The roots of a quartic polynomial can be obtained "analytically", or more precisely as expressions involving radicals, but these are usually too complicated to be of much use.  For example, try

> solve(x^4 + x - 3 = 0, explicit);

If there is only one variable, and you want numerical values for the roots, use fsolve instead of solve.  Thus

> fsolve(x^4 + x - 3 = 0);

I'm curious about how determinant=0 corresponds to a unique solution.  Usually
it's the opposite: determinant=0 means the solution, if any, is nonunique.

Your post seems to consist of image tags pointing to files that you have on your own computer (e.g.
C:/Users/JeanPhi/AppData/Local/Temp/moz-screenshot-29.png), but which we have no access to.

If you want to upload files, you can use the green up-arrow button.  But the best thing to do is simply type your question, including any equations or Maple code, in ordinary text.  Use the Preview comment button to see what it your posting will look like before you post it.  If you see html code such as <p> and &nbsp;, click on "Input format" and change it to "Filtered HTML" instead of "Plain text".

Your post seems to consist of image tags pointing to files that you have on your own computer (e.g.
C:/Users/JeanPhi/AppData/Local/Temp/moz-screenshot-29.png), but which we have no access to.

If you want to upload files, you can use the green up-arrow button.  But the best thing to do is simply type your question, including any equations or Maple code, in ordinary text.  Use the Preview comment button to see what it your posting will look like before you post it.  If you see html code such as <p> and &nbsp;, click on "Input format" and change it to "Filtered HTML" instead of "Plain text".

One thing to note about Eigenvalues: if your B has rational entries, it will look for an exact expression for the eigenvalues by solving the characteristic polynomial.  This may not be what you want.  If you want numerical approximations for the eigenvalues, you should either make sure that B contains at least one float or use

> Eigenvalues(evalf(M));

One thing to note about Eigenvalues: if your B has rational entries, it will look for an exact expression for the eigenvalues by solving the characteristic polynomial.  This may not be what you want.  If you want numerical approximations for the eigenvalues, you should either make sure that B contains at least one float or use

> Eigenvalues(evalf(M));

I don't think Maple "guesses" or ":chooses" a vertical range per se.  It calculates points using its adaptive plotting algorithm (see the section on adaptive in the help page ?plot,options). The vertical range is then taken to fit all the values that have been calculated.

I don't think Maple "guesses" or ":chooses" a vertical range per se.  It calculates points using its adaptive plotting algorithm (see the section on adaptive in the help page ?plot,options). The vertical range is then taken to fit all the values that have been calculated.

This is two separate loops, not a nested loop.  Read again what I wrote.

This is two separate loops, not a nested loop.  Read again what I wrote.

Another example of why it's best to avoid 2D math mode.

Another example of why it's best to avoid 2D math mode.

I think you mean this.

> S:=simplify(sum(x^k/product(2*j+1,j=0..k),k=0..infinity));
S := 1/2*2^(1/2)*exp(1/2*x)*Pi^(1/2)*erf(1/2*2^(1/2)*x^(1/2))/x^(1/2)

> series(S, x);
1+1/3*x+1/15*x^2+1/105*x^3+1/945*x^4+1/10395*x^5+O(x^(11/2))

I think you mean this.

> S:=simplify(sum(x^k/product(2*j+1,j=0..k),k=0..infinity));
S := 1/2*2^(1/2)*exp(1/2*x)*Pi^(1/2)*erf(1/2*2^(1/2)*x^(1/2))/x^(1/2)

> series(S, x);
1+1/3*x+1/15*x^2+1/105*x^3+1/945*x^4+1/10395*x^5+O(x^(11/2))

The region between the two circles |z-1|=1 and |z-2|=2 corresponds, by the conformal mapping w = 1/z, to the region between the two straight lines Re(w) = 1/4 and Re(w) = 1/2.

> with(plots): 
  conformal(1/w, w=1/4 - 10*I .. 1/2 + 10*I, scaling=constrained, numxy=[30,1000],
     grid=[5,400],colour=[red,blue]);

The region between the two circles |z-1|=1 and |z-2|=2 corresponds, by the conformal mapping w = 1/z, to the region between the two straight lines Re(w) = 1/4 and Re(w) = 1/2.

> with(plots): 
  conformal(1/w, w=1/4 - 10*I .. 1/2 + 10*I, scaling=constrained, numxy=[30,1000],
     grid=[5,400],colour=[red,blue]);