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        riddles >> medium >> prove the sum =pi/3
        
(Message started by: tony123 on Jan 25th, 2008, 1:15am)
Title: prove the sum =pi/3
Post by tony123 on Jan 25th, 2008, 1:15am
prove the sum  of this series =pi/3

1 + 1/5 - 1/7 - 1/11 +1/13 + 1/17 - ...
Title: Re: prove the sum =pi/3
Post by cool_joh on Jan 25th, 2008, 3:51am
What is the general rule for Un?
Title: Re: prove the sum =pi/3
Post by towr on Jan 25th, 2008, 3:58am
I think it's http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf (-1)k [1/(6k+1) + 1/(6k+5)]
Title: Re: prove the sum =pi/3
Post by temporary on Jan 25th, 2008, 6:49am
Would remnian zeta function work?
Title: Re: prove the sum =pi/3
Post by ThudanBlunder on Jan 25th, 2008, 7:31am

on 01/25/08 at 06:49:32, temporary wrote:
Would remnian zeta function work?

Learn to spell names before dropping them.


on 01/25/08 at 03:51:12, cool_joh wrote:
What is the general rule for Un?

un = 2/[6n + (-1)n - 3]

Edit: Nope, forgot about the signs - better express as two alternating series.
Title: Re: prove the sum =pi/3
Post by Joe Fendel on Jan 25th, 2008, 7:56am

on 01/25/08 at 03:58:01, towr wrote:
I think it's http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf (-1)k [1/(6k+1) + 1/(6k+5)]


This just looks like (4/3) * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf (-1)k/(2k+1)
Title: Re: prove the sum =pi/3
Post by Joe Fendel on Jan 25th, 2008, 9:20am

on 01/25/08 at 07:56:57, Joe Fendel wrote:
This just looks like (4/3) * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf (-1)k/(2k+1)


Oh, and this is actually just the same as (1 - i/2) *(4/3) * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf ik/(k+1)
Title: Re: prove the sum =pi/3
Post by Hippo on Jan 25th, 2008, 10:38am

on 01/25/08 at 09:20:45, Joe Fendel wrote:
Oh, and this is actually just the same as (1 - i/2) *(4/3) * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf ik/(k+1)



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.
Title: Re: prove the sum =pi/3
Post by temporary on Jan 25th, 2008, 5:52pm
[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)?
Title: Re: prove the sum =pi/3
Post by Icarus on Jan 25th, 2008, 5:53pm
Riemann
Title: Re: prove the sum =pi/3
Post by mikedagr8 on Jan 25th, 2008, 6:07pm

on 01/25/08 at 17:53:38, Icarus wrote:
Riemann

At least he had the correct letters.
Did you know that this topic (Reimann zeta (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1191331733)) is what put srn347 over the edge of whether he should stay in the forum?

I'm sure you knew this though.
Title: Re: prove the sum =pi/3
Post by Icarus on Jan 25th, 2008, 6:51pm
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.)
Title: Re: prove the sum =pi/3
Post by Eigenray on Jan 25th, 2008, 7:08pm

on 01/25/08 at 09:20:45, Joe Fendel wrote:
Oh, and this is actually just the same as (1 - i/2) *(4/3) * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf ik/(k+1)

I think you mean

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif (-1)k/(2k+1)  =   http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif ik/(k+1) - i/2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif (-1)k/(k+1),

or just http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/re.gif[ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif ik/(k+1) ].
Title: Re: prove the sum =pi/3
Post by Hippo on Jan 27th, 2008, 3:55am

on 01/25/08 at 19:08:13, Eigenray wrote:
I think you mean

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif (-1)k/(2k+1)  =   http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif ik/(k+1) - i/2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif (-1)k/(k+1),

or just http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/re.gif[ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif ik/(k+1) ].


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 01/25/08 at 10:38:59, 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)
Title: Re: prove the sum =pi/3
Post by temporary on Jan 27th, 2008, 11:20am

on 01/25/08 at 18:07:25, mikedagr8 wrote:
At least he had the correct letters.
Did you know that this topic (Reimann zeta (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1191331733)) is what put srn347 over the edge of whether he should stay in the forum?

I'm sure you knew this though.


You mean that they banned him for misspelling riemann?! Something like that is not worth banning someone.
Title: Re: prove the sum =pi/3
Post by mikedagr8 on Jan 27th, 2008, 3:29pm

on 01/27/08 at 11:20:21, temporary wrote:
You mean that they banned him for misspelling riemann?! Something like that is not worth banning someone.

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.
Title: Re: prove the sum =pi/3
Post by temporary on Jan 27th, 2008, 4:07pm
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.
Title: Re: prove the sum =pi/3
Post by FiBsTeR on Jan 27th, 2008, 4:09pm

on 01/27/08 at 15:29:33, mikedagr8 wrote:
They banned him for being imcompetent [...]


Incompetent*.  ::)
Title: Re: prove the sum =pi/3
Post by mikedagr8 on Jan 27th, 2008, 4:10pm

on 01/27/08 at 16:09:35, FiBsTeR wrote:
Incompetent*.  ::)

Typo. Laziness.
Title: Re: prove the sum =pi/3
Post by Icarus on Jan 27th, 2008, 8:21pm

on 01/27/08 at 16:07:24, 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.


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.
Title: Re: prove the sum =pi/3
Post by temporary on Jan 27th, 2008, 8:47pm

on 01/27/08 at 20:21:23, 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.


