -1, (2) = 0}), (25, 1) = Vector(2, {(1) = 6, (2) = 3}), (25, 2) = Vector(2, {(1) = -1, (2) = 2}), (25, 3) = Vector(2, {(1) = -1, (2) = -1}), (26, 1) = Vector(2, {(1) = 5, (2) = 5}), (26, 2) = Vector(2, {(1) = -2, (2) = 1}), (26, 3) = Vector(2, {(1) = -1, (2) = -1}), (27, 1) = Vector(2, {(1) = 3, (2) = 6}), (27, 2) = Vector(2, {(1) = -3, (2) = 0}), (27, 3) = Vector(2, {(1) = -1, (2) = -1}), (28, 1) = Vector(2, {(1) = 0, (2) = 6}), (28, 2) = Vector(2, {(1) = -3, (2) = -1}), (28, 3) = Vector(2, {(1) = 0, (2) = -1}), (29, 1) = Vector(2, {(1) = -3, (2) = 5}), (29, 2) = Vector(2, {(1) = -2, (2) = -2}), (29, 3) = Vector(2, {(1) = 1, (2) = -1}), (30, 1) = Vector(2, {(1) = -5, (2) = 3}), (30, 2) = Vector(2, {(1) = -1, (2) = -3}), (30, 3) = Vector(2, {(1) = 1, (2) = -1}), (31, 1) = Vector(2, {(1) = -6, (2) = 0}), (31, 2) = Vector(2, {(1) = 0, (2) = -3}), (31, 3) = Vector(2, {(1) = 1, (2) = 0}), (32, 1) = Vector(2, {(1) = -6, (2) = -3}), (32, 2) = Vector(2, {(1) = 1, (2) = -2}), (32, 3) = Vector(2, {(1) = 1, (2) = 1}), (33, 1) = Vector(2, {(1) = -5, (2) = -5}), (33, 2) = Vector(2, {(1) = 2, (2) = -1}), (33, 3) = Vector(2, {(1) = 1, (2) = 1}), (34, 1) = Vector(2, {(1) = -3, (2) = -6}), (34, 2) = Vector(2, {(1) = 3, (2) = 0}), (34, 3) = Vector(2, {(1) = 1, (2) = 1}), (35, 1) = Vector(2, {(1) = 0, (2) = -6}), (35, 2) = Vector(2, {(1) = 3, (2) = 1}), (35, 3) = Vector(2, {(1) = 0, (2) = 1}), (36, 1) = Vector(2, {(1) = 3, (2) = -5}), (36, 2) = Vector(2, {(1) = 2, (2) = 2}), (36, 3) = Vector(2, {(1) = -1, (2) = 1}), (37, 1) = Vector(2, {(1) = 5, (2) = -3}), (37, 2) = Vector(2, {(1) = 1, (2) = 3}), (37, 3) = Vector(2, {(1) = -1, (2) = 1}), (38, 1) = Vector(2, {(1) = 6, (2) = 0}), (38, 2) = Vector(2, {(1) = 0, (2) = 3}), (38, 3) = Vector(2, {(1) = -1, (2) = 0}), (39, 1) = Vector(2, {(1) = 6, (2) = 3}), (39, 2) = Vector(2, {(1) = -1, (2) = 2}), (39, 3) = Vector(2, {(1) = -1, (2) = -1}), (40, 1) = Vector(2, {(1) = 5, (2) = 5}), (40, 2) = Vector(2, {(1) = -2, (2) = 1}), (40, 3) = Vector(2, {(1) = -1, (2) = -1}), (41, 1) = Vector(2, {(1) = 3, (2) = 6}), (41, 2) = Vector(2, {(1) = -3, (2) = 0}), (41, 3) = Vector(2, {(1) = -1, (2) = -1}), (42, 1) = Vector(2, {(1) = 0, (2) = 6}), (42, 2) = Vector(2, {(1) = -3, (2) = -1}), (42, 3) = Vector(2, {(1) = 0, (2) = -1}), (43, 1) = Vector(2, {(1) = -3, (2) = 5}), (43, 2) = Vector(2, {(1) = -2, (2) = -2}), (43, 3) = Vector(2, {(1) = 1, (2) = -1}), (44, 1) = Vector(2, {(1) = -5, (2) = 3}), (44, 2) = Vector(2, {(1) = -1, (2) = -3}), (44, 3) = Vector(2, {(1) = 1, (2) = -1}), (45, 1) = Vector(2, {(1) = -6, (2) = 0}), (45, 2) = Vector(2, {(1) = 0, (2) = -3}), (45, 3) = Vector(2, {(1) = 1, (2) = 0}), (46, 1) = Vector(2, {(1) = -6, (2) = -3}), (46, 2) = Vector(2, {(1) = 1, (2) = -2}), (46, 3) = Vector(2, {(1) = 1, (2) = 1}), (47, 1) = Vector(2, {(1) = -5, (2) = -5}), (47, 2) = Vector(2, {(1) = 2, (2) = -1}), (47, 3) = Vector(2, {(1) = 1, (2) = 1}), (48, 1) = Vector(2, {(1) = -3, (2) = -6}), (48, 2) = Vector(2, {(1) = 3, (2) = 0}), (48, 3) = Vector(2, {(1) = 1, (2) = 1}), (49, 1) = Vector(2, {(1) = 0, (2) = -6}), (49, 2) = Vector(2, {(1) = 3, (2) = 1}), (49, 3) = Vector(2, {(1) = 0, (2) = 1}), (50, 1) = Vector(2, {(1) = 3, (2) = -5}), (50, 2) = Vector(2, {(1) = 2, (2) = 2}), (50, 3) = Vector(2, {(1) = -1, (2) = 1}), (51, 1) = Vector(2, {(1) = 5, (2) = -3}), (51, 2) = Vector(2, {(1) = 1, (2) = 3}), (51, 3) = Vector(2, {(1) = -1, (2) = 1}), (52, 1) = Vector(2, {(1) = 6, (2) = 0}), (52, 2) = Vector(2, {(1) = 0, (2) = 3}), (52, 3) = Vector(2, {(1) = -1, (2) = 0}), (53, 1) = Vector(2, {(1) = 6, (2) = 3}), (53, 2) = Vector(2, {(1) = -1, (2) = 2}), (53, 3) = Vector(2, {(1) = -1, (2) = -1}), (54, 1) = Vector(2, {(1) = 5, (2) = 5}), (54, 2) = Vector(2, {(1) = -2, (2) = 1}), (54, 3) = Vector(2, {(1) = -1, (2) = -1}), (55, 1) = Vector(2, {(1) = 3, (2) = 6}), (55, 2) = Vector(2, {(1) = -3, (2) = 0}), (55, 3) = Vector(2, {(1) = -1, (2) = -1}), (56, 1) = Vector(2, {(1) = 0, (2) = 6}), (56, 2) = Vector(2, {(1) = -3, (2) = -1}), (56, 3) = Vector(2, {(1) = 0, (2) = -1}), (57, 1) = Vector(2, {(1) = -3, (2) = 5}), (57, 2) = Vector(2, {(1) = -2, (2) = -2}), (57, 3) = Vector(2, {(1) = 1, (2) = -1}), (58, 1) = Vector(2, {(1) = -5, (2) = 3}), (58, 2) = Vector(2, {(1) = -1, (2) = -3}), (58, 3) = Vector(2, {(1) = 1, (2) = -1}), (59, 1) = Vector(2, {(1) = -6, (2) = 0}), (59, 2) = Vector(2, {(1) = 0, (2) = -3}), (59, 3) = Vector(2, {(1) = 1, (2) = 0}), (60, 1) = Vector(2, {(1) = -6, (2) = -3}), (60, 2) = Vector(2, {(1) = 1, (2) = -2}), (60, 3) = Vector(2, {(1) = 1, (2) = 1})})

