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On Mon, 27 Jul 2015 18:47:22 +0100, Orchid Win7 v1 wrote:
> It's a little like... what is b^0.5? How do you multiply something by
> itself half a time?
Isn't that called a square root?
Jim
--
"I learned long ago, never to wrestle with a pig. You get dirty, and
besides, the pig likes it." - George Bernard Shaw
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>> But according to the wikipedia page below the -1/12 thing does have some
>> practical uses? I couldn't find any actual information about those
>> practical uses though.
>
> Well, as "practical" as the Riemann zeta function I guess...
Yes I suppose "complex analysis, quantum field theory, and string
theory" are all quite theoretical non-practical things (from an
Engineering point of view). I was hoping it would be something like
complex numbers, that do actually have real world proper practical uses
(like analysing AC circuits or mechanical vibrations).
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scott <sco### [at] scott com> wrote:
> Maybe I'm a bit late to the party here, probably because I'm an Engineer
> rather than a Mathematician, but this seemed a pretty crazy "proof" of
> what you get if you sum all the natural numbers up:
> s= 1+2+3+4+5+6+...
> 4s= 4+8+12+16+...
> (s-4s) = 1+2+3+4+5+ 6+...
> -4 -8 -12-...
> -3s = 1-2+3-4+5-6+...
> -3s-3s = 1-2+3-4+5-6+...
> +1-2+3-4+5-6+...
> -6s = 1-1+1-1+1-1+1-...
Which is equal to:
-6s = (1-1)+(1-1)+(1-1)+...
= 0+0+0+0+... = 0
s = 0/-6 = 0
Therefore:
1+2+3+4+5+6+... = 0
Crazy, huh?
--
- Warp
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>> -6s = 1-1+1-1+1-1+1-...
>
> Which is equal to:
>
> -6s = (1-1)+(1-1)+(1-1)+...
> = 0+0+0+0+... = 0
I'm no mathematician, but to do that you must make the assumption that
there are an even number of terms in the infinite sum (ie every +1 has a
-1 to pair with it). You could have assumed an odd number of terms and
got a sum of 1 instead.
Writing the sum equals 1 minus the sum seems to avoid the need to make
such an assumption.
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scott <sco### [at] scottcom> wrote:
> >> -6s = 1-1+1-1+1-1+1-...
> >
> > Which is equal to:
> >
> > -6s = (1-1)+(1-1)+(1-1)+...
> > = 0+0+0+0+... = 0
> I'm no mathematician, but to do that you must make the assumption that
> there are an even number of terms in the infinite sum (ie every +1 has a
> -1 to pair with it). You could have assumed an odd number of terms and
> got a sum of 1 instead.
> Writing the sum equals 1 minus the sum seems to avoid the need to make
> such an assumption.
An infinite sum can't have an "odd" or an "even" number of terms.
But you bring a good point. If you pair the elements differently, you get:
-6s = 1+(-1+1)+(-1+1)+(-1+1)+...
= 1+0+0+0+0+... = 1
Therefore s = -1/6.
In fact, you can get basically any integer value you want for -6s when
you group the elements appropriately.
This goes to show that when you are dealing with infinities, you can
"prove" anything you want.
I think the original "proof" is bogus.
The "proof" using Riemann's zeta function is also bogus in a sense.
Riemann's zeta function is the infinite sum of 1/n^s, but only for
values of s so that Real(s) > 1. (The sum would give 1+2+3+4+... when
s = -1, but the zeta function is not 1/n^s for values of s < 1. It's
something a lot more complicated.)
--
- Warp
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>> I'm no mathematician, but to do that you must make the assumption that
>> there are an even number of terms in the infinite sum (ie every +1 has a
>> -1 to pair with it). You could have assumed an odd number of terms and
>> got a sum of 1 instead.
>
>> Writing the sum equals 1 minus the sum seems to avoid the need to make
>> such an assumption.
>
> An infinite sum can't have an "odd" or an "even" number of terms.
Yes that was my point, "grouping" in any way is invalid because you must
make assumptions about the total number of terms, which you can't for an
infinite list.
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scott <sco### [at] scottcom> wrote:
> >> I'm no mathematician, but to do that you must make the assumption that
> >> there are an even number of terms in the infinite sum (ie every +1 has a
> >> -1 to pair with it). You could have assumed an odd number of terms and
> >> got a sum of 1 instead.
> >
> >> Writing the sum equals 1 minus the sum seems to avoid the need to make
> >> such an assumption.
> >
> > An infinite sum can't have an "odd" or an "even" number of terms.
> Yes that was my point, "grouping" in any way is invalid because you must
> make assumptions about the total number of terms, which you can't for an
> infinite list.
I don't think that's how it works. (If it were, then that original "proof"
would be invalid from the get-go, because it's grouping elements and
summing those groups.)
--
- Warp
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>> Yes that was my point, "grouping" in any way is invalid because you must
>> make assumptions about the total number of terms, which you can't for an
>> infinite list.
>
> I don't think that's how it works. (If it were, then that original "proof"
> would be invalid from the get-go, because it's grouping elements and
> summing those groups.)
There's no grouping like you did in the original proof. Which part of
the original proof assumes the length of the summation is anything other
than infinite?
Welcome back BTW :-)
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Am 27.07.2015 um 19:54 schrieb Orchid Win7 v1:
> This latter type of shenanigans is mostly used in cryptography and
> number theory, but does also pop up in places like error-correcting
> codes. (If you've ever tried to scan a bar code or play a CD, you care
> about error-correcting codes.)
Nobody cares about error-correcting codes when playing an audio CD.
Unlike DVD or even data CDs (aka CD-ROMs), Sony's audio CD format
doesn't waste any data capacity on bit error recovery.
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Am 27.07.2015 um 11:19 schrieb scott:
> https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF
You may also like the proof that all triangles are equilateral:
https://youtu.be/Yajonhixy4g
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