| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1393.1 | hopefully helpful | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Thu Feb 28 1991 09:46 | 70 |
| 1393.2 | | ALLVAX::JROTH | The ships of state sail on mirage... | Thu Feb 28 1991 11:58 | 28 |
| This is a geometric series that collapses, much like compound
interest (where I first saw this as a kid).
A heuristic way to think about it is to formally perform long
division:
1 + z + z^2 + ...
____________________________
1-z ) 1
1-z
---
z
z-z^2
-----
z^2
.......
Breaking this off after n terms we have a remainder of z^n/(1-z)
so loosely speaking we might hope that if z is less than 1 this
remainder will approach zero.
The full theory of convergence Andrew mentions actually appeared centuries
after mathematicians had started playing with series in a formal way.
It was suspicion about the validity of these manipulations that brought
on the seemingly involved theorems we take for granted today.
- Jim
|
| 1393.3 | formal rules. | CADSYS::COOPER | Topher Cooper | Thu Feb 28 1991 13:47 | 41 |
| RE: .1 (Andrew)
Without disagreeing with your conclusions, I have a minor nit to pick
with your argument.
>= lim N->oo (1-u^(N+1))/(1-u) [...]
>
>= 1/1-u - lim N->oo (u^(N+1))/(1-u).
>
> Now the question is, what is the limit of this second term? It
>depends on the modulus of u. ... If |u| > 1, then the term will go to
>infinity and so your series doesn't converge.
You got from the first step above to the second by applying to
algebraic rulse for limits:
(1) lim x->a (f(x) - g(x)) = (lim x->a (f(x))) - (lim x->a (g(x)))
(2) lim x->a (f) = f when f is independent of x.
These are quite correct but there is a restriction on (1) -- it is
only applicable when both the limits on the right converge. It is
possible for them to fail to converge and yet the expression on the
right to converge. For example:
lim N->oo (0) = lim N->oo (N-N) = (lim N->oo (N)) - (lim N->oo (N))
obviously (by the second rule above) the limit on the left converges
while neither of the two on the right do. You cannot, therefore,
simply conclude from the non-convergence of the expression on the
right that the original expression will not converge.
You might be able to show that for the particular case of
h = lim x->a (f - g(x)) {x not in f}
f - g(x) converges iff g(x) converges,
but I haven't seen that anywhere.
As I said, just a nit.
Topher
|
| 1393.4 | | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Fri Mar 01 1991 11:57 | 4 |
| Sheesh, if I can't even take a constant out of a limit without being jumped on!
Regards,
Andrew.
|
| 1393.5 | Apology | CADSYS::COOPER | Topher Cooper | Fri Mar 01 1991 15:40 | 14 |
| RE: .4 (Andrew)
Sorry. My note came across as much more critical than I intendend --
written too quickly without having read it over. Of course you knew
what you were doing. What I was trying to do is to warn people that
what you did is not, *in general*, correct. The reason I thought
it was worth "nitting" you about is that it is a common error for
beginners -- they partition some perfectly reasonable, convergent
expression into non-convergent expressions and then give up on the
assumption that the original was not convergent. I wanted to put
in a "watch it" for the less experienced.
Topher
|
| 1393.6 | Other collapsing operations :-) | CHOVAX::YOUNG | Bulldozer & Buzzsaw Methodology | Sun Mar 03 1991 19:48 | 8 |
| Re .2:
> This is a geometric series that collapses, much like compound
> interest ...
and much like many of those institutions that deal in compund interest.
-- Barry
|