M := Matrix(60, 3, {(1, 1) = Vector(2, {(1) = -3, (2) = 5}), (1, 2) = Vector(2, {(1) = -2, (2) = -2}), (1, 3) = Vector(2, {(1) = 0, (2) = 0}), (2, 1) = Vector(2, {(1) = -5, (2) = 3}), (2, 2) = Vector(2, {(1) = -1, (2) = -3}), (2, 3) = Vector(2, {(1) = 1, (2) = -1}), (3, 1) = Vector(2, {(1) = -6, (2) = 0}), (3, 2) = Vector(2, {(1) = 0, (2) = -3}), (3, 3) = Vector(2, {(1) = 1, (2) = 0}), (4, 1) = Vector(2, {(1) = -6, (2) = -3}), (4, 2) = Vector(2, {(1) = 1, (2) = -2}), (4, 3) = Vector(2, {(1) = 1, (2) = 1}), (5, 1) = Vector(2, {(1) = -5, (2) = -5}), (5, 2) = Vector(2, {(1) = 2, (2) = -1}), (5, 3) = Vector(2, {(1) = 1, (2) = 1}), (6, 1) = Vector(2, {(1) = -3, (2) = -6}), (6, 2) = Vector(2, {(1) = 3, (2) = 0}), (6, 3) = Vector(2, {(1) = 1, (2) = 1}), (7, 1) = Vector(2, {(1) = 0, (2) = -6}), (7, 2) = Vector(2, {(1) = 3, (2) = 1}), (7, 3) = Vector(2, {(1) = 0, (2) = 1}), (8, 1) = Vector(2, {(1) = 3, (2) = -5}), (8, 2) = Vector(2, {(1) = 2, (2) = 2}), (8, 3) = Vector(2, {(1) = -1, (2) = 1}), (9, 1) = Vector(2, {(1) = 5, (2) = -3}), (9, 2) = Vector(2, {(1) = 1, (2) = 3}), (9, 3) = Vector(2, {(1) = -1, (2) = 1}), (10, 1) = Vector(2, {(1) = 6, (2) = 0}), (10, 2) = Vector(2, {(1) = 0, (2) = 3}), (10, 3) = Vector(2, {(1) = -1, (2) = 0}), (11, 1) = Vector(2, {(1) = 6, (2) = 3}), (11, 2) = Vector(2, {(1) = -1, (2) = 2}), (11, 3) = Vector(2, {(1) = -1, (2) = -1}), (12, 1) = Vector(2, {(1) = 5, (2) = 5}), (12, 2) = Vector(2, {(1) = -2, (2) = 1}), (12, 3) = Vector(2, {(1) = -1, (2) = -1}), (13, 1) = Vector(2, {(1) = 3, (2) = 6}), (13, 2) = Vector(2, {(1) = -3, (2) = 0}), (13, 3) = Vector(2, {(1) = -1, (2) = -1}), (14, 1) = Vector(2, {(1) = 0, (2) = 6}), (14, 2) = Vector(2, {(1) = -3, (2) = -1}), (14, 3) = Vector(2, {(1) = 0, (2) = -1}), (15, 1) = Vector(2, {(1) = -3, (2) = 5}), (15, 2) = Vector(2, {(1) = -2, (2) = -2}), (15, 3) = Vector(2, {(1) = 1, (2) = -1}), (16, 1) = Vector(2, {(1) = -5, (2) = 3}), (16, 2) = Vector(2, {(1) = -1, (2) = -3}), (16, 3) = Vector(2, {(1) = 1, (2) = -1}), (17, 1) = Vector(2, {(1) = -6, (2) = 0}), (17, 2) = Vector(2, {(1) = 0, (2) = -3}), (17, 3) = Vector(2, {(1) = 1, (2) = 0}), (18, 1) = Vector(2, {(1) = -6, (2) = -3}), (18, 2) = Vector(2, {(1) = 1, (2) = -2}), (18, 3) = Vector(2, {(1) = 1, (2) = 1}), (19, 1) = Vector(2, {(1) = -5, (2) = -5}), (19, 2) = Vector(2, {(1) = 2, (2) = -1}), (19, 3) = Vector(2, {(1) = 1, (2) = 1}), (20, 1) = Vector(2, {(1) = -3, (2) = -6}), (20, 2) = Vector(2, {(1) = 3, (2) = 0}), (20, 3) = Vector(2, {(1) = 1, (2) = 1}), (21, 1) = Vector(2, {(1) = 0, (2) = -6}), (21, 2) = Vector(2, {(1) = 3, (2) = 1}), (21, 3) = Vector(2, {(1) = 0, (2) = 1}), (22, 1) = Vector(2, {(1) = 3, (2) = -5}), (22, 2) = Vector(2, {(1) = 2, (2) = 2}), (22, 3) = Vector(2, {(1) = -1, (2) = 1}), (23, 1) = Vector(2, {(1) = 5, (2) = -3}), (23, 2) = Vector(2, {(1) = 1, (2) = 3}), (23, 3) = Vector(2, {(1) = -1, (2) = 1}), (24, 1) = Vector(2, {(1) = 6, (2) = 0}), (24, 2) = Vector(2, {(1) = 0, (2) = 3}), (24, 3) = Vector(2, {(1) = -1, (2) = 0}), (25, 1) = Vector(2, {(1) = 6, (2) = 3}), (25, 2) = Vector(2, {(1) = -1, (2) = 2}), (25, 3) = Vector(2, {(1) = -1, (2) = -1}), (26, 1) = Vector(2, {(1) = 5, (2) = 5}), (26, 2) = Vector(2, {(1) = -2, (2) = 1}), (26, 3) = Vector(2, {(1) = -1, (2) = -1}), (27, 1) = Vector(2, {(1) = 3, (2) = 6}), (27, 2) = Vector(2, {(1) = -3, (2) = 0}), (27, 3) = Vector(2, {(1) = -1, (2) = -1}), (28, 1) = Vector(2, {(1) = 0, (2) = 6}), (28, 2) = Vector(2, {(1) = -3, (2) = -1}), (28, 3) = Vector(2, {(1) = 0, (2) = -1}), (29, 1) = Vector(2, {(1) = -3, (2) = 5}), (29, 2) = Vector(2, {(1) = -2, (2) = -2}), (29, 3) = Vector(2, {(1) = 1, (2) = -1}), (30, 1) = Vector(2, {(1) = -5, (2) = 3}), (30, 2) = Vector(2, {(1) = -1, (2) = -3}), (30, 3) = Vector(2, {(1) = 1, (2) = -1}), (31, 