## 593 Reputation

16 years, 268 days

## Thanks, next step...

Thank you to both of you (Joe Riel and Robert Israel).

So, how would I translate the following pseudocode:

for each permutation, g, in the set (not group) of permutations { [1,4], [1,5], [3,5,4]}

do proess which depends upon g([1,2,3,4,5]) as parameters end do;

?

Second problem = same question, but for the group of permutations = the full Symmetric group on {1,2,3,4,5}?

## Thanks, next step...

Thank you to both of you (Joe Riel and Robert Israel).

So, how would I translate the following pseudocode:

for each permutation, g, in the set (not group) of permutations { [1,4], [1,5], [3,5,4]}

do proess which depends upon g([1,2,3,4,5]) as parameters end do;

?

Second problem = same question, but for the group of permutations = the full Symmetric group on {1,2,3,4,5}?

## Wow! Thank you, Alec!...

I will check our answer to make certain it works with my particular problem. I was not expecting an explicit solution from anyone!

I already knew that the inverse of  b(n) = sum of s(n,k) * d(k) from k=0 to j where s(n,k) = the Stirling number of the Second Kind

is d(k) = sum of S(k,n) from n=0 to j where S(k,n) = the Stirling number of the First Kind. This relates to my problem if one defines

b(n) = sum of k^n * c(k) from k=1 to j.  Then d(k) = k! * sum of c(j)*(j-choose-k) from k=0 to j. Then another relabelling brought us close to, but not exactly, to my original problem.  I sensed my problem was not quite one of any of the "textbook" answers.

Anyway, this is why I am still a skeptic, until I check your answer! :)

## Wow! Thank you, Alec!...

I will check our answer to make certain it works with my particular problem. I was not expecting an explicit solution from anyone!

I already knew that the inverse of  b(n) = sum of s(n,k) * d(k) from k=0 to j where s(n,k) = the Stirling number of the Second Kind

is d(k) = sum of S(k,n) from n=0 to j where S(k,n) = the Stirling number of the First Kind. This relates to my problem if one defines

b(n) = sum of k^n * c(k) from k=1 to j.  Then d(k) = k! * sum of c(j)*(j-choose-k) from k=0 to j. Then another relabelling brought us close to, but not exactly, to my original problem.  I sensed my problem was not quite one of any of the "textbook" answers.

Anyway, this is why I am still a skeptic, until I check your answer! :)

## Clean it up first...

Why not start by cleaning it up a little first?

If I read this expression's parentheses correctly, you have

y = ( 2522 * (1 - exp(-x)) / (18*x+19) ) / ( ( 1- exp(-x))/x  + (2522/3)/(18*x+19) - 1 )

Then use the Lagrange Inversion Formula, which I thought Maple has built in (after checking analyticity at (x,y)=(0,0), if necessary making the proper transformation first).

## Clean it up first...

Why not start by cleaning it up a little first?

If I read this expression's parentheses correctly, you have

y = ( 2522 * (1 - exp(-x)) / (18*x+19) ) / ( ( 1- exp(-x))/x  + (2522/3)/(18*x+19) - 1 )

Then use the Lagrange Inversion Formula, which I thought Maple has built in (after checking analyticity at (x,y)=(0,0), if necessary making the proper transformation first).

## Thanks, again....

Thank you, again, for sticking with me on this thread, Robert Israel.

I will have to try an interlibrary loan with my local county library to take out Rudin's book (I am not employed at a university or anywhere).

Basically, what I do not understand is:

how do the countably infinitely many solutions, z=z(k), of z^a + c*z = w "match up with" the countably infinitely many solutions, z=z(k'), of z^a = w ?

In other words, for which k and k' does z(k) -> z(k') as c->0 ?

where you and I have defined z(k') above, as exp( (ln(w) + 2*Pi*i*k')/a), but I have not yet successfully defined z(k), but have considered ways of defining z(k) involving limits and combinations of solutions of z^p + c*z = w for rational numbers, p, since I consider all M roots of z^p + c*z =w known, where p=N/M, gcd(M,N)=1, integers M & N.

## Thanks, again....

Thank you, again, for sticking with me on this thread, Robert Israel.

I will have to try an interlibrary loan with my local county library to take out Rudin's book (I am not employed at a university or anywhere).

Basically, what I do not understand is:

how do the countably infinitely many solutions, z=z(k), of z^a + c*z = w "match up with" the countably infinitely many solutions, z=z(k'), of z^a = w ?

