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        riddles >> putnam exam (pure math) >> harmonic
        
(Message started by: srn347 on Oct 1st, 2007, 8:26pm)
Title: harmonic
Post by srn347 on Oct 1st, 2007, 8:26pm
What is 1-1/2 +1/3 -1/4...
what about 1/9 +1/19+1/29...(all that have a nine anywhere in the denominator)
Title: Re: harmonic
Post by Sameer on Oct 1st, 2007, 8:43pm
First is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/eta.gif(1) :)

Second is complemented depleted/Kempner series.
Title: Re: harmonic
Post by srn347 on Oct 1st, 2007, 9:06pm
I meant what are their numerical values.
Title: Re: harmonic
Post by Sameer on Oct 1st, 2007, 9:19pm

on 10/01/07 at 21:06:10, srn347 wrote:
I meant what are their numerical values.


First one is another way of saying what is its numerical value!! That's the way how we do it!!

For second I would say [hide] 80 http://www.qbyte.org/puzzles/p072s.html [/hide]
Title: Re: harmonic
Post by srn347 on Oct 1st, 2007, 9:24pm
The n thing doesn't tell me the number. the 80 was on the right track, but out of the loop(the track shouldn't be shaped like a loop anyway). Go back to line 10(in binary).
Title: Re: harmonic
Post by Sameer on Oct 1st, 2007, 9:42pm

on 10/01/07 at 21:24:12, srn347 wrote:
The n thing doesn't tell me the number


It does if you know Dirichlet!!
Title: Re: harmonic
Post by srn347 on Oct 2nd, 2007, 6:28am
was that supposed to be the remnien zeta? regardless, I require an actual answer.
Title: Re: harmonic
Post by pex on Oct 2nd, 2007, 6:50am

on 10/02/07 at 06:28:05, srn347 wrote:
was that supposed to be the remnien zeta? regardless, I require an actual answer.

Not many people write eta when they mean zeta. Sameer's answer is correct: the Dirichlet Eta function evaluated at 1.
Title: Re: harmonic
Post by srn347 on Oct 2nd, 2007, 7:21am
Nope. I don't know how you got an answer of 1...
Title: Re: harmonic
Post by pex on Oct 2nd, 2007, 7:26am

on 10/02/07 at 07:21:48, srn347 wrote:
Nope. I don't know how you got an answer of 1...

He doesn't, he gets eta(1).
Title: Re: harmonic
Post by Grimbal on Oct 2nd, 2007, 7:27am
The Dirichlet Eta function evaluated at 1 evaluates as ln(2).
Title: Re: harmonic
Post by srn347 on Oct 2nd, 2007, 7:27am
Does that mean zeta 1? That is incorrect.
Title: Re: harmonic
Post by pex on Oct 2nd, 2007, 7:28am

on 10/02/07 at 07:27:41, srn347 wrote:
Does that mean zeta 1? That is incorrect.

No, it means exactly what it says.
Title: Re: harmonic
Post by ThudanBlunder on Oct 2nd, 2007, 10:06am
347, your series 1 - 1/2 + 1/3 - 1/4 +..... = loge2 = 0.63452380952380952380952380952381...

Edit: loge2 = 0.69314718055994530941723212145818...
Title: Re: harmonic
Post by srn347 on Oct 2nd, 2007, 4:32pm
Nope. there is fallacy in your statement.
Title: Re: harmonic
Post by pex on Oct 2nd, 2007, 4:40pm

on 10/02/07 at 16:32:19, srn347 wrote:
Nope. there is fallacy in your statement.
Until "loge2", ThudanBlunder is correct (although you may be more familiar with seeing it written as ln 2), but I don't know where he got his decimal expression. Maple tells me it's 0.693147180559945309417232121458176568...

