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        riddles >> medium >> 1 - 1 + 1 - 1 + 1 - 1 ...
        
(Message started by: william wu on Oct 29^th, 2003, 12:42am)

Title: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by william wu on Oct 29^th, 2003, 12:42am

What is 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 ... ?

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by towr on Oct 29^th, 2003, 1:41am

::[hide]undefined[/hide]::

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by Sir Col on Oct 29^th, 2003, 2:50am

I know that's the 'answer', but why isn't it [hide]0 or 1[/hide]? For example, [hide]the answer to, |x–0.5|=0.5, is x=0,1, but we wouldn't say it is undefined[/hide].

In fact, could we reason as follows...
::[hide]
Let S=1–1+1–1+...
-S=–1+1–1+1–...

Adding both sides, 0=?, and the only way for this to remain consistent is if the RHS is also zero. Hence there are as many +1's as -1's; so the sum, S=0.
[/hide]::

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by towr on Oct 29^th, 2003, 3:20am

on 10/29/03 at 02:50:32, Sir Col wrote:

but why isn't it <hidden>

::[hide]infinity is neither odd, nor even, nor both.. so while for any finite n number of terms it is 1 when n is odd, and 0 when n is even, you can't say what it is 'at infinity'. Taking 0 or 1 or the average, 0.5 is just as (un)reasonable as taking 0.1, or 0.7 or sqrt(1/2).[/hide]::

Quote:

In fact, could we reason as follows...

we could, but ::[hide]
If S-S =0, that doesn't mean S=0
besides S is also 1 + (-S), so 2S = 1 which means S = 0.5.
But we also have S = 2S, since S = 1 -1 + 1 - 1 + 1 -1 + 1 - 1 + ... = 1 + 1 - 1 - 1 + 1 + 1 - 1 - 1 + ... = 2*1 +2*-1 + 2*1 + 2*-1 ... = 2S
So S = 0
And let's not forget S = 1 + S, since 1 -1 +1 - 1 +1 = 1 + 1 - 1 + 1 - 1 ...
You can push any number of +1's, or -1's to the front if you want, as long as it's a finite number.. The only problem is that with an infinite sum, the same number of -1's or 1's respectively 'drop off at the other end' (you'll never get to add them back to the sum)[/hide]::

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by Dudidu on Oct 29^th, 2003, 3:36am

on 10/29/03 at 02:50:32, Sir Col wrote:

I know that's the 'answer', but why isn't it ...

The problem for my opinion is that
[hide]the [sum]_i=0[smiley=supinfty.gif] (-1)ⁱ does not converges :-/, thus the result is undefined !
[/hide]

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by william wu on Oct 29^th, 2003, 9:25am

Some mathematicians would argue that "in some sense", the sum converges to [hide]0.5[/hide] :o

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by Sir Col on Oct 29^th, 2003, 10:03am

Actually, I've just found a flaw in my reasoning (now there's a surprise):

Consider S=1–1+1=1, -S=–1+1–1. Obviously S+-S=0, but that tells us nothing about S.

[e]Edited to apologise to towr: Yep, you caught my nonesene post before I realised what a load of rubbish I was talking and 'corrected' it.[/e]

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by towr on Oct 29^th, 2003, 10:11am

[e]meh, you allready changed your post..
anyway, this is still a valid point I think..[/e]

It doesn't matter that there's a one to one mapping.
if you take s = 1 +1 -1 +1 +1 -1... there is also a one-to-one mapping here,
every 1 at position 3n+1 can be paired with a -1 at 6n,
and each 1 at 3n+2 has his companian -1 at 6n+3..
So there's equally many 1's and -1's , infinitely many.. Yet the sum clearly goes to infinity regardless..

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by Icarus on Oct 29^th, 2003, 7:00pm

on 10/29/03 at 09:25:46, william wu wrote:

Some mathematicians would argue that "in some sense", the sum converges to [hide]0.5[/hide] :o

Not just "in some sense", but rather "in the sense of Cesaro".

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by Dudidu on Oct 30^th, 2003, 12:26am

on 10/29/03 at 19:00:49, Icarus wrote:

Not just "in some sense", but rather "in the sense of Cesaro".

Icarus, just for general knowledge what is "the sense of Cesaro" ?
(A reference might be also helpful :P)

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by towr on Oct 30^th, 2003, 1:21am

I can't find anything on it at either mathworld, nor wikipedia. But it is caually mentioned at a few pages on the web

Basicly the Cesaro limit seem to be the limit of the average of the partial sums..
lim [ a₁ + a₂ + ... + a_n ]/n
Which is the same as the lim a_n, if the sequence a_n converges..

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by Icarus on Oct 30^th, 2003, 3:34pm

A sequence {a_n} is called "Cesaro summable" if the sequence of averages {1/n [sum]_k=1ⁿ a_k} converges.

As towr as said, if {a_n} converges, then it is Cesaro summable, and the Cesaro limit is the same as the ordinary limit. But some sequences (such as the sequence of partial sums of 1 - 1 + 1 - 1 + ...) are Cesaro summable even though they do not converge.

The Cesaro limit is the most common of many possible extensions of the concept of limit to non-convergent sequences.

These extensions are not really all that useful. Generally, if a sequence fails to converge in the normal sense, that is exactly the information you need from it. But occasionally the application requires a value to put out even though it does not require convergence. In this case, some extension is needed (which one depends on the application), and Cesaro's is the most likely to work (as might be expected).

Title: Re: 1 - 1 + 1 - 1 + 1 - 1 ...
Post by Dudidu on Oct 31^st, 2003, 11:23am

Icarus and towr,
Thank you... every day you learn something new ;).