In other words, for which k and k' does z(k) -> z(k') as c->0 ?

where you and I have defined z(k') above, as exp( (ln(w) + 2*Pi*i*k')/a), but I have not yet successfully defined z(k), but have considered ways of defining z(k) involving limits and combinations of solutions of z^p + c*z = w for rational numbers, p, since I consider all M roots of z^p + c*z =w known, where p=N/M, gcd(M,N)=1, integers M & N.

## Searching references...

I emailed the author of the arxiv paper you mentioned. I have no idea if they will email me back. Nevertheless, I need to know soon whether they have fully developed their method or if their ideas are mere sketches, like mine are. That is why I pushed so hard this summer with Maple to crank out at least one fully computed example, namely, to the Abel-type ODE dy/dx = g(x)*y^m + h(x)*y^n. I need to know now if I am re-inventing the wheel, or if there is sufficient difference in our methods to warrant me presenting my ideas to the Differential Algebra & Related Topics workshop in Newark, NJ this year in November.

Even if my work is not original (but independently discovered), there still remains the task of computerizing it and also relating these formal solutions to my work in differential resolvents. I am guessing that these Russians' formal solutions are in the form of power series. If I can connect them to my work on resolvents, I wonder about getting the formal solutions more neatly/compactly in terms of finite Picard-Vessiot equations, including quadratures.

## Searching references...

I emailed the author of the arxiv paper you mentioned. I have no idea if they will email me back. Nevertheless, I need to know soon whether they have fully developed their method or if their ideas are mere sketches, like mine are. That is why I pushed so hard this summer with Maple to crank out at least one fully computed example, namely, to the Abel-type ODE dy/dx = g(x)*y^m + h(x)*y^n. I need to know now if I am re-inventing the wheel, or if there is sufficient difference in our methods to warrant me presenting my ideas to the Differential Algebra & Related Topics workshop in Newark, NJ this year in November.

Even if my work is not original (but independently discovered), there still remains the task of computerizing it and also relating these formal solutions to my work in differential resolvents. I am guessing that these Russians' formal solutions are in the form of power series. If I can connect them to my work on resolvents, I wonder about getting the formal solutions more neatly/compactly in terms of finite Picard-Vessiot equations, including quadratures.

## Exactly! And now.....

Exactly, Joe Riel! Thank you!
And now, one more level of complication: I need to make the vars list itself be of indefinite lengths.  Plus, I have a lot of R's to compute, so I'll have to put your code into its own loop.  More specifically, I am computing c[i,j,k,l] for all i,j,k,l in terms of (using my recursion, as linear combinations of) just those of the form c[0,j',0,l'] and c[i',0,k',0] where i',j',k',l' run from 0 to about 2*i,2*j,k,l. Hence, the number of variables c[i',0,k',0] and c[0,j',0,l'] needed increases with i,j,k,l.  I need to compute a sufficient number of c[i,j,k,l], as I mentioned earlier, about 0<=i,j,k,l<=5, to visualize the pattern and hopefully guess a general formula.

## Thanks to all for your help: next proble...

My next problem is to COLLECT all the like powers of c[0,j,0,l] and c[i,0,k,0]. Actually, there are no powers. My output (all it R) is a LINEAR combination of assorted TABLE elements c[0,j,0,l] and c[i,0,k,0] for i,j,k,l in the range 1 through 5. 5 is about the largest range I can choose before my computer completely hangs up. It has to set up storage space for 5^4 c[i,0,k,0]'s and c[0,j,0,l]'s.
R = Linear combinations of these c[ ] elements over polynomials in indeterminate table elements d[] and e[] and rational in d[0].
For example, I now need to turn R:= (1/d[0])*(c[0,1,0,2] - c[3,0,4,0]) + (5*c[0,1,0,2] - e[1]*c[3,0,4,0]) into (1/d[0] + 5)*c[0,1,0,2] + (-1/d[0] - e[1])*c[3,0,4,0].

## Thank you, Edgar!...

THANK you for recreating the error message that I get all the time.
The "evaluates to 6=1..5" thing.

## "Resetting" a counter loop...

# Desired output 1, 2, 3 s[i]
# Actual output: 1, 2, 3, s[4]

for i fro 1 to 3 do i end do;
s[i];

I have had more complicated examples than this,
such as when the counter loop is used again as a counter loop
in which I have had to "reset" the "i". I can't dig those up now.
(They're too long to type in.)

## Need to clear loop counters...

One does indeed need to clear variables used as loop counters. I have experimented with that and proven that is necessary many times.

Thank you for the evaln function. I will try it.

There is no way I would have been able to look in the Help menu for the "evaln" function unless I already knew that it was able to clear array elements.

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