## simplification of arguments...

If I have an expression like this

f:=ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))

maple has trouble to simplify the argument.

In particular is it possible to apply expand() only to the denominator?

This is meant in general, so if I have many terms with expressions like this (possibly of products with other functions in each term), I want this simplification to be done termwise for the arguments of the functions.

Expanding the fraction doesn't work as in frontend(expand, [f]).

## polylog vs. dilog...

This is not a problem per se, but more to understand the background.

restart;

f := polylog(2, -x);

int(f/(x+1), x);

convert(f, dilog);

int(%/(x+1), x)

The integration of the polylog maple is not capable of doing, but after converting to dilog it finds an anti derivative.

That leads to the question, why is dilog as a separate to polylog(2,*) implemented anyway? Why couldn't it all be done with the more general polylog function?

I'm also wondering why maple has difficulties to integrate

int(dilog(x+1)/(x+a),x)

for general a.

## How do I solve an propositions such as this (exist...

How would I go about getting true or false returned on these propositions?
I have tried just about every eval and various syntax methods, but nothing has worked so far.

I know most can easilly be done by hand/thinking, but I'm sure Maple should have a way to do this as well.

∀n∈Z:2n>n+2   ,   ∃n∈Z:2|(3n+1)    ,   ∃k∈Z:∀n∈Z:n=kn   ,   ∃k∈Z:∀n∈Z:2|(n+k)   ,   ∀n∈Z:∀k∈Z:(n>k∨k≥n)

## ImportMatrix Kernel Lost...

Using Maple 2018.2.1, I'm receiving a lost kernel message when importing the attached data file with ImportMatrix. I traced the issue to a "*" symbol at the end of the file but would have expected this to cause an error message (if any error at all) instead of the connection to the kernel to be lost. Is this a bug or am I misunderstanding the usage of ImportMatrix?

test.mw

test2.txt

## PDE_and_BC_during_2018_err...

Hello, to all,
On my computer I have installed Windows 7 Professional, Maple 2018.2.1 and
Physics:-Version()[2];
2019, January 5, 13:32 hours, version in the MapleCloud: 276,

version installed in this computer: 276
When I try to compute some examples from your poste "PDE_and_BC_during_2018.mw", I get an error in Example 8:
Example 8: This problem represents the temperature distribution in a thin circular plate whose lateral surfaces are insulated (Articolo example 6.9.2):
pde__8 := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r);
/
|
|
|
d                    1   |/ d                \
pde__8 := --- u(r, theta, t) = ---- ||--- u(r, theta, t)|
dt                  25 r |\ dr               /
\

2                  \
d                   |
-------- u(r, theta, t)|
/  2                \          2               |
| d                 |    dtheta                |
+ r |---- u(r, theta, t)| + -----------------------|
|   2               |              r           |
\ dr                /                          /
iv__8 := D[1]*u(1, theta, t) = 0, u(r, 0, t) = 0, u(r, Pi, t) = 0, u(r, theta, 0) = (r-(1/3)*r^3)*sin(theta);
iv__8 := D[1] u(1, theta, t) = 0, u(r, 0, t) = 0,

/    1  3\
u(r, Pi, t) = 0, u(r, theta, 0) = |r - - r | sin(theta)
\    3   /
pdsolve([pde__8, iv__8], u(r, theta, t), HINT = boundedseries(r = [0]));
Error, (in dsolve) cannot determine if this expression is true or false: not 0 <= -Pi

or I get no answer as in Example 10:
Example 10: A Laplace PDE with one homogeneous and three non-homogeneous conditions:
pde__10 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0;
/  2         \   /  2         \
| d          |   | d          |
pde__10 := |---- u(x, y)| + |---- u(x, y)| = 0
|   2        |   |   2        |
\ dx         /   \ dy         /
iv__10 := u(0, y) = 0, u(Pi, y) = sinh(Pi)*cos(y), u(x, 0) = sin(x), u(x, Pi) = -sinh(x);
iv__10 := u(0, y) = 0, u(Pi, y) = sinh(Pi) cos(y),

u(x, 0) = sin(x), u(x, Pi) = -sinh(x)
pdsolve([pde__10, iv__10]);

There are also no answer as in Examle 10 in the Examples 15, 18, 19
Can you give me a hint,  what could be wrong?
With kind regards
Wolfgang Gellien

## Maintain Sequence Iterator In Same Line...

So I was trying to create a shorthand for creating a plot of multiple arrows, with the arrow colour dependent on the magnitude of the vector.
I currently have a set of vectors, v, I want to display, and v[4] is the largest.