1) = Vector(2, {(1) = -6, (2) = 0}), (31, 2) = Vector(2, {(1) = 0, (2) = -3}), (31, 3) = Vector(2, {(1) = 1, (2) = 0}), (32, 1) = Vector(2, {(1) = -6, (2) = -3}), (32, 2) = Vector(2, {(1) = 1, (2) = -2}), (32, 3) = Vector(2, {(1) = 1, (2) = 1}), (33, 1) = Vector(2, {(1) = -5, (2) = -5}), (33, 2) = Vector(2, {(1) = 2, (2) = -1}), (33, 3) = Vector(2, {(1) = 1, (2) = 1}), (34, 1) = Vector(2, {(1) = -3, (2) = -6}), (34, 2) = Vector(2, {(1) = 3, (2) = 0}), (34, 3) = Vector(2, {(1) = 1, (2) = 1}), (35, 1) = Vector(2, {(1) = 0, (2) = -6}), (35, 2) = Vector(2, {(1) = 3, (2) = 1}), (35, 3) = Vector(2, {(1) = 0, (2) = 1}), (36, 1) = Vector(2, {(1) = 3, (2) = -5}), (36, 2) = Vector(2, {(1) = 2, (2) = 2}), (36, 3) = Vector(2, {(1) = -1, (2) = 1}), (37, 1) = Vector(2, {(1) = 5, (2) = -3}), (37, 2) = Vector(2, {(1) = 1, (2) = 3}), (37, 3) = Vector(2, {(1) = -1, (2) = 1}), (38, 1) = Vector(2, {(1) = 6, (2) = 0}), (38, 2) = Vector(2, {(1) = 0, (2) = 3}), (38, 3) = Vector(2, {(1) = -1, (2) = 0}), (39, 1) = Vector(2, {(1) = 6, (2) = 3}), (39, 2) = Vector(2, {(1) = -1, (2) = 2}), (39, 3) = Vector(2, {(1) = -1, (2) = -1}), (40, 1) = Vector(2, {(1) = 5, (2) = 5}), (40, 2) = Vector(2, {(1) = -2, (2) = 1}), (40, 3) = Vector(2, {(1) = -1, (2) = -1}), (41, 1) = Vector(2, {(1) = 3, (2) = 6}), (41, 2) = Vector(2, {(1) = -3, (2) = 0}), (41, 3) = Vector(2, {(1) = -1, (2) = -1}), (42, 1) = Vector(2, {(1) = 0, (2) = 6}), (42, 2) = Vector(2, {(1) = -3, (2) = -1}), (42, 3) = Vector(2, {(1) = 0, (2) = -1}), (43, 1) = Vector(2, {(1) = -3, (2) = 5}), (43, 2) = Vector(2, {(1) = -2, (2) = -2}), (43, 3) = Vector(2, {(1) = 1, (2) = -1}), (44, 1) = Vector(2, {(1) = -5, (2) = 3}), (44, 2) = Vector(2, {(1) = -1, (2) = -3}), (44, 3) = Vector(2, {(1) = 1, (2) = -1}), (45, 1) = Vector(2, {(1) = -6, (2) = 0}), (45, 2) = Vector(2, {(1) = 0, (2) = -3}), (45, 3) = Vector(2, {(1) = 1, (2) = 0}), (46, 1) = Vector(2, {(1) = -6, (2) = -3}), (46, 2) = Vector(2, {(1) = 1, (2) = -2}), (46, 3) = Vector(2, {(1) = 1, (2) = 1}), (47, 1) = Vector(2, {(1) = -5, (2) = -5}), (47, 2) = Vector(2, {(1) = 2, (2) = -1}), (47, 3) = Vector(2, {(1) = 1, (2) = 1}), (48, 1) = Vector(2, {(1) = -3, (2) = -6}), (48, 2) = Vector(2, {(1) = 3, (2) = 0}), (48, 3) = Vector(2, {(1) = 1, (2) = 1}), (49, 1) = Vector(2, {(1) = 0, (2) = -6}), (49, 2) = Vector(2, {(1) = 3, (2) = 1}), (49, 3) = Vector(2, {(1) = 0, (2) = 1}), (50, 1) = Vector(2, {(1) = 3, (2) = -5}), (50, 2) = Vector(2, {(1) = 2, (2) = 2}), (50, 3) = Vector(2, {(1) = -1, (2) = 1}), (51, 1) = Vector(2, {(1) = 5, (2) = -3}), (51, 2) = Vector(2, {(1) = 1, (2) = 3}), (51, 3) = Vector(2, {(1) = -1, (2) = 1}), (52, 1) = Vector(2, {(1) = 6, (2) = 0}), (52, 2) = Vector(2, {(1) = 0, (2) = 3}), (52, 3) = Vector(2, {(1) = -1, (2) = 0}), (53, 1) = Vector(2, {(1) = 6, (2) = 3}), (53, 2) = Vector(2, {(1) = -1, (2) = 2}), (53, 3) = Vector(2, {(1) = -1, (2) = -1}), (54, 1) = Vector(2, {(1) = 5, (2) = 5}), (54, 2) = Vector(2, {(1) = -2, (2) = 1}), (54, 3) = Vector(2, {(1) = -1, (2) = -1}), (55, 1) = Vector(2, {(1) = 3, (2) = 6}), (55, 2) = Vector(2, {(1) = -3, (2) = 0}), (55, 3) = Vector(2, {(1) = -1, (2) = -1}), (56, 1) = Vector(2, {(1) = 0, (2) = 6}), (56, 2) = Vector(2, {(1) = -3, (2) = -1}), (56, 3) = Vector(2, {(1) = 0, (2) = -1}), (57, 1) = Vector(2, {(1) = -3, (2) = 5}), (57, 2) = Vector(2, {(1) = -2, (2) = -2}), (57, 3) = Vector(2, {(1) = 1, (2) = -1}), (58, 1) = Vector(2, {(1) = -5, (2) = 3}), (58, 2) = Vector(2, {(1) = -1, (2) = -3}), (58, 3) = Vector(2, {(1) = 1, (2) = -1}), (59, 1) = Vector(2, {(1) = -6, (2) = 0}), (59, 2) = Vector(2, {(1) = 0, (2) = -3}), (59, 3) = Vector(2, {(1) = 1, (2) = 0}), (60, 1) = Vector(2, {(1) = -6, (2) = -3}), (60, 2) = Vector(2, {(1) = 1, (2) = -2}), (60, 3) = Vector(2, {(1) = 1, (2) = 1})})

