Re: [obm-l] Risch algorithm

["Luis Lopes" <llopes@xxxxxxxxxxxxx>]

Sauda,c~oes,

Não sabia como obter f(x)=x^{x+1}  do email abaixo.
Então escrevi pro prof. Rousseau novamente.
Como havia um engano na resposta dele, mando
este email somente para fazer o registro.

Para os que gostam da transformada de Laplace,
mais um exemplo do uso desta ferramenta.

Lamento a notação exótica.

===
Dear Luis:

It seems that  I made a mistake.  The result I get now seems
to be slightly different from the one that I quoted.  The way that
I took may be the long way around, but this is how I proceeded.
Start with the Laplace transform of t^n:

\int_0^{\infty} t^n e^{-st} dt = n!/s^{n+1}.

Replace n by n-1 and set s = n+1.  Thus

\int_0^{\infty} t^{n-1} e^{-(n+1)t} dt = \frac{(n-1)!}{(n+1)^n}.

Thus (formally) the series in question is given by

1 + \sum_{n=1}^{\infty} \frac{1}{(n+1)^n}
= 1 + \int_0^{\infty} \sum_{n=1}^{\infty} \frac{(t e^{-t})^{n-1}}{(n-1)!}
e^{-2t} dt  = 1 + \int_0^{\infty} exp(t exp(-t)) e^{-2t} dt.

Now set u = e^{-t} so the integral becomes

\int_0^{\infty} exp(t exp(-t)) e^{-t} dt = \int_1^0 \frac{1}{u^u} u (-du)
= \int_0^1 \frac{u du}{u^{u}}.

Thus the sum of the series is

1 + \int_0^1 \frac{u du}{u^u}.

This checks numerically using Maple.  It follows that there is an exact
formula for the sum of the series if and only if there is one for the
integral.  I'll stick by my conviction that this is highly unlikely.  I
haven't done it, but I believe that the Risch algorithm will show that the
antiderivative of u/u^u is not an elementary function.  This doesn't
complete the story since there are definite integrals that one can
evaluate even though you can't express the indefinite integral as
an elementary function (for example \int_0^{\infty} exp(-x^2) dx).

Cheers,

Cecil
===

[]'s
Luis

===
As for the other question, I would be exceedingly surprised if
the series in question has closed form sum.  Of course, one can
re-express the series sum as an integral; a quick calculation gives

\int_0^1 x^{x+1} dx,

and I am confident that one prove (using the Risch algorithm) that
x^{x+1} has no antiderivative in elementary terms.   While this
doesn't completely settle the issue, it comes close.
===

Para registrar, o problema 2 era

2) Calcule S = 1 / (1+n)^n =
= 1 + 1/2 + 1/3^2 + 1/4^3 + ....

Agora uma pergunta: alguém conhece esse algoritmo
de Risch? Nunca ouvi falar disso. E então aquela outra
soma que apareceu por aqui - S = \sum 1 / n^n  -
recentemente deve ter o mesmo tratamento e conclusão:
nada de forma fechada.

[]'s
Luís


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