## 312 Reputation

18 years, 348 days

## Maple 2016 exited with code 158...

Maple 2016

This is simplified from a larger example.  I never saw Maple exit from a numercal calculation.

> 123456789 ^ 987654321:
Execution stopped: Stack limit reached.

Process Maple R2016 exited abnormally with code 158

> kernelopts(maxdigits);
38654705646

> lprint(123456789. ^ 987654321.);
.4339566080e7991619731

kernelopts(memusage) shows about 2~3 meg bytes used.

The system has 16G Ram.  16G swap.

Seems like something maple should have caught.

## Limit Does Not Eval...

Maple

restart;
eq1 := T - M*g = 0;
eq2 := T1*sin(theta1) + T2*sin(theta2) - T = 0;
eq3 := -T1*cos(theta1) + T2*cos(theta2) = 0;
soln := solve({eq1,eq2,eq3},{T,T1,T2});

limit(subs(theta2=theta1,soln[3]),theta1=0);

This returns a limit -> ... expression.  Since there is a sin(theta1) in the denominator, this should be undefined or infinity.

## Maplev-mode Key Bindings...

Maple

I don't know where to ask this.  Sorry if this is the wrong forum.

I use maplev-mode

;; Version:    2.155
;; X-RCS:      \$Id: maplev.el,v 1.14 2006-06-02 14:02:38 joe Exp \$

I am attempting to modify it so
f1 invokes maplev-cmaple-send-line
f2 invokes maplev-cmaple-send-region
f3 invokes maplev-cmaple-send-buffer

(define-key map [(control c) (control c) ?l] 'maplev-cmaple-send-line)
(define-key map [(f1)]  'maplev-cmaple-send-line)

but, emacs seems to not recognize this.  I can change the global keymap to get this behavior.
Is there a way to do this in maplev.el?

Tom Dean

## How do I get larger icons...

Maple 2016

I use maple 2016 on linux with X.  When minimized, the maple icon is very small.  Also, the help icon is very small.

Are larger icons available?

The maple icon is ./bin/Maple2016.png.  I can not find the help icon.  Any ideas?

## How do I express bc for PDE...

Maple 2016

I have a PDE with boundary conditions, from a NASA paper.  I always seem to have problems expressing the bc.

PDE := diff(u(x, t), t) = (1/4)*exp(2)*exp(-u(x, t))*(diff(diff(u(x, t), x), x))/(x^2+2);

The initial/boundary conditions are

@t=0, u(x, t) = 2-2*ln(-x^2+2)

@x=0, diff(u(x,t),x)=0  ## this is the bc I have problem expressing

@x=1, u(x,t) = 2+ln(1+t)

The exact solution given in the paper:

2 + ln(1+t) - 2*ln(2-x^2)

I tried

ics := u(x, 0) = 2*(1-ln(2-x^2));

bcs := D[1](u(0,t))=0, u(1,t)=2+ln(1+t);

PDEtools[Solve]([PDE,ics,bcs]); ## no solution

How do I do this?

Tom Dean

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