taro

440 Reputation

12 Badges

6 years, 23 days
Maple is to me difficult. The first version I bought was Maple9, and it was more than 15 years ago. But, I couldn't use it, feeling it too difficult. But, three years ago, I thought Maple might be helpful to my study, and since then, I have continued to learn Maple. As I got able to read the Maple help, I think that I could get to use maple better now than before. But, I feel that I am a beginner yet.

MaplePrimes Activity


These are questions asked by taro

Hello people in mapleprimes,
I have a question.

I use maple2019 with mac os 10.14.6.

$ sw_vers
ProductName:	Mac OS X
ProductVersion:	10.14.6
BuildVersion:	18G95

With maple2019, errors appears.

> kernelopts(version);
        Maple 2019.1, APPLE UNIVERSAL OSX, Jun 6 2019, Build ID 1403154

> assume(a>-1,b>0);
> additionally(a<=1);
> about(a);
Originally a, renamed a~:
  is assumed to be: FAIL

> assume(tau<1,tau>0,s<1,s>0):
> a_e1:=tau*s*(1+tau)<tau*s+tau+s-1:
> b_e2:=expand(lhs(a_e1)-rhs(a_e1))<0:
>  b_e3:=collect(b_e2,s,factor):
> solve(b_e2,s) assuming tau<1;
Error, (in assuming) when calling 'property/ConvertProperty'. Received: 'FAIL
is an invalid property'

On the other hand, with maple2018, they do not.

> kernelopts(version);
       Maple 2018.2, APPLE UNIVERSAL OSX, Nov 16 2018, Build ID 1362973

> assume(a>-1,b>0);
> additionally(a<=1);
> about(a);
Originally a, renamed a~:
  is assumed to be: RealRange(Open(-1),1)

> assume(tau<1,tau>0,s<1,s>0):
> a_e1:=tau*s*(1+tau)<tau*s+tau+s-1:
> b_e2:=expand(lhs(a_e1)-rhs(a_e1))<0:
> b_e3:=collect(b_e2,s,factor):
> solve(b_e2,s) assuming tau<1;
                                    1
                               [{-------- < s~}]
                                 1 + tau~

Are errors due to some bugs in maple2019 when being used with mac os 10.14.6?

Take care.

taro

Sorry,

I have been away from Maple for a year.
Then, when I used op command, I am puzzed to notice the results were different from those I know.

>op((x+5)^2*(x+y));
      _EXPSEQ((x+5)^2, x+y)

Result I know is 

        (x+5)^2, x+y

Has a modifire such as _EXPSEQ   automatically come to be attatched?
Or, can I have maple express it  in the form I know:  (x+5)^2, x+y?

Thank you in advance.

taro

 

 

 

 

 


 

Hello people in mapleprimes,

I installed maple 2018 Japanese version.
And, with solve(x^2-1,x), its solution is expressed as _EXPSEQ(1, -1)

I know this expression is an internal represantation.
How can I have maple answer as 1, -1?

Thanks in advance.

Addition: my pc is mac osx 10.13.6.
 

 

 

Hello people in mapleprimes,

I am writing this sentences, hoping to be given any answer from you, to the 
question I have, about conversion of Array to List.

restart;
a:=plot3d(theta*z,theta=0..2,z=0..1,coords=cylindrical,grid=[3,2]);
A:=op([1,1],a);

brings a response of Array as, which I call BB,

Array(1 .. 3, 1 .. 2, 1 .. 3, {(1, 2, 3) = HFloat(1.), (2, 2, 1) = HFloat(.540302305868139765), (2, 2, 2) = HFloat(.841470984807896505), (2, 2, 3) = HFloat(1.), (3, 2, 1) = HFloat(-.832293673094284814), (3, 2, 2) = HFloat(1.81859485365136342), (3, 2, 3) = HFloat(1.)}, datatype = float[8])

And, I converted it to List, with 

convert(A,list,nested);

The response of this is, which I call CC,

[[[0., 0., 0.], [0., 0., 1.]], [[0., 0., 0.], [.540302305868140, .841470984807897, 1.]], [[-0., 0., 0.], [-.832293673094285, 1.81859485365136, 1.]]]

, which is composed of three lists:
[[0., 0., 0.], [0., 0., 1.]]

[[0., 0., 0.], [.540302305868140, .841470984807897, 1.]]

[[-0., 0., 0.], [-.832293673094285, 1.81859485365136, 1.]]

Then, my question is following:

Seeing CC, I can see three segments are shown.
The first one is  the segment between [0., 0., 0.] and [0., 0., 1.].
But, I can't find any elements of Array BB corresponding to [0,0,0] above in CC.
And as for 0, 0 in [0, 0, 1] in CC above, the first 0 should be (1,2,1) and the second one should be (1,2,2).
But, there are no such (1,2,1) and (1,2,2) in BB. 
Despite of these, BB is converted to CC, from any reason I want to know and ask this question for.

As for [[0., 0., 0.], [.540302305868140, .841470984807897, 1.]] in BB, the circumstances are the same.
I don't know from where [0,0,0] appeared. Surely, as MESH requires three points, you might say that
such things might occur. But, it is only after I converted Array BB to List CC, with plot3d or MESH then having no relation 
to the present circumstances. So, I think that it is not possible to say that Maple added [0,0,0] or such things to Array BB

 as it is about plot 3D or about MESH.

So, my question is how the conversion of BB to CC could have occurred.

Thanks in advance.

May_16_2018_question.mw

Hello people in mapleprimes,

I am writing this question, hoping you to teach me about the meanig of the Array in a plot3d object.

c:=plot3d(x*y,x=0..2,y=0..1,coords=cylindrical,grid=[3,2]);
lprint(c);

makes 

PLOT3D(MESH(Array(1 .. 3, 1 .. 2, 1 .. 3, {(1, 2, 3) = HFloat(2.), (2, 2, 1) = HFloat(1.08060461173627953), (2, 2, 2) = HFloat(1.68294196961579301), (2, 2, 3) = HFloat(2.), (3, 2, 1) = HFloat(-1.66458734618856963), (3, 2, 2) = HFloat(3.63718970730272684), (3, 2, 3) = HFloat(2.)}, datatype = float[8])))

I thought I had known the cylindrical coords.
And, I think that (1, 2, 3) in  (1, 2, 3) = HFloat(2.) means the value of x*y when x, the angle, =0, y, the hight, =1, z, distance from the origin, =3. And, it should become 0, as the value of x, times 1, the value of y, equals 0. But, it takes HFloat(2.) as
shown in (1, 2, 3) = HFloat(2.).

And, as for (3, 2, 1) = HFloat(-1.66458734618856963), which means x=2 and y=1, I think the value of x*y=2. But, 
it is wrong as seen in  (3, 2, 1) = HFloat(-1.66458734618856963). 

I hope you will teach me how to consider about sample points in the cylindrical plot.

Thank you advance.

 

 

 

 

1 2 3 4 5 6 7 Last Page 1 of 19