How is what I just said an insult?
Title: Re: prove the sum =pi/3
Post by Icarus on Jan 27th, 2008, 9:20pm
Good grief. Please stop being intentionally obtuse.
Title: Re: prove the sum =pi/3
Post by temporary on Jan 28th, 2008, 4:52pm
Comparing me to an angle? I'm not sure if that comment deserves response, but for some reason, it's getting it anyway.
Title: Re: prove the sum =pi/3
Post by JiNbOtAk on Jan 28th, 2008, 7:32pm

on 01/27/08 at 21:20:20, Icarus wrote:
Please stop being intentionally obtuse.



on 01/28/08 at 16:52:52, temporary wrote:
Comparing me to an angle?


I don't think it's intentional, more like inherent.
Title: Re: prove the sum =pi/3
Post by Sir Col on Jan 29th, 2008, 2:59pm

on 01/28/08 at 16:52:52, temporary wrote:
Comparing me to an angle?

Quite honestly, I wouldn't worry myself about that.



on 01/28/08 at 16:52:52, temporary wrote:
I'm not sure if that comment deserves response, but for some reason, it's getting it anyway.

In any case, could you ever be certain? I'm not sure if it could.
Title: Re: prove the sum =pi/3
Post by temporary on Jan 29th, 2008, 6:11pm
It just did.
Title: Re: prove the sum =pi/3
Post by Sir Col on Jan 30th, 2008, 12:01am
Is that so.
Title: Re: prove the sum =pi/3
Post by Ghost Sniper on Jan 30th, 2008, 1:42pm
Look, Temporary. What you are doing right now is completely reflective of what srn347 did in his day. You've already aggravated enough people right now, including me. If I was a moderator, I would have banned you by now. However, I believe that the moderators are giving you a chance to undo your wrongs right now, so please change before it is too late.
Title: Re: prove the sum =pi/3
Post by FiBsTeR on Jan 30th, 2008, 5:20pm
I think we've drifted off topic long enough.

I started by trying to express the infinite sum using sigma notation. I began by summing all the positive terms and subtracting all the negative terms, which gave:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif 4 / (12k+1)(12k+5) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif 4 / (12k+7)(12k+11)

This boiled down to the very ugly looking:

4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif (67k + 36) / (10368k4 + 21456k3 + 14760k2 + 3697k + 385)

This appears to be a dead end...

On another track, I was reading through one of my problem solving books last night and I saw a proof of how
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=1 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif k-2 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif2/6, using the Taylor series expansion of the sine function. Could something similar be used here?
Title: Re: prove the sum =pi/3
Post by Icarus on Jan 30th, 2008, 7:48pm
Looking at it, I would suspect that it might work better to use a fourier series instead of a power series. It looks to me like the fourier expansion of some inverse trig function (for example, tan-1(q(x)), where q(x) is some elementary function).
Title: Re: prove the sum =pi/3
Post by SWF on Jan 30th, 2008, 8:17pm
= 1 + (1/3 - 1/3) +1/5 -1/7 + (1/9-1/9) -1/11 + 1/13 +(1/15-1/15) +1/17 - 1/19 +(1/21-1/21) - ...

= (1-1/3+1/5-1/7+1/9-1/11+1/13-1/15+1/17-1/19+1/21-...) + (1/3-1/9+1/15-1/21+...)

=atan(1) + (1/3)*atan(1) = 4/3*(pi/4) = pi/3
Title: Re: prove the sum =pi/3
Post by pex on Jan 31st, 2008, 3:00am
SWF: it works here, but can we justify this trick for a conditionally convergent series?

Actually, you used a conversion to integrals here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1183124601;start=21#21) that also works in this case.
Title: Re: prove the sum =pi/3
Post by FiBsTeR on Jan 31st, 2008, 6:01am

on 01/30/08 at 20:17:04, SWF wrote:
= 1 + (1/3 - 1/3) +1/5 -1/7 + (1/9-1/9) -1/11 + 1/13 +(1/15-1/15) +1/17 - 1/19 +(1/21-1/21) - ...

= (1-1/3+1/5-1/7+1/9-1/11+1/13-1/15+1/17-1/19+1/21-...) + (1/3-1/9+1/15-1/21+...)

=atan(1) + (1/3)*atan(1) = 4/3*(pi/4) = pi/3


Beautiful!  :)
Title: Re: prove the sum =pi/3
Post by Eigenray on Jan 31st, 2008, 4:12pm

on 01/31/08 at 03:00:02, pex wrote:
SWF: it works here, but can we justify this trick for a conditionally convergent series?

If we have two convergent series, we can 'merge' them however we like and the sums will add, as long as the terms of each sequence stay in the same order relative to each other.  I.e., if f : {1,2} x N http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif N is a bijection with f(i,j) < f(i,k) whenever j<k, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A(f-1(n)) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A(1,n) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A(2,n) as long as both sums exist.

And I think Joe had the same proof in mind:

on 01/25/08 at 07:56:57, Joe Fendel wrote:
This just looks like (4/3) * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifk=0..inf (-1)k/(2k+1)
Title: Re: prove the sum =pi/3
Post by pex on Feb 1st, 2008, 1:24am

on 01/31/08 at 16:12:34, Eigenray wrote:
If we have two convergent series, we can 'merge' them however we like and the sums will add, as long as the terms of each sequence stay in the same order relative to each other.  I.e., if f : {1,2} x N http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif N is a bijection with f(i,j) < f(i,k) whenever j<k, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A(f-1(n)) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A(1,n) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A(2,n) as long as both sums exist.

Ah, thanks. I agree with FiBsTeR then: beautiful!


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