@Joe Riel 

Thank you!

I've changed the case there. My both procedures there, ordered and A049525 could be, certainly, written better - but they work fast enough up to the point where the sequence calculation makes sense, so I didn't even try to improve them.

Alec

@Joe Riel 

Joe, it is not that long, and it is quite useful. Perhaps, you could post it here, or in a separate post (or in Maplesoft blog)?

It would have more chances to survive that way.

Alec

@Alejandro Jakubi 

That's great! I've just used it to construct a sequence in the OEIS.

Alec

I think, you typed typsetting instead of typesetting.

In a conditional statement, that can be done without is, just as

if interface(typesetting)=standard then

Alec

I think, you typed typsetting instead of typesetting.

In a conditional statement, that can be done without is, just as

if interface(typesetting)=standard then

Alec

As the title says. It returns standard if it is standard, or extended if it is extended.

Alec

As the title says. It returns standard if it is standard, or extended if it is extended.

Alec

@hirnyk 

I am sure that using Maple for that is not sufficient, not necessary, and also not desirable.

The asymptotics themselves are not enough for the convergence proof - one has to prove that the difference between the series of original terms and the series of their asymptotics converges, too. For example, the series sum((-1)^k*f(k),k=1..infinity) is divergent with the following f,

f:=k->k^(-1/2)+(-1)^k/k:
asympt(f(k),k,0);

                                    1/2
                             O((1/k)   )

but the main term of the asymptotic is the same as in this example (well, with a coefficient 1/2 which could be obtained by division of f(k) by 2).

Two terms may be good though, because the sum of the second terms of the asymptotic in the original example converges (absolutely), but the sum that you used is already a deformed one - it misses floor and ceil.

sum(1/n,n=1..exp(1))=add(1/n,n=1..floor(exp(1)));

                    Psi(exp(1) + 1) + gamma = 3/2

evalf(%);

                      1.750021380 = 1.500000000

Again, one doesn't need Maple to find the asymptotic that you cited (which actually, is too long for that purpose - 2 terms would be enough, or even one term and a correct O-term,)

asympt(sum(1/n , n = exp(sqrt(k)) .. exp(sqrt(k+1))), k,10); 

       1/2        3/2        5/2          7/2          9/2
  (1/k)      (1/k)      (1/k)      5 (1/k)      7 (1/k)
  -------- - -------- + -------- - ---------- + ----------
     2          8          16         128          256

                   11/2           13/2            15/2
           21 (1/k)       33 (1/k)       429 (1/k)
         - ------------ + ------------ - -------------
               1024           2048           32768

                  (17/2)
         + O((1/k)      )

Compare that with

asympt(sqrt(k+1)-sqrt(k),k,8);

       1/2        3/2        5/2          7/2          9/2
  (1/k)      (1/k)      (1/k)      5 (1/k)      7 (1/k)
  -------- - -------- + -------- - ---------- + ----------
     2          8          16         128          256

                   11/2           13/2            15/2
           21 (1/k)       33 (1/k)       429 (1/k)
         - ------------ + ------------ - -------------
               1024           2048           32768

                  (17/2)
         + O((1/k)      )

which is just a binomial series minus 1 and multiplied by sqrt(k), after rewriting sqrt(k+1)-sqrt(k) as sqrt(k)*(sqrt(1+1/k)-1).

Maple finds that asymptotic the same way as I did - it expresses the harmonic sum through Psi, then the main term of the asymptotic of Psi is ln, and other terms don't matter because they become exponentially small if the limits of summation are exponential,

Alec

@hirnyk 

I am sure that using Maple for that is not sufficient, not necessary, and also not desirable.

The asymptotics themselves are not enough for the convergence proof - one has to prove that the difference between the series of original terms and the series of their asymptotics converges, too. For example, the series sum((-1)^k*f(k),k=1..infinity) is divergent with the following f,

f:=k->k^(-1/2)+(-1)^k/k:
asympt(f(k),k,0);

                                    1/2
                             O((1/k)   )

but the main term of the asymptotic is the same as in this example (well, with a coefficient 1/2 which could be obtained by division of f(k) by 2).

Two terms may be good though, because the sum of the second terms of the asymptotic in the original example converges (absolutely), but the sum that you used is already a deformed one - it misses floor and ceil.

sum(1/n,n=1..exp(1))=add(1/n,n=1..floor(exp(1)));

                    Psi(exp(1) + 1) + gamma = 3/2

evalf(%);

                      1.750021380 = 1.500000000

Again, one doesn't need Maple to find the asymptotic that you cited (which actually, is too long for that purpose - 2 terms would be enough, or even one term and a correct O-term,)

asympt(sum(1/n , n = exp(sqrt(k)) .. exp(sqrt(k+1))), k,10); 

       1/2        3/2        5/2          7/2          9/2
  (1/k)      (1/k)      (1/k)      5 (1/k)      7 (1/k)
  -------- - -------- + -------- - ---------- + ----------
     2          8          16         128          256

                   11/2           13/2            15/2
           21 (1/k)       33 (1/k)       429 (1/k)
         - ------------ + ------------ - -------------
               1024           2048           32768

                  (17/2)
         + O((1/k)      )

Compare that with

asympt(sqrt(k+1)-sqrt(k),k,8);

       1/2        3/2        5/2          7/2          9/2
  (1/k)      (1/k)      (1/k)      5 (1/k)      7 (1/k)
  -------- - -------- + -------- - ---------- + ----------
     2          8          16         128          256

                   11/2           13/2            15/2
           21 (1/k)       33 (1/k)       429 (1/k)
         - ------------ + ------------ - -------------
               1024           2048           32768

                  (17/2)
         + O((1/k)      )

which is just a binomial series minus 1 and multiplied by sqrt(k), after rewriting sqrt(k+1)-sqrt(k) as sqrt(k)*(sqrt(1+1/k)-1).

