## Surface fit with boundary conditions...

Hello dear Maple users! I have the following problem that I am trying to solve with maple. I have a set of data points (i made these up for now, the real set has more points) x:=( 1,2,3,4,5,6,7,8 ) y:=( 1,2,3,4,5,6,7,8 ) z:=( 3,1,3,2,4,7,7,9 ) Now, I know how to surface fit this and get maple to formulate an equation for z=f(x,y) with the following statement: w:=(Statistics:-Fit(a + b*x + c*y + d*x^2 + e*x*y+f*y^2,

## Using plots(inequal)...

Hi Maple's experts, I would like to draw inequalities in R^3. Is there a similar command (to "inequal") for more than two unknown variables?. Thanks in advance, JJ

## genvecs in maple...

Hi. There is a command called genvecs in mathcad and i wonder if there is something similar in maple. Or can somebody tell me how to make the same in maple. Regards M

## Sheet for percolation...

has someone a sheet about percolation theory ? I put a very simple one on my blog: percolation

## First order difference equation !...

Can maple do this ? and how .... I know you can do the graph, but I need to know if maple can do my problem. ex: Y(t+1) + Y(t) = (-1)^t and y(0) = 2

## problem with max.....

Hello, How to select the maxima of the list H1:= [[8,9,11],[8,10,11],[5,10,6]]? I do mean, I would need to select the maxima among the first components (8,8,5), second components (9,10,10) and third components (11,11,6). I would also need the index of this elements that are in the maxima set. Leika

## How to draw it?...

I want to draw the set of points that satisfies the following sets of inequalities: x+y>=50; #(1) x+z>=50; # (2) y+z>=50; # (3) subject to x+y+z=100; I tried: >with(plots): >implicitplot3d( {x+y+z=100, x+y>=50,x+z>=50,y+z>=50}, x=0..120, >y=0..110, z=0..115, axes=normal); 1. How could I do to see exactly the required set in a different colour? It would be possible to give different colours to each constraint (1-> yellow, 2->blue, 3 ->red)? 2. It would be possible to avoid the three dimension space and to represent everything on the plane x+y+z=100 (that is, to represent the inequalities 1, 2 and 3 only on this plane)?.

## assuming and procedure parameters...

Hello,

I am looking for an explanation for the following behavior :

> f := proc(x) somefunc(x) end proc:
> a := table([variable=z]):
> f(a);
somefunc(a)
> f(a) assuming z::real;
somefunc(table([variable=z]))

Why is 'a' replaced with its corresponding table in the second case? And is it possible to use the name 'a' in the return value?

Franky.

## Plotting with units?...

Is it possible to use units on functions/data that will be plotted? I have tried but it seems to flag an error. If it is not allowed, what is the best way to strip the units before plotting? Thanks, Dgiznya

## pointplot a family of curves...

I can pointplot a single curve if I select a single element from the list below. Is it possible to use all the elements of the list and plot several curves on the same graph?

## Units...

Hi! How do you compare values with units attached: eg. this error message displays x := 4(m) y := 5(m) if x <>

## Undefined values...

hi again! Is it possible to disregard part of an equation if the denominator is zero? Thanks.

## Unit simplification...

Hi, I was just wondering whether it is possible to limit the simplification of units to display N and mm rather than kg/s^2. Also when i square the units and then square root them is it possible for it to simplify the square root without right clicking and selecting simplify? Thanks!

## Numerical integration...

Hello!

I´m Trying to do numerical integration for a function defined by a product of characteristic functions. I either receive the message " Error, (in unknown) too many levels of recursion " or "error, (in evalf/int) unable to handle singularity". What should I do?

f_1:=(x,y,z,w)->piecewise(x+w > 0 , 1);
f_2:=(x,y,z,w)->piecewise(x*w - y*z > 0,1);
f_3:=(x,y,z,w)->piecewise((x+w)^2 - 4*(x*w - y*z)<0,1);
prod:=(x,y,z,w)->f_1(x,y,z,w)*f_2(x,y,z,w)*f_3(x,y,z,w);

I want to integrate prod, with all four variables ranging from -1 to 1.

Thank you, Bernardo