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Topic: prove the sum =pi/3 (Read 2003 times) |
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tony123
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prove the sum =pi/3
« on: Jan 25th, 2008, 1:15am » |
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prove the sum of this series =pi/3
1 + 1/5 - 1/7 - 1/11 +1/13 + 1/17 - ...
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cool_joh
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What is the general rule for Un?
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towr
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Re: prove the sum =pi/3
« Reply #2 on: Jan 25th, 2008, 3:58am » |
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I think it's  k=0..inf (-1)k [1/(6k+1) + 1/(6k+5)]
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| « Last Edit: Jan 25th, 2008, 3:58am by towr » |
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temporary
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Re: prove the sum =pi/3
« Reply #3 on: Jan 25th, 2008, 6:49am » |
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Would remnian zeta function work?
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ThudnBlunder
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Re: prove the sum =pi/3
« Reply #4 on: Jan 25th, 2008, 7:31am » |
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on Jan 25th, 2008, 6:49am, temporary wrote:| Would remnian zeta function work? |
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Learn to spell names before dropping them.
on Jan 25th, 2008, 3:51am, cool_joh wrote:| What is the general rule for Un? |
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un = 2/[6n + (-1)n - 3]
Edit: Nope, forgot about the signs - better express as two alternating series.
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| « Last Edit: Jan 25th, 2008, 9:01am by ThudnBlunder » |
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Joe Fendel
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Re: prove the sum =pi/3
« Reply #5 on: Jan 25th, 2008, 7:56am » |
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on Jan 25th, 2008, 3:58am, towr wrote:| I think it's k=0..inf (-1)k [1/(6k+1) + 1/(6k+5)] |
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This just looks like (4/3) * k=0..inf (-1)k/(2k+1)
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Joe Fendel
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Re: prove the sum =pi/3
« Reply #6 on: Jan 25th, 2008, 9:20am » |
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on Jan 25th, 2008, 7:56am, Joe Fendel wrote:
This just looks like (4/3) * k=0..inf (-1)k/(2k+1) |
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Oh, and this is actually just the same as (1 - i/2) *(4/3) * k=0..inf ik/(k+1)
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Hippo
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Re: prove the sum =pi/3
« Reply #7 on: Jan 25th, 2008, 10:38am » |
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on Jan 25th, 2008, 9:20am, Joe Fendel wrote:
Oh, and this is actually just the same as (1 - i/2) *(4/3) * k=0..inf ik/(k+1) |
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I didn't check the thread ...
May be you know the sum exactly ... otherwise this may help ... , but I hope \sum_{k\ge 1} i^k/(k+1) can be calculated using lim_{x\to 1-}F(x) where F(x)=\sum_{k\ge 1} x^{k+1}i^k/(k+1). F'(x)=\sum_{k\ge 0} (xi)^k=1/(1-xi) so F(x)=c+\int dx/(1-xi) ...
= c-i ln (1-ix) 0=F(0)=c-0 ... F(x)=-i ln (1-ix) and \sum=-i ln (1-i)
Unfortunately I don't understand your derivation.
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| « Last Edit: Jan 25th, 2008, 12:27pm by Hippo » |
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temporary
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Re: prove the sum =pi/3
« Reply #8 on: Jan 25th, 2008, 5:52pm » |
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[quote author=ThudanBlunder link=board=riddles_medium;num=1201252512;start=0#4 date=01/25/08 at 07:31:10]
Learn to spell names before dropping them.
Learn how to correct someone before doing it. What is the correct way to spell it( and don't say i-t)?
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Icarus
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Re: prove the sum =pi/3
« Reply #9 on: Jan 25th, 2008, 5:53pm » |
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Riemann
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mikedagr8
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Re: prove the sum =pi/3
« Reply #10 on: Jan 25th, 2008, 6:07pm » |
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on Jan 25th, 2008, 5:53pm, Icarus wrote:
At least he had the correct letters.
Did you know that this topic (Reimann zeta) is what put srn347 over the edge of whether he should stay in the forum?
I'm sure you knew this though.
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| « Last Edit: Jan 25th, 2008, 6:24pm by mikedagr8 » |
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Icarus
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Re: prove the sum =pi/3
« Reply #11 on: Jan 25th, 2008, 6:51pm » |
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Actually, I haven't bothered to read what was posted while I was away, unless it was to a thread that is still active.
Personally, I prefer to take people at face-value, even if they think they are fooling me with hidden insults or sly references. But I don't care for sloppy posts. If someone makes a mistake because they mis-typed, or even don't realize that they've got it wrong, this doesn't bother me. I do the same. But if someone makes a mistake because they can't be bothered to find out the correct spelling or are too lazy to even type it out, and just leave it to the reader to figure out what they really meant, that I find presumptive and rude. (This is why I hate posts with "u r" for "you are" and the like. I tend to ignore them.)
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And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Eigenray
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Re: prove the sum =pi/3
« Reply #12 on: Jan 25th, 2008, 7:08pm » |
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on Jan 25th, 2008, 9:20am, Joe Fendel wrote:| Oh, and this is actually just the same as (1 - i/2) *(4/3) * k=0..inf ik/(k+1) |
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I think you mean
(-1)k/(2k+1) = ik/(k+1) - i/2 (-1)k/(k+1),
or just  [ ik/(k+1) ].
