@Kitonum 

@Preben Alsholm

@Mac Dude

 @acer

Thank you for your kindness of teaching me a lot.

Kitonum's method is to remove terms with I, isn't it.

As far as it is limited to the solution of  y=x^3, surd(y,3) is easiest.

But, considering other cases, surely, the way of removing terms with I is actually useful.

I didn't know well how to use remove or select actually. But, owing to Kitonum, I know it well now.

Thank you for the code of

b:=solve(x^3-3*x+1=0, x):

map(t->Re(t), remove(t->is(Im(t)<>0), map(evalc,[b])));

All coefficients of terms with I of the solution of solve(x^3-3*x+1=0, x); being real is difficult to see at first sight, isn't it?

And, thank you acer for pointing out that y has to be positive.

as far as y>0(y<0) is concerned,

surd(y,3) assuming y::positive;

and

surd(y,3) assuming y::negative;

brings y^(1/3) and -(-y)^(1/3)

And, thank you for

for s in s1 do your code:
  print( s, radnormal(eval(s, y=-8)), evalf[5](evalf(eval(s, y=-8))) );
end do:

And, thank you for showing me the code:

solve( y-x^3, x, parametric, real );
It is new to me. Thank you for teaching it.

Thank you.

taro

@Carl Love 

Thank you again, Carl,

Owing to you, I could show strings from 1 to the number I designated, with substring.

>substring("a := 4; a^2; a^3;",1..7);
                         "a := 4;"

>substring("a := 4; a^2; a^3;",1..12);
                         "a := 4; a^2;"
substring("a := 4; a^2; a^3;",1..17);
                      "a := 4; a^2; a^3;"

Thank you.

taro

@Carl Love 

Hello, Carl, thank you for teaching me a lot. I could understand well.

As the string has characters in the following order, values pointed at by lastread are 8, 13, and 17, and then

each statement beginning from that position are offseted by the option offset of parse.

> seq( i, i="a := 4; a^2; a^3;");
"a", " ", ":", "=", " ", "4", ";", " ", "a", "^", "2", ";", " ",  "a", "^", "3", ";"

By the way, is there way to show the number of the position 8, 13, and 17, not through counting with the above

command?

Thank you.

taro

@Christopher2222 @peter2108

Description in help is

Each command in the geometry package can be accessed by using either the long form or the short form of the command name in the command calling sequence.

@peter2108 

Following Thomas Richard's code, it will.

latex(%);
\left(\cos \left(\sqrt{2}~t \right)-\cos \left(\omega ~t \right)
\right)~\frac{1}{\omega ^{2}-2}

Hello Carl,

@Carl Love 

b double _ and 12, that is b__12 works completely well. Thank you for telling me it.

Best wishes.

taro

@vv 

Thank you.

I could obtain the result owing to you.

Thank you.

taro

@vv 

Thank you vv,

The assumption you added: c>0, e>0, corresponds to theta >1,m > 1 in my case, which is from

>solve(-theta/(1-theta)>0,theta);solve(m-1>0,m);
  RealRange(-infinity, Open(0)), RealRange(Open(1), infinity)
                  RealRange(Open(1), infinity)

And, though in your example, calculation is successful, in my case it does not go well as before they are added:

a:=int(((beta/beta[1,2])^(-theta/(1-theta))-kappa[1]^(-theta/(1-theta)))*m*beta^(m-1),beta=0 .. kappa[1]*beta[1,2]) assuming (theta >1,m > 1);

Can I ask for more advice of yours?

Best wishes

taro

@acer 

I just tried to keep the code of Corentin alive:

f(x) = x  + 6 x + 12
>> g := x -> f(x)*exp(x) ;
                          x -> f(x) exp(x)
g(2);
                          f(2) exp(2)
was his code, for which I thought he didn't realize that f(x)=x*6*x+12, which is the result of dsolve, does not have

any functions of assign the rhs to f(x) in the lhs.

I though that if he used assign after f(x) = x  + 6 x + 12, g := x -> f(x)*exp(x) would have worked well except that 

-> should be replaced with unapply.

I couldn't understand your statement of "unreasonably unnecessary assignment."

taro

@Rouben Rostamian  

I think that hhoang1 5 had just forgotten to write as plot(ln(sqrt((x-12)/(-x^2+15*x)))); 

15x is not an adequate expression, while 15*x is.

taro

@Scot Gould 

Have you tried currentdir command?

currentdir();

shows the current dir. And, if you write as currentdir(\Users\me), then you can change the current directory to

\Users\me.

And, I think it might be good for you to write it in Maple.init file in your PC.

taro

Hello @Markiyan Hirnyk 

Thank you for suggesting the command for lagrange method. After googling, I did know it.

As for the reason I wanted to make the code myself is that

I thought it is convenient to write procedures if I use them frequently, but I hadn't written procedures except for practices from any textbook.

I think I will call and use commands from a file in which I write commands, when I need them.

Of course, I can have the code of the LagrangeMultipliers on my screen and try to understand it, with a code of

showstat(Student[MultivariateCalculus][LagrangeMultipliers]);

But, I thought it might be a tough work to a person who haven't understood even local variables well,

(I included x, y, obj, expre of my codes into the category of local variable at first.)

 But, I thought that the procedure I made should be right basically, having only some parts of correction,

sorts of which I would need to do by myself from now but which I couldn't understand and needed any help from others then for. 

Best wishes.

taro

@vv 

Thank you for teaching me a lot.

From your code, I could understand three variables are required to solve for a system of three equations.

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,lambda}); %this place is the key point.
end proc;

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

The above codes works well, which I could understand from your code.

And, thank you for teaching other parts in your code as well.

As for lagrange multipriers as such, next code works.

Student[MultivariateCalculus][LagrangeMultipliers] (a*b,[z-a-b],[a,b]);
http://www.maplesoft.com/support/help/Maple/view.aspx?path=Student/MultivariateCalculus/LagrangeMultipliers

But, what I wanted to know was not to use the code in the above link, but how to write code with procedure.

Thank you very much for telling me enough.

Best wishes.

taro

@acer 

Thank you for telling me how Int works.

I could understand your explanation well and your examples.

And, I read

http://facstaff.cbu.edu/~wschrein/media/ANA/Maplenumint.pdf

along with your maple code and output, and Help of evalf,Int.

And, I read a pdf about some rules of numerical integration only with Simpthon's rule most advanced one.

I could get a lot of knowledge about numerical integration owing to you.

Thank you.

taro

@Jean-Michel Collard 

Yes. codes I cited are from essential maple7 by Dr. Corless.