Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

princess snowwhite
2012-11-07 18:54:34

Thank you all. You have been most helpful. Mitu

noelevans
2012-11-07 17:35:49

Hi!

I assume that we are dealing with the PRINCIPAL roots of -1 (when k=0) since for each n there are n
distinct roots of -1 equally spaced about the unit circle.  Fistfiz's example using the clock gives a good
illustration of that sequence progressing counterclockwise from e^ipi to 1 around the top of the circle.

smile

bobbym
2012-11-07 08:51:51

Hi;

or use the composition law for limits to treat it according to this rule:





Fistfiz
2012-11-07 07:28:14

noelevans wrote:

How does this look? :0)
                        i*180                    i*(180/n)                                  i0
(-1)^(1/n) = (1*e       )^(1/n) = 1*e              so this approaches 1*e   = 1 as n goes to infinity.

(The angles are in degrees.)

I have to admit that at first sight this looked funny; but after being (maybe) less superficial i'm seeing a meaning behind this:
look it geometrically (i write polar coordinates for complex numbers)...

the (first) square root for -1 is    (1,pi/2)       (midnight)
the (first) 3rd root for  -1        (1,pi/3)          (one o'clock)
the (first) 4th root for -1 is    (1,pi/4)            (half past one)
.....
.....                                                             (...some time passes...)
.....
the (first) nth root for -1 tends to (1,0)         (almost three o' clock)

so it seems to me that your limit is what the first nth root of (-1) tends to.


EDIT: I want to add something:

where k=0,1,2...,n-1. In particular, the integer part of (n+1)/2 (which is n/2 if n is even and (n+1)/2 if odd) belongs to the list of k's;
If we accept your and my proceeding then we get:



(where i put n/2 or n+1/2 as k)

so one of us (or eventually both dizzy) must be wrong.

noelevans
2012-11-07 07:05:47

How does this look? :0)
                        i*180                    i*(180/n)                                  i0
(-1)^(1/n) = (1*e       )^(1/n) = 1*e              so this approaches 1*e   = 1 as n goes to infinity.

(The angles are in degrees.)

Fistfiz
2012-11-07 06:49:10

anonimnystefy wrote:

What do you mean by a succession from N to C?




You see that, for example

anonimnystefy
2012-11-07 04:00:00

What do you mean by a succession from N to C?

Fistfiz
2012-11-07 03:55:47

hi mitu, I would say it does not exist if your succession is from N to R;
here's a short proof of non-existence:
if LIM[a(n)]=L, then for all a(n(k)) LIM[a(n(k))]=L
you see that LIM[a(2k)]!=L since a(2k) is not defined for each k. But maybe someone would argue that for each n in dom(a(n)) a(n)=-1, so LIMa(n)=-1... i see it just as a formal problem, maybe someone can be more precise.

While writing my post i realized that if your succession is from N to C it is not even a function, so i don't know if it has any meaning to talk about limit...

anonimnystefy
2012-11-07 01:05:34

I would say it is either 1 or non-existent...

mitu
2012-11-06 20:45:47

what is the limit of (-1)^(1/n) when n tends to infinity? does it exist?

Board footer

Powered by FluxBB