I know this could be done by creating an arrow plot for each vector seperately and then by combining them using display:
`arrow1 := arrow( v[1], width=0.15,length=20,color=ColorTools:-Color( (norm(v[1])/norm(v[4]))*[0,0,1] ) );`
```arrow2 := arrow( v[2], width=0.15,length=20,color=ColorTools:-Color( (norm(v[2])/norm(v[4]))*[0,0,1] ) ); ... print(plots:-display([arrow1, arrow2, ...]));```

But I was wondering if it could be done in a fashion similar to this:
```arrows := arrow([seq(v[i], i=1..4)],width=0.15,length=20,color=ColorTools:-Color((norm(v[i])/norm(v[4]))*[0,0,1])); print(arrows);```

(btw: it works fine replacing the last i with 1, which draws all arrows nearly black, or with 4, which as you guessed, draws all arrows blue...)

Help appreciated!

## pdetest fails when called after another pdetest...

I found a case where pdetest fails when called after another call to pdetest. I am using Physics version 272 from the clould. Using Maple 2018.2.1 on windows 10

```restart;
u:='u';x:='x';t:='t';
pde := diff(u(x, t), t) + diff(u(x, t),x) =0;
sol:=pdsolve(pde,u(x,t));
pdetest(sol,pde);
#0  OK

#restart;
u:='u';x:='x';t:='t';
pde:=diff(u(x,t),t)+diff(u(x,t),x)=0;
bc:=u(0,t)=0;
ic:=u(x,0)=sin(x);
sol:=pdsolve([pde,ic,bc],u(x,t)) assuming x>0;
pdetest(sol,pde);

#sol := u(x, t) = -sin(-x+t)*Heaviside(-t+x)   #OK

#-Dirac(-t+x)*sin(x)*cos(t)+Dirac(-t+x)*cos(x)*sin(t)+
Dirac(-x+t)*sin(x)*cos(t)-Dirac(-x+t)*cos(x)*sin(t)   #WRONG should be 0
```