By the way, also eta(1) = ln 2.
Title: Re: harmonic
Post by pex on Oct 2nd, 2007, 5:02pm
Just for completeness, the second series diverges. Proof:

Its "complement", the series summing the reciprocals of all integers containing no nines, is
(1/1 + 1/2 + ... + 1/8)
+ (1/10 + 1/11 + ... + 1/18 + 1/20 + ... + 1/88)
+ (1/100 + ... + 1/888)
+ ...
The n-th bracketed term contains 8*9n-1 numbers which are reciprocals of n-digit integers. Thus, the sum is less than
8*(1/1) + 8*9*(1/10) + 8*92*(1/100) + ...
= 8*sum{n=0 to infinity} (9/10)n
= 8 * 1/(1-9/10) = 80,
which means that it converges.

We know that the sum of reciprocals of all integers (1/1 + 1/2 + 1/3 + ...) diverges, and we know that the sum of the reciprocals of all integers that contain no nines converges. This means that the part that is left over, the sum of the reciprocals of all integers that do contain nines, must diverge.
Title: Re: harmonic
Post by srn347 on Oct 2nd, 2007, 8:46pm
It is not log or ln of whatever. should I tell you what it is?
Title: Re: harmonic
Post by Obob on Oct 2nd, 2007, 10:19pm
Its very arrogant of you to tell all these people who are much smarter than you that they are wrong.  Especially since they are in fact correct, and you don't even know the math involved.
Title: Re: harmonic
Post by srn347 on Oct 2nd, 2007, 11:38pm
The first one is zero, because it is the harmonic series(infinity) minus itself. 1-1/2+1/3...=1-(1-1/2)+1/3-(1/2-1/4)...=(1-1)+(1/2-1/2)+(1/3)-1/3... If you think it's arrogant, try taking ritalin. if you are offended, blame the one whom the phase originated from.
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 12:40am

on 10/02/07 at 23:38:41, srn347 wrote:
The first one is zero, because it is the harmonic series(infinity) minus itself. 1-1/2+1/3...=1-(1-1/2)+1/3-(1/2-1/4)...=(1-1)+(1/2-1/2)+(1/3)-1/3... If you think it's arrogant, try taking ritalin. if you are offended, blame the one whom the phase originated from.

Wow. This could just be one of the most stupid things I've ever seen.

One. "Infinity minus infinity" is not well-defined; by no means does it have to be zero.

Two. You are trying to use the commutative property (look it up) of addition in a conditionally convergent series (look that one up too). This generally does not work.

Three. Any decent textbook or website that treats limits will tell you that the value is ln 2.

Four. So did multiple math-inclined (to say the least) people above.
Title: Re: harmonic
Post by towr on Oct 3rd, 2007, 2:25am
Five. Try the start of the sequence on your trusty calculator, and you'll see that in no way does it tend to 0.
Six. Ritalin does not work against other people's arrogance, so no matter how much ritalin any of us were to take, you'd still be an arrogant little bugger.
Title: Re: harmonic
Post by srn347 on Oct 3rd, 2007, 6:55am
Try solving the second series.
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 7:20am

on 10/03/07 at 06:55:06, srn347 wrote:
Try solving the second series.

Look further up in the thread - it diverges. But you're not going to believe this anyway.

Edit: perhaps it's nice if I mention that "my" proof of this fact was inspired rather heavily by NickH's solution in the link Sameer posted. Night-time plagiarism...
Title: Re: harmonic
Post by Sameer on Oct 3rd, 2007, 7:58am
*sigh* how come you didn't tell us you didn't want the solution that everyone knows but the one that you concocted up... it would have made our thinking so much simpler than these god knows what eta, zeta, etc..

Edit: I tried giving benefit of doubt and encouraging good posts, but it seems it is pointless.
Title: Re: harmonic
Post by Grimbal on Oct 3rd, 2007, 8:21am
Srn347,

it seems to me that you are very curious about mathematics, you want to explore it beyond what you are taught at school, and you already know much more than typical for your age.

But your approach is way too careless.  You grab all the concepts you can get your hands on and apply them without any precaution.  And in the process you take all the wrong turns.

