In general, consider a DE diff(y(r),r,r) = - b * diff(y(r),r)/r + f(y(r)) where b > 0. 

Note that diff(y(r),r,r) + b*diff(y(r),r)/r = r^(-b) * diff(r^b * diff(y(r),r), r).  If there is a constant A such that  f(y(r)) < A for sufficiently large r, then for such r we have

diff(r^b*diff(y(r),r), r) < A*r^b

, so for some constants C1 and C2,

y(r) < A*r^2/(2*(b+1)) - C1*r^(1-b)/(1-b) + C2 [ in the case b=1, replace the r^(1-b)/(1-b) by ln(r) ], and similarly if the inequalities are reversed.  In particular, the only way for y(r) to have a finite limit L as r -> infinity with f continuous at L would be to have f(L) = 0.

Note that the residue will not exist in this case: 1/sqrt(1+x^4) has branch points, not poles.

The problem seems to be this.  residue substitutes x+(1+I)/sqrt(2) for x, and expands the resulting expression (more precisely, it
uses normal(..., 'expanded'))..  Then it tries to take a power series for this.
Since this series will involve fractional powers, it will not be of type 'series', so residue keeps trying to higher and higher
orders, eventually to order 27.  If the original expression had not been expanded, that would not be so bad:

series(1/sqrt(1+(t+(1+I)/sqrt(2))^4), t, 27) has length 2289, and is returned quickly.

But because the expression has been expanded, the sizes of the un-simplified coefficients get rapidly larger.  I'd guess that the
length of the order-27 series would be in the billions.  So it seems to take forever (presumably it would eventually either finish or use up
all available memory).

I'm submitting an SCR.

I don't understand the question.  What strings do you want to suppress, in what output?

I don't understand the question.  What strings do you want to suppress, in what output?

I don't understand your question.  Why should the frequency of the cosine have an effect on s?

s = (j-0.9)*deltat/sampling.  You have j going from 1 to 20, deltat = 0.0025 and sampling = 1, so s goes from .00025 to .04775

I don't understand your question.  Why should the frequency of the cosine have an effect on s?

s = (j-0.9)*deltat/sampling.  You have j going from 1 to 20, deltat = 0.0025 and sampling = 1, so s goes from .00025 to .04775

There are several errors here, I think.  The "too many levels of recursion" comes from the fact that when i=j=2 you solve the equation 9.*A[2,0]+49.50000000*A[2,1]+333. = 0, obtaining {A[2,0] = -5.500000000*A[2,1]-37., A[2,1] = A[2,1]}, and then assign that result to A[2,0]. 
Assigning a variable a value containing the name of that variable is a bad idea.

Not quite.  for plot, this will generally get you a list of points, but for implicitplot you'll have an Array.  You may want to use

convert(op([1,1],p), listlist);

Also there will be more than one Array if the curve has more than one component.

Not quite.  for plot, this will generally get you a list of points, but for implicitplot you'll have an Array.  You may want to use

convert(op([1,1],p), listlist);

Also there will be more than one Array if the curve has more than one component.

You have ln*x all through these equations.  ln is a function, not a variable: it needs parentheses, not a multiplication sign.  Also, in each equation you have a pi that should be Pi.

After making those corrections, fsolve still finds no real solutions.

You have ln*x all through these equations.  ln is a function, not a variable: it needs parentheses, not a multiplication sign.  Also, in each equation you have a pi that should be Pi.

After making those corrections, fsolve still finds no real solutions.

Very few equations have "closed form" solutions, i.e. solutions you can write down a formula for in terms of "well-known" functions.
That would not be "due to some approximations".  In fact, approximations generally produce simpler equations, which are more likely to have closed form solutions.

Your eqn2 is incorrect, because you left out the multiplication sign after the b.  With that corrected, eqn2 becomes

.8e11*polylog(3/2,exp(b*(u-4300))) = 1

If we substitute x for exp(b*(u-4300)), eqn2 says polylog(3/2,x) = .125e-10. This has only one real solution, approximately
x = .125e-10.  Substituting that into eqn1, we get

1.750000000/b^(5/2)+2508.333334/b^(3/2) = -1290

which clearly has no real solutions (for b > 0 the left side is positive and the right side is negative).

There are, however, complex solutions.  For example, fsolve gives me a solution of b = -.7784682642-1.349152491*I.
That would correspond to u = 4308.055177-13.96031445*I.

Very few equations have "closed form" solutions, i.e. solutions you can write down a formula for in terms of "well-known" functions.
That would not be "due to some approximations".  In fact, approximations generally produce simpler equations, which are more likely to have closed form solutions.

Your eqn2 is incorrect, because you left out the multiplication sign after the b.  With that corrected, eqn2 becomes

.8e11*polylog(3/2,exp(b*(u-4300))) = 1

If we substitute x for exp(b*(u-4300)), eqn2 says polylog(3/2,x) = .125e-10. This has only one real solution, approximately
x = .125e-10.  Substituting that into eqn1, we get

1.750000000/b^(5/2)+2508.333334/b^(3/2) = -1290

which clearly has no real solutions (for b > 0 the left side is positive and the right side is negative).

There are, however, complex solutions.  For example, fsolve gives me a solution of b = -.7784682642-1.349152491*I.
That would correspond to u = 4308.055177-13.96031445*I.

There is some of that.  The help system now contains an "Error Message Guide" which has pages explaining some of the more common error messages.  And in Maple 13 (Standard GUI) error messages are hyperlinks that take you to the corresponding help page in the on-line Maple help.

There is some of that.  The help system now contains an "Error Message Guide" which has pages explaining some of the more common error messages.  And in Maple 13 (Standard GUI) error messages are hyperlinks that take you to the corresponding help page in the on-line Maple help.