## 450 Reputation

6 years, 309 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.

## Procedure of constrained optimum...

Maple 2015

Hello people in mapleprime

maxi:=proc(obj,expre,x,y)
local eq1,eq2,eq3,lagrangean;
lagrangean:=obj+lambda*expre;
eq1:=diff(lagrangean,x)=0:
eq2:=diff(lagrangean,y)=0:
eq3:=diff(lagrangean,lambda)=0:
solve({eq1,eq2,eq3},{x,y});
end proc;

Using the above procedure, next code will not return the values of a and b, or a[2] and b[2].

Can you teach me the reason why?

maxi(a*b,z-a-b,a,b);

maxi(a[2]*b[2],z-a[2]-b[2],a[2],b[2])

taro

## How does evalf@Int choose points?...

Maple 2015

Hello people in mapleprimes,

I have a question about how Int does.

The following function spy returns 0, of course, with a side effect of listing the value of x one by one to secrets.

secrets := NULL:

spy:=proc(x::{name,numeric})
global secrets;
if type(x,name) then
return 'procname'(args)
else
secrets:=x,secrets;
return 0;
end if;
end proc;

And, with this function, calculation of the Int, that is, following brings a sequence of numbers:

evalf(Int(spy,0..1));

secrets;

.7506605773, .2493394227, .9118140517, 0.881859483e-1, .9970470440, 0.29529560e-2, 1.0000000000, 2.2449529449*10^(-11), .5000000000

The question I have is why the number of this sequence is not from smaller( or greater) to greater (smaller) in order,

but in random order. And, numerical calculation of Int can be done with only 9 points extracted?

Best wishes.

taro

## Is there any use of the specific 1-D mat...

Hello those in mapleprimes,

I have a question.

I opened Help and entered diff, for example, to have the explanation of how to use diff.

Whether it is diff or not, is not the problem I want to ask about now. But, what I want to ask is the following.

I cklicked the WS botton at the tool bar of the Help, so that I created the worksheet of the same contents of the help of the diff.

Then, I chose an expression in Calling Sequence. That is,

the first line shows

diff(f,x1, ..., xj)     on the left, and on the right \frac{d^j}{dx_j \cdots dx_I} f (I wrote this in the TeX ways, though

it is written in 2-D math input, actually)

I chose an expression on the right side, and from context menu, which appeared though right clicking of my mouse,

I selected convert to and 1-D math input, through which I changed 2-D math of the worksheet of the help file to 1-D math input.

Then, the expression shown is a strange one:

((&DifferentialD;)^j)/(&DifferentialD;x[j] ... &DifferentialD;x[1]) f;

This is still easier one. As for the third line in the Help, it has a more complicated form as

((&DifferentialD;)^r)/(&DifferentialD;x[k]^m &DifferentialD;x[j] ... &DifferentialD;x[3] &DifferentialD;x[2]^n &DifferentialD;x[1]^n) f;

Do you know how this way to notate is applied in maple worksheet?

Why does maplesoft use this way to notate expressions limited to the Help pages, if it is limited to the help pages.

taro

## Why does (-8)^(1/3) become 1+Complex(1)*...

Hello people in mapleprimes,

I have a question about why what is shown by maple by simplify(-8)^(1/3) is 1+ Complex(1)* (3)^(1/2)?

Solutions of x^3=-8 are -2, 1+Complex(1)*(3)^(1/2) and 1- Complex(1)*(3)^(1/2). And, as for the last one, 1- Complex(1)*(3)^(1/2), it is

the conjugate of the second, so it might not need to be written, because of it being easily seen so.

Is it the same reason why just -2 is not shown as the result of simplify((-8)^(1/3))?

PS. I know the instruction to use surd in such a case.

the reason I asked this question is this:

I am reading Essential Maple, where

ln(z) = ln(rho*exp(Complex(1)*theta));

ln(rho*exp(Complex(1)*theta))=ln(rho)+Complex(1)*theta;

ln(rho)+Complex(1)*theta;=ln(rho)+Complex(1)*arctan(y,x);

and

z^a=exp(a*ln(z));

and

"Because of

exp(w*Pi*Complex(1)*k)=cos(2*Pi*k) + Complex(1)*sin(2*Pi*k);

and

cos(2*Pi*k) + Complex(1)*sin(2*Pi*k)=1;

we could equally well have chosen

ln_k z = ln(z) + 2*Pi*Complex(1)*k"

are written.

Supposing these, there is a sentence that

"we choose k=0, and thus -Pi<=theta<=Pi to be the one (that for our canonical logarithm).

Every computer algebra language and numerical language follows this standard and takes the

complex logarithm to have its imaginary part in this range.

With this definition, (-8)^(1/3)=1 + Complex(1)*sqrt(3), and not -2. (the end of quotation)"

And, I can't understand the last sentence"With this definition", so I asked the above question.

I hope someone give an answer to the above question.

taro

## Why does the -> operator behaves like ...

Maple

Hello people in mapleprimes,

I have a question.

f1:=1+x:

unapply(f1,x);

returns x->1+x , with maple realizing 1+x is assigned to f1.

But, its short form

x-> f1;

returns

x->f1;

Why does this occur, and how can I have maple return the former result with x->f1 being input?