If I run the above again, but with restart call in between active, so that all is cleared, then pdetest gives 0 as expected on the second pde, with the same solution

```restart;
u:='u';x:='x';t:='t';
pde := diff(u(x, t), t) + diff(u(x, t),x) =0;
sol:=pdsolve(pde,u(x,t));
pdetest(sol,pde);
#0   #OK

restart;
u:='u';x:='x';t:='t';
pde:=diff(u(x,t),t)+diff(u(x,t),x)=0;
bc:=u(0,t)=0;
ic:=u(x,0)=sin(x);
sol:=pdsolve([pde,ic,bc],u(x,t)) assuming x>0;
pdetest(sol,pde);

#0   #OK !! ```

can any one explain why this happens?  Is this a bug? It seems like pdetest caching problem. it remembers something from the last call and this affects the result it gives for the next call.

any work around?

## evaluate a list of polynomials over a list of valu...

Hello everybody.

This is my question. I tried to evaluate a list of polynomial over a list of values. Something like this:

eval([a*x, b*x], x = [p, q, t])

to get something like this:

[[a*p, a*q, a*t], [b*p, b*q, b*t]]

I know this method: eval~(a*x,x=~[p,q,t])  though this works for one polynomial over a list of values. Not precisely, what I am looking for.

I figured out a method that worked defining functions and with ‘apply’ and ‘map’. Here an example:

m:=t->3*2^t:

n:=t->(t+4)^2:

map(apply~,[m,n],[1,2,3]);

[[6, 12, 24], [25, 72, 147]]

However, how can I get this result using the ‘eval’ function.

Thank you all in advanced for any contribution.

## Evaluation of 'combined' constants...

Might be a beginners trap...

How can I evaluate a constant expression or an expression/formula that contains several scientific constants without "Constantin'g' them out like in sample 2)??

1) Example:

Rb := (1/4)*alpha/(Pi*R[infinity])

'real' value of (Rb)??

1) This example works:

evalf(1-Constant(m[e])/Constant(m[p])) -- but it's awful with respect to the simple demand to just divide two known constants..

## why can not collect(m+f(xi))...

restart;
T := K+F(xi)*F(xi);
2
K + F(xi)
U := alpha[0]+alpha[1]*(m+F(xi))+beta[1]/(m+F(xi))+alpha[2]*(m+F(xi))*(m+F(xi))+beta[2]/(m+F(xi))^2;
beta[1]
alpha[0] + alpha[1] (m + F(xi)) + ---------
m + F(xi)

2     beta[2]
+ alpha[2] (m + F(xi))  + ------------
2
(m + F(xi))
diff(U, xi);
/ d        \
beta[1] |---- F(xi)|
/ d        \           \ dxi      /
alpha[1] |---- F(xi)| - --------------------
\ dxi      /                  2
(m + F(xi))

/ d        \
2 beta[2] |---- F(xi)|
/ d        \             \ dxi      /
+ 2 alpha[2] (m + F(xi)) |---- F(xi)| - ----------------------
\ dxi      /                   3
(m + F(xi))
d := alpha[1]*T-beta[1]*T/(m+F(xi))^2+2*alpha[2]*(m+F(xi))*T-2*beta[2]*T/(m+F(xi))^3;
/         2\
/         2\   beta[1] \K + F(xi) /
alpha[1] \K + F(xi) / - --------------------
2
(m + F(xi))

/         2\
/         2\   2 beta[2] \K + F(xi) /
+ 2 alpha[2] (m + F(xi)) \K + F(xi) / - ----------------------
3
(m + F(xi))
diff(d, xi);
/ d        \
2 beta[1] F(xi) |---- F(xi)|
/ d        \                   \ dxi      /
2 alpha[1] F(xi) |---- F(xi)| - ----------------------------
\ dxi      /                      2
(m + F(xi))

/         2\ / d        \
2 beta[1] \K + F(xi) / |---- F(xi)|
\ dxi      /
+ -----------------------------------
3
(m + F(xi))

/ d        \ /         2\
+ 2 alpha[2] |---- F(xi)| \K + F(xi) /
\ dxi      /

/ d        \
+ 4 alpha[2] (m + F(xi)) F(xi) |---- F(xi)|
\ dxi      /

/ d        \
4 beta[2] F(xi) |---- F(xi)|
\ dxi      /
- ----------------------------
3
(m + F(xi))

/         2\ / d        \
6 beta[2] \K + F(xi) / |---- F(xi)|
\ dxi      /
+ -----------------------------------
4
(m + F(xi))
collect(%, diff);
/                                               /         2\
|                   2 beta[1] F(xi)   2 beta[1] \K + F(xi) /
|2 alpha[1] F(xi) - --------------- + ----------------------
|                               2                     3
\                    (m + F(xi))           (m + F(xi))

/         2\
+ 2 alpha[2] \K + F(xi) / + 4 alpha[2] (m + F(xi)) F(xi)

/         2\\
4 beta[2] F(xi)   6 beta[2] \K + F(xi) /| / d        \
- --------------- + ----------------------| |---- F(xi)|
3                     4     | \ dxi      /
(m + F(xi))           (m + F(xi))      /
S := (2*alpha[1]*F(xi)-2*beta[1]*F(xi)/(m+F(xi))^2+2*beta[1]*(K+F(xi)^2)/(m+F(xi))^3+2*alpha[2]*(K+F(xi)^2)+4*alpha[2]*(m+F(xi))*F(xi)-4*beta[2]*F(xi)/(m+F(xi))^3+6*beta[2]*(K+F(xi)^2)/(m+F(xi))^4)*T;
