Hi there,

is it possible to have the legend of a pointplot accepting arguments?

Say I have an array of 3 values (e.g. p:=[1.5,2,2.5]) representing a given parameter for a function. When representing this function by means of a plot, I would like the legend (legend="Parameter " + value in array; legend=["Parameter", "Parameter", "Parameter") to accept the contents of the array, without entering the values manually:

Parameter 1.5

Parameter 2

Parameter 2.5

How can this be achieved?

Thanks,

jon

When I plot an expresion I would the graphics be framed. What I am getting, instead, is the curve and the axis, whith no frame at all. How can I force Maple to make the graphic surrounded by a frame? 

@ecterrab 

I figured I'd start a new thread for odd things I come across whilst using the new physics package. 

I have found this, and am not sure if it is expected. 

Download Simplify2.mw

Archimedes supposedlly, was asked to determine whether a crown made for the king consisted of pure gold. According to 

legend, he solved this problem by weighing the crown first in air and then in water. Suppose the scale read 7.84 N when the 

crown was in the air and 6.84 N when it was in water.

What should Archimedes have told the king ?

Hi! I have a matlab worksheet that i need to convert it to Maplesoft. Is it possible?

Regards

Hikmat

Take a look at below. I was expecting maple to give me "g'(1)"! :)

hi.i am a problem with following dsolve.please help me....thanks alot

dsys3 := {10*f2(x)+12*(diff(f1(x), x))+14*f3(x) = 0, 2*(diff(f1(x), x, x))+4*(diff(f2(x), x))+6*(diff(f3(x), x)) = 0, 16*(diff(f3(x), x, x, x, x))+19*(diff(f3(x), x, x))+22*(diff(f1(x), x))+25*f2(x)+27*f3(x)+29*f3(x)+31+32 = 0, f1(0) = 0, f1(1) = 0, f2(0) = 0, f2(1) = 0, f3(0) = 0, f3(1) = 0, ((D@@1)(f1))(0) = 0, ((D@@1)(f1))(1) = 0, ((D@@1)(f2))(0) = 0, ((D@@1)(f2))(1) = 0, ((D@@1)(f3))(0) = 0, ((D@@1)(f3))(1) = 0}; dsol5 := dsolve(dsys3, 'maxmesh' = 500, numeric, range = 0 .. 1, abserr = .1, output = listprocedure); fy3 := eval(f3(x), dsol5); fy2 := eval(f2(x), dsol5); fy1 := eval(f1(x), dsol5)ERROR.mw

Let N be an integer. 

For each pair of integers (n,m) where 1<= n,m <= N, we have a variable f_{n,m}(t). 

Then for these we have a system of ODEs 

d/dt f_{n,m}(t) = \sum_{n', m'} f_m'n' * f_m''n'' * (m'n'' - m''n') 

where m''=m-m', n''=n=n', and the sum is simply over for all pairs (n',m'). 

I simply do not know how to put these set of equations into Maple in a nice way. 

I will really appreciate any help!

1. Sketch the two vectors listed after the formula for r (t).
2. Sketch, on the same plane, the curve C dterminated by r (t), and indicate the orientation for the given values of t.
   
   r (t)=t*i +4*cos (t)j +9*sin (t)k,
   
   r (0), r(Pi/4),  t>=0
   
   

Cause CodeTools:-Test will not eval the paramater:

```
f := x -> x+1;

y := 2;

CodeTools:-Test( f(1), 2); # this will work as normal

CodeTools:-Test( f(1), y); # this will not work because `y` will not eval as 2

```

So current my solution is

```
test := subs(y=2, () -> CodeTools:-Test( f(1), y));

test();

```

I don't know is there any more proper solution for this.

More general, is there a way to force evaluate an `uneval` parameter?

the final result I want to get should be b*x

Thank you.

How to convert an  algebraic equation to a parametrized one ?

How to sketch the curve of a circular cylinder , x^2+y^3^=a^2 with the parametric equations lying on it ?

x=a*cos't) , y=a*sin(t) , z=b*t , if t varies from 0 to 2*Pi, the point P starts at(a,0,0) and moves upward.

How to show that the parametric equations are indeed lying on the cylinder ?

How to plot the whole things ?

test.mw

Hello. As you can see the minimum of set X is 2.

the time pair in set W whose difference is 2 corresponds to elements 54 and 55. How do i locate and print these automatically?

Download test.mw

Good afternoon sir.

I request your kind suggestion to my cited query.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

I am a research student and I am writing up my thesis. I was reading one of the paper written earlier by authors in 2003.

The authors calculated a symbolic rank of a matrix A, and got 9, using Maple 6.

Then new methods developed in 2012 and proved that the "true" rank should be 8. The later paper that "Most  recent versions of Maple have this simplification built in and are able to return a rank of 8.".

I just noticed that in Maple 18.01, this matrix was still evaluted for a symbolic rank 9, if no simplifcaiton was done before using Rank(), see attached.

I didnt explore a lot, but just as a notice. I am a bit concerned as most of my research was trying to deal with exponentials.

Is that something to be fixed in future versions?

Thanks!

 Bug.mw