 (1)

M[1, 1]; for i from 2 while M[i, 1] <> M[1, 1] and i < 25 do print(i, M[i, 1]) end do

24, Vector[column](%id = 18446745366646139710)

 (2)

``


 

Download Test_While_do_loop.mw

Cut of Errorfunction...

Asked by: digerdiga 390
error-function
June 22 2018
0  2

I'm currently wondering about the cut I'm looking for in the following worksheet.

I evaluate it in 2 ways but get different answers. Any idea what the problem here is?

Thanks


 

> 

restart; dIs := sqrt(Pi/(I*s))*exp(I*s*t-I*s*omega0^2); Is1 := `assuming`([simplify(int(dIs, s))], [s > 0]); dIs := `assuming`([int(exp(-I*(omega^2+omega0^2-t)*s), omega = -infinity .. infinity)], [s > 0]); Is2 := int(%, s); plot3d(Im(eval(Is1, [t = x+I*y, s = 1, omega0 = 1])), x = -3 .. 3, y = -3 .. 3)

(-I*Pi/s)^(1/2)*exp(I*s*t-I*s*omega0^2)

  

(1/2-(1/2)*I)*Pi*2^(1/2)*erf(s^(1/2)*(I*(omega0^2-t))^(1/2))/(I*(omega0^2-t))^(1/2)

  

exp(I*s*t-I*s*omega0^2)*Pi^(1/2)/(I*s)^(1/2)

  

-I*Pi*erf((omega0^2-t)^(1/2)*(I*s)^(1/2))/(omega0^2-t)^(1/2)

  

 
> 

``


 

Download CutErrorFunction.mw

HELP ME TO SOLVE THIS ERROR (USING DSOLVE)...

Asked by: farah adanan 15
dsolve numeric differential-equations
June 22 2018
2  5

restart; with(plots);
[animate, animate3d, animatecurve, arrow, changecoords, 

  complexplot, complexplot3d, conformal, conformal3d, 

  contourplot, contourplot3d, coordplot, coordplot3d, 

  densityplot, display, dualaxisplot, fieldplot, fieldplot3d, 

  gradplot, gradplot3d, implicitplot, implicitplot3d, inequal, 

  interactive, interactiveparams, intersectplot, listcontplot, 

  listcontplot3d, listdensityplot, listplot, listplot3d, 

  loglogplot, logplot, matrixplot, multiple, odeplot, pareto, 

  plotcompare, pointplot, pointplot3d, polarplot, polygonplot, 

  polygonplot3d, polyhedra_supported, polyhedraplot, rootlocus, 

  semilogplot, setcolors, setoptions, setoptions3d, spacecurve, 

  sparsematrixplot, surfdata, textplot, textplot3d, tubeplot]


fixedparameter1 := [n = .3, W[e] = .3, M = .2, gamma = 1, delta = -1, N[r] = .8, Pr = .72, Nb = .5, Nt = .5, Bi = 2, Pr = .72, Le = 5];
[n = 0.3, W[e] = 0.3, M = 0.2, gamma = 1, delta = -1, N[r] = 0.8, 