/                                               /         2\
|                   2 beta[1] F(xi)   2 beta[1] \K + F(xi) /
|2 alpha[1] F(xi) - --------------- + ----------------------
|                               2                     3
\                    (m + F(xi))           (m + F(xi))

/         2\
+ 2 alpha[2] \K + F(xi) / + 4 alpha[2] (m + F(xi)) F(xi)

/         2\\
4 beta[2] F(xi)   6 beta[2] \K + F(xi) /| /         2\
- --------------- + ----------------------| \K + F(xi) /
3                     4     |
(m + F(xi))           (m + F(xi))      /
expand((2*w*k*k)*beta*S-(2*A*k*k)*d-2*w*U+k*U*U);
2                   2               2
-2 A k  alpha[1] K - 2 A k  alpha[1] F(xi)

2               3
- 4 A k  alpha[2] F(xi)  - 4 w alpha[2] F(xi) m

+ 2 k alpha[0] alpha[1] m + 2 k alpha[0] alpha[1] F(xi)

2 k alpha[0] beta[1]                          2
+ -------------------- + 2 k alpha[0] alpha[2] m
m + F(xi)

2   2 k alpha[0] beta[2]
+ 2 k alpha[0] alpha[2] F(xi)  + --------------------
2
(m + F(xi))

2                         3
+ 2 k alpha[1]  m F(xi) + 2 k alpha[1] m  alpha[2]

3            2 k beta[1] beta[2]
+ 2 k alpha[1] F(xi)  alpha[2] + -------------------
3
(m + F(xi))

2  3                     2  2      2
+ 4 k alpha[2]  m  F(xi) + 6 k alpha[2]  m  F(xi)

2      3                              2
+ 4 k alpha[2]  F(xi)  m - 2 w alpha[0] + k alpha[0]

2                    3        2                2
+ 4 w k  beta alpha[1] F(xi)  + 4 w k  beta alpha[2] K

2
2                    4   2 A k  beta[1] K
+ 12 w k  beta alpha[2] F(xi)  + ----------------
2
(m + F(xi))

2              2
2 A k  beta[1] F(xi)         2
+ --------------------- - 4 A k  alpha[2] m K
2
(m + F(xi))

2                 2        2
- 4 A k  alpha[2] m F(xi)  - 4 A k  alpha[2] F(xi) K

2                  2              2
4 A k  beta[2] K   4 A k  beta[2] F(xi)
+ ---------------- + ---------------------
3                    3
(m + F(xi))          (m + F(xi))

2 k alpha[1] m beta[1]
+ 4 k alpha[0] alpha[2] F(xi) m + ----------------------
m + F(xi)

2
+ 6 k alpha[1] m  alpha[2] F(xi)

2   2 k alpha[1] m beta[2]
+ 6 k alpha[1] m alpha[2] F(xi)  + ----------------------
2
(m + F(xi))

2 k alpha[1] F(xi) beta[1]   2 k alpha[1] F(xi) beta[2]
+ -------------------------- + --------------------------
m + F(xi)                              2
(m + F(xi))

2                             2
2 k beta[1] alpha[2] m    2 k beta[1] alpha[2] F(xi)
+ ----------------------- + ---------------------------
m + F(xi)                   m + F(xi)

2                             2
2 k alpha[2] m  beta[2]   2 k alpha[2] F(xi)  beta[2]
+ ----------------------- + ---------------------------
2                           2
(m + F(xi))                 (m + F(xi))

2
2  2             2      2    k beta[1]
+ k alpha[1]  m  + k alpha[1]  F(xi)  + ------------
2
(m + F(xi))

2
2  4             2      4    k beta[2]
+ k alpha[2]  m  + k alpha[2]  F(xi)  + ------------
4
(m + F(xi))

2 w beta[1]
- 2 w alpha[1] m - 2 w alpha[1] F(xi) - -----------
m + F(xi)

2                     2   2 w beta[2]
- 2 w alpha[2] m  - 2 w alpha[2] F(xi)  - ------------
2
(m + F(xi))

2                             2                     2
4 w k  beta beta[1] F(xi) K   8 w k  beta beta[1] K F(xi)
- --------------------------- + ----------------------------
2                              3
(m + F(xi))                    (m + F(xi))

2
2                           8 w k  beta beta[2] F(xi) K
+ 8 w k  beta alpha[2] F(xi) m K - ---------------------------
3
(m + F(xi))

2                     2
24 w k  beta beta[2] K F(xi)         2
+ ----------------------------- + 4 w k  beta alpha[1] F(xi) K
4
(m + F(xi))

2                   3        2               2
4 w k  beta beta[1] F(xi)    4 w k  beta beta[1] K
- -------------------------- + ----------------------
2                          3
(m + F(xi))                (m + F(xi))

2                   4
4 w k  beta beta[1] F(xi)          2                      2
+ -------------------------- + 16 w k  beta alpha[2] K F(xi)
3
(m + F(xi))

2                   3
2                    3     8 w k  beta beta[2] F(xi)
+ 8 w k  beta alpha[2] F(xi)  m - --------------------------
3
(m + F(xi))

2               2         2                   4
12 w k  beta beta[2] K    12 w k  beta beta[2] F(xi)
+ ----------------------- + ---------------------------
