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jaap
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njperrone wrote:
jaap wrote:I have already given an example of a function that is neither even nor odd that has a zero sum:
f(x) = { 1-cos(2pi x) for 0<x<1; 0 elsewhere}

this actually does not sum up to either 0 or infinity. So is excluded, by definition, from the theorem.

f'(x) = { -2pi sin(2pi x) for 0<x<1; 0 elsewhere}
f'(x) = 0 for all integers x
so your sum will be 0.

njperrone wrote:
jaap wrote:A worse example, here is an odd function with a zero sum.
f(x) = { 1-cos(2pi x) for x>=0; cos(2pi x)-1 for x<0 }

and this one actually works with my theorem.

f'(x) = { 2pi sin(2pi x) for x>=0; -2pi sin(2pi x) for x<0 }
f'(x) = 0 for all integers x
so your sum will be 0, even though the function is odd.

njperrone
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jaap wrote:f'(x) = { 2pi sin(2pi x) for x>=0; -2pi sin(2pi x) for x<0 }
f'(x) = 0 for all integers x
so your sum will be 0, even though the function is odd.

Please tell me you know what |x| means.

It means the absolute value of x. Look at my hypothesis and you will see that the absolute value is present.

jaap
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njperrone wrote:
jaap wrote:f'(x) = { 2pi sin(2pi x) for x>=0; -2pi sin(2pi x) for x<0 }
f'(x) = 0 for all integers x
so your sum will be 0, even though the function is odd.

Please tell me you know what |x| means.

It means the absolute value of x. Look at my hypothesis and you will see that the absolute value is present.

For that odd function f, the derivative happens to satisfy f'(x)=0 for all integers x, so we get:

$\sum_{x=0}^{\infty}{|f'(\pm x)|} = \sum_{x=0}^{\infty}{|f'(x)+f'(-x)|} = \sum_{x=0}^{\infty}{|0+0|} = \sum_{x=0}^{\infty}{0} = 0$

njperrone
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jaap wrote:
For that odd function f, the derivative happens to satisfy f'(x)=0 for all integers x, so we get:

$\sum_{x=0}^{\infty}{|f'(\pm x)|} = \sum_{x=0}^{\infty}{|f'(x)+f'(-x)|} = \sum_{x=0}^{\infty}{|0+0|} = \sum_{x=0}^{\infty}{0} = 0$

njperrone
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jaap wrote:
njperrone wrote:
jaap wrote:
f'(x) = 0 for all integers x

What exactly do you mean by this?

jaap
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njperrone wrote:
jaap wrote:
njperrone wrote:
jaap wrote:
f'(x) = 0 for all integers x

What exactly do you mean by this?

The example was
f(x) = { 1-cos(2pi x) for x>=0; cos(2pi x)-1 for x<0 }
The cosine here uses radians of course.

Differentiating gives
f'(x) = { 2pi sin(2pi x) for x>=0; -2pi sin(2pi x) for x<0 }

The sine function is zero at all multiples of 2pi, i.e. sin(0) = sin(2pi) = sin(4pi) = ... = 0.
Therefore, sin(2pi x) = 0 for every integer value of x.
Therefore f'(x)=0 for every integer value of x, including negative ones.
Therefore f'(x)+f'(-x)=0 for every integer value of x.
Therefore your formula, which is a sum, will also be 0.

njperrone
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jaap wrote:The example was
f(x) = { 1-cos(2pi x) for x>=0; cos(2pi x)-1 for x<0 }
The cosine here uses radians of course.

Differentiating gives
f'(x) = { 2pi sin(2pi x) for x>=0; -2pi sin(2pi x) for x<0 }

The sine function is zero at all multiples of 2pi, i.e. sin(0) = sin(2pi) = sin(4pi) = ... = 0.
Therefore, sin(2pi x) = 0 for every integer value of x.
Therefore f'(x)=0 for every integer value of x, including negative ones.
Therefore f'(x)+f'(-x)=0 for every integer value of x.
Therefore your formula, which is a sum, will also be 0.

Creative argument. But at the same time you could argue that the sine function should never leave a value other than zero, and the cosine function should never have a value other than one. Because pi/2 is a multiple of 2pi because 2pi*1/4=pi/2 but sin(pi/2)=1. Creative but graphing and an example such that i have given have proven it wrong. on a side note, I just used the reductio ad absurdom metheod; my first time too. haha. Nice argument, but wrong.

jaap
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njperrone wrote:
jaap wrote:The example was
f(x) = { 1-cos(2pi x) for x>=0; cos(2pi x)-1 for x<0 }
The cosine here uses radians of course.