  Pr = 0.72, Nb = 0.5, Nt = 0.5, Bi = 2, Pr = 0.72, Le = 5]


eq1 := (1-n)*(diff(f(eta), eta, eta, eta))+f(eta)*(diff(f(eta), eta, eta))-M*(diff(f(eta), eta))+n*W[e]*(diff(f(eta), eta, eta, eta))*(diff(f(eta), eta, eta)) = 0;
        /  d   /  d   /  d         \\\
(1 - n) |----- |----- |----- f(eta)|||
        \ deta \ deta \ deta       ///

            /  d   /  d         \\     /  d         \
   + f(eta) |----- |----- f(eta)|| - M |----- f(eta)|
            \ deta \ deta       //     \ deta       /

            /  d   /  d   /  d         \\\ /  d   /  d         \\   
   + n W[e] |----- |----- |----- f(eta)||| |----- |----- f(eta)|| = 
            \ deta \ deta \ deta       /// \ deta \ deta       //   

  0
deq1; eval(eq1, fixedparameter1);
    /  d   /  d   /  d         \\\
0.7 |----- |----- |----- f(eta)|||
    \ deta \ deta \ deta       ///

            /  d   /  d         \\       /  d         \
   + f(eta) |----- |----- f(eta)|| - 0.2 |----- f(eta)|
            \ deta \ deta       //       \ deta       /

          /  d   /  d   /  d         \\\ /  d   /  d         \\   
   + 0.09 |----- |----- |----- f(eta)||| |----- |----- f(eta)|| = 
          \ deta \ deta \ deta       /// \ deta \ deta       //   

  0
eq2 := (1+(4/3)*N[r])*(diff(theta(eta), eta, eta))+Pr*f(eta)*(diff(theta(eta), eta))+Nb*(diff(phi(eta), eta))*(diff(theta(eta), eta))+Nt*(diff(theta(eta), eta))*(diff(theta(eta), eta)) = 0;
          /    4     \ /  d   /  d             \\
          |1 + - N[r]| |----- |----- theta(eta)||
          \    3     / \ deta \ deta           //

                         /  d             \
             + Pr f(eta) |----- theta(eta)|
                         \ deta           /

                  /  d           \ /  d             \
             + Nb |----- phi(eta)| |----- theta(eta)|
                  \ deta         / \ deta           /

                                    2    
                  /  d             \     
             + Nt |----- theta(eta)|  = 0
                  \ deta           /     
deq2; eval(eq2, fixedparameter1);
                      /  d   /  d             \\
          2.066666667 |----- |----- theta(eta)||
                      \ deta \ deta           //

                           /  d             \
             + 0.72 f(eta) |----- theta(eta)|
                           \ deta           /

                   /  d           \ /  d             \
             + 0.5 |----- phi(eta)| |----- theta(eta)|
                   \ deta         / \ deta           /

                                     2    
                   /  d             \     
             + 0.5 |----- theta(eta)|  = 0
                   \ deta           /     
eq3 := diff(phi(eta), eta, eta)+Pr*Le*f(eta)*(diff(phi(eta), eta))+Nt*(diff(theta(eta), eta, eta))/Nb = 0;
    /  d   /  d           \\                /  d           \
    |----- |----- phi(eta)|| + Pr Le f(eta) |----- phi(eta)|
    \ deta \ deta         //                \ deta         /

            /  d   /  d             \\    
         Nt |----- |----- theta(eta)||    
            \ deta \ deta           //    
       + ----------------------------- = 0
                      Nb                  
deq3 := eval(eq3, fixedparameter1);
    /  d   /  d           \\               /  d           \
    |----- |----- phi(eta)|| + 3.60 f(eta) |----- phi(eta)|
    \ deta \ deta         //               \ deta         /

                     /  d   /  d             \\    
       + 1.000000000 |----- |----- theta(eta)|| = 0
                     \ deta \ deta           //    
bcs1 := f(0) = 0, D(f)(0) = 1+gamma*(D@D)(F)(0)+delta*(D@D@D)(f)(0), D(f)(8) = 0;
 f(0) = 0, 

   D(f)(0) = 1 + gamma @@(D, 2)(F)(0) + delta @@(D, 3)(f)(0), 

   D(f)(8) = 0
bc1 := eval(bcs1, fixedparameter1);
   f(0) = 0, D(f)(0) = 1 + @@(D, 2)(F)(0) - @@(D, 3)(f)(0), 

     D(f)(8) = 0
bcs2 := D(theta)(0) = Bi*(theta(0)-1), theta(8) = 0;
         D(theta)(0) = Bi (theta(0) - 1), theta(8) = 0
bc2 := eval(bcs2, fixedparameter1);
           D(theta)(0) = 2 theta(0) - 2, theta(8) = 0
bcs3 := Nb*D(phi)(0)+Nt*D(theta)(0) = 0, Nb*D(phi)(0)+Nt*D(theta)(0) = 0, phi(8) = 0;
        Nb D(phi)(0) + Nt D(theta)(0) = 0, 

          Nb D(phi)(0) + Nt D(theta)(0) = 0, phi(8) = 0
bc3 := eval(bcs3, fixedparameter1);
       0.5 D(phi)(0) + 0.5 D(theta)(0) = 0, 

         0.5 D(phi)(0) + 0.5 D(theta)(0) = 0, phi(8) = 0
R := dsolve({bc1, bc2, bc3, deq1, deq2, deq3}, [f(eta), theta(eta), phi(eta)], numeric, output = listprocedure);
Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations


 

How to code an if statement with a variable not eq...

Asked by: amycormier 5
syntax conditional if-then
June 22 2018
0  2

Hello! I am trying to make an if statement that is IF a bound is not equal to NULL, it does things, and if it IS equal to NULL, the bounds are set to zero. When a bound is null, they say 

bound1:=()

My first if statement will not work, please help!

 

bound1:=solve(tau(x)=(Intv||j)[1],x,useassumptions) assuming (Intv||i)[1]<=x<=(Intv||i)[2] ;  

bound2:=solve(tau(x)=(Intv||j)[2],x,useassumptions) assuming (Intv||i)[1]<=x<=(Intv||i)[2];

if bound1<>NULL;bound2<>NULL;  then

if bound1<=bound2   then  

lower:=bound1;  upper:=bound2  

else lower:=bound2;   upper:=bound1 end if;

else lower:=0; upper:=0 end if;

Another way to find the length of edges of a tetra...

