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These are answers submitted by vv

You define u(x,t) := ...,  but u is already defined as a table u[i]:=...
This generates a total mess. Use another "variable", e.g. U(x,t) instead.

How can you believe that the expressions could be equal? Not only the exponents of sin(...) are different (2/n  versus 2/(n-1)) but the constants are not the same, one of them is piecewise, ...

C:=(m::nonnegint,n::integer) -> 
  coeff(coeff(product((1-q^i)*(1-q^i/z)*(1-q^(i-1)*z), i=1..m+1),q,m),z,n):

C(6,2),  C(6,4); 

      0,  1

(k is an arbitrary integer)

for c to N do  for d from c+1 to N do
  for x1 to c-1 do
    for F in [f,g] do
      if nops(factor(F(x,c,d)-F(x1,c,d)))=4 then lprint(F(x+k,c,d),m=F(x1,c,d)) fi;
od od:

(x+k-4)*(x+k+3)*(x+k-5)*(x+k+4), m = 180
(x+k-4)*(x+k+3)*(x+k-6)*(x+k+5), m = 360
(x+k-4)*(x+k+4)*(x+k-7)*(x+k+7), m = 720
(x+k-5)*(x+k+4)*(x+k-7)*(x+k+6), m = 504
(x+k-5)*(x+k+4)*(x+k-9)*(x+k+8), m = 1260
(x+k-5)*(x+k+4)*(x+k-10)*(x+k+9), m = 1800
(x+k-5)*(x+k+5)*(x+k-10)*(x+k+10), m = 2016
(x+k-6)*(x+k+5)*(x+k-7)*(x+k+6), m = 1260
(x+k-6)*(x+k+6)*(x+k-7)*(x+k+7), m = 1440
(x+k-6)*(x+k+5)*(x+k-9)*(x+k+8), m = 1080
(x+k-7)*(x+k+7)*(x+k-9)*(x+k+9), m = 2880
(x+k-7)*(x+k+6)*(x+k-10)*(x+k+9), m = 3780
(x+k-8)*(x+k+8)*(x+k-9)*(x+k+9), m = 5040
(x+k-9)*(x+k+8)*(x+k-10)*(x+k+9), m = 5544
(x+k-9)*(x+k+8)*(x+k-10)*(x+k+9), m = 2520

1. In Maple 2020 it can be done because a (statement)  is an expression

x:=10: str:="A":
str:=cat(str,  `if`(x=10,  [(x:=11)," it was 10"][2], [(x:=8)," it was not 10"][2]));  x;


2. `if`  is alias for ifelse  and a link appears in the help page of if. Or, use ?ifelse 


`%+`(sort([op(p)], key=abs@coeffs)[]);

Use value(%)  to  go back.

This isthe well known  Pell's equation.

for _Z1 from 0 to 6 do

0, 1
226153980, 1766319049
798920165762330040, 6239765965720528801
2822295814832482312327709940, 22042834973108102061352541449
9970149719303180503641083029374964080, 77869358613928486808166555366140995201
35220930741174421456911021812718768924061809900, 275084262906388245923976756042747916825335226249
124422801783292138491822391332416163557158135530198606120, 971773147303355325052564141449134520779147876502526039201

It is easy to obtain such equalities. Here is a simple example. It can be extended to a generic one.
Why do you consider them so important?

   local m:=subs(sqrt(3)=0,z),
   m^2+n^2-1+ 2*m*n-sqrt(3)
G:= n -> (F@@n)(g@@n)(1+sqrt(3))=1+sqrt(3):

You do not have a value (true or false) for print_table
Note also that a better  header for your proc is:

proc(f::procedure, a::realcons, b::realcons, N::posint, print_table::truefalse)


You must use parallel substitutions in ex:
ex := simplify(subs([ X = cos(theta)*X-sin(theta)*Y, Y = sin(theta)*X+cos(theta)*Y ], eq)); 

(You have defined the procedure f. It is then easier to use f(X-9/7, Y+8/7)  and similar for ex, instead of subs) .

It is interesting that for diff, the first argument can be almost anything, including a mathematical nonsense. For sets, lists, rtables it acts as expected (elementwise), but even strings are accepted. 
Note that series may produce errors.

f:=sin(x) + x^3*"a";
g:=sin(x) + [cos(x), x^3];
h:=sin(x) + [cos(x)*"a"] + {x^2*"b"^3 + "c"};

You can reduce the time to 0 because if a,b,... are >0 (as you have tried) then g must be a square.


The problem is generated by simplify.

(1 + (203808*exp(-(342569*t)/506))/131537)^(131537/203808);

simplify(%, sqrt);  # or simplify(%)

For example, with a smaller exponent:

simplify((1 + (203808*exp(-(342569*t)/506))/131537)^(3/7), sqrt); 




You should try to understand and explain what happens in the next lines:

local S:=LinearAlgebra:-RandomMatrix(3, generator=rand(-10..10),shape=skewsymmetric),
      V0:=LinearAlgebra:-RandomVector(3, generator=rand(-10..10))^+,
      A:=(S-1)^(-1) . (S+1) . <a,0,0;0,b,0;0,0,c>;
<V0, V0+A[1],V0+A[2],V0+A[3]>
seq(P(3,5,14), 1..10);

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