Newsgroup sci.math 156292

Directory

In article <5961lq$m0d@srv4-poa.nutecnet.com.br>,
Gustavo Freitas Reis  wrote:
>mao@wuchem.wustl.edu (Dean Mao) wrote:
>
>>Is there a proof somewhere on the 'net proving that 0^0=1?
This question is discussed in the sci.math FAQ.
>0^0 is NOT 1, it is an indetermined expression. If you consider that
>0^x, for every x != 0, equals 0, and x^0, for every x != 0, equals 1,
>you realize that we can't determine the value for 0^0.
Except that the question does not mention anything about limits, but merely
concerns the value of 0^0.  For reasons mentioned in the FAQ, the value of
this expression is 1 (certainly if the operands are viewed as integers).
>The same thing happens to 0/0. What do you think: the answer is 0, 1
>or doesn't exist because of the "division by zero"? :) Any
>suggestions?
It's not the same at all.   0^0 has a value, but 0/0 does not.
-- 
Dave Seaman			dseaman@purdue.edu
      ++++ stop the execution of Mumia Abu-Jamal ++++
    ++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++

In article <59dvp9$lrk@news-central.tiac.net>, numtheor@tiac.net (Bob Silverman) writes:
> mab@dst17.wdl.loral.com (Mark A Biggar) wrote:
> 
>>What is the genreal closed form solution for the recurrence relation:
> 
>>X(n+1) = a*X(n) + b;
> 
> Let   alpha,Beta be the two roots   of   x^2 - ax - b = 0.
> Then  X(n) =  c1 alpha^n + c2 Beta^n,    where c1,c2 are given
> by the initial conditions.
> 
Is this correct?  The given recurrence is first order, hence should
only need one initial condition and therefore only one "arbitrary"
constant, c1.
Since it is a first order recurrence, the solution can be given quite
easily in terms of a and b:
X(n) = C a^n + b/(1-a), where C is a constant determined by an initial
condition (such as X(0) or X(1)).
Manley Perkel

In article <32B9AB32.383A@EM.AGR.CA>, JEANDITBAILLEULP@EM.AGR.CA wrote:
> Hi
> 
> Does any one know how to integrate the following function :
> a*exp(-b*exp(c*exp(d*t)))*dt
> where a, b, c and d are fixed parameters, t a variable and exp the
> natural exponation
The function is continuous, so the indefinite integral exists.
It may be computed numerically.
Unless one of the parameters a,b,c,d is zero, that indefinite
integral is not an elementary function.

Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) wrote:
>
>  The Successor axiom of Peano , as it stands by itself is none other
>than the very definition of a p-adic. Both are Series additions. In
>fact, you can replace the Successor (endless adding of ...0001) axiom
>with p-adics in the Peano axiom system. When you do this replacement
>you are left with some other axioms of which two of them contridict the
>p-adic series. The Mathematical Induction axiom contradicts the
>Successor axiom and the no predecessor to 0 axiom contradicts the
>Successor axiom.
>
You are right in one thing: We don't understand you. Let's analyze this
more precisely:
>
>  The Successor axiom of Peano , as it stands by itself is none other
>than the very definition of a p-adic. 
Let's rewrite what you write here: Anything that keeps to the Successor
axiom is a p-adic. Somehow I don't believe you...
>Both are Series additions.
Fair enough. There is an infinitude of Series additions, so that doesn't
surprise me in the least.
>In fact, you can replace the Successor (endless adding of ...0001) axiom
>with p-adics in the Peano axiom system.
'the Successor' is a function, p-adics are a kind of number. How can you
replace the one by the other? Probably you mean something else, but I'm in
the dark as to WHAT you mean.
>When you do this replacement
>you are left with some other axioms of which two of them contridict the
>p-adic series. 
Well, if you replace one axiom by something else, yes, then you have the
chance that the system becomes contradicting.
>The Mathematical Induction axiom contradicts the
>Successor axiom and the no predecessor to 0 axiom contradicts the
>Successor axiom.
>
No, they don't contradict the Successor axiom, but the definition of p-adics.
But that's because Peano's axioms were never meant to define the p-adics.
Andre Engels

"Capt. Nemo"  wrote:
>For flexibility and "user comfort", Matlab is very nice, and it's
>available on Windows. 
Does Matlab do symbolic manipulation, e.g. integration and
differentiation?
Regards,
-Karl
-Karl
+========================================+
| Karl F. Bloss, Senior Systems Engineer |
| Air Products & Chemicals, Inc.         |
| blosskf@apci.com                       |
| http://www.airproducts.com/            |
| < PGP encrypted mail preferred >       |
| key: http://www.enter.net/~bloss       |
+========================================+
   I speak for myself.  Public Affairs speaks for Air Products...

Might anyone know how to effectively compute the monodromy group of
an n-th order, linear, homogeneous, ODE with polynomial coefficients
in the independent variable?
What is the relation between the monodromy group of an ODE and the
differential Galois group?

I'm interest about the best sources to find out what has been
written about the kissing numbers?
During this Xmas time, please if possible, also via email.
Jorma Kyppo
Finland
jorma@jytko.jyu.fi