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Heart

I think we need a new math thread! (Thanks, Heart, for pointing me at this thread...)

Firstly, the focus on the RH is the location of the zeros of the Riemann Zeta function. There are trivial and non-trivial zeros. Let us look at one zeta equation:

ddcbd58d5a228364ab9da20f5cd59495.png

We have some Pi's in the equation, and we can see that this will go to zero whenever the sine function equals zero, and that is when "s" (and "s" is a complex number) is equal to -2, -4, -6, -8, etc. That is, whenever the numerator in the sine function is a multiple of minus Tau.

These are the "trivial zeros" of the function. They need Tau's, not just Pi's.

Further, if we take the derivative of the zeta function and evaluate them at these trivial zeros, we get results with Tau's. For instance:

taus.jpg

If I didn't make an error the general version of the first two equations is

zeta'(-2n) = (-1)^n(2n)!/(2*tau^(2n)) zeta(1+2n) for integer n > 0

(sorry for not doing Latex).

This follows from the reflection formula that Kelly wrote down earlier.

Calculating zeta'(0) seems a bit trickier but I think it follows from the alternating series Kelly wrote down later along with the Wallis product for pi.

Positive even numbers also give values determined from Tau (due to the two pi in the equation):

1ac2fe4a57481a891fff2dfffe2cb19c.png

If anyone wants to try and prove this, here's a hint. Start with the equation

cot x = 1/x + sum_(r=1...inf) [1/(x-r*pi) + 1/(x+r*pi)]

(This formula follows by logarithmically differentiating Euler's sine product.)

From here you just need to differentiate. Differentiating both sides and analysing the x->0 behaviour gives you, with some algebra, zeta(2) = pi^2/6.

To get the general relation for zeta(2n) that Kelly wrote down you might want to make use of the definition of the Bernoulli numbers as the following generating function:

x/(e^x - 1) = sum_r B_r x^r/r!

Two further exercises: can you derive zeta(-2n-1) - i.e. the value of zeta at the negative odd integers? Hint: Kelly's reflection formula again! Also can you get the values of:

1/1^3 - 1/3^3 + 1/5^3 - ...

1/1^5 - 1/3^5 + 1/5^5 - ...

?

One final small puzzle: as Kelly writes

Firstly, the focus on the RH is the location of the zeros of the Riemann Zeta function. There are trivial and non-trivial zeros. Let us look at one zeta equation:

ddcbd58d5a228364ab9da20f5cd59495.png

We have some Pi's in the equation, and we can see that this will go to zero whenever the sine function equals zero, and that is when "s" (and "s" is a complex number) is equal to -2, -4, -6, -8, etc. That is, whenever the numerator in the sine function is a multiple of minus Tau.

These are the "trivial zeros" of the function. They need Tau's, not just Pi's.

Why doesn't this show that zeta has trivial zeroes at the *positive* even integers too? After all the sine function in that formula does go to 0 at these values too...

Oooo oooo, more of the puzzle! With you two around, how can I not get better at maths? :D

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Kelly

If anyone wants to try and prove this, here's a hint. Start with the equation

cot x = 1/x + sum_(r=1...inf) [1/(x-r*pi) + 1/(x+r*pi)]

(This formula follows by logarithmically differentiating Euler's sine product.)

From here you just need to differentiate. Differentiating both sides and analysing the x->0 behaviour gives you, with some algebra, zeta(2) = pi^2/6.

.

Fascinating!

Why doesn't this show that zeta has trivial zeroes at the *positive* even integers too? After all the sine function in that formula does go to 0 at these values too...

.

Good point. The equation is:

ddcbd58d5a228364ab9da20f5cd59495.png

The sine function does go to zero at s = even integers. However, this is the wrong equation to use for even integers. One of the formulae that I used and plotted would be better. Easiest of all for even integers is the one that I showed using Bernoulli numbers

The problem is the gamma function. For instance, let us use s = 2. The equation becomes:

zetagamma.jpg

We have gamma of -1, which really cannot be utilized. For example, let us plot the gamma function on the real axis. Using Mathematica, I have:

gamma.jpg

We have a zero from the sine function, but gamma at the negative integers give a plus or minus infinity (depending on from which direction we approach it). Using the reflection formula for even integers would be difficult. One can use limits, but I can save that for another day.

Calculating zeta'(0) seems a bit trickier but I think it follows from the alternating series Kelly wrote down later along with the Wallis product for pi.

.

I agree that it can be tricky. But, yes, using the Wallis formula can make it easier and I was not thinking of it. We see it here:

http://mathworld.wolfram.com/WallisFormula.html

But what I did instead is use a formula that I saw in Apostol's 1985 article, Formulas for Higher Derivatives of the Riemann Zeta Function, in the Mathematics of Computation journal.

