You can set the relative error for an entire session by the setting the Digits environment variable. It's set to 10 by default. But what you want is akin to setting the absolute error for an entire session. There's no easy way to do that (maybe no way at all?). However, you can apply fnormal to any individual expression containing floating-point constants, and it'll turn the small ones into zero.

If you use the hardware-floating-point environment, certain very small numbers will underflow and automatically become 0. This environment is used in the background by default by many (most?) numerical commands when you have Digits set to 15 or less. For example, I noticed that you used exp(-10000) in a call to pdsolve in a recent Question. We can see below that this will be treated as zero:

evalhf(exp(-10000));

     0.

No. That is, I believe, well beyond the capability of any current CAS. Even curves can only be parameterized when f(x,y) has certain special forms, such as polynomial.

The value under the square root sign is negative for any real rho. Therefore both h(rho) and its integral are purely imaginary for any real range of rho. We could plot these imaginary values if you want: Simply replace v(rho) with Im(v(rho)) in the plot command.

In Maple 2015, I tried with the default method= rkf45, and with method= rosenbrock and I had no trouble. Did you try the default method? I confirm that dverk78 and lsode don't work for this problem.

If you'd followed today's discussion in the followups to your prior Question, you would've tried is. It's the most basic command for testing the mathematical equality (over the complex numbers) of expressions. Unfortunately, testeq is very limited in the expressions that it can handle.

is(C45=C54);

     false

(I'm a bit disturbed that you ignored Kitonum's advice (in your prior Question) that you not use I as a variable. I had to substitute J for I below to get meaningful results.)

Here's an explicit proof that the two expressions are fundamentally different: We evaluate each one setting all derivatives and variables to Pi/4, then simplifying.

C45a:= subs(I= J, C45):
C45b:= evalindets(convert(C45a, rational), specfunc(diff), freeze):
C45c:= evalindets(C45b, name, ()-> Pi/4):
simplify(C45c);

C54a:= subs(I= J, C54):
C54b:= evalindets(convert(C54a, rational), specfunc(diff), freeze):
C54c:= evalindets(C54b, name, ()-> Pi/4):
simplify(C54c);

RootFinding:-NextZero performs a job similar to fsolve (for real functions only). It's a little trickier to use because you have to carefully adjust its options or else it'll miss roots. But, if you want all the roots in a given interval, it may be a better choice than fsolve. Here's the first 20 positive roots:

K:= N-> KummerM(1/2-sqrt(2*N)/4, 1, sqrt(2*N)):

R:= table([0=0]):
for k to 20 while R[k-1]<10000 do
     R[k]:= RootFinding:-NextZero(K, R[k-1], maxdistance= 10000, guardDigits= 3)
od:

convert(R, list)[2..];

[3.65679345776329, 22.3047305506807, 56.9605153816721, 107.620271629881, 174.282057717514, 256.945030311759, 355.608766328753, 470.273028271534, 600.937671278859, 747.602601263178, 910.267754050297, 1088.93308412430, 1283.59855815621, 1494.26415108509, 1720.92984364050, 1963.59562071823, 2222.26147028210, 2496.92738260118, 2787.59334970751, 3094.25936500266]

Another root-finding command that's supposed to find all the roots in an interval is Student:-Calculus1:-Roots. However, in this case, it only finds 4 of the first 20.

I can't tell what you're doing wrong since you didn't post your code or upload a worksheet. I'll assume that by "deriving" you mean finding the derivative. (The verb form of derivative in mathematical English is, oddly enough, differentiate rather than derive.) To differentiate L with respect to x, enter the command

diff(L, x)

Here's a procedure for it:

A:= M->
     <<Matrix(M) | -<seq(1/k, k= M..1, -1)> | Matrix(M, (i,j)-> `if`(i+j=M+1, 1/j, 0))>,
      Vector[row]([1$M, Pi, 1$M]),
      <Matrix(M, (i,j)-> `if`(i+j=M+1, 1/i, 0)) | <seq(1/k, k= 1..M)> | Matrix(M)>
     >/Pi:

There is a serious bug that VectorCalculus integrals consistenly ignore the sign, even in simple cases. For example:

VectorCalculus:-PathInt(x^2, [x,y]= Line(<0,0>, <1,1>));

VectorCalculus:-PathInt(x^2, [x,y]= Line(<1,1>, <0,0>));

Your ODE has a singularity at r=0. Note the 1/r factor. So you can't specify boundary or initial conditions at 0.

If your worksheet is short, please include it in your Question as well as attaching it. I think that that facilitates the discussion.

Where you have length(v), you should have numelems(v).

Making that correction, your code still produces incorrect results. But I don't understand your algorithm so I can't correct it. One thing I find surprising, and probably erroneous, is for j to 26. Why 26?

To plot the curves, do

plot([seq(<v|P[..,k]>, k= 1..10)]);

But, as it stands, the curves are constant.

The source of your problem is incorrect 2D input. To see it and correct it, select with the mouse your assignment statement that defines B[22]. Right click to bring up the context menu, then select 2-D Math -> Convert To -> 1-D Math Input. Then you will that this assignment statement is

B[22] := Q^%T . `#mi("M")` . `#mi("Q")`+`#mi("m")`[L1]*T[b]^%T . S[L1]^%T . S[L1] . T[b]+Q^%T . `#msubsup(mi("R"),mi("L1",fontstyle = "normal"),mi("b",fontstyle = "normal"))` . I__L1 . `#msubsup(mi("R"),mi("L1",fontstyle = "normal"),mi("b",fontstyle = "normal"))`^%T . Q+m[L2]*T[b]^%T . S[L2]^%T . S[L2] . T[b]+Q^%T . `#msubsup(mi("R"),mi("L2",fontstyle = "normal"),mi("b",fontstyle = "normal"))` . I__L2 . `#msubsup(mi("R"),mi("L2",fontstyle = "normal"),mi("b",fontstyle = "normal"))`^%T . Q;

The first term needs to be changed from

Q^%T . `#mi("M")` . `#mi("Q")`

to simply

Q^%T.M.Q

After making that change, the IsMatrixShape command (with the mapped expand) will instantly return true.

To avoid making mistakes like this in the future, I suggest that you switch to using 1D input.

Maple makes a distinction between pairs of expressions that are manifestly equal (identically equal) and those pairs that can be proved to be equal after some simplification. For example, sin(1)^2+cos(1)^2 is not identically equal to 1, as can be seen simply by entering the former expression and noting that it does not automatically and immediately appear as 1. IsMatrixShape(..., symmetric) is checking whether the matrix's entries are manifestly equal to those of the transpose. So,

M:= <0, 1; sin(1)^2+cos(1)^2, 0>:
IsMatrixShape(M, symmetric);

     false

but

IsMatrixShape(simplify~(M), symmetric);

     true

You haven't said whether your matrices have symbolic or numeric entries. If they have symbolic or exact entries, then you need to apply expand~ to the matrix before the symmetric check, as TomLeslie noted. If your matrices have floating-point entries, then it is a little bit trickier to check for symmetry, but it can still be done. Let me know.

You should not use the same variable, i, as both the index of summation and the index of a for loop. At the end of the for loop, the value of i is 7. Thus, the subsequent sum command

sum(c[i], i= 1..6)

is interpretted as if you'd entered

sum(c[7], 7= 1..6)

Since c has only six elements, the 7 is an invalid subscript. Using a different variable for the for loop index will solve both your problem 1 and problem 2.

The evalc command is the "natural" command for evaluating complex expressions whose variables are assumed to be real. So, what first occurred to me is

convert(evalc(conjugate(exp(I*x))), exp);