| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1510.1 | your forgot the :-) | BUZON::BELDIN_R | Pull us together, not apart | Thu Oct 24 1991 09:12 | 1 |
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| 1510.2 | | ZFC::deramo | Dan D'Eramo, zfc::deramo | Thu Oct 24 1991 17:59 | 3 |
| re .1, why do you believe that the base note is not serious?
Dan
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| 1510.3 | Its a good joke! | CORREO::BELDIN_R | Pull us together, not apart | Fri Oct 25 1991 17:24 | 17 |
| The aforesaid institute doesn't exist.
They don't do math.
Physicians don't do math.
The title of the paper is in English, not Portuguese.
Any operator which is wielded by a physician and is self-destructive
will only be used once, and if used on itself, negates its future
existence.
Pick a reason, any reason.
:-)
Dick
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| 1510.4 | re.3 | VAXSPO::WANDERLEY | | Thu Nov 21 1991 06:04 | 57 |
| >The aforesaid institute doesn't exist.
I think that the fact of you don't know something outside your
country, doesn't mean that it doesn't exist. You should read the news.
>They don't do math.
If you don't know the institute, how can you do what they do or
not?
>Physicians don't do math.
No comments. Its too jerk.
>The title of the paper is in English, not Portuguese.
I think the guy should be a little bit smart do know that I wrote
the title in English, just to make the things easy.
Let's call R$ the Self-Destructive Operator.
where: R$ * R$ = 0
R$ * N = N (N = any natural number)
Let's suppose the function:
2
S*(t) = 3-2t+4t that be operated with R$.
R$S*(t) = S*(t+R$)
2
R$S*(t) = 3-2(t+R$) + 4(t+R$)
2
R$S*(t) = 3-2t-R$2+4t + R$8t + 4 * R$R$
2
R$S*(t) = 3-2t+4t - R$2 + R$8t + 0
2
R$S*(t) = 3-2t+4t + R$(-2+8t)
2
R$S*(t) = 3-2t+4t - 2+8(t+R$)
2
R$S*(t) = 3-2t+4t - 2 + 8t + R$8
2
R$S*(t) = 3-2t+4t - 2 + 8t + 8
------- ------- ---
S(t) V(t) a(t)
That's it.
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| 1510.5 | | ZFC::deramo | Dan D'Eramo | Thu Nov 21 1991 11:56 | 19 |
| > 2
> R$S*(t) = 3-2t+4t - 2 + 8t + 8
> ------- ------- ---
> S(t) V(t) a(t)
Where V(t) is the derivative of S(t), and a(t) the derivative of V(t),
like in distance, velocity, and acceleration.
It seems you mean for R to change the polynomial p(t) into
the polynomial p(t + R), where you evaluate terms with an
R as R^n -> 0 for n>1, and terms*R*terms -> R*terms*terms
as another R of q(t) -> q(t + R).
For example, R(t^3) -> (t+R)^3 -> t^3 + 3t^2R + 3tR^2 + R^3 -> t^3 + R(3t^2)
and it already appears that in general we have R(p(t)) -> p(t) + R(p'(t))
-> p(t) + p'(t) + R(p''(t)) -> etc., stopping when the derivative
becomes a constant, as R(n) -> n.
Dan
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| 1510.6 | I still don't get it | PULPO::BELDIN_R | Pull us together, not apart | Thu Nov 21 1991 12:20 | 13 |
| > Let's call R$ the Self-Destructive Operator.
> where: R$ * R$ = 0
> R$ * N = N (N = any natural number)
I'm the one who thought it was a joke. I still have my doubts.
What kinds of things can R$ multiply? Reals, Complex numbers,
Functions, ...?
Your definition shows what it means when multiplied by itself (an operator)
and by a natural number. How can I get from that to the multiplications of
polynomials that you go on to describe? I
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| 1510.7 | on further study... | PULPO::BELDIN_R | Pull us together, not apart | Tue Nov 26 1991 12:29 | 10 |
|
On examining the self destructive operator, it becomes clear that it is not
associative since
(R$ * R$) * N = 0 and R$ * ( R$ * N) = R$ * N = N
I guess that's what is meant by self destruction. Does it have any
specific applications that anyone knows of?
Dick
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| 1510.8 | re.7 | VAXSPO::WANDERLEY | | Fri Nov 29 1991 06:32 | 146 |
| The self-destructive operator can be operated on so called
super-functions (any mathematical object wich can generate one or more
functions through a certain mathematical transformation).
To extract the information from the super-function we use the
self-destructive uncopling operator, wich presents the following
operational behavior:
R$F(x) = F(x + R$) (I)
R$ R$ = 0 (II)
R$ N = N (III)
R$ R$ = 1 (IV)
x y
So, we can use the self-destructive operator to:
- Super-functions
- Construction of Pascal's Triangle
- Trigonometrical Super-functions
- Construction of Derivatives and Integral's
- Resolution of Differential Equations
- Non-linear Differential Equation Solutions
- Partial Differential Equations
- Lagrange's Mechanics
- etc.
METHOD OF FLUXIONS - Newton (1966) - England
--------------------------------------------
2
Step 1 : S(t) = 3 + 2t + 4t
.
