| (1) Morale
> this must have been too obvious to the author, as they dont mention
> slightest clue how it is so. which makes me feel even more depressed,
> never mind other worst feelings..
Hey, don't beat yourself up! I know how easy it is to take it
all personally, but really You're OK. We've all been there!
(2) As simple as A,B,C=(B�/A)...:-)
> any one could say why this series has this closed form?
> oo
> ----
> \ j(2*Pi/6) -1 n
> / { e Z }
> ---
> n=0
> 1
> = -----------------------
> j(2*Pi/6) -1
> 1- e Z
>
>
> where j= sqrt(-1)
> Z is general complex number (a+jb)
OK, let's write u = e^(2*j*PI/6)/z. Then you want
sum(n=0...oo) u^n.
= lim N->oo sum(n=0...N) u^n [*]
= lim N->oo (1-u^(N+1))/(1-u) [multiply * by 1-u to see this]
[assuming u <> 1 for the moment]
= 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, I hope you can see that this
term will go to zero. If |u| > 1, then the term will go to infinity and
so your series doesn't converge.
Lastly, let's look at what happens when |u| = 1. If u = 1, then
the series diverges and we could never get the expression * above. For
other u, we can express u = exp^(2*j*PI*�), where � lies in (0,1). u^N
does not converge to 0, it just whizzes round with modulus always 1, so there
is no way that your series could converge to anything finite.
This is all actually a very common situation. We say that your series
has a *radius of convergence*, which equals 1 in this case, and is here
centred on the origin. While u is inside the circle, the series converges to
1/1-u. While u is outside, the series diverges to oo. What happens *on*
the circumference can vary, and in this case it's always divergent.
If you substitute back for u:
|u| = |e^(2*j*PI/6)/z|
= |e^(2*j*PI/6)|/|z|
= 1/|z|
So, your series converges not for all z, but for those z such that
|z| > 1, ie *outside* the unit circle. "general z" don't work.
I hope this is helpful...
Regards,
Andrew.
|
| 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
|
| 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
|