@Carl Love 

Owing to your code, I could understand at last the meanings of k:=[[1,2,10],[1,1,8],[0,1,5],[0,0,6]].
I could understand the meaning of your passage, "The listlist k is a specification of 12 values of z; the (x,y)-values are determined implicit."

Thank you.

Taro yamada

@Kitonum

Thank you. I had read that part of help. But I still had my question unresolved.
After reading carefully your passage:  "The Array A may be replaced by a list of the form [[z11,...z1n], [z21,...z2n],...[zm1...zmn]] where zij is the function value at grid point (i, j), I should have understood its meanings.
But, I couldn't then understand what was written there.

Taro yamada

@acer

Thank you for your explanation about the distinction between operator-form and expression-form.  ribbonplot5 is a little strange program which makes it OK even if x=-Pi..Pi is used instead of -Pi..Pi, 
in so far as procedure names are used in the place of Flist. If caring for the distinction between the two, 
why simmilar care isn't been payed to the distinction between only the names of procedure and 
expressions is the place I think any people might wonder.

Taro yamada

@tomleslie 

Thank you. Owing to your teaching, I could notice that it should be Flist(lhs(r1)) because unapply is used.
And, thank you for showing your short and more understandable code.

Taro yamada

@acer

Thank you.
I could understand what things are going owing to you.
And, putting K as the third term is the way which make it so easy to understand the meaning 
of the third argument, isn't it. Thank you for telling me the way,

Best wishes.

Taro yamada

Best wishes. 

@acer 

Hello, Happy New Year.

Thank you for your giving me a splendid answer in the other day.
As for you answer cord, I have a question. So, if you don't mind, I will be very happy to get your answer again.
My question this time is about the denominator of a in the previous question, that is,

c:=I__11*I__22 - I__11*X__2 - I__12*I__21 - I__22*X__1 + X__1*X__2;

Following modification works well, which you used in the previous answer. 

> c:=I__11*I__22 - I__11*X__2 - I__12*I__21 - I__22*X__1 + X__1*X__2;
> collect(c,[I__12],simplify);

              (X__2 - I__22)*(X__1 - I__11) - I__12*I__21

 But, if I do this as in the following, the outcome is not good.

>c:=I__11*I__22 - I__11*X__2 - I__12*I__21 - I__22*X__1 + X__1*X__2;

>collect(c,[I__12]);
                      I__11 I__22 - I__11 X__2 - I__12 I__21 - I__22 X__1 + X__1 X__2
>simplify(%);
                     (I__22 - X__2)*I__11 + X__1*X__2 - I__22*X__1 - I__12*I__21

The part of X__1*X__2 - I__22*X__1 is not collected with X__1.

So, what I want  to ask you this time is how the code of "collect(c,[I__12],simplify);" works well.

Thank in advance.

taro yamada

@acer 

Dear Mr. acer,

Thank you very much for teaching,
I had forgotten the use of collect with option, and that of sort with option.
Thank you for teaching me how to use them to solve the present problem of mine.
taro yamada
 

@nm

Dear Mr. nm,

Thank you very much. Owing to you, I could resolve the problem.
I have forgotten a lot, about maple.
But, I will regain a knack of using maple now.

Thank you .

taro yamada

@Christian Wolinski 

I, at last, understood that {Non}(a) as is in your code.
Thank you for showing me the new code bringing the outcome in the form with O, to show  the role of frontend.

I haven't understand the last code yet. I will try to do.

taro

@vv 

Thank you for teaching me  that just factor after expand is enough.

taro

@Kitonum 

Thank you. At last, I have just understood  what you mean:  the use of expand being crucial.

taro

@Christian Wolinski 

Thank you for your instruction.

I am writing this message hoping that you would teach me a little further about your code.
Please see the appended file.
In the modification of from e_n_1b2 to e_n_1b3, convert/parfrac is needed.
But, with w^sigma being in e_n_1b2, using just convert brings out an error message.
So, I think you used frontend, to regard w^sigma as a single variable.
And, as seen in e_n_1b3, with frontend, the modified equation I wanted appear.

But, on the other hand, I can't understand why identical(w^sigma) should be contained in the option 
of frontend, for I think that the option of frontend means variables which are not regarded as a single variable.

 So, I wonder why it is there. And, I can't understand why option is like [{Non}(identical(w^sigma)), {}]);

I am very glad if you explain them to me. I'm sorry.

Dec_23_b.mw

taro

@acer 
@Kitonum

Thank you for kindly teaching.
If I couldn't have borrowed your hand here, I would be astrayed at how to begin maple again.

Thanks a lot.

taro.
 

As seen in

https://www.maplesoft.com/products/roadmap.aspx

osx 10.15 is not supported by maple2018 or 2019, yet.

Hello
This page might be useful.

https://www.facebook.com/Maplesoft/posts/10157599572441042

Take care