The general formula is:

apos1.jpg

But to find zeta'(0), we set k = 1 and s on the LHS to 1, but that gives us a zeta(1) on the RHS, so we need help (because zeta(1) is not defined; it is a danger zone and should be avoided unless for exploring it for fun). Further down, we have:

apos2.jpg

But we still need help for the zeta(1) on the RHS. So, he addresses that and we come to:

apos3.jpg

The coefficients really are not shown in (7), but they are listed at the end. Nonetheless, for n = 1 (first derivative) the sum term vanishes, and we have:

apos4.jpg

Or:

apos5.jpg

Leaving the imaginary, we have:

apos6.jpg

Two further exercises: can you derive zeta(-2n-1) - i.e. the value of zeta at the negative odd integers? Hint: Kelly's reflection formula again! Also can you get the values of:

1/1^3 - 1/3^3 + 1/5^3 - ...

1/1^5 - 1/3^5 + 1/5^5 - ...

?

.

Perhaps next time. ;)

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michaeld

If anyone wants to try and prove this, here's a hint. Start with the equation

cot x = 1/x + sum_(r=1...inf) [1/(x-r*pi) + 1/(x+r*pi)]

(This formula follows by logarithmically differentiating Euler's sine product.)

From here you just need to differentiate. Differentiating both sides and analysing the x->0 behaviour gives you, with some algebra, zeta(2) = pi^2/6..

Fascinating!

Anyone care to try this calculation? If you do it will give a huge hint for evaluating the alternating odd-degree sums:

1/1^3 - 1/3^3 + 1/5^3 - 1/7^3 + ...

1/1^5 - 1/3^5 + 1/5^5 - 1/7^5 + ...

1/1^7 - 1/3^7 + 1/5^7 - 1/7^7 + ...

...

Why doesn't this show that zeta has trivial zeroes at the *positive* even integers too? After all the sine function in that formula does go to 0 at these values too...

.

Good point. The equation is:

ddcbd58d5a228364ab9da20f5cd59495.png

The sine function does go to zero at s = even integers. However, this is the wrong equation to use for even integers. One of the formulae that I used and plotted would be better. Easiest of all for even integers is the one that I showed using Bernoulli numbers

The problem is the gamma function. [...]

We have gamma of -1, which really cannot be utilized. For example, let us plot the gamma function on the real axis. Using Mathematica, I have:

gamma.jpg

We have a zero from the sine function, but gamma at the negative integers give a plus or minus infinity (depending on from which direction we approach it). Using the reflection formula for even integers would be difficult. One can use limits, but I can save that for another day.

Yep! The answer to this puzzle is that the gamma function has simple poles at the non-positive integers.

One way of seeing this is to remember that for non-negative integers Gamma(n+1) = n! that is n factorial. The factorial property is that

n! = n(n-1)!

The corresponding formula for the gamma function is

Gamma(z+1) = z*Gamma(z)

and this is required to hold everywhere.

But then as

Gamma(z) = Gamma(z+1)/z

if you let z->0 then the numerator Gamma(z+1) -> Gamma(1) = 1 (remember this is 0 factorial = 1) but the denominator goes to 0. So Gamma(z) blows up as z->0.

Similarly by iterating the above, for a positive integer n:

Gamma(z) = Gamma(z+n+1)/[z(z+1)...(z+n)]

and if you let z->-n, then the numerator Gamma(z+n+1) -> 1 and the denominator -> 0 due to the z+n factor, so Gamma(z) blows up at -n.

In fact Gamma is holomorphic at every point of the complex plane except for 0,-1,-2,... where it has a simple pole. Similarly Zeta is holomorphic except at the point 1, where it also has a simple pole.

Calculating zeta'(0) seems a bit trickier but I think it follows from the alternating series Kelly wrote down later along with the Wallis product for pi.

.

I agree that it can be tricky. But, yes, using the Wallis formula can make it easier and I was not thinking of it. We see it here:

Thanks for the reference! Interesting stuff indeed.

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littlepersonparadox

I'm in awe of the matheness going on here. I feel like small potatoes compared to everyone else since i don't know calculus.

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Heart

This makes me so glad I did this thread :D I have enough going on right now to not have time to properly digest this, but it has gone on the list of interesting things to ponder on on my free time. Kelly, Mic, you are so awesome!

I'm loving how we've gone from IB troubles and high school musings, to post-PhD/Masters people math musings. It makes me excited to see where this goes in the future!

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Kelly

Two further exercises: can you derive zeta(-2n-1) - i.e. the value of zeta at the negative odd integers? Hint: Kelly's reflection formula again!

.

