Dear Maple community,

I was going to export my math assignment about vector functions as an PDF, when I noticed that it butchered every column vector printed out in blue in the document as math-output. I only have this issue when exporting my work.

How do I make Maple display both coordinates of vector in the blue output field?

I produced the below image as an illustration of my problem:

Thank you in advance.

The flag of Germany on the strip of the German mathematician August Ferdinand Möbius. Basically, it's just one way to represent flags of a certain type. It seemed that the flag looked good on the Mobius strip.
FLAG.mw

Why can't the variables in my procedure be called? Omega1, Omega2, Q1, Q2 are all expressed in the form of local variables in the loop, why can't we solve the equation because there are too many variables? At the same time, I would like to ask how to output the results of Q1 and Q2?

equation915.mw

Why can the coeff operator return the coefficient of 1/b, namely -1/x in the example below correctly,

but cannot return the coefficient of 1/x , which should be -1/b?

dummy := asympt(x*1/(1 - a*x - b*x^2), x);
coeff(dummy, 1/x);
coeff(dummy, 1/b);

What one generally wants is to be able to return the coefficients of the orders of an asymptotic expansion, but coeff seems unable to do that as soon as you want the coefficients of 1/x^n

Maple help pages is silent about this.

This is my file
 

restart;
F := proc(ee, LL) Typesetting:-mrow(InertForm:-Typeset(InertForm:-Display(eval(eval(InertForm:-MakeInert(factor(ee)), [`%*` = `*`]) = InertForm:-MakeInert(map(sort, algsubs(a*x = InertForm:-MakeInert(a*x), ee), order = plex(b))), `=`~([a, b], LL)), inert = false)), Typesetting:-mo("="), InertForm:-Typeset(eval(ee, `=`~([a, b], LL)))); end proc;
p := a^3*x^3 + 3*a^2*b*x^2*y + 3*a*b^2*x*y^2 + b^3*y^3;
L := [[2, 3], [1, 2], [1/3, -sqrt(2)]];
ans := F~(p, L);
print~(ans);

I get 

I want to the order of terms of the polynomial, e.g, (2 x + 3y)^3 like this

(2 x + 3 y)^3 = (2 x)^3 + 3 (2 x)^2 (3 y) + 3 (2 x) (3 y)^2 + (3 y )^ 3 = 8*x^3 + 36*x^2*y + 54*x*y^2 + 27*y^3. 

How can I get that? 

Solving differential equations, Maple sometimes unfortunately returns the solution in an utterly  unusable form I never encountered a use for.

As an example a solution was found assuming separation of variables. Maple returns the following solution,

Is there a command where I can just get the argument of the solution, namely the differential equation diff(_Y(ts),ts,ts) + etc  ?  or even better as diff(X(ts),ts,ts) + etc   as I intended and expected it to be without the silly and unnecessary proxy variable _Y(ts) ?

It is such a pity that Maple return these results not as a differential equation in F2(ts), but gives the result in a proxy variable which isutterly unnecessary.

Anyway, if anyone knows a command just to get the argument of the solution above to get rid of all the unnecessary and proxy structure, I will appreciate it as I currently copy the solution and redefine it, which can introduce errors and destroys the generality of the document.

A classic result states that the equation x³＋px²＋qx＋r＝0 with real coefficients p, q, r has positive roots iff p＜0, q＞0, r＜0 and -27r² - 2p(2p² - 9q)r + q²(p² - 4q) ⩾ 0 (see for example this question). 
However, Maple appears unable to find the condition: 

a, b, c := allvalues(RootOf(x^3 + p*x^2 + q*x + r, x), 'implicit'):
RealDomain:-solve({a, b, c} >~ 0, [p, q, r]);
 = 
Warning, solutions may have been lost
                               []

Is there a way to get the above conditions in Maple with as little human intervention as possible (I mean, without a priori knowledge of the theory of polynomials)? 