4                           4
(m + F(xi))                 (m + F(xi))

4 k beta[1] alpha[2] F(xi) m   4 k alpha[2] F(xi) m beta[2]
+ ---------------------------- + ----------------------------
m + F(xi)                                2
(m + F(xi))
value(%);
2                   2               2
-2 A k  alpha[1] K - 2 A k  alpha[1] F(xi)

2               3
- 4 A k  alpha[2] F(xi)  - 4 w alpha[2] F(xi) m

+ 2 k alpha[0] alpha[1] m + 2 k alpha[0] alpha[1] F(xi)

2 k alpha[0] beta[1]                          2
+ -------------------- + 2 k alpha[0] alpha[2] m
m + F(xi)

2   2 k alpha[0] beta[2]
+ 2 k alpha[0] alpha[2] F(xi)  + --------------------
2
(m + F(xi))

2                         3
+ 2 k alpha[1]  m F(xi) + 2 k alpha[1] m  alpha[2]

3            2 k beta[1] beta[2]
+ 2 k alpha[1] F(xi)  alpha[2] + -------------------
3
(m + F(xi))

2  3                     2  2      2
+ 4 k alpha[2]  m  F(xi) + 6 k alpha[2]  m  F(xi)

2      3                              2
+ 4 k alpha[2]  F(xi)  m - 2 w alpha[0] + k alpha[0]

2                    3        2                2
+ 4 w k  beta alpha[1] F(xi)  + 4 w k  beta alpha[2] K

2
2                    4   2 A k  beta[1] K
+ 12 w k  beta alpha[2] F(xi)  + ----------------
2
(m + F(xi))

2              2
2 A k  beta[1] F(xi)         2
+ --------------------- - 4 A k  alpha[2] m K
2
(m + F(xi))

2                 2        2
- 4 A k  alpha[2] m F(xi)  - 4 A k  alpha[2] F(xi) K

2                  2              2
4 A k  beta[2] K   4 A k  beta[2] F(xi)
+ ---------------- + ---------------------
3                    3
(m + F(xi))          (m + F(xi))

2 k alpha[1] m beta[1]
+ 4 k alpha[0] alpha[2] F(xi) m + ----------------------
m + F(xi)

2
+ 6 k alpha[1] m  alpha[2] F(xi)

2   2 k alpha[1] m beta[2]
+ 6 k alpha[1] m alpha[2] F(xi)  + ----------------------
2
(m + F(xi))

2 k alpha[1] F(xi) beta[1]   2 k alpha[1] F(xi) beta[2]
+ -------------------------- + --------------------------
m + F(xi)                              2
(m + F(xi))

2                             2
2 k beta[1] alpha[2] m    2 k beta[1] alpha[2] F(xi)
+ ----------------------- + ---------------------------
m + F(xi)                   m + F(xi)

2                             2
2 k alpha[2] m  beta[2]   2 k alpha[2] F(xi)  beta[2]
+ ----------------------- + ---------------------------
2                           2
(m + F(xi))                 (m + F(xi))

2
2  2             2      2    k beta[1]
+ k alpha[1]  m  + k alpha[1]  F(xi)  + ------------
2
(m + F(xi))

2
2  4             2      4    k beta[2]
+ k alpha[2]  m  + k alpha[2]  F(xi)  + ------------
4
(m + F(xi))

2 w beta[1]
- 2 w alpha[1] m - 2 w alpha[1] F(xi) - -----------
m + F(xi)

2                     2   2 w beta[2]
- 2 w alpha[2] m  - 2 w alpha[2] F(xi)  - ------------
2
(m + F(xi))

2                             2                     2
4 w k  beta beta[1] F(xi) K   8 w k  beta beta[1] K F(xi)
- --------------------------- + ----------------------------
2                              3
(m + F(xi))                    (m + F(xi))

2
2                           8 w k  beta beta[2] F(xi) K
+ 8 w k  beta alpha[2] F(xi) m K - ---------------------------
3
(m + F(xi))

2                     2
24 w k  beta beta[2] K F(xi)         2
+ ----------------------------- + 4 w k  beta alpha[1] F(xi) K
4
(m + F(xi))

2                   3        2               2
4 w k  beta beta[1] F(xi)    4 w k  beta beta[1] K
- -------------------------- + ----------------------
2                          3
(m + F(xi))                (m + F(xi))

2                   4
4 w k  beta beta[1] F(xi)          2                      2
+ -------------------------- + 16 w k  beta alpha[2] K F(xi)
3
(m + F(xi))

2                   3
2                    3     8 w k  beta beta[2] F(xi)
+ 8 w k  beta alpha[2] F(xi)  m - --------------------------
3
(m + F(xi))

2               2         2                   4
12 w k  beta beta[2] K    12 w k  beta beta[2] F(xi)
+ ----------------------- + ---------------------------
4                           4
(m + F(xi))                 (m + F(xi))

4 k beta[1] alpha[2] F(xi) m   4 k alpha[2] F(xi) m beta[2]
+ ---------------------------- + ----------------------------
m + F(xi)                                2
(m + F(xi))
simplify(%);
2                   2               2
-2 A k  alpha[1] K - 2 A k  alpha[1] F(xi)

2               3
