taro

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5 years, 361 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 replies submitted by taro

@tomleslie 

The first way is

x:={b[1]=alpha*a[1]^6, b[2]=alpha*a[2]^6, c[1]=beta*a[1]^4+gamma*d[1], c[2]=beta*a[2]^4+gamma*d[2]};

y:=a[1]^2+b[1]^2+c[1]^2+d[1]^2+a[2]^2+b[2]^2+c[2]^2+d[2]^2=0;

z:=subs(x,y);

solve(z,a[1]);

which brings a solution, though including RootOf.

 

The second way is

k:= solve({op(x),y},a[1]);

But, this does not bring any answer, which I can't understand why.

 

@tomleslie 

Hello

Two ways to solve

P:=(a,b,c,d,e,f)->a^2+b^2+c^2+d^2+e^2+f^2;

that is
b:=x->x^2;
c:=x->x^2;
d:=x->x^2;
e:=x->x^2;
P(a,b(a),c(a),d(f),e(f),f)=0;
solve(%,a);

 

and

solve(P(a,b,c,d,e,f),[a,b,c,d]);

bring different solutions.

So, I couldn't understand your answer.

 

@vv 

evalf(151/11);
convert(%,rational,3);

which brings 41/3 of OP.

@tomleslie 

Thank you for your reply.

with the change of preferences of maple in Precision, Round screen display, to 3
decimal places, I can get the view of an only 3 digit below a decimal point like this.

The meaning of my question is how to do the same without changing the preference
so that without exerting this modification on other calculation on the same worksheet, that is to limit this effect on only a part of various calculation.

I thought of this question seeing the difference of the expression of the original questioner's expression of the numbers, and your result of exact expression of the numbers.

 

@tomleslie 

N := [7.5000000000, 2.5000000000, 4.0000000000]

is different from

M:={7.5, 2.5, 4}

printf("%g %g %g\n", op(N) );
outputs 7.5 2.5 4, but this cannot be used for the continuing calculation.
How can I use the outputs for the later calculation?

I'm not the questioner, but I will be very glad if you will give me an answer to this question.

 

@Kitonum 

Thank you for your detailed explanation.
I could understand the limit of inequality in maple and how to deal with inequalities owing to you.

 taro

@ThU 
Surely, in this case, it is easier to rewrite the expression manually.
But, I think I could improve my skill much in using Maple, through asking that question here and reading the code by Kitonum. I owe a lot to people who gave me answers to my questions.



 

@Kitonum 

Thank you. I owe you a lot.
I found from your answer that the basic idea, when a variable overlaps between the part already collected and the part not yet collected, is to replace the part already collected with another variable and to collect the part not yet collected.

And, thank you for showing me the use of freeze and thaw, and how to chooze the already
collected part. Your code is very helpful.
Thanks a lot.

@vv 

I'm sorry. I couldn't understand what you wrote. What is some subexpression which you wrote? and what can I do if I had defined that subexpression. You posted your
comment while I was modifying what I wrote, that is, in the part of mentioning that algsubs could not be used for appropriately getting the answer. And, if my
remove(has, c1, delta) means your writing that I have to obtain 'that subexpression,'
what can I do after doing it?

@vv 

Thank you for your comment. But, this is not for typesetting, though looking so.
My question is how to do additional collect in the case that the variable with which I want to collect the rest of expression has already been used for another collected part.


More writing to explain my problem, as op([5,2..3],a1) is not -epsilon+z, I cannot use applyop for collecting -epsilon*z + z to 
z*(1-epsilon).
   On the other hand, I can identify -epsilon*z+z with

But, this identification does not look helpful, as algsubs(d1=collect(d1,z), c1) does not play the role which I expected.

@nidojan 

I'm sorry. I think I might have lacked of enough explanation about my question.
What I want to do is to collect the part of -epsilon*z+z in a1 to z*(1-epsilon) in b1.

 

 

@Markiyan Hirnyk 

I think that your advice is to divide both of numerator and denominator with a specific expression. In that meaning, through doing so, I could simplify the expression which I wanted to simplify. Thank you.

 

@Kitonum 
Thank you for your teaching.
From your answer, I found substitution plays well.

Though a similar one, dividing first works well.

e1:=n = (theta-1)*(z-1)*(-k*psi*L*(-(epsilon-1)*psi/epsilon)^epsilon*k^epsilon+K*(-(epsilon-1)*psi*k/epsilon)^epsilon)/(gamma*((theta-1)*(z-1)*(-(epsilon-1)*psi*k/epsilon)^epsilon-(-(epsilon-1)*psi/epsilon)^epsilon*k^epsilon*theta*z*psi));
b:=(-(epsilon-1)*psi*k/epsilon)^epsilon;
c:=simplify(numer(rhs(e1))/b)/simplify(denom(rhs(e1))/b) assuming (k>0);

The problem in this case to me migh have been the use of assuming(k>0).

Thank you.

 

 

 @sand15 

Doesn't the way to ascertain whether all eigenvalues are positive become an answer?

Vector[row]([r,s]).Matrix([[1,-1/2],[-1/2,1]]).Vector[column]([r,s]);

expand(%)

a:=Matrix([[1,-1/2],[-1/2,1]]);

LinearAlgebra[Eigenvalues](a);

 

@Kitonum 

In my environment: maple2016, osx EL Capitan, the result of assignment is different from yours. A and B are reverse in my results. Could I ask you to show the way to correct this?

 

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