Asked by: minhthien2016 385
geometry angle tetrahedron
June 21 2018
0  0

Let be given tetrahedron ABCD, where AB = BC = AC = a, AD = d, AD = e, CD = f. I know that, If the measure of angle of AB and CD equal to Pi/3, then we have d^2 - e^2 - a*f = 0. I tried:
ListTools[Categorize];
L := []; 
for a to 30 do for d to 30 do
for e to 30 do for f to 30 do
if abs(d-e) < a and a < d+e and abs(a-e) < d and d < a+e and abs(d-a) < e and e < d+a and abs(d-f) < a and a < d+f and abs(a-f) < d and d < a+f and abs(d-a) < f and f < d+a and abs(e-f) < a and a < e+f and abs(a-f) < e and e < a+f and abs(a-e) < a and a < a+e and -a*f+d^2-e^2 = 0 and igcd(a, d, e, f) = 1 and nops({a, d, e, f}) = 4
then L := [op(L), [a, d, e, f]] end if end do end do end do end do; 
nops(L); 
L;

Another way to find the length of edges of a tetrahedron knowing that the mesure angle of two opposite?

Improvements in solving PDE & BC in Maple 2018.1

by: Katherina von Bulow 140 Maple 2018
pde pdsolve differential-equations
June 21 2018
8  11


 

For Maple 2018.1, there are improvements in pdsolve's ability to solve PDE with boundary and initial conditions. This is work done together with E.S. Cheb-Terrab. The improvements include an extended ability to solve problems involving non-homogeneous PDE and/or non-homogeneous boundary and initial conditions, as well as improved simplification of solutions and better handling of functions such as piecewise in the arguments and in the processing of solutions. This is also an ongoing project, with updates being distributed regularly within the Physics Updates.

Solving more problems involving non-homogeneous PDE and/or non-homogeneous boundary and initial conditions

 

 

Example 1: Pinchover and Rubinstein's exercise 6.17: we have a non-homogenous PDE and boundary and initial conditions that are also non-homogeneous:

> 

pde__1 := diff(u(x, t), t)-(diff(u(x, t), x, x)) = 1+x*cos(t)
iv__1 := (D[1](u))(0, t) = sin(t), (D[1](u))(1, t) = sin(t), u(x, 0) = 1+cos(2*Pi*x)

> 

pdsolve([pde__1, iv__1])

u(x, t) = 1+cos(2*Pi*x)*exp(-4*Pi^2*t)+t+x*sin(t)

(1)

How we solve the problem, step by step:

   

 

Example 2: the PDE is homogeneous but the boundary conditions are not. We solve the problem through the same process, which means we end up solving a nonhomogeneous pde with homogeneous BC as an intermediate step:

> 

pde__2 := diff(u(x, t), t) = 13*(diff(u(x, t), x, x))
iv__2 := (D[1](u))(0, t) = 0, (D[1](u))(1, t) = 1, u(x, 0) = (1/2)*x^2+x

> 

pdsolve([pde__2, iv__2])

u(x, t) = 1/2+Sum(2*(-1+(-1)^n)*cos(n*Pi*x)*exp(-13*Pi^2*n^2*t)/(Pi^2*n^2), n = 1 .. infinity)+13*t+(1/2)*x^2

(12)

How we solve the problem, step by step:

   

 

Example 3: a wave PDE with a source that does not depend on time:

> 

pde__3 := (diff(u(x, t), x, x))*a^2+1 = diff(u(x, t), t, t)
iv__3 := u(0, t) = 0, u(L, t) = 0, u(x, 0) = f(x), (D[2](u))(x, 0) = g(x)

> 

`assuming`([pdsolve([pde__3, iv__3])], [L > 0])

u(x, t) = (1/2)*(2*(Sum(sin(n*Pi*x/L)*(2*L*(Int(sin(n*Pi*x/L)*g(x), x = 0 .. L))*sin(a*Pi*t*n/L)*a-Pi*(Int(sin(n*Pi*x/L)*(-2*f(x)*a^2+L*x-x^2), x = 0 .. L))*cos(a*Pi*t*n/L)*n)/(Pi*n*a^2*L), n = 1 .. infinity))*a^2+L*x-x^2)/a^2

 (23)

How we solve the problem, step by step:

   

 

Example 4: Pinchover and Rubinstein's exercise 6.23 - we have a non-homogenous PDE and initial condition:

> 

pde__4 := diff(u(x, t), t)-(diff(u(x, t), x, x)) = g(x, t)
iv__4 := (D[1](u))(0, t) = 0, (D[1](u))(1, t) = 0, u(x, 0) = f(x)

> 

pdsolve([pde__4, iv__4], u(x, t))

u(x, t) = Int(f(tau1), tau1 = 0 .. 1)+Sum(2*(Int(f(tau1)*cos(n*Pi*tau1), tau1 = 0 .. 1))*cos(n*Pi*x)*exp(-Pi^2*n^2*t), n = 1 .. infinity)+Int(Int(g(x, tau1), x = 0 .. 1)+Sum(2*(Int(g(x, tau1)*cos(n1*Pi*x), x = 0 .. 1))*cos(n1*Pi*x)*exp(-Pi^2*n1^2*(t-tau1)), n1 = 1 .. infinity), tau1 = 0 .. t)

 (30)

If we now make the functions f and g into specific mappings, we can compare pdsolve's solutions to the general and specific problems:

> 

f := proc (x) options operator, arrow; 3*cos(42*x*Pi) end proc
g := proc (x, t) options operator, arrow; exp(3*t)*cos(17*x*Pi) end proc

 

Here is what pdsolve's solution to the general problem looks like when taking into account the new values of f(x) and g(x,t):