The example you gave is one of the pitfalls of infinite series: sometimes, by reorganizing the terms you can change the result.
See Conditional Convergence on Mathworld (http://mathworld.wolfram.com/ConditionalConvergence.html).

I can understand that you want to explore and try more "exotic" concepts in mathematics.  I do because I did the same at your age.  But you have to know that each concept has a "user's manual" attached that tells you when and how the concept can be used.  You cannot just pick it up and use it as your intuition tells you it should be used.  To use it properly, you have to take the time to study the subject.  If you don't, you are moving in a minefield.  You will get funny results and you won't know why.  If you are lucky, you will stumble over a paradox, like a "proof" that 1=0.  These are a sure sign that you haven't been careful enough in one or the other concept you used.  It is something you shouldn't ignore.  Get back and understand what you did wrong.

I don't just want to criticize you.  (In fact, yes, when you make bold claims about things you know only superficially.)  But I am worried about your approach of mathematics.  You cannot learn mathematics by just trying things and see if they works.  It's not going to work in the long term.  Eventually, you have to take a good book and learn the concepts thoroughly, chapter by chapter.
Title: Re: harmonic
Post by towr on Oct 3rd, 2007, 8:32am

on 10/02/07 at 23:38:41, srn347 wrote:
The first one is zero, because it is the harmonic series(infinity) minus itself. 1-1/2+1/3...=1-(1-1/2)+1/3-(1/2-1/4)...=(1-1)+(1/2-1/2)+(1/3)-1/3...

Roughly you can say http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifni=1 1/i http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/approx.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif + ln n (the approximation improves with n) where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif is the Euler constant.
And so the alternating harmonic series up to the nth term (for large n) gives http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif + ln n - (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif + ln n/2) = ln n -  ln n/2 =  ln n/[n/2] = ln 2
So as a formality we take n to infinity, and get lim n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif ln 2 = ln 2

What you seem to be doing on the other hand, is
 lim n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif (ln n  -   ln n/2) = lim n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif ln n  -   lim n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif ln n/2 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif, but you shouldn't haphazardly treat limits like that.



Who wants to try 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + ... ?
Title: Re: harmonic
Post by Grimbal on Oct 3rd, 2007, 8:41am

on 10/03/07 at 08:32:55, towr wrote:
Who wants to try 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + ... ?

Easy!  It is (1+1/2+1/3+...) - (2/3+2/6+2/9+...)
= (1+1/2+1/3+...) - 2/3·(1+1/2+1/3+...)
= infinity - 2/3·infinity = 1/3·infinity.
It is not infinity, but quite big anyway.  ::)
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 8:53am

on 10/03/07 at 08:41:14, Grimbal wrote:
Easy!  It is (1+1/2+1/3+...) - (2/3+2/6+2/9+...)
= (1+1/2+1/3+...) - 2/3·(1+1/2+1/3+...)
= infinity - 2/3·infinity = 1/3·infinity.
It is not infinity, but quite big anyway.  ::)

LOL! ;D

More seriously, I find [hide]divergence to positive infinity[/hide].
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 9:46am
One that I just came up with, solved, and found interesting:

1/1 + 1/2 - 1/3 - 1/4 + 1/5 + 1/6 - 1/7 - 1/8 + 1/9 + 1/10 - 1/11 - 1/12 + ...
Title: Re: harmonic
Post by Obob on Oct 3rd, 2007, 10:07am
Without even knowing that the alternating harmonic series sums to ln 2, it is trivial to see that at least it is positive or divergent:  regroup the terms like this:

(1-1/2)+(1/3-1/4)+(1/5-1/6)+...

Then the even partial sums S2, S4, S6, ... are increasing, since each of the parenthesized terms above is positive, and S2 is 1/2.  So, if the series converges, its limit is at least 1/2.
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 10:26am

on 10/03/07 at 09:46:58, pex wrote:
One that I just came up with, solved, and found interesting:

1/1 + 1/2 - 1/3 - 1/4 + 1/5 + 1/6 - 1/7 - 1/8 + 1/9 + 1/10 - 1/11 - 1/12 + ...

