Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: Sequence limit
Replies: 72   Last Post: Nov 26, 2013 12:07 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
quasi

Posts: 10,327
Registered: 7/15/05
Re: Sequence limit
Posted: Nov 9, 2013 6:29 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

quasi wrote:
>quasi wrote:
>>Bart Goddard wrote:
>>>Paul wrote:
>>>>Bart Goddard wrote:
>>>>> Timothy Murphy wrote:
>>>>> >
>>>>> > It's clear that the upper limit is 1,

>>
>>Rephrased:
>>
>> "It's clear that
>>
>> lim_sup {|sin(n)|^(1/n) | n in N} = 1."

>
>In meant:
>
> "It's clear that lim sup of the sequence
>
> |sin(n)|^(1/n), n = 1,2,3, ...
>
> is equal to 1."
>

>>I agree -- it's obvious, or at least elementary.
>>
>>The fact that the lim sup equals 1 can be easily shown without
>>invoking the stronger, more advanced result that the actual
>>limit exists and is equal to 1.
>>

>>>>> > since n pi mod Z will be distributed evenly in [0,1),
>>
>>Rephrased:
>>
>> "The set {(n*Pi) mod 1 | n in N} is dense in (0,1)."
>>

>>>>> > and so will infinitely often be in the range (1/3,2/3).
>>
>>Yes, true.
>>

>>>>> I don't think this is true.
>>
>>I think you misunderstand Tim's claim.
>>
>>Tim wasn't claiming that the sequence
>>
>> |sin(n)|^(1/n), n = 1,2,3, ...
>>
>>has infinitely many elements in the interval (1/3,2/3).
>>
>>That would be false since the limit is 1.
>>
>>Rather, he was claiming that the sequence
>>
>> (n*Pi) mod 1, n = 1,2,3, ...
>>
>>has infinitely many elements in the interval (1/3,2/3).
>>
>>That claim is true, and, as an easy consequence, it follows
>>that lim_sup |sin(n)|^(1/n) | n in N} = 1.

>
>I meant:
>
>lim sup of the sequence
>
> |sin(n)|^(1/n), n = 1,2,3, ...
>
>is equal to 1.
>

>>Thus, the lim sup is easy.
>>
>>By contrast, the limit is hard.
>>

>>>>>|Sin(n)| is probably distributed evenly,
>>
>>Yes, in the sense that the set
>>
>> {|sin(n)| : n in N}
>>
>>is dense in the interval (0,1).
>>

>>>>>but raising to (1/n) power is going to crowd things toward 0.
>>
>>No, it will crowd things toward 1.
>>

>>>> This sounds ambiguous to me.
>>>>
>>>> Do you mean that you don't think it's true that the
>>>> upper limit is 1.

>>
>>We're way past that.
>>
>>In this thread, it's already been established that the limit is
>>known to be equal to 1, so of course the lim sup is also 1.
>>

>>>or do you mean that you don't think it's true that it's clear
>>>that the upper limit is 1?
>>>
>>>Neither.
>>>
>>>Obviously the thing I don't think is true is the
>>>sentence to which I'm responding:
>>>
>>>That the values of |sin n|^(1/n) are evenly distributed.

>>
>>Tim never made that claim.
>>

>>>You could infer this by my comment about the (1/n) power
>>>pushing things toward zero.
>>>
>>>It has already been proven that the limit is 1.

>>
>>Right, but Tim is focusing just on the lim sup, using only
>>elementary reasoning.
>>
>>To me, his reasoning seems correct.
>>
>>But the argument can be made even simpler ...
>>
>>Let L = lim_sup {|sin(n)|^(1/n) | n in N}.

>
>I meant:
>
>Let L = lim sup of the sequence
>
> |sin(n)|^(1/n), n = 1,2,3, ...
>

>>We wish to show L = 1.
>>
>>It's clear that |sin(n)|^(1/n) < 1 for all n in N, hence L <= 1.
>>
>>lemma:
>>
>>For any two consecutive terms of the sequence
>>
>> |sin(n)|, n = 1,2,3, ...
>>
>>at least one of them exceeds 2/5.
>>
>>Proof:
>>
>>Suppose n is a positive integer such that |sin(n)| <= 2/5.
>>
>>We will show that |sin(n+1)| > 2/5.
>>
>>Firstly,
>>
>> => |sin(n)| <= 2/5
>>
>> => sin^2(n) <= 4/25
>>
>> => cos^2(n) >= 21/25
>>
>>Then
>>
>> sin^2(n + 1)
>>
>> = (sin(n)cos(1) + cos(n)sin(1))^2
>>
>> = sin^2(n)cos^2(1)
>> + cos^2(n)sin^2(1)
>> - 2sin(n)cos(n)sin(1)cos(1)


Of course that last line above should be

+ 2sin(n)cos(n)sin(1)cos(1)

but that correction doesn't affect what follows.

