## 5640 Reputation

7 years, 189 days

## I didn't know this syntax...

@Kitonum
I use to use  sort(L, key=(x=x[1]))  instead.
Is your syntax docummented elsewhere ?

## PS : see Mariusz Iwaniuk's answ...

This will explain why CDF(burr_distribution, x), or Mean(burr_distribution), or any other statistics has no closed form for general values of the parameters

Maybe this ?

burr.mw

## @Les  The fit you propose seems od...

@Les

The fit you propose seems odd ...

Maybe a typo error in your f:= t->439e^-2.9t function ?

 > Pts:= [[0, 438.912], [0.0295, 402.336], [0.125, 365.76], [0.2009, 341.376]]:
 > with(Statistics):
 > fit := ExponentialFit(Pts, t);
 (1)
 > FIT := 439*exp(-2.9*t)
 (2)
 > plot(
 > [Pts, fit, FIT], style= [point, line, line], symbol= diamond, symbolsize= 12,
 > color=[blue, red, gold], gridlines, view= [0..0.25, 0..500]
 > );
 >
 >

## Here is your complete worksheet in&...

Here is your complete worksheet in WORKSHEET-MODE style

EulerLagrange.mw

Don't be afraid if the copy-paste operation transforms the first line into the second one ... the results are the same

## If you want to build the solution "...

If you want to build the solution "sol2" from t=1000 to t=3000, the simplest way is to set the IC equal to sol(1000) and solve in the range 1000..3000.
If you do this (brown code in the attached file), you will see that sol2 is equal to sol in the range 1000..3000

aa_problem_MWE2.mw

## I did not analyze your code for solution...

I did not analyze your code for solution "sol2".

But, if you try to solve the same problem from t=1000 to t=3000, the only thing tou have to do is to set the new IC at t=1000 equal to sol(1000) and solve in the range 1000..3000
If you do this, sol2 returns the solution sol gave in the range 1000..3000 (see the code in brown in the attached file).

By the way, I added legend=... in the graphs for a smarter plot and change method=rkf45 (default choice) ny method=rosenbrock (to hande potential stiff problems).

So I guess your code for "sol2" could contain some mistake(s) ?

aa_problem_MWE2.mw

## I did not analyze your code for solution...

I did not analyze your code for solution "sol2".

But, if you try to solve the same problem from t=1000 to t=3000, the only thing tou have to do is to set the new IC at t=1000 equal to sol(1000) and solve in the range 1000..3000
If you do this, sol2 returns the solution sol gave in the range 1000..3000 (see the code in brown in the attached file).

By the way, I added legend=... in the graphs for a smarter plot and change method=rkf45 (default choice) ny method=rosenbrock (to hande potential stiff problems).

So I guess your code for "sol2" could contain some mistake(s) ?

aa_problem_MWE2.mw

## @acer  (from sand15's home-alia...

@acer
(from sand15's home-alias)

The example given in the help pages, even with a legend on the first plot, works perfectly at home

Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895
(imac)

## Thank you Preben...

@Preben Alsholm

I'll will contact you again as soon as possible.

Thanks for the help

## @vv    Do you mean: why do I...

Do you mean: why do I do this

```ff:=x->2*x+3;
M := module()
option package;
export f;
f:=eval(:-ff);
end module:```

```M := module()
option package;
export f;
f:=eval(:-ff);
end module:
ff:=x->2*x+3:```

?

The answer is: just because I found this method somewhere here, years ago.

However I'll be unavaliable for two days to investigate this point further.
I'll will contact you again as soon as possible to clarify this.

Thanks for the help

## @vv  (different lognames but the sa...

@vv
(different lognames but the same person, now out from my office)

I have tested your coding and I understand the different results it gives.
But it seems (maybe I did not examine the results with enough attention) that

```ff:=x->2*x+3;
M := module()
option package;
export f;
f:=eval(:-ff);
end module:
#ff:=x->2*x+3:
```

returns the same things  ...

More of this, if the definitions of ff and M are done in a specific worksheet ending with the creation on an archive M.mla, and if M is used in a separate worksheet through the command with(M) (libname having been instanciated correctly), which is my case, should I have to expect

```#ff:=x->2*x+3;
M := module()
option package;
export f;
f:=eval(:-ff);
end module:
ff:=x->2*x+3:```

and

```ff:=x->2*x+3;
M := module()
option package;
export f;
f:=eval(:-ff);
end module:
#ff:=x->2*x+3:
```

to behave differently ?

## Thank you for your interest in my proble...

(different lognames but the same person, now out from my office)

I proceed exactly as I understand you do.

More precisely :

1/ I open an interactive Maple session (Maple 2015, Windows 7)

2/  I open the file MyModule.mw in worksheet W1

and the file Test.mw in a separate worksheet W2

W1 is aimed to develop the package MyModule

The code in W2 contains  the call to the procedure MyProc included in  MyModule

3/ W2 begins this way :

restart:

MyLib := ….                           # the directory which contains  MyModule.mla

libname := MyLib, libname;    # should I have write lib name := libname, MyLib,  ?

with(MyModule):

4/ W1 begins with a « restart » command and is organized this way :

* the codes corresponding to the N procedures MyModule contains (each of them in a separate block)

* a new block where the module MyModule is defined

* a last block where the archive MyModule.mla  is created in the directory Mylib (the same name used in W2)

5.0/ I execute W1 up to the command MyProc(…), where MyProc is one of the N procedures MyModule contains.

I’m not happy with the result it returns, so:

5.1/ I go back to W1

5.2/ I modify MyProc

5.3/ I execute (!!!) the whole W1 worksheet

If MyLib already contains MyModule.mla, which is generally the case, this archive is destroyed

before a new might be created

I verified in a third worksheet that MyModule.mla contains the modified procedure MyProc

5.4/ I go back to W2.

I insert a new block just after the one which contains the call to MyProc.

This block contains

unwith(MyModule);

package();             # to insure that MyModule has been « unwithed » … which is the case (***, see below)

with(MyModule);

The command showstat(MyProc) keeps displaying the unmodified MyProc loaded by the first with(MyModule)

(point 3/ above).

It all goes as if unwith(MyModule) + with(MyModule) was ineffective.

*** More surprisingly, if I  execute the command MyProc(…) after unwith(MyModule) , I do not obtain the output

MyProc(…) which would signify that  MyProc is unknown, but the same result I obtained at point 5.0 !?!?

PS : it is of course not a blocking situation for I can always execute the W2 worksheet from its first « restart » command to the MyProc(…).

It’ is more something which is troubling me...

Thank you,

best regards

## Shorter...

S := (a,d) -> sum((a+d*k)^(r), k=1..infinity):
S(a, d) - S(f0, f1)

Nevertheless I'm surprised by the opposite signs of d*n in

`Zeta(0, -r, (d*n+a+d)/d)`

between Mathematica and I : probably some mistake on my side ???

Best regards

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