I think that I might. I will use -2n+1 and start with 1. To establish this expression to "s" and to 2n and others (which I will use), I made a table:

table.jpg

Here is the reflection formula:

ref1.jpg

We see that s = -2n+1 (the odd negative numbers that give s), s-1 = -2n, and 1-s = 2n. This gives:

ref2.jpg

Further, the sine term will be evaluated at [negative] odd integers, and multiplied by pi/2, that gives plus and minus one, with a negative one at s = -1, -5, -9..., and a positive one at -3, -5, -7....

We have established on the table that this reduces to (-1)^n.

Putting negative exponents in the denominator and substituting the sine function as described, we now have:

ref3.jpg

Gamma of a positive integer is given by: Gamma(x) = (x-1)! Our 1-s is 2n so we want Gamma(2n) and that is (2n-1)!

Lastly, we have the zeta of even integers. We already know the equation for even positive integers:

bern.jpg

Putting all of those together gives:

new.jpg

We can simplify. The (2n-1)!/(2n)! gives 1/2n. The (-1)^n times (-1)^n+1 gives (-1)^2n-1. But 2n-1 is always an odd number. This simply becomes -1. Other terms that cancel can be eliminated.

We have:

new2.jpg

Now we substitute back in s (2n = 1-s):

new3.jpg

That looks pretty simple. Does it work? Here is a check. And in the check, I used negative integers, including even integers. It still works (and that is because, besides B(1), all Bernoulli numbers for odd integers are zero; and in this case, that translates to negative even integers for s):

table2.jpg

So, for odd negative integers (and any negative integer), zeta(s) is given by:

new3.jpg

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Dodecahedron314

This makes me so glad I did this thread :D I have enough going on right now to not have time to properly digest this, but it has gone on the list of interesting things to ponder on on my free time. Kelly, Mic, you are so awesome!

I'm loving how we've gone from IB troubles and high school musings, to post-PhD/Masters people math musings. It makes me excited to see where this goes in the future!

If it progresses anything like my IB troubles and high school musings did IRL, probably topology. Which according to the IBL script for this week, we're learning about soon in my math class...this'll be interesting.

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michaeld

Yep! Nicely done Kelly. And thanks Heart for starting this thread. :cake:

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Kelly

In other news, the New Horizons spacecraft reached Pluto (flyby) today. We will get pictures soon. It is the fastest spacecraft yet.

*celebrates the journey to Pluto* Remember when Pluto was a planet? :)

*also celebrates Heart's awesome work at the LHC*

In 1998, Homer Simpson appears to have determined the mass of the Higgs Boson (to within an order of magnitude: ~775GeV as opposed to ~125GeV):

o-SIMPSONS-570.jpg

This is many years before Heart and her colleagues finally found the Higgs Boson at CERN's LHC. Impressive.

Below that, he gives an example of a contradiction to Fermat's Last Theorem. Type that into your calculator or whatever you have. Is he wrong? ;)

Below that, he has an equation showing that the density parameter Omega of the universe is more than unity. This suggests that the universe is closed and may come back together in a big crunch.

That was a possibility in 1998. However, years later, sciency-type cosmologist peeps determined that with so much dark energy, the universe's expansion is not slowing; indeed, it is even expanding at an increasing rate!

Since Homer is rather prescient in these matters, he later, in this episode, erases the inequality sign and replaces it with the correct "<" sign.

Below that is a topological puzzle. He shows how a torus (such as a donut with one hole) is homeomorphic to a sphere (or, rather, a solid ball). This is similar to saying that a beer mug with a handle is topologically equivalent to a wine glass. We know that a torus and a sphere are not topologically homeomorphic. But Homer solves this ingeniously. If you take a donut and take a few bites out of it,

HomerDonut.jpghomer_eating_Donut.jpg

the hole disappears.

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Heart

Below that is a topological puzzle. He shows how a torus (such as a donut with one hole) is homeomorphic to a sphere (or, rather, a solid ball). This is similar to saying that a beer mug with a handle is topologically equivalent to a wine glass. We know that a torus and a sphere are not topologically homeomorphic. But Homer solves this ingeniously. If you take a donut and take a few bites out of it,

HomerDonut.jpghomer_eating_Donut.jpg

the hole disappears.

I don't know if I believe your assertion there Kelly. As an experimental physicist, I think we need a few more experiments to reach statistical significance.

Anyone up for helping me bite a few donuts? ;)

This makes me so glad I did this thread :D I have enough going on right now to not have time to properly digest this, but it has gone on the list of interesting things to ponder on on my free time. Kelly, Mic, you are so awesome!

I'm loving how we've gone from IB troubles and high school musings, to post-PhD/Masters people math musings. It makes me excited to see where this goes in the future!