> 

value(simplify(evalindets(u(x, t) = Int(f(tau1), tau1 = 0 .. 1)+Sum(2*(Int(f(tau1)*cos(n*Pi*tau1), tau1 = 0 .. 1))*cos(n*Pi*x)*exp(-Pi^2*n^2*t), n = 1 .. infinity)+Int(Int(g(x, tau1), x = 0 .. 1)+Sum(2*(Int(g(x, tau1)*cos(n1*Pi*x), x = 0 .. 1))*cos(n1*Pi*x)*exp(-Pi^2*n1^2*(t-tau1)), n1 = 1 .. infinity), tau1 = 0 .. t), specfunc(Int), proc (u) options operator, arrow; `PDEtools/int`(op(u), AllSolutions) end proc)))

u(x, t) = 3*cos(42*Pi*x)*exp(-1764*Pi^2*t)+cos(Pi*x)*(65536*cos(Pi*x)^16-278528*cos(Pi*x)^14+487424*cos(Pi*x)^12-452608*cos(Pi*x)^10+239360*cos(Pi*x)^8-71808*cos(Pi*x)^6+11424*cos(Pi*x)^4-816*cos(Pi*x)^2+17)*(exp(289*Pi^2*t+3*t)-1)*exp(-289*Pi^2*t)/(289*Pi^2+3)

 (31)

 

Here is pdsolve's solution to the specific problem:

> 

pdsolve([pde__4, iv__4], u(x, t))

u(x, t) = ((867*Pi^2+9)*cos(42*Pi*x)*exp(-1764*Pi^2*t)+cos(17*Pi*x)*(exp(3*t)-exp(-289*Pi^2*t)))/(289*Pi^2+3)

(32)

 

And the two solutions are equal:

> 

simplify(combine((u(x, t) = 3*cos(42*x*Pi)*exp(-1764*Pi^2*t)+cos(x*Pi)*(65536*cos(x*Pi)^16-278528*cos(x*Pi)^14+487424*cos(x*Pi)^12-452608*cos(x*Pi)^10+239360*cos(x*Pi)^8-71808*cos(x*Pi)^6+11424*cos(x*Pi)^4-816*cos(x*Pi)^2+17)*(exp(289*Pi^2*t+3*t)-1)*exp(-289*Pi^2*t)/(289*Pi^2+3))-(u(x, t) = ((867*Pi^2+9)*cos(42*x*Pi)*exp(-1764*Pi^2*t)+cos(17*x*Pi)*(exp(3*t)-exp(-289*Pi^2*t)))/(289*Pi^2+3)), trig))

0 = 0

(33)
> 

f := 'f'; g := 'g'

 

Improved simplification in integrals, piecewise functions, and sums in the solutions returned by pdsolve

 

 

Example 1: exercise 6.21 from Pinchover and Rubinstein is a non-homogeneous heat problem. Its solution used to include unevaluated integrals and sums, but is now returned in a significantly simpler format.

> 

pde__5 := diff(u(x, t), t)-(diff(u(x, t), x, x)) = t*cos(2001*x)
iv__5 := (D[1](u))(0, t) = 0, (D[1](u))(Pi, t) = 0, u(x, 0) = Pi*cos(2*x)

> 

pdsolve([pde__5, iv__5])

u(x, t) = (1/16032024008001)*(4004001*t+exp(-4004001*t)-1)*cos(2001*x)+Pi*cos(2*x)*exp(-4*t)

(34)
> 

pdetest(%, [pde__5, iv__5])

[0, 0, 0, 0]

(35)

 

Example 2: example 6.46 from Pinchover and Rubinstein is a non-homogeneous heat equation with non-homogeneous boundary and initial conditions. Its solution used to involve two separate sums with unevaluated integrals, but is now returned with only one sum and unevaluated integral.

> 

pde__6 := diff(u(x, t), t)-(diff(u(x, t), x, x)) = exp(-t)*sin(3*x)
iv__6 := u(0, t) = 0, u(Pi, t) = 1, u(x, 0) = phi(x)

> 

pdsolve([pde__6, iv__6], u(x, t))

u(x, t) = (1/8)*(8*(Sum(2*(Int(-(-phi(x)*Pi+x)*sin(n*x), x = 0 .. Pi))*sin(n*x)*exp(-n^2*t)/Pi^2, n = 1 .. infinity))*Pi-Pi*(exp(-9*t)-exp(-t))*sin(3*x)+8*x)/Pi

 (36)
> 

pdetest(%, [pde__6, iv__6])

[0, 0, 0, (-phi(x)*Pi^2+Pi*x+2*(Sum((Int(-(-phi(x)*Pi+x)*sin(n*x), x = 0 .. Pi))*sin(n*x), n = 1 .. infinity)))/Pi^2]

(37)

 

More accuracy when returning series solutions that have exceptions for certain values of the summation index or a parameter

 

 

Example 1: the answer to this problem was previously given with n = 0 .. infinity instead of n = 1 .. infinity as it should be:

> 

pde__7 := diff(v(x, t), t, t)-(diff(v(x, t), x, x))

iv__7 := v(0, t) = 0, v(x, 0) = -(exp(2)*x-exp(x+1)-x+exp(1-x))/(exp(2)-1), (D[2](v))(x, 0) = 1+(exp(2)*x-exp(x+1)-x+exp(1-x))/(exp(2)-1), v(1, t) = 0
> 

pdsolve([pde__7, iv__7])

v(x, t) = Sum(-2*sin(n*Pi*x)*((Pi^2*(-1)^n*n^2-Pi^2*n^2+2*(-1)^n-1)*sin(Pi*t*n)-(-1)^n*cos(Pi*t*n)*Pi*n)/(Pi^2*n^2*(Pi^2*n^2+1)), n = 1 .. infinity)

 (38)

 

Example 2: the answer to exercise 6.25 from Pinchover and Rubinstein is now given in a much simpler format, with the special limit case for w = 0 calculated separately:

> 

pde__8 := diff(u(x, t), t) = k*(diff(u(x, t), x, x))+cos(w*t)
iv__8 := (D[1](u))(L, t) = 0, (D[1](u))(0, t) = 0, u(x, 0) = x