- 4 A k  alpha[2] F(xi)  - 4 w alpha[2] F(xi) m

+ 2 k alpha[0] alpha[1] m + 2 k alpha[0] alpha[1] F(xi)

2 k alpha[0] beta[1]                          2
+ -------------------- + 2 k alpha[0] alpha[2] m
m + F(xi)

2   2 k alpha[0] beta[2]
+ 2 k alpha[0] alpha[2] F(xi)  + --------------------
2
(m + F(xi))

2                         3
+ 2 k alpha[1]  m F(xi) + 2 k alpha[1] m  alpha[2]

3            2 k beta[1] beta[2]
+ 2 k alpha[1] F(xi)  alpha[2] + -------------------
3
(m + F(xi))

2  3                     2  2      2
+ 4 k alpha[2]  m  F(xi) + 6 k alpha[2]  m  F(xi)

2      3                              2
+ 4 k alpha[2]  F(xi)  m - 2 w alpha[0] + k alpha[0]

2                    3        2                2
+ 4 w k  beta alpha[1] F(xi)  + 4 w k  beta alpha[2] K

2
2                    4   2 A k  beta[1] K
+ 12 w k  beta alpha[2] F(xi)  + ----------------
2
(m + F(xi))

2              2
2 A k  beta[1] F(xi)         2
+ --------------------- - 4 A k  alpha[2] m K
2
(m + F(xi))

2                 2        2
- 4 A k  alpha[2] m F(xi)  - 4 A k  alpha[2] F(xi) K

2                  2              2
4 A k  beta[2] K   4 A k  beta[2] F(xi)
+ ---------------- + ---------------------
3                    3
(m + F(xi))          (m + F(xi))

2 k alpha[1] m beta[1]
+ 4 k alpha[0] alpha[2] F(xi) m + ----------------------
m + F(xi)

2
+ 6 k alpha[1] m  alpha[2] F(xi)

2   2 k alpha[1] m beta[2]
+ 6 k alpha[1] m alpha[2] F(xi)  + ----------------------
2
(m + F(xi))

2 k alpha[1] F(xi) beta[1]   2 k alpha[1] F(xi) beta[2]
+ -------------------------- + --------------------------
m + F(xi)                              2
(m + F(xi))

2                             2
2 k beta[1] alpha[2] m    2 k beta[1] alpha[2] F(xi)
+ ----------------------- + ---------------------------
m + F(xi)                   m + F(xi)

2                             2
2 k alpha[2] m  beta[2]   2 k alpha[2] F(xi)  beta[2]
+ ----------------------- + ---------------------------
2                           2
(m + F(xi))                 (m + F(xi))

2
2  2             2      2    k beta[1]
+ k alpha[1]  m  + k alpha[1]  F(xi)  + ------------
2
(m + F(xi))

2
2  4             2      4    k beta[2]
+ k alpha[2]  m  + k alpha[2]  F(xi)  + ------------
4
(m + F(xi))

2 w beta[1]
- 2 w alpha[1] m - 2 w alpha[1] F(xi) - -----------
m + F(xi)

2                     2   2 w beta[2]
- 2 w alpha[2] m  - 2 w alpha[2] F(xi)  - ------------
2
(m + F(xi))

2                             2                     2
4 w k  beta beta[1] F(xi) K   8 w k  beta beta[1] K F(xi)
- --------------------------- + ----------------------------
2                              3
(m + F(xi))                    (m + F(xi))

2
2                           8 w k  beta beta[2] F(xi) K
+ 8 w k  beta alpha[2] F(xi) m K - ---------------------------
3
(m + F(xi))

2                     2
24 w k  beta beta[2] K F(xi)         2
+ ----------------------------- + 4 w k  beta alpha[1] F(xi) K
4
(m + F(xi))

2                   3        2               2
4 w k  beta beta[1] F(xi)    4 w k  beta beta[1] K
- -------------------------- + ----------------------
2                          3
(m + F(xi))                (m + F(xi))

2                   4
4 w k  beta beta[1] F(xi)          2                      2
+ -------------------------- + 16 w k  beta alpha[2] K F(xi)
3
(m + F(xi))

2                   3
2                    3     8 w k  beta beta[2] F(xi)
+ 8 w k  beta alpha[2] F(xi)  m - --------------------------
3
(m + F(xi))

2               2         2                   4
12 w k  beta beta[2] K    12 w k  beta beta[2] F(xi)
+ ----------------------- + ---------------------------
4                           4
(m + F(xi))                 (m + F(xi))

4 k beta[1] alpha[2] F(xi) m   4 k alpha[2] F(xi) m beta[2]
+ ---------------------------- + ----------------------------
m + F(xi)                                2
(m + F(xi))

collect(%, m+F(xi));
Error, (in collect) cannot collect m+F(xi)

## why Maple sometimes uses _C1, _C2, etc.. for const...

I thought in Maple the standard was to use _C1, and _C2, etc... for constants in the solutions returned.

Sometimes Maple mixes _C1 and c[2] in the same result. Is this common, to be expected sometimes and is OK? I noticed this only recently.

I was thinking may be some part of Maple code still was not updated to use _C1 notation? Here is an example