Differentiating gives
f'(x) = { 2pi sin(2pi x) for x>=0; -2pi sin(2pi x) for x<0 }

The sine function is zero at all multiples of 2pi, i.e. sin(0) = sin(2pi) = sin(4pi) = ... = 0.
Therefore, sin(2pi x) = 0 for every integer value of x.
Therefore f'(x)=0 for every integer value of x, including negative ones.
Therefore f'(x)+f'(-x)=0 for every integer value of x.
Therefore your formula, which is a sum, will also be 0.

Creative argument. But at the same time you could argue that the sine function should never leave a value other than zero, and the cosine function should never have a value other than one. Because pi/2 is a multiple of 2pi because 2pi*1/4=pi/2 but sin(pi/2)=1. Creative but graphing and an example such that i have given have proven it wrong. on a side note, I just used the reductio ad absurdom metheod; my first time too. haha. Nice argument, but wrong.

You do understand that [imath]\sum_{x=0}^{\infty}[/imath] means summing over integer values of x, that it does not include any term with say, x=1/2 ?
Last edited by jaap on Sun Sep 27, 2009 8:43 pm UTC, edited 1 time in total.

njperrone
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jaap wrote:...integer value of x.

Also another reason to make things i assumed to be known more explicit. I assumed that it was common knowlege that x implied all real numbers and n implied all real integers.

jestingrabbit
Factoids are just Datas that haven't grown up yet
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njperrone wrote:
jaap wrote:The example was
f(x) = { 1-cos(2pi x) for x>=0; cos(2pi x)-1 for x<0 }
The cosine here uses radians of course.

Differentiating gives
f'(x) = { 2pi sin(2pi x) for x>=0; -2pi sin(2pi x) for x<0 }

The sine function is zero at all multiples of 2pi, i.e. sin(0) = sin(2pi) = sin(4pi) = ... = 0.
Therefore, sin(2pi x) = 0 for every integer value of x.
Therefore f'(x)=0 for every integer value of x, including negative ones.
Therefore f'(x)+f'(-x)=0 for every integer value of x.
Therefore your formula, which is a sum, will also be 0.

Creative argument. But at the same time you could argue that the sine function should never leave a value other than zero, and the cosine function should never have a value other than one. Because pi/2 is a multiple of 2pi because 2pi*1/4=pi/2 but sin(pi/2)=1. Creative but graphing and an example such that i have given have proven it wrong. on a side note, I just used the reductio ad absurdom metheod; my first time too. haha. Nice argument, but wrong.

njperrone, you are wrong. All that jaap has done is use the periodicity of the sine function. There is no way that you could use that periodicity to show that sine or cosine are constant, but it is elementary to observe that if you only look at values separated by an integer multiple of the period you only get one value. Jaap has used this to create a counterexample to your theorem.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

jaap
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njperrone wrote:
jaap wrote:...integer value of x.

Also another reason to make things i assumed to be known more explicit. I assumed that it was common knowlege that x implied all real numbers and n implied all real integers.

And I thought is was common knowledge that sums can only be taken over a countable number of things, and Sigma notation implies that it is over integers. Summing over an uncountably infinite number of things is impossible, unless you use infinitesimals and that just results in integration.
Last edited by jaap on Sun Sep 27, 2009 8:54 pm UTC, edited 1 time in total.

njperrone
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Then all that needs to be done is to imply that there is a set, the real numbers, that values of x come from, denote this as xi and this will take into account fractional values of x. and it has been shown that sin(x) is 0 for k+pi where k is an integer, and 1 for k+pi/2 where k is an integer.

njperrone
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jaap wrote:And I thought is was common knowledge that sums can only be taken over a countable number of things, and Sigma notation implies that. Summing over an uncountably infinite number of things is impossible, unless you use infinitesimals and that just results in integration.

That statement is itself a contradiction. Because infinity itself is uncountable.

jaap
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njperrone wrote:
jaap wrote:And I thought is was common knowledge that sums can only be taken over a countable number of things, and Sigma notation implies that. Summing over an uncountably infinite number of things is impossible, unless you use infinitesimals and that just results in integration.