Maple finds that asymptotic the same way as I did - it expresses the harmonic sum through Psi, then the main term of the asymptotic of Psi is ln, and other terms don't matter because they become exponentially small if the limits of summation are exponential,

Alec

f1 is not a solution of motion, so it is not surprising that Maple can't find it.

If you substitute f1 in the lhs of motion[], you get

pdetest(h(x1,x2)=f1,motion);
           2   2        2            2  2
  {- (12 x1  x2  + 12 x1  x2 c + 3 x1  c  - 8 x1 x2 - 4 c x1

                                                               2   2
         + 8 ln(2) - 16 ln(2) x1 x2 - 8 ln(2) c x1 + 8 ln(2) x1  x2

                     2                  2  2
         + 8 ln(2) x1  x2 c + 2 ln(2) x1  c  - 8 %1 + 16 %1 x1 x2

                              2   2          2               2  2
         + 8 %1 c x1 - 8 %1 x1  x2  - 8 %1 x1  x2 c - 2 %1 x1  c

               3          3         4    /                       2
         + 8 x1  x2 + 4 x1  c + 4 x1 )  /  ((-2 + 2 x1 x2 + c x1)
                                       /

          3
        x1 )}

  %1 := ln(2 - 2 x1 x2 - c x1)

which is not 0,

eval(%,[x1=1,x2=1]);
                             2            2             2
              16 + 12 c + 3 c  + 2 ln(2) c  - 2 ln(-c) c
           {- -------------------------------------------}
                                   2
                                  c

Alec

f1 is not a solution of motion, so it is not surprising that Maple can't find it.

If you substitute f1 in the lhs of motion[], you get

pdetest(h(x1,x2)=f1,motion);
           2   2        2            2  2
  {- (12 x1  x2  + 12 x1  x2 c + 3 x1  c  - 8 x1 x2 - 4 c x1

                                                               2   2
         + 8 ln(2) - 16 ln(2) x1 x2 - 8 ln(2) c x1 + 8 ln(2) x1  x2

                     2                  2  2
         + 8 ln(2) x1  x2 c + 2 ln(2) x1  c  - 8 %1 + 16 %1 x1 x2

                              2   2          2               2  2
         + 8 %1 c x1 - 8 %1 x1  x2  - 8 %1 x1  x2 c - 2 %1 x1  c

               3          3         4    /                       2
         + 8 x1  x2 + 4 x1  c + 4 x1 )  /  ((-2 + 2 x1 x2 + c x1)
                                       /

          3
        x1 )}

  %1 := ln(2 - 2 x1 x2 - c x1)

which is not 0,

eval(%,[x1=1,x2=1]);
                             2            2             2
              16 + 12 c + 3 c  + 2 ln(2) c  - 2 ln(-c) c
           {- -------------------------------------------}
                                   2
                                  c

Alec

@hirnyk 

Why make things more complicated than they are?

I think that one doesn't need Maple to evaluate ln(exp(A))=A for a positive A and to find the asymptotic of sqrt(k+1)−sqrt(k) by putting sqrt(k) out and using the series for sqrt(1+1/k) = 1 + 1/2/k + ... which one actually doesn't need here because the decreasing follows from the derivative calculation (done mentally) and limit 0 follows from rewriting sqrt(k+1)−sqrt(k) as 1/(sqrt(k+1)+sqrt(k)), which could be also used for the decreasing proof instead of a derivative.

Using Maple for proofs is dangerous - I posted recently on this site Maple proofs that 0 = 1, that Pi ≥ 11 and e ≥ 200,000 (assuming that they are positive) and that Psi(t) = −∞ for positive t, generalizing Preben's Alsholm's example of a divergent integral which Maple evaluates to gamma.

Alec

@hirnyk 

Why make things more complicated than they are?

I think that one doesn't need Maple to evaluate ln(exp(A))=A for a positive A and to find the asymptotic of sqrt(k+1)−sqrt(k) by putting sqrt(k) out and using the series for sqrt(1+1/k) = 1 + 1/2/k + ... which one actually doesn't need here because the decreasing follows from the derivative calculation (done mentally) and limit 0 follows from rewriting sqrt(k+1)−sqrt(k) as 1/(sqrt(k+1)+sqrt(k)), which could be also used for the decreasing proof instead of a derivative.

Using Maple for proofs is dangerous - I posted recently on this site Maple proofs that 0 = 1, that Pi ≥ 11 and e ≥ 200,000 (assuming that they are positive) and that Psi(t) = −∞ for positive t, generalizing Preben's Alsholm's example of a divergent integral which Maple evaluates to gamma.

Alec

@hirnyk 

I don't exactly understand you. What generalities?

It is a proof. No code is needed.

Alec

@hirnyk 

I don't exactly understand you. What generalities?

It is a proof. No code is needed.

Alec