```restart;
pde:=diff(u(x,t),t)+ diff( u(x,t),x )^3 + 6 * u(x,t)* diff(u(x,t),x) = 0;
sol:=pdsolve(pde,u(x,t));
```

which gives

`sol := u(x, t) = -(3/2)*_C1^2+3*(t*_c[2]+x)*_C1-(3/2)*(t*_c[2]+x)^2-(1/6)*_c[2]`

## how to make pdsolve return other solutions to nonl...

Is there an option, like AllSolutions used with solve, so that pdsolve would return all solutions to a PDE when it is nonlinear?

I looked at pdsolve help and do not see a HINT that looks like might do this.

For example, this PDE, Maple returns one solution. But Mathematica returns 2 solutions

```restart;
pde:= diff(u(x,t),t) = diff(u(x,t),x\$5)+10*diff(u(x,t),x\$3)*u(x,t)+25*diff(u(x,t),x\$2)*diff(u(x,t),x)+
20*u(x,t)^2*diff(u(x,t),x);
sol:=pdsolve(pde,u(x,t));

#sol := u(x, t) = -12*tanh(176*_C2^5*t+_C2*x+_C1)^2*_C2^2+8*_C2^2
```

But there is another solution

```sol1:=u(x,t)=-(1/2)* _C1^2*(-2 + 3*tanh(x*_C1+ t*_C1^5 + _C2)^2)
pdetest(sol1,pde)
#0
```

Here is another example. Maple returns one solution and Mathematica 7 solutions

