Carl Love

Carl Love

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8 years, 251 days
Natick, Massachusetts, United States
My name was formerly Carl Devore.

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These are answers submitted by Carl Love

The integral is simply infinity. It doesn't depend on any variable. The x is gone after you integrate.

Preben's answer is great, and it will work for procedures in general. I just want to direct you to a great command that is made especially for your kind of function: It's called ?piecewise.

Maple is case sensitive, and Return is not the same thing as return, which is what Maple uses. (Indeed, all of Maple's ?keywords are in lowercase.) One of the procedures must use the capitalized version.

Just guessing here, but was one of the sets of procedures translated automatically from Mathematica?

You need to expand (i.e., distribute sin(z) over the rest) the first expression; otherwise, it won't contain the subexpression that you told applyrule to look for. But simply using the command expand will do too much: It will also apply a trig identity expand cos(2*z). So do this

frontend(expand, [exprsn1]);

Then use applyrule on the result of this.

First, get rid of the line k:= 0, 1..npts/2. Do a restart to eliminate the effect of that command. Then, at the end, do

plot(Zuncomp(k), k= 0..npts/2, labels= ["k", "Zuncomp"])

Another workaround is to set Digits:= 20.

Assumptions via assume on one variable are not cumulative. Your second assume command erases the effect of the first. The second one is unnecessary anyway: It is implied by the first. If you actually want to make multiple assumptions on a single variable, put them together in one assume command or use the additionally command. See the examples at ?assume for details. If you remove the second assume, you will get your expected result.

By the way, you misspelled alpha as alph, but that doesn't change the result.

I agree with Preben about assuming. I think that it makes for cleaner, easier-to-read, less-error-prone, and more-flexible worksheets. Although there is still a place for assume if you're trying to get Maple to do a symbolic proof.

This is a bug in the Standard GUI. Your commands run fine in the command-line Maple. If you have access to the Classic GUI (I don't), try that also.

A warning to anyone trying to investigate this: If you try to probe this, say with printlevel:= 100, the GUI will lose connection with the kernel (without any message), and you'll be stuck. Setting interface(prettyprint= 0) does not help. Another curiosity: The output, i.e. the returned value, of the OP's commands is literally the symbol `Non-fatal error while reading data from kernel.`; that is what soltn is set to; it is not an error message in the usual sense. Therefore, it must the kernel that is sending the message!

There are two independent variables, t and x. So they need to be 3D plots. I don't think two 3D plots, together, of functions that are so close will be very useful. But it can be done if that's what you really want. You can have two side-by-side 3D plots on different axes. Another option is a plot of the difference between the two functions.

There are two immediate problems:

First, the initial condition P(0) = P uses P in two different ways. You need to use a different letter.

Second, the exponential function is represented in Maple as exp(...), not e^(...).  If you correct this to exp(-1.15*10^12), the resulting number is so incredibly small that it is difficult (not impossible) to work with. So, before we go further, I have to ask if this is what you really meant.

As a statement, to return from a procedure:

if `and`(0<z, abs(y)<=6, 0<x,x<1) then return else return 0 fi

As an expression, which may or may not appear after return:

`if`(`and`(0<z, abs(y)<=6, 0<x,x<1), NULL, 0)

NULL is equivalent to [][]; I prefer the latter, which is easier to type and doesn't clutter the code with words in ALL CAPS. In several contexts in Maple, a NULL is implied by the absence of anything else; such is the case with the first return statement above. The if-statement bracket fi is if backwards, and is equivalent to end if. And `and`(cond1, cond2, ..., condN) is equivalent to cond1 and cond2 and ... and condN, with the former being less to type when there are more than two conditions.

Another way --- one of many.

restart;
r:= [3,3,3,3,4,4,4,3,3,3,3,3,3,3,3,2,4,3,1]:
T:= ListTools:-Classify(nops, [ListTools:-Categorize](`=`, r));

T := TABLE([1 = {[1], [2]}, 4 = {[4, 4, 4, 4]}, 13 = {[3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3]}])

m:= max(indices(T));

m := 13

op ~ (T[m])[];

3

I wrote the above to cleanly handle the multi-modal case (as Kitonum also did), which adds a bit of complexity.

Yes, dsolve with type= series says that the positive-term series that you posted is the solution. I think that I found the sign bug in your code.  You have a line

u[m,k+1]:= -R[m](seq(u[i,k],i=1..nEq))-A[m,k];

I think that the last term should be +A[m,k]. If I make this change, then I get the positive-term series that you want.

Maple is saying that there's a singularity near x = 2.5. Do you have reason to doubt that? Here's the code and plot. You can see that it's approaching an asymptote.

restart;
dq:= diff(y(x),x) = x^2 + y(x)^2 - 1:
DEtools[DEplot](
    dq, y(x), x= 0..2.5, {y(0)=0}
   ,view= [default, -1..5]
   ,arrows= line
   ,title= 'Slope*Field'
);

Just apply coeff twice:

S:= A*sin(x)*cos(z) + B*sin(2*x)*cos(z) + C*sin(3*x)*cos(2*z):

coeff(coeff(S, sin(2*x)), cos(z));

                                                                 B


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