Carl Love

## 25831 Reputation

10 years, 358 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

## Need to add in the 1s...

@brian bovril Do you mean all the factorings of 36 into 3 parts? The program under discussion in this thread does not consider 1 to be a part of any factoring, but the results could be easily modified to include 1. Just take all the factorings, Factorings(36,1), Factorings(36,2), and Factorings(36,3), and pad the results with the appropriate number of 1s.

## Not trying to prove...

@Markiyan Hirnyk I'm not try to prove anything about the expression F in the mathematical sense of proof. I'm trying to prove, if you could call it that, or verify, that plot was unable to evaluate F at any of the values that it chose, except for k = 0. It may well be that there are other values of k at which F can be evaluated.

Now I am convinced that this situation---where plot can only evaluate the expression at one value---should be considered a bug, because it should give the same warning message that one gets if it can't evaluate the expression at any values. Displaying empty axes with no warning is not helpful.

Compare plot(1+x*I, x= 0..1) with plot(1+x*I, x= 1..2).

## Not trying to prove...

@Markiyan Hirnyk I'm not try to prove anything about the expression F in the mathematical sense of proof. I'm trying to prove, if you could call it that, or verify, that plot was unable to evaluate F at any of the values that it chose, except for k = 0. It may well be that there are other values of k at which F can be evaluated.

Now I am convinced that this situation---where plot can only evaluate the expression at one value---should be considered a bug, because it should give the same warning message that one gets if it can't evaluate the expression at any values. Displaying empty axes with no warning is not helpful.

Compare plot(1+x*I, x= 0..1) with plot(1+x*I, x= 1..2).

## PDEtools:-Solve just calls dsolve for od...

PDEtools:-Solve just passes the problem on to dsolve if it detects that they're ODEs. If dsolve is used for this problem without the numeric option, it will spend a huge amount of time and memory trying to find a symbolic solution. I didn't let it finish, as I was getting close to my memory limit.

## PDEtools:-Solve just calls dsolve for od...

PDEtools:-Solve just passes the problem on to dsolve if it detects that they're ODEs. If dsolve is used for this problem without the numeric option, it will spend a huge amount of time and memory trying to find a symbolic solution. I didn't let it finish, as I was getting close to my memory limit.

Plain text is easiest to work with. Use one of the three clipboard icons on the toolbar in MaplePrimes to cut-and-paste directly into your post. Maple understands plaintext, but not Word AFAIK. Or upload a worksheet. Why take the extra step of going through Word?

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Plain text is easiest to work with. Use one of the three clipboard icons on the toolbar in MaplePrimes to cut-and-paste directly into your post. Maple understands plaintext, but not Word AFAIK. Or upload a worksheet. Why take the extra step of going through Word?

.

## Should be enough, with correct initial c...

That definition of T(x) should be enough. But your initial conditions are ill-posed:

`R(x) = 0, X(x) = 10;Shall I assume that that should beR(0) = 0, X(0) = 10?`

## Should be enough, with correct initial c...

That definition of T(x) should be enough. But your initial conditions are ill-posed:

`R(x) = 0, X(x) = 10;Shall I assume that that should beR(0) = 0, X(0) = 10?`

## Polyominoes...

Another great algorithm, Kitonum. As a point of reference, these structures are formally called (in English) "polyominoes" (rather than "polygons of the match"), and there is much literature on them.

Could you please edit the first paragraph of your post?  The right sides of the lines are cut off; they did not word wrap.

And how did you export an array of plots from Maple? Perhaps by using an external plotter? When I try (in Standard GUI), it only lets me select one cell.

## But that's the only point plot can evalu...

@Markiyan Hirnyk Well, of course, that's just -ln(2). But that is the only point in the plot range that plot was able to evaluate. You can see that in the plot structure:

P:= plot(F, k= 0..2):  # F cut-and-paste from OP's warning msg.
op(P);

CURVES(Matrix(2, 2, {(1, 2) = -.693147180559945, (2, 1) = HFloat(HFloat(undefined)), (2, 2) = HFloat(HFloat(undefined))}, datatype = float[8], storage = rectangular, order = Fortran_order, shape = []), COLOUR(RGB, .47058824, 0., 0.54901961e-1, _ATTRIBUTE("source" = "mathdefault"))), AXESLABELS(k, ""), VIEW(0. .. 2., DEFAULT, _ATTRIBUTE("source" = "mathdefault"))

For every other value of k that plot tried, it got the error about abs not being differentiable, but it traps those errors in a try ... end try block. It won't draw a curve with only one point, so it just displays empty axes.

## But that's the only point plot can evalu...

@Markiyan Hirnyk Well, of course, that's just -ln(2). But that is the only point in the plot range that plot was able to evaluate. You can see that in the plot structure:

P:= plot(F, k= 0..2):  # F cut-and-paste from OP's warning msg.
op(P);

CURVES(Matrix(2, 2, {(1, 2) = -.693147180559945, (2, 1) = HFloat(HFloat(undefined)), (2, 2) = HFloat(HFloat(undefined))}, datatype = float[8], storage = rectangular, order = Fortran_order, shape = []), COLOUR(RGB, .47058824, 0., 0.54901961e-1, _ATTRIBUTE("source" = "mathdefault"))), AXESLABELS(k, ""), VIEW(0. .. 2., DEFAULT, _ATTRIBUTE("source" = "mathdefault"))

For every other value of k that plot tried, it got the error about abs not being differentiable, but it traps those errors in a try ... end try block. It won't draw a curve with only one point, so it just displays empty axes.

## I realize that...

@Markiyan Hirnyk I realize that my plot is not what the OP was trying to plot. I was trying to illustrate that the warning message was unrelated to the failure to get a plot. Rather, the issue is as Preben suggested: the inability to evaluate the derivative of abs that appears in the OP's function.

## I realize that...

@Markiyan Hirnyk I realize that my plot is not what the OP was trying to plot. I was trying to illustrate that the warning message was unrelated to the failure to get a plot. Rather, the issue is as Preben suggested: the inability to evaluate the derivative of abs that appears in the OP's function.

## Being Socratic?...

@Markiyan Hirnyk If you were asking the question to be Socratic with the OP, then I am sorry for interfering here. But if you really were asking because you didn't understand the OP's problem, then C = Limit(z(t), t= infinity).

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