Step 2 : S(t) = S(t + T) - S(t)
---------------
T
.
Step 3 : S(t) = 2 + 8t + 4T
Step 4 : "T" being very small, it is negligible.
.
Step 5 : S(t) = 2 + 8t
DERIVATIVE THEORY - Cauchy (1823) - France
------------------------------------------
2
Step 1 : S(t) = 3 + 2t + 4t
S(t + T) - S(t)
Step 2 : S'(t) = lim ----------------
T->0 T
Step 3 : S'(t) = lim 2 + 8t + 4T
T->0
Step 4 : For the point tangent T->0, so
Step 5 : S'(t) = 2 + 8t
WEIERSTRASS EPSILON-DELTA METHOD - Heine (1872) - Germany
---------------------------------------------------------
2
Step 1 : S(t) = 3 + 2t + 4t
dS S(t+dT) - S(t)
Step 2 : --- = ------------------
dt dt
dS
Step 3 : -- = 2 + 8t + 4dt
dt
Step 4 : Given a positive real number very small
dS
Step 5 : -- = 2 + 8t
dt
NONSTANDARD ANALYSIS - Robinson (1966) - USA
--------------------------------------------
2
Step 1 : S(t) = 3 + 2t + 4t
dS S(t + dt) - S(t)
Step 2 : -- = St ----------------
dt dt
dS
Step 3 : -- = St 2 + 8t + 4dt
dt
Step 4 : Hiper Real = Standard + Nonstandard
dS
Step 5 : -- = 2 + 8t
dt
SELF-DESTRUCTIVE OPERATOR - A.P.Ricieri (1984) - Brazil
-------------------------------------------------------
2
Step 1 : S(t) = 3 + 2t + 4t
Step 2 : R$S(t) = S(t+R$)
2
Step 3 : R$S(t) = 3 + 2t + 4t + R$(2+8t) + 4R$R$
Step 4 : R$R$ = 0
Step 5 : R$R$(t) = S(t) + R$S'(t)
---------------#-------------
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| 1510.9 | | SSDEVO::LARY | Laughter & hope & a sock in the eye | Sun Dec 01 1991 01:47 | 11 |
| > R$F(x) = F(x + R$) (I)
>
> R$ R$ = 0 (II)
>
> R$ N = N (III)
>
> R$ R$ = 1 (IV)
> x y
I may be confused by the notation, but isn't (III) a special case of (I)
when f(x) = N?
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| 1510.10 | re .9 | VAXSPO::WANDERLEY | | Mon Dec 02 1991 06:18 | 1 |
| R$ N = N , where N is a Natural Number.
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| 1510.11 | order of operations? | CORREO::BELDIN_R | Pull us together, not apart | Tue Dec 03 1991 09:44 | 16 |
| re <<< Note 1510.10 by VAXSPO::WANDERLEY >>>
-< re .9 >-
> R$ N = N , where N is a Natural Number.
The problem that .9 had, and I still have, is that a statement like
R$ f(t) = ....
appears to be a general statement of which R$ N = N is a special case.
I get the impression that one must postpone evaluation of f() until all
applications of R$ have been resolved, but I see nothing that compels that
discipline.
Dick
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| 1510.12 | | ZFC::deramo | Dan D'Eramo, zfc::deramo | Tue Dec 03 1991 12:45 | 14 |
| >> > R$ N = N , where N is a Natural Number.
>>
>> The problem that .9 had, and I still have, is that a statement like
>>
>> R$ f(t) = ....
>>
>> appears to be a general statement of which R$ N = N is a special case.
No no :-) you are confusing, say, R$ 9 = 9, with the case of
`9':R->R given by `9'(t)=9, when R$ `9'(t) = `9'(t + R$) = 9
and so R$ `9' = `9'.
Dan
:-)
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| 1510.13 | back to .1? | CORREO::BELDIN_R | Pull us together, not apart | Tue Dec 03 1991 15:14 | 5 |
| re .12
Are you telling me my original reaction was appropriate?
Dick
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| 1510.14 | | ZFC::deramo | Dan D'Eramo | Tue Dec 03 1991 19:51 | 14 |
| >Are you telling me my original reaction was appropriate?
No, it's just that I thought it was neat that mathematicians
will use the same notation for a number as for a function.
In one sense I agree with .9 and .11, but in another I can
see how someone might consider R$ 9 = 9 not to have been
established by earlier rules that considered functions and
operators. But just push the symbols around in the "obvious"
way and you get the "right" results.
It would be interesting to see a rigorous development of
this.
Dan
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| 1510.15 | nullary functions. | CADSYS::COOPER | Topher Cooper | Wed Dec 04 1991 10:47 | 18 |
| RE: .14 (Dan)
I believe that the lambda calculus treats constants as nulary
(no-argument) functions. This provides a pretty clean
recursion/induction termination because an n-ary function is treated
recursively as a 1-ary function which processes the first argument and
returns an (n-1)-ary function. So "add(1 2)" is interpretted as
follows:
add(1(), 2())
(add(1()))(2())
add-1(2())
3()
And the interpretation stops at nullary functions, which are,
nevertheless functions with all the properties of functions.
Topher
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