That statement is itself a contradiction. Because infinity itself is uncountable.

The integers are countably infinite, which is what I meant with countable in the above. The reals are uncountably infinite. The latter means roughly speaking that you cannot use an index number xi to count off all the reals, even if you could count for an infinite amount of time.

njperrone
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Thank you again, just requires further investigation.

majikthise
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I know it's not really my place to say this, but the above discussion seems to have diverged from the purpose of this thread. Perhaps njperrone should eat humble pie and admit he(?) was wrong, or instead create a new thread for addressing the problems with his approach?
Is this a wok that you've shoved down my throat, or are you just pleased to see me?

njperrone
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majikthise wrote:I know it's not really my place to say this, but the above discussion seems to have diverged from the purpose of this thread. Perhaps njperrone should eat humble pie and admit he(?) was wrong, or instead create a new thread for addressing the problems with his approach?

Well, it kinda did. I posted something I thought to have been right. And I was defending it under the conditions I thought I knew to be right. But when they mentioned that the summation cannot have non-integer values that is when it fell apart. Because they were arguing that from the start, and I thought they were arguing something else. So it was a divergence, yes. But any more aruing about it I feel is over because I have realised what they were pointing out (finally). So thanks jaap, I just wish either you had mentioned that before, or that I picked up on it.

toastedcrumpets
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Ok, to introduce myself to the board and to try and bring this thread back on topic... I have an Erdos number of 4, at least that's the best I can do using the AMS search engine.

M.Bannerman & William G. Hoover, paper
William G. Hoover & William Moran paper
William Moran & Harold S. Shapiro book
P. Erdős & Harold S. Shapiro paper

That seems to be about par for the course though.

SidMason
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Yep my Erdos number is 1 yep.

atomictown
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Ok, I have a Bacon number of 2 (I worked on the national parks documentary, and appeared on camera in one scene, which was narrated by tom hanks, who was in apollo 13 with Bacon). So if anyone here has an erdos number of 1 we could joint publish something and have an erdos-bacon number of 4, which would be a new record.

Starx
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Klotz
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My first paper has been accepted, and it will give me an Erdos number of 7. I'm not sure that is the lowest I can get; my coauthor is not in the collaboration distance calculator so I had to investigate and find a link.

I have an Erdos number of 5 if you count my undergrad thesis, or the footnote in my old supervisor's paper that mentions me.

jstnice
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Note that in most cases Erdös numbers are an upper bound. Particularly for people with higher numbers, there's always the possibility an obscure publication or unexplored path will reduce their numbers. An individual's number may decrease if any of the authors in their path publishes a paper with anybody whose number is less than their previous predecessor, or if a new shorter path is created when an individual's Erdös number is reduced. silverhammermba
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I recently finished my undergraduate thesis. I'm not entirely sure if it will be published by anyone and technically it doesn't have a coauthor, but if you count my adviser then that would give me a number of 4 (I think).

makc
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I did not published anything, so I guess my number is +∞

vatar
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My paper has been accepted to Congress on Evolutionary Computation 2010. That puts my Erdos number at 5.
Erdos -> Oved Shisha -> Nehad Morsi -> Etienne Kerre -> Martine De Cock -> me

Something Awesome
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I'm a coauthor of a paper that was just accepted to Fractals, which will give me an Erdös Number of 3! (that's 3 excited, not 3 factorial)

Aardvarki
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makc wrote:I did not published anything, so I guess my number is +∞

Makc, your number is undefined, not infinity. Even an infinite number of links is insufficient to link yourself to Erdos.

.. As is mine. Though, my UG senior project was with a professor with Erdos number 4, however we did not publish anything, so I lost out on the chance there to define my Erdos number.
-Aa
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jaap
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Aardvarki wrote:
makc wrote:I did not published anything, so I guess my number is +∞

Makc, your number is undefined, not infinity. Even an infinite number of links is insufficient to link yourself to Erdos.

According to wikipedia
A person with no such coauthorship chain connecting to Erdős has an Erdős number of infinity (or an undefined one).

Infinity is the most common convention. For example The Erdös Number project uses that.

hashbrowns
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### Re:

HenryS wrote:This paper, on the "100 prisoners and a lightbulb" problem, gets me an Erdős number of 4 You know, it is absolutely cruel that you posted this... I spent like 3 hours of research time toying with the problem! Nice work by the way. Very clever set of solutions so far.
also: Yeah, I know I'm a drunk Irish cowboy ninja...

