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scott <sco### [at] scott com> 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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On 31/07/2015 11:22 PM, clipka wrote:
> 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.
In fact, the audio CD format uses cross-interleaved Reed-Solomon codes
for error recovery.
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Am 01.08.2015 um 11:24 schrieb Orchid Win7 v1:
> On 31/07/2015 11:22 PM, clipka wrote:
>> 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.
>
> In fact, the audio CD format uses cross-interleaved Reed-Solomon codes
> for error recovery.
Damn, I hate to stand corrected.
But there was /something/ with regards to error recovery that CD-ROMs
have but CD-DAs don't.
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On 8/1/2015 4:36 PM, clipka wrote:
>> In fact, the audio CD format uses cross-interleaved Reed-Solomon codes
>> for error recovery.
>
> Damn, I hate to stand corrected.
Write this Date in our calenders.
On the first of August two thousand and fourteen. Clipka admitted he was
wrong. :-P
--
Regards
Stephen
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On 01/08/2015 04:36 PM, clipka wrote:
> Am 01.08.2015 um 11:24 schrieb Orchid Win7 v1:
>> In fact, the audio CD format uses cross-interleaved Reed-Solomon codes
>> for error recovery.
>
> Damn, I hate to stand corrected.
>
> But there was /something/ with regards to error recovery that CD-ROMs
> have but CD-DAs don't.
Yes, I have that vague recollection as well. I should think ISO-9660
probably has block-level checksums or similar, to allow corruption to be
detected. For audio CDs, the player is supposed to just fill in any
unreadable chunks with silence.
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