```restart;
pde:= diff(u(x,t),t)= u(x,t)*(1-u(x,t))+ diff(u(x,t),x\$2);
sol:=pdsolve(pde,u(x,t));

#sol := u(x, t) = (1/4)*tanh(-5*t*(1/12)+(1/12)*sqrt(6)*x+_C1)^2-
(1/2)*tanh(-5*t*(1/12)+(1/12)*sqrt(6)*x+_C1)+1/4```

But there are other solutions

```pde = D[u[x, t], t] == u[x, t] (1 - u[x, t]) + D[u[x, t], {x, 2}];
DSolve[pde, u[x, t], {x, t}]```

I've tested some (not all) of these 7 solutions in Maple using pdetest and Maple agrees they are solutions:

```restart;
pde:= diff(u(x,t),t)= u(x,t)*(1-u(x,t))+ diff(u(x,t),x\$2);
sol:=pdsolve(pde,u(x,t));
with(MmaTranslator);
sol2:=FromMma(`-(1/4) (-3 + Tanh[(5 t)/12 - (I x)/(2 Sqrt[6]) - C[3]]) (1 +
Tanh[(5 t)/12 - (I x)/(2 Sqrt[6]) - C[3]])`);
pdetest(u(x,t)=sol2,pde);
#0```

I tried setting

_AllSolutions:=true

But it had no effect. Is there other options?

## Not return result of an integration...

Hi experts,

I want to compute the following formula using Maple but It returns the formula of integration only.

int(sin(x)/(a*b+a^2*sin(x)^2-d^2*cos(x)^2)(c+cos(x)), x = 0 .. x)

Thank a lot.

## how to view source code in a .maple file (workbook...

I'd like to view the Maple source code in .maple file. I am not able to find how to do this.

I tried commands in 136697-Can-One-Get-The-Source-Code-Back-From

which applies to .mla using the march command. But it did not work for .maple. It gives an error when trying it on .maple file.

I can open the .maple file in the Maple user interface OK using File->Open

but other than documentation, I see no Maple code.

https://maple.cloud/app/5137472255164416

Since most Maple code can be viewed, I assume one can also view the source code in the above.

Any idea how to do this?

## combine/polylog error...

Hello,

I have this error I'm not sure how to solve

restart;

`assuming`([simplify(int(ln(1+x)^3/(x+a), x = 0 .. 1))], [a > 0]);

combine(expand((eval(%, a = I)+eval(%, a = -I))*(1/2)))

What is the precise problem here?

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