And, a bit more difficult:

1/1 + 1/2 + 1/3 - 1/4 - 1/5 - 1/6 + 1/7 + 1/8 + 1/9 - 1/10 - 1/11 - 1/12 + ...
Title: Re: harmonic
Post by ThudanBlunder on Oct 3rd, 2007, 10:43am

on 10/02/07 at 23:38:41, srn347 wrote:
The first one is zero, because it is the harmonic series(infinity) minus itself. 1-1/2+1/3...=1-(1-1/2)+1/3-(1/2-1/4)...=(1-1)+(1/2-1/2)+(1/3)-1/3... If you think it's arrogant, try taking ritalin. if you are offended, blame the one whom the phase originated from.

IIRC, one can re-bracket terms of an infinite series and get the right answer iff the series is absolutely convergent. Your series is conditionally convergent only. You simply don't know the answers to you own problems.  
Title: Re: harmonic
Post by Eigenray on Oct 3rd, 2007, 11:08am
1/1 + 1/2 + ... + 1/r - 1/(r+1) - 1/(r+2) - ... - 1/(2r) + 1/(2r+1) + ... + 1/(3r) - 1/(3r+1) - ... - 1/(4r) + 1/(4r+1) + ....
= 1/r * { log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/(1-coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) * arctan[sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/(1-coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)] }

where the sum is over http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif = (2k+1)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/r, k going from 1 to r.  The first few are

log(2)
1/2 [ log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2 ]
1/3 [ log 2 + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 ]
1/4 [ log 2 +  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif(1/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2) ]
1/5 [ log 2 +  2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{1+2/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5} ]
1/6 [ log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/6 (15 + 4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3) ]
1/7 [ log 2 + ?? ]
1/8 [ log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif(1/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2})
...
1/r [ log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif * http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gifr ].

Is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gifr always algebraic?
Title: Re: harmonic
Post by Obob on Oct 3rd, 2007, 11:09am
If a series is convergent, then you can "rebracket" the terms however you want, in the sense that if your series is a0+a1+a2+... then it will equal say (a0+a1)+(a2+...).  That is, you can add parentheses however you wish, so long as you don't rearrange terms.  Adding parentheses corresponds to passing to a subsequence of the sequence of partial sums of the series.  Since a subsequence of a convergent sequence is convergent and converges to the same limit, adding brackets doesn't change the limit.

However, you can rearrange the terms of the series if and only if the series is absolutely convergent.  If the series converges, but not absolutely, then the terms of the series can be rearranged to make the sum come out to any value you want.  So, by suitable rearrangement of the terms of the alternating harmonic series, we can make the limit be 0, say.  This does not mean that the limit of the alternating harmonic series is 0, though!  It is undeniably ln 2.
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 11:22am

on 10/03/07 at 11:08:40, Eigenray wrote:
1/1 + 1/2 + ... + 1/(r-1) - 1/r - 1/(r+1) - ... - 1/(2r-1) + 1/(2r) + ... + 1/(3r-1) - 1/(3r) - ... - 1/(4r-1) + 1/(4r) + ....
= 1/r * { log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/(1-coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) * arctan[sinhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/(1-coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)] }

where the sum is over http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif = (2k+1)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/r, k going from 1 to r.

Wow! Mind if I don't see where this comes from straight away? (Anyway, your first three values match the ones I found - so I guess my messy integrals were correct.)
Title: Re: harmonic
Post by Eigenray on Oct 3rd, 2007, 11:26am
If you already have them as integrals, try [hide]partial fractions[/hide].
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 11:31am

on 10/03/07 at 11:08:40, Eigenray wrote:
Is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gifr always algebraic?