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Hippo
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Re: prove the sum =pi/3
« Reply #13 on: Jan 27th, 2008, 3:55am » |
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on Jan 25th, 2008, 7:08pm, Eigenray wrote:
I think you mean
(-1)k/(2k+1) = ik/(k+1) - i/2 (-1)k/(k+1),
or just [ ik/(k+1) ]. |
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In that case I must generalize ... for a not equal 1 (may be absolute value at least 1 is required ... yes I am using lim ax^n=0 for n going to infty and x going to 1).
on Jan 25th, 2008, 10:38am, Hippo wrote:
May be you know the sum exactly ... otherwise this may help ... , but I hope \sum_{k\ge 1} a^k/(k+1) can be calculated using lim_{x\to 1-}F(x) where F(x)=\sum_{k\ge 1} x^{k+1}a^k/(k+1). F'(x)=\sum_{k\ge 0} (ax)^k=1/(1-ax) so F(x)=c+\int dx/(1-ax) ...
= c-a ln (1-ax); 0=F(0)=c-0 ... F(x)=-a ln (1-ax) and \sum=-a ln (1-a)
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| « Last Edit: Jan 27th, 2008, 4:06am by Hippo » |
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temporary
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Re: prove the sum =pi/3
« Reply #14 on: Jan 27th, 2008, 11:20am » |
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on Jan 25th, 2008, 6:07pm, mikedagr8 wrote:
At least he had the correct letters.
Did you know that this topic (Reimann zeta) is what put srn347 over the edge of whether he should stay in the forum?
I'm sure you knew this though. |
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You mean that they banned him for misspelling riemann?! Something like that is not worth banning someone.
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mikedagr8
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Re: prove the sum =pi/3
« Reply #15 on: Jan 27th, 2008, 3:29pm » |
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on Jan 27th, 2008, 11:20am, temporary wrote:
You mean that they banned him for misspelling riemann?! Something like that is not worth banning someone. |
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No. They banned him for being incompetent and abusive. It was in this thread and the thread after it (also created by srn347) that got him booted. People were sick of his antics and assumptions.
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| « Last Edit: Jan 27th, 2008, 4:48pm by mikedagr8 » |
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temporary
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Re: prove the sum =pi/3
« Reply #16 on: Jan 27th, 2008, 4:07pm » |
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TB? Thud and blunder, or as srn347 used to call him: thunderblunder. Lol. Although I still object to his ban being constitutional, let's get back on topic.
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FiBsTeR
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Re: prove the sum =pi/3
« Reply #17 on: Jan 27th, 2008, 4:09pm » |
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on Jan 27th, 2008, 3:29pm, mikedagr8 wrote:
They banned him for being imcompetent [...] |
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Incompetent*. 
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mikedagr8
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Re: prove the sum =pi/3
« Reply #18 on: Jan 27th, 2008, 4:10pm » |
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on Jan 27th, 2008, 4:09pm, FiBsTeR wrote:
Incompetent*. |
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Typo. Laziness.
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| « Last Edit: Jan 27th, 2008, 4:49pm by mikedagr8 » |
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Icarus
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Re: prove the sum =pi/3
« Reply #19 on: Jan 27th, 2008, 8:21pm » |
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on Jan 27th, 2008, 4:07pm, temporary wrote:| TB? Thud and blunder, or as srn347 used to call him: thunderblunder. Lol. Although I still object to his ban being constitutional, let's get back on topic. |
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I'm sure it was insults, such as that one, that got him banned. That and other abuse of the forum. Incompetence is not a banning offense, or surely most of us would be gone by now for stupid things like dropping i's from our calculations.
But regardless, this is not a constitutional issue. The US constitution right to free speech does not extend to allowing you to force yourself into places where your nastiness is not appreciated. Even in the public venues the right was intended to cover, editors and organizers still have the right to demand decorum from those speaking.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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temporary
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Re: prove the sum =pi/3
« Reply #20 on: Jan 27th, 2008, 8:47pm » |
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on Jan 27th, 2008, 8:21pm, Icarus wrote:
I'm sure it was insults, such as that one, that got him banned. That and other abuse of the forum. Incompetence is not a banning offense, or surely most of us would be gone by now for stupid things like dropping i's from our calculations.
But regardless, this is not a constitutional issue. The US constitution right to free speech does not extend to allowing you to force yourself into places where your nastiness is not appreciated. Even in the public venues the right was intended to cover, editors and organizers still have the right to demand decorum from those speaking. |
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How is what I just said an insult?
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Icarus
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Re: prove the sum =pi/3
« Reply #21 on: Jan 27th, 2008, 9:20pm » |
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Good grief. Please stop being intentionally obtuse.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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temporary
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Re: prove the sum =pi/3
« Reply #22 on: Jan 28th, 2008, 4:52pm » |
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Comparing me to an angle? I'm not sure if that comment deserves response, but for some reason, it's getting it anyway.
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JiNbOtAk
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Re: prove the sum =pi/3
« Reply #23 on: Jan 28th, 2008, 7:32pm » |
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on Jan 27th, 2008, 9:20pm, Icarus wrote:| Please stop being intentionally obtuse. |
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on Jan 28th, 2008, 4:52pm, temporary wrote:| Comparing me to an angle? |
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I don't think it's intentional, more like inherent.
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Sir Col
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Re: prove the sum =pi/3
« Reply #24 on: Jan 29th, 2008, 2:59pm » |
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on Jan 28th, 2008, 4:52pm, temporary wrote:| Comparing me to an angle? |
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Quite honestly, I wouldn't worry myself about that.
on Jan 28th, 2008, 4:52pm, temporary wrote:| I'm not sure if that comment deserves response, but for some reason, it's getting it anyway. |
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In any case, could you ever be certain? I'm not sure if it could.
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