>> > cos^2(n)sin^2(1) - |2sin(n)cos(n)sin(1)cos(1)|
>>
>> >= (21/25)sin^2(1) - 2(2/5)(1)sin(1)cos(1)
>>
>> > 4/25
>>
>>Hence |sin(n+1)| > 2/5, as was to be shown.
>>
>>It follows that
>>
>> (2/5)^(1/n) < |sin(n)|^(1/n) < 1
>>
>>holds for infinitely many positive integers n.
>>
>>Since (2/5)^(1/n) -> 1 as n -> oo, it follows that L = 1.


quasi


Date Subject Author
10/3/13
Read Sequence limit
Bart Goddard
10/3/13
Read Re: Sequence limit
Karl-Olav Nyberg
10/3/13
Read Re: Sequence limit
quasi
10/3/13
Read Re: Sequence limit
quasi
10/3/13
Read Re: Sequence limit
Karl-Olav Nyberg
10/3/13
Read Re: Sequence limit
quasi
10/4/13
Read Re: Sequence limit
Roland Franzius
10/4/13
Read Re: Sequence limit
quasi
10/5/13
Read Re: Sequence limit
Roland Franzius
10/5/13
Read Re: Sequence limit
quasi
10/26/13
Read Re: Sequence limit
Roland Franzius
10/26/13
Read Re: Sequence limit
karl
10/26/13
Read Re: Sequence limit
Roland Franzius
10/26/13
Read Re: Sequence limit
gnasher729
10/27/13
Read Re: Sequence limit
karl
10/3/13
Read Re: Sequence limit
quasi
10/4/13
Read Re: Sequence limit
Leon Aigret
10/4/13
Read Re: Sequence limit
William Elliot
10/4/13
Read Re: Sequence limit
quasi
10/4/13
Read Re: Sequence limit
William Elliot
10/4/13
Read Re: Sequence limit
quasi
10/4/13
Read Re: Sequence limit
David C. Ullrich
10/4/13
Read Re: Sequence limit
Robin Chapman
10/5/13
Read Re: Sequence limit
Bart Goddard
10/4/13
Read Re: Sequence limit
GoogleOnly@mpClasses.com
10/4/13
Read Re: Sequence limit
Bart Goddard
10/4/13
Read Re: Sequence limit
GoogleOnly@mpClasses.com
10/4/13
Read Re: Sequence limit
Peter Percival
10/5/13
Read Re: Sequence limit
Virgil
10/4/13
Read Re: Sequence limit
Bart Goddard
10/6/13
Read Re: Sequence limit
David Bernier
10/6/13
Read Re: Sequence limit
Virgil
10/6/13
Read Re: Sequence limit
Bart Goddard
10/7/13
Read Re: Sequence limit
Mohan Pawar
10/7/13
Read Re: Sequence limit
Bart Goddard
10/7/13
Read Re: Sequence limit
gnasher729
10/7/13
Read Re: Sequence limit
Richard Tobin
10/7/13
Read Re: Sequence limit
Robin Chapman
10/7/13
Read Re: Sequence limit
Michael F. Stemper
10/7/13
Read Re: Sequence limit
Michael F. Stemper
10/7/13
Read Re: Sequence limit
David Bernier
10/7/13
Read Re: Sequence limit
fom
10/8/13
Read Re: Sequence limit
Virgil
10/8/13
Read Re: Sequence limit
fom
10/8/13
Read Re: Sequence limit
Virgil
10/8/13
Read Re: Sequence limit
fom
10/4/13
Read Re: Sequence limit
fom
10/4/13
Read Re: Sequence limit
quasi
10/4/13
Read Re: Sequence limit
quasi
10/9/13
Read Re: Sequence limit
Shmuel (Seymour J.) Metz
10/10/13
Read Re: Sequence limit
Bart Goddard
11/5/13
Read Re: Sequence limit
Shmuel (Seymour J.) Metz
11/6/13
Read Re: Sequence limit
Bart Goddard
11/11/13
Read Re: Sequence limit
Shmuel (Seymour J.) Metz
11/12/13
Read Re: Sequence limit
Bart Goddard
11/15/13
Read Re: Sequence limit
Shmuel (Seymour J.) Metz
11/15/13
Read Re: Sequence limit
Bart Goddard
11/6/13
Read Re: Sequence limit
Timothy Murphy
11/8/13
Read Re: Sequence limit
Bart Goddard
11/8/13
Read Re: Sequence limit
Paul
11/8/13
Read Re: Sequence limit
Bart Goddard
11/9/13
Read Re: Sequence limit
Paul
11/9/13
Read Re: Sequence limit
quasi
11/9/13
Read Re: Sequence limit
quasi
11/9/13
Read Re: Sequence limit
quasi
11/13/13
Read Re: Sequence limit
Timothy Murphy
11/13/13
Read Re: Sequence limit
quasi
11/14/13
Read Re: Sequence limit
Timothy Murphy
11/14/13
Read Re: Sequence limit
Virgil
11/14/13
Read Re: Sequence limit
Roland Franzius
11/26/13
Read Re: Sequence limit
Shmuel (Seymour J.) Metz
11/9/13
Read Re: Sequence limit
Roland Franzius
11/9/13
Read Re: Sequence limit
Paul

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.