If we let k go from 0 to r-1 (which makes no difference), so that all thetas are between zero and 2pi, and we recall sin(theta)/(1-cos(theta)) = cot(theta/2), we find that the summands equal

sin(theta)/(1-cos(theta)) * arctan(sin(theta)/(1-cos(theta))
= cot(theta/2) * arctan(cot(theta/2))
= cot(theta/2) * (pi - theta)/2.

Cotangents of such angles are always algebraic (IIRC) and (pi-theta) is a rational multiple of pi. So I'd say the answer is yes.
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 11:32am

on 10/03/07 at 11:26:21, Eigenray wrote:
If you already have them as integrals, try [hide]partial fractions[/hide].

Yes, I did that before I posted the problems - I had found them in exactly the same forms you posted.
Title: Re: harmonic
Post by Eigenray on Oct 3rd, 2007, 12:53pm

on 10/03/07 at 11:31:52, pex wrote:
we recall sin(theta)/(1-cos(theta)) = cot(theta/2)

Aha, yes.  I suck at trig.


on 10/03/07 at 11:31:52, pex wrote:
Cotangents of such angles are always algebraic (IIRC) and (pi-theta) is a rational multiple of pi. So I'd say the answer is yes.

Yes; they can be expressed in terms of a 2r-th root of unity, so http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gifr has degree no more than http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(2r) over http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif.  But it looks like we should be able to improve this bound to just http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(r).


on 10/03/07 at 11:32:32, pex wrote:
Yes, I did that before I posted the problems - I had found them in exactly the same forms you posted.

Do you have the general solution as an integral?


We can show also that if

S(r) = 1/r [ log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifcot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/2)(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/2 ]

is the sum in question, then

S(r) - log(2r/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0

as r http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif.  Do you see how?
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 1:41pm

on 10/03/07 at 12:53:38, Eigenray wrote:
Do you have the general solution as an integral?

Isn't it [hide]integral(0..1) (1-xr) / [(1-x)(1+xr)] dx[/hide]? (Of course, we could [hide]cancel (1-x), but it just doesn't look as nice[/hide].)
Title: Re: harmonic
Post by pex on Oct 3rd, 2007, 1:57pm

on 10/03/07 at 12:53:38, Eigenray wrote:
We can show also that if

S(r) = 1/r [ log 2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifcot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/2)(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/2 ]

is the sum in question, then

S(r) - log(2r/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0

as r http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif.  Do you see how?

Looking at it for a while: no, I don't. Unfortunately, I don't have any more time to spend on the problem now. :(
Title: Re: harmonic
Post by Eigenray on Oct 3rd, 2007, 2:49pm

on 10/03/07 at 13:41:02, pex wrote:
Isn't it [hide]integral(0..1) (1-xr) / [(1-x)(1+xr)] dx[/hide]? (Of course, we could [hide]cancel (1-x), but it just doesn't look as nice[/hide].)

And what happens when you expand this using partial fractions?


on 10/03/07 at 13:57:24, pex wrote:
Looking at it for a while: no, I don't. Unfortunately, I don't have any more time to spend on the problem now. :(

It can be done using the fact that

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif02http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif { cot(t/2)(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif-t)/2 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif[1/t+1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif-t)] } dt = -2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.giflog(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif).

(I just used Mathematica for this.)  But maybe there is a nicer way.
Title: Re: harmonic
Post by Sir Col on Oct 3rd, 2007, 3:30pm
srn347, this is the first occasion I've had to say anything negative towards you, but your posts in this thread are entirely rude. You have received pertinent response to your original questions, so your subsequent responses are clear evidence that you are out of your depth with the mathematics. Moreover, in reading your responses here and in other threads I suspect that you have not quite appreciated how lucky you are to be a member on this forum. There are few other places on the planet where so many talented, knowledgeable, and skilled expositors gather. And what is more, they are always willing and entirely generous in sharing their knowledge.

I suggest you take serious stock of your actions and consider carefully if you wish to remain a member. Patience is wearing very thin, and in most cases you have already gone too far.

