limit

mitu · 2012-11-05 21:45:47

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

anonimnystefy · 2012-11-06 02:05:34

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

Fistfiz · 2012-11-06 04:55:47

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 · 2012-11-06 05:00:00

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

Fistfiz · 2012-11-06 07:49:10

anonimnystefy wrote:

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

You see that, for example

Last edited by Fistfiz (2012-11-06 07:51:38)

noelevans · 2012-11-06 08:05:47

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 · 2012-11-06 08:28:14

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 ) must be wrong.

Last edited by Fistfiz (2012-11-06 09:00:00)

bobbym · 2012-11-06 09:51:51

Hi;

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

noelevans · 2012-11-06 18:35:49

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.

princess snowwhite · 2012-11-06 19:54:34

Thank you all. You have been most helpful. Mitu

Last edited by princess snowwhite (2012-11-06 19:55:03)

Math Is Fun Forum

#1 2012-11-05 21:45:47

limit

#2 2012-11-06 02:05:34

Re: limit

#3 2012-11-06 04:55:47

Re: limit

#4 2012-11-06 05:00:00

Re: limit

#5 2012-11-06 07:49:10

Re: limit

#6 2012-11-06 08:05:47

Re: limit

#7 2012-11-06 08:28:14

Re: limit

#8 2012-11-06 09:51:51

Re: limit

#9 2012-11-06 18:35:49

Re: limit

#10 2012-11-06 19:54:34

Re: limit

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