Incompetent
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masakatsu
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The abstract I wrote for my master’s thesis:

Mourfield and Medcalfe, from Capella University and Augusta State University, used a quantitative analysis approach in the post positivist worldview in a survey to determine the mean time nurses spend reviewing evidence-based research (EBR) per week. Out of 125 respondents who completed the survey, it is likely that the population of the studied hospital spends less than an hour and less than half an hour at work reviewing evidence based research. There was evidence that “knowledge of EBR may be coming from other sources, such as a specific person in a unit or the attending physician” (p.16). There is a lack of correlation between a nurse’s role, education, and evidence in determining time spends reviewing EBR. The conclusion is that the “current level of understanding of the value and use of EBR in current research is incorrect” p. 16). The strength of this work is its examination of a Magnet facility, considered an excellent indicator of EBR use by nurses, even if it was just one facility. The value of this piece is the understanding the lack of use of EBR by nurses and possibly why.

The editors cut out most of the statistical analysis. Let us face it, nurses suck at math. They also changed the title, origionally being Mean time nurses review EBR per week.

I think some mathematician is writing his graduate thesis on calculating my Erdős number.

References
Mourfield, N. M., & Medcalfe, S. (2009). How much time do you spend reviewing research per week? Nursing Management, 40(12), 13-18. doi:10.1097/01.NUMA.0000365464.94098.a7

You think you have it bad, I teach Intro to Project Management to Undergrads.

Sagekilla
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My first actual published article! My prof emailed me and told me that the referees for Physical
Review Letters only had two minor issues with formatting, but it was otherwise good to go.

Not quite sure if I have an Erdos number yet though.

(Semi old) Pre-print: http://arxiv.org/abs/1008.1258
http://en.wikipedia.org/wiki/DSV_Alvin#Sinking wrote:Researchers found a cheese sandwich which exhibited no visible signs of decomposition, and was in fact eaten.

ImTestingSleeping
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I just did a bit of research (https://files.oakland.edu/users/grossman/enp/Erdos1.html) and it turns out:

My Erdos number is 3!

And I'm a third year undergrad! This is really cool! I have to admit though, the coauthor of my paper who was listed on that site is someone I have never met. My advisor and me collaborated with him through email though, so I think it's legit!

user2.0
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http://www.math.ksu.edu/~pinner/RevisedHeilbronn3.pdf
Published in Journal of Number Theory Volume 125, Issue 2, August 2007, Pages 289-297. This gives me Erdős 3 via Pinner.

http://www.nsc.ru/EMIS/journals/INTEGERS/papers/j34/j34.pdf
Published in INTEGERS 9 (2009), 435-440.

Zach 739085133
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vatar wrote:My paper has been accepted to Congress on Evolutionary Computation 2010.

That's cool. I was published there too, in 2004.

V. Gordon and Z. Matley (2004). Evolving Sparse Direction Maps for Maze Pathfinding. IEEE Congress on Evolutionary Computation (CEC2004), Portland, OR, pp 835-838.

One path goes: Erdös → Blass → Mitavskiy → Rowe → Whitley → Gordon → Matley (me)

Jyrki
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My Erdös number is 3. Just in case somebody is building stats on this ex-kgb
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I'm just out of highschool and I'll soon have an erdős number of 4 for my paper in graph theory. I won't post my paper right now, but I'll say a main result is proving that [imath]\lceil\Delta\rceil(G) \geq Z(G)[/imath], where [imath]\lceil a\rceil(G) = sup\{ a(H) | H \preceq G \}[/imath] for any graph parameter [imath]a(G)[/imath].
Last edited by ex-kgb on Fri Sep 02, 2011 1:29 am UTC, edited 1 time in total.

silverhammermba
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My paper was published in the Journal of Number Theory. Unfortunately there doesn't seem to be a free way to access it... all I have is the reference info.

J. Number Theory
131 (2011), 2442-2460.

rukhin
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### Re:

3.14159265... wrote:Anyone here working on the 3x+1 problem? I love it, and want to know some good articles written on it already.

I recommend Lagarias' recent release on "The 3x+1 Conjecture". Great starting point