A sincere apology would be a starting point. From there I would strongly advise a more measured approach to any future posts. If you genuinely want help, then ask; if someone is explaining something and is using terms/concepts above your current level of mathematics then say so. Otherwise, if you have nothing useful to say, then please say nothing.
Title: Re: harmonic
Post by ThudanBlunder on Oct 3rd, 2007, 7:30pm
Well said, Grimbal and Sir Col!
Title: Re: harmonic
Post by ima1trkpny on Oct 3rd, 2007, 7:35pm

on 10/03/07 at 19:30:55, ThudanBlunder wrote:
Well said, Grimbal and Sir Col!


True. However it seems he is bent on being defiant and won’t take a hint even though he has been banned.


on 10/03/07 at 19:11:42, sm347 wrote:
But if you ban me, because i am so clever, I will find many more ways of showing you that I am smarter than u.
Title: Re: harmonic
Post by temporary on Jan 26th, 2008, 5:55pm

on 10/03/07 at 19:35:56, ima1trkpny wrote:
True. However it seems he is bent on being defiant and won’t take a hint even though he has been banned.


http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1201066684

They are not the same person.
Title: Re: harmonic
Post by ima1trkpny on Jan 26th, 2008, 7:37pm

on 01/26/08 at 17:55:43, temporary wrote:
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1201066684

They are not the same person.


Prove it. ::)
Title: Re: harmonic
Post by temporary on Jan 27th, 2008, 11:25am
Compare what they say.
Title: Re: harmonic
Post by mikedagr8 on Jan 27th, 2008, 3:40pm

on 01/27/08 at 11:25:59, temporary wrote:
Compare what they say.

Look at how sm347 writes his age. Everyone who was used to srn347 (sad to say) knows he was obsessed with infinity. Maybe he was an imposter sm347. It really doesn't matter. He was booted before then. Someone was just having a joke. It hasn't harmed anyone as srn347 was already banned. I'm sorry you feel that way and you are certain they are not the same. I understand your opinions but I think this subject of srn347 should be put behind us. I don't want the problems he caused arrising again.
Title: Re: harmonic
Post by temporary on Jan 27th, 2008, 4:09pm
Yet sm347 posted only spam, while srn347 at least tried to solve the riddles. For further discussion, http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1201066684
Title: Re: harmonic
Post by Obob on Jan 27th, 2008, 5:48pm
To be honest, nobody cares if srn347 and sm347 are/were the same person.  Whether you and srn347 are the same person (and I am 99% sure you are) is much more pertinent.
Title: Re: harmonic
Post by Eigenray on Jan 27th, 2008, 7:02pm
In any case, temporary and srn347 both certainly tend to sully any thread they touch.

Getting back on topic, it was still to be shown that

Sr = 1/1 + 1/2 + ... + 1/r - 1/(r+1) - 1/(r+2) - ... - 1/(2r)
  + 1/(2r+1) + ... + 1/(3r) - 1/(3r+1) - ... - 1/(4r)
  + 1/(4r+1) + ....
= 1/r [ log 2 + 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif tanhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif ]
~ log(2r/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif,

where the sum is over http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif= khttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2r,  k= r-1, r-3, r-5, ..., k http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1.

In fact it looks like

Sr ~ log(2r/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif + log(2)/r - 0.137078/r2

Any takers?
Title: Re: harmonic
Post by temporary on Jan 27th, 2008, 7:31pm

on 01/27/08 at 17:48:50, Obob wrote:
To be honest, nobody cares if srn347 and sm347 are/were the same person.  Whether you and srn347 are the same person (and I am 99% sure you are) is much more pertinent.


Wasn't srn347 ip banned also, or is that just a myth. Also, if what you are saying is true, then

1/1 + 1/2 + ... + 1/r - 1/(r+1) - 1/(r+2) - ... - 1/(2r)
  + 1/(2r+1) + ... + 1/(3r) - 1/(3r+1) - ... - 1/(4r)
  + 1/(4r+1) + .... ~ log(2r/) +  + log(2)/r - 0.137078/r2


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