SandorSzabo

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These are answers submitted by SandorSzabo

Which version do you use?

Sandor

(Maple)

 

(Maple)

Thanks for detailed explanation. Unfortunately Maple says inner is undocumented protected name. So the question arises, what does inner do with matrices and vectors?

.

The "solution" is that Maple not says anything the x-range of the hypergeometric identity in the answer. It is very annoying, working with identities myself have to guess the range or to browse the help or search on the net.

Sorry for disturbing.

                                    Sandor

Hi Robert,

Something is wrong with your solution, coeff(Q,diff(w(x),x$2))  works well, but

coeff(Q,diff(w(x),x$1)) or coeff(Q,diff(w(x),x)) not. I absolutely don't understand why.

jakubi

Luckily, jakubi's solution works well. In fact I don't understand every detail, but very strange for me, that in Help there is  coeff - extract a coefficient of a polynomial.

I suspect that in jacubi's solution dop is somehow a polynomial differential operator that is the reason that Maple can apply coeff.

Thanks,  Sandor

 

Earlier I never used table and tried to follow your suggestion but it didn't succeed.

I have no idea at all to solve (without cut, copy, paste by hand) the problem.

Sandor

Using the composition operator @, or @@ the repeated composition operator, you can use/build more complex structure, if I don't misunderstand your second question.

Hmm i'm not sure if this is allowed: take the sequence and replace it by a continuous function

restart:

f:=2^x/x;

int(f,x);

asympt(%,x,2);

convert(%,polynom);

 

on a sidenote, strange that Maple considers the last term as a polynom, but it works to remove the O-term.

I don't know well enough hypergeometric functions, but I know there are a lot of identities.

So my question is:  what identities have to apply  to prove that

 

[;\frac{253}{108} hypergeom\left(\left[ -\frac{1}{2},-\frac{1}{6},\frac{1}{6}\right], \left[\frac{4}{3},\frac{5}{3}\right],-\frac{1}{729\right);][;-\frac{911}{6298560} hypergeom\left(\left[\frac{1}{2},\frac{5}{6},\frac{7}{6}\right],\left[\frac{7}{3},\frac{8}{3}\right],-\frac{1}{729}\right);]

[;+\frac{73}{9795520512} hypergeom\left(\left[ \frac{3}{2},\frac{11}{6},\frac{13}{6}\right],\left[\frac{10}{3},\frac{11}{3}\right],-\frac{1}{729}\right);][;-\frac{2}{27};] [;= ;] [;\frac{2}{27} (10\sqrt{10}-1);]  ?

I'm interested in by-hand or/and  by-Maple proof,  because I'm sure there exists a general identity involving hypergeom.

Maybe this identity is new discovered by Maple ;-)

Sandor

Thanks for your answers.

Joe: the problem is not that the result of Maple is "approximately" equal or not by the correct result.

jakubi: your solution is correct.

However I still don't understand the Maple's result.

You writes

VectorCalculus:-ArcLength(<x,(x/2)^(2/3)>, x=0..u) assuming u>0;

So the problem is, the number 2>0 so why Maple gives me hypergeom?

I tried to convert hypergeom into elementary but it doesn't help.

Sandor

 

 

Sorry, I made a misprint.

Since I can't  type math2d I type it by hand.

[;simplify\left( ArcLength \left( \left< x,\left( \frac{x}{2} \right)^{\frac{2}{3}} \right>,x=0..2\right)\right);]

The correct answer is  [;\frac{2}{27} (10\sqrt{10}-1);] (by elementary calculations).

Maple's answer is

[;\frac{253}{108} hypergeom\left(\left[ -\frac{1}{2},-\frac{1}{6},\frac{1}{6}\right], \left[\frac{4}{3},\frac{5}{3}\right],-\frac{1}{729\right);][;-\frac{911}{6298560} hypergeom\left(\left[\frac{1}{2},\frac{5}{6},\frac{7}{6}\right],\left[\frac{7}{3},\frac{8}{3}\right],-\frac{1}{729}\right);]

[;+\frac{73}{9795520512} hypergeom\left(\left[ \frac{3}{2},\frac{11}{6},\frac{13}{6}\right],\left[\frac{10}{3},\frac{11}{3}\right],-\frac{1}{729}\right);][;-\frac{2}{27};].

How could derive this result the Maple?

Sandor

 

I haven't found exact formula.

I don't know what is your assumption about  w, small, or large positive numbers.

Hopefully, using the following calculation you will be able to derive a useful asymptotic formula. If not, ask our, and we try.

Earlier I knew only the simple regularization. Thanking to jakubi now I know the double version. Using his idea I was able (hopefully)

to give a correct derivation. Download 3807_partitionfunction.pdf
View file details

Sandor

See unapply in the help.

I hope it helps.

Sandor

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