> 

`assuming`([pdsolve([pde__8, iv__8], u(x, t))], [L > 0])

u(x, t) = piecewise(w = 0, (1/2)*L+Sum(2*L*(-1+(-1)^n)*cos(n*Pi*x/L)*exp(-Pi^2*n^2*k*t/L^2)/(n^2*Pi^2), n = 1 .. infinity)+t, (1/2)*(L*w+2*(Sum(2*L*(-1+(-1)^n)*cos(n*Pi*x/L)*exp(-Pi^2*n^2*k*t/L^2)/(n^2*Pi^2), n = 1 .. infinity))*w+2*sin(w*t))/w)

 (39)

 

Improved handling of piecewise, eval/diff in the given problem

 

 

Example 1: this problem, which contains a piecewise function in the initial condition, can now be solved:

> 

pde__9 := diff(f(x, t), t) = diff(f(x, t), x, x)
iv__9 := f(0, t) = 0, f(1, t) = 1, f(x, 0) = piecewise(x = 0, 0, 1)

> 

pdsolve([pde__9, iv__9])

f(x, t) = Sum(2*sin(n*Pi*x)*exp(-Pi^2*n^2*t)/(n*Pi), n = 1 .. infinity)+x

(40)

 

Example 2: this problem, which contains a derivative written using eval/diff, can now be solved:

> 

pde__10 := -(diff(u(x, t), t, t))-(diff(u(x, t), x, x))+u(x, t) = 2*exp(-t)*(x-(1/2)*x^2+(1/2)*t-1)

iv__10 := u(x, 0) = x^2-2*x, u(x, 1) = u(x, 1/2)+((1/2)*x^2-x)*exp(-1)-((3/4)*x^2-(3/2)*x)*exp(-1/2), u(0, t) = 0, eval(diff(u(x, t), x), {x = 1}) = 0
> 

pdsolve([pde__10, iv__10], u(x, t))

u(x, t) = -(1/2)*exp(-t)*x*(x-2)*(t-2)

(41)

 

References:

 

Pinchover, Y. and Rubinstein, J.. An Introduction to Partial Differential Equations. Cambridge UP, 2005.


 

Download What_is_New_after_Maple_2018.mw

Katherina von Bülow

is this another evalf bug?...

Asked by: Adam Ledger 360
evalf frac
June 21 2018
3  19


evalf[100](frac(exp(19*Pi)-19*Pi));

0.32853457802957784855876405976954586639886249604033514784046998713819112593


evalf[10](frac(exp(19*Pi)-19*Pi));

 

0.

Likelihood function and finding some constants...

Asked by: Dark Energy 100
statistics
June 21 2018
1  3

Hello Friends

I have a critical problem that I wish to solve it with maple

suppose we have a list like following: y_obs=(2,4,8,7,9,52,35,478,52) and corresponding variance σy=(.2,.3,.5,.87,.1.2,.22,.78,.99,1.5)
we know y as the function of x described such as y_theoric=x+p and minimizing X is

X=Sigma [(y_theoric-y_obs)^2]/σy which includes the sum of nine numbers...

the question is:

How we can find p from likelihood function and plot general behavior of y versus of x through two above series?

for example this solution used in article under the names Hubble parameter data constraints on dark energy by Yun Chen and Bhatra Ratra (Physics Letters B)

Thank you

 

Removal of Periodic, Salt and Pepper Noises from...

by: stan2018 100 Maple
June 21 2018
7  0

In an attempt to explore the field of image processing, @Samir Khan and I created an application (download here) that demonstrates the removal of two types of noises from an image through frequency and spatial filtering.

Periodic noises and salt & pepper noises are two common types of image noises, usually caused by errors during the image capturing or data transmission process. Periodic noises result in repetitive patterns being added onto the original image, while salt & pepper noises are the irregular appearance of dark pixels in the bright area and bright pixels in the dark area of the image. In this application, we artificially generate these noises and pollute a clean picture in order to demonstrate the removal techniques.

(Fig 1: Picture of Waterloo Office taken by Sophie Tan            Fig 2: Converted to greyscale for processing, added two noises)

In order to remove periodic noises from the image, we apply a 2D Fourier Transform to convert the image from spatial domain to frequency domain, where periodic noises can be visually detected as separate, discrete spikes and therefore easily removed.

(Fig 3 Frequency domain of the magnitude of the image)

One way to remove salt and pepper noises is to apply a median filter to the image. In this application, we run a 3 by 3 kernel across the image matrix that sorts and places the median among the 9 elements as the new matrix entry, thus resulting in the whole image being median-filtered.

Comparison of the image before and after noise removal:

Please refer to the application for more details on the implementation of the two removal techniques.

 

How do I choose two coordinates points over four p...

Asked by: Sabrina Kamal 15
solve substitution
June 21 2018
1  2

Hello everyone! I am currently solving on a basic coordinates points for my final year project. This is a part of my coding in maple.


ans := solve({eq5, eq6}, {P2, Q2});
             {P2 = 3.222860033, Q2 = 3.170614592}, 

               {P2 = 1.572224939, Q2 = 5.670614592}

 

I am finding the points of P2 and Q2. From there after solving for eq5 and 6 it will gives two points for P2 and two points for Q2. So how am i going to choose the points using maple coding without copy paste the answer?

Really appreciate if any of us can help me. Thank you in advanced :)

extracting coefficients of exponential functions...

Asked by: guras 100
solve exponential coefficients
June 20 2018
1  2

 

Consider for instance the following equation:

Eq:=(a-4)*exp(4*x)+(b+1)*exp(2*x)+(c-2)=0

How can I list the coefficients of the exponential functions and also solve the equation for the constant parameters 

a, b, and c?

I tried 

[coeffs(collect(lhs(EQ), exp), exp)] =~ 0;

but it did not work. Thank you for your help.

What does this plot look like?...

Asked by: Markiyan Hirnyk 7809
June 20 2018
1  22

I mean 

restart;
 plots:-implicitplot(sqrt(b)*sqrt(1-4*p/b)-2*arctan(sqrt((9*p/b-22201/10000)/(9/4-9*p/b))) = 0, b = 0 .. 5,
 p = 0 .. 5, gridrefine = 2, rational);

I find the above result unsatisfactory.

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