Edit. An interesting problem is when these three positive roots can further be the lengths of sides of a triangle. For reference, here are some (unenlightening) results from some other software: 

I don't know how make my graph be beter for real part and imaginary part and abs part which part how work with parameter can any one explain on this example?

G.mw

I have this file
 

restart:
F := proc(ee,LL)
  uses InertForm, Typesetting;
  mrow(Typeset(Display(eval(eval(MakeInert(factor(ee)),[`%*`=`*`])
                                 =MakeInert(subs(b=MakeInert(b*y)/y,
                        a=MakeInert(a*x)/x,p)),[a,b]=~LL),
                       inert=false)),
       mo("="),Typeset(eval(ee,[a,b]=~LL)))
end proc:
p := (a*x)^2 - 2*a*x*b*y + (b*y)^2:
L := [[sqrt(2),3],[2,5],[3,12],[1/3,5/7]];
ans := F~(p, L):
print~(ans):

How can I put the results like this
 

\documentclass[12pt,a4paper]{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{enumitem}
\theoremstyle{definition}
\newtheorem{ex}{Exercise}
\begin{document}
\begin{ex}
\[(\sqrt{2} x-3 y)^2=(\sqrt{2} \cdot x)^2-2 \cdot(\sqrt{2} \cdot x) \cdot(3 \cdot y)+(3 \cdot y)^2=2 x^2-6 \sqrt{2} x y+9 y^2.\]
\end{ex}

\begin{ex}
\[(2 x-5 y)^2=(2 \cdot x)^2-2 \cdot(2 \cdot x) \cdot(5 \cdot y)+(5 \cdot y)^2=4 x^2-20 x y+25 y^2. \]
\end{ex}

\begin{ex}
\[(3 x-12 y)^2=(3 \cdot x)^2-2 \cdot(3 \cdot x) \cdot(12 \cdot y)+(12 \cdot y)^2=9 x^2-72 x y+144 y^2. \]
\end{ex}

\begin{ex}
\[\left(\frac{x}{3}-\frac{5 y}{7}\right)^2=\left(\frac{1}{3} \cdot x\right)^2-2 \cdot\left(\frac{1}{3} \cdot x\right) \cdot\left(\frac{5}{7} \cdot y\right)+\left(\frac{5}{7} \cdot y\right)^2=\frac{1}{9} x^2-\frac{10}{21} x y+\frac{25}{49} y^2. \]
\end{ex}
 
\end{document} 

The attached sheet contains equations H1 to H6 and K1 to K3. What data do I need to modify to ensure that the values of H1 to H6 fall between 0 and 1, and K1 to K3 are negative? The parameter ranges are also given in the sheet. Is there a method to achieve this?

rouhg.mw

i did two case of this equation and odetest is worked good but in this case the odetest is not worked well anyone can determine what is mistake ?

F_P_Correct_case_three.mw

Dear Maple user I am facing error while running the codes  to plot the graph for two data sets .

I am attaching the files.

Error_in_Display_1.mw

Why does dsolve not call odetest by default before a solution is returned?

I mean, why do I have test each result separately. dsolve could have an odetest option (default=true).

In case of discrepancies dsolve could inform the user and suggest to call dsolve with odetest=false and run odetest separately to analyse the problem.

Set up this way, dsolve would never return potentially incorrect solutions that do not pass odetest.

HTR.mw

In above problem, Additionally How to  plot  heat transfer rate  Q versus L^2  for distinct porosity parmeters(Sh) , using  heat transfer rate formula, Q = (q*L)/(k*A*T[b])=theta'(1).

using  [Sh = 0.1, L^2 = 0.1, Nr =0 .1, Ha =0 .1, Pe = 0.1],  [Sh = 0.3, L^2 = 0.3, Nr = 0.1, Ha = 0.1, Pe =0 .1],   [Sh = 0.5, L^2 =0 .5, Nr =0 .1, Ha = 0.1, Pe =0 .1].