If it progresses anything like my IB troubles and high school musings did IRL, probably topology. Which according to the IBL script for this week, we're learning about soon in my math class...this'll be interesting.

Your wish is Kelly's command apparently :D

Though who else could ever have done with with donuts? Kelly is my new hero. Well, that's a lie, she's been a hero for a long time now :P

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Kelly

I don't know if I believe your assertion there Kelly. As an experimental physicist, I think we need a few more experiments to reach statistical significance.

Anyone up for helping me bite a few donuts? ;)

I would love to collaborate on that project. 8)

Below that, he gives an example of a contradiction to Fermat's Last Theorem. Type that into your calculator or whatever you have. Is he wrong? ;)

:)

Anyone?

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michaeld

Below that, he gives an example of a contradiction to Fermat's Last Theorem. Type that into your calculator or whatever you have. Is he wrong? ;)

:)

Anyone?

You don't need a calculator. :P You can look at the remainders when divided by 4. The remainder of 3987 is 3 or -1 so raising it to the power of 12 we get 1 mod 4. Similarly 4365^any power leaves a remainder of 1 when divided by 4. So their sum leaves a remainder of 2. But 4472^12 is divisible by 4. So the sum cannot be true.

But someone who has a calculator should try it. :-p I think I can guess what will happen...

Oh and btw.... Heart! Kelly! No mention at all of the possible pentaquark discovery at the LHC? :huh:

EDIT: OK I don't have a normal calculator but I do have the google calculator. If you put in either

3987^12 + 4365^12

or

4472^12

then the google calculator gives you 6.3976656e+43!

But if you put in 3987^12 + 4365^12 - 4472^12 you get 1.2118938e+33...

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Kelly

You don't need a calculator. :P You can look at the remainders when divided by 4. The remainder of 3987 is 3 or -1 so raising it to the power of 12 we get 1 mod 4. Similarly 4365^any power leaves a remainder of 1 when divided by 4. So their sum leaves a remainder of 2. But 4472^12 is divisible by 4. So the sum cannot be true.

:) That is the smart way to prove Homer wrong. 8)

EDIT: OK I don't have a normal calculator but I do have the google calculator. If you put in either

3987^12 + 4365^12

or

4472^12

then the google calculator gives you 6.3976656e+43!

But if you put in 3987^12 + 4365^12 - 4472^12 you get 1.2118938e+33...

:) Excel can handle that, too (as can Mathematica, but Mathematica us too uber awesome to use as a simple calculator). This seems to be a near miss for a FLT contradiction. Can we think of closer ones?

Oh and btw.... Heart! Kelly! No mention at all of the possible pentaquark discovery at the LHC? :huh:

:cake: to Heart and her colleagues! (again!) *celebrates*

I read about that and was hoping that someone would post about it, else I would have tomorrow (I almost used that as the hangman word). It had been predicted years ago. Now I need to think of what properties it would have. And I wonder how we could use it.

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nerdperson777

I don't have my math degree yet! *whines because not able to understand yet*

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michaeld

If you use a system that can handle large integers, e.g. Python, you can calculate it exactly.

3987^12 + 4365^12 - 4472^12 = 1211886809373872630985912112862690.

Yep the theory of pentaquarks has been around for years. I remember quite well attending a research seminar on them over a decade ago, but their theoretical prediction goes way back to the 60s or 70s I think.

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Heart

Below that, he gives an example of a contradiction to Fermat's Last Theorem. Type that into your calculator or whatever you have. Is he wrong? ;)

:)

Anyone?

You don't need a calculator. :P You can look at the remainders when divided by 4. The remainder of 3987 is 3 or -1 so raising it to the power of 12 we get 1 mod 4. Similarly 4365^any power leaves a remainder of 1 when divided by 4. So their sum leaves a remainder of 2. But 4472^12 is divisible by 4. So the sum cannot be true.

But someone who has a calculator should try it. :-p I think I can guess what will happen...

Oh and btw.... Heart! Kelly! No mention at all of the possible pentaquark discovery at the LHC? :huh:

Dang. I didn't think to use remainders. I really should have thought of that, I know this trick. It just didn't register :P

Too much physics in my life recently!

And yes, the pentaquark is exciting!! It's so hard to measure and show experimentally, this is yet another triumph for the theoretical particle physics understanding of the world! The various groups at the LHC are picking away at so many exciting things. Every time something like this is given in a press release I get so excited to see what's next. I was just out for beers with some people from ATLAS, and they have some stuff in their back pocket too it sounds like. LHCb is the one that published about the pentaquark, and unfortunately I don't know as many people from that collaboration. But regardless, it's exciting news :3

For those who don't know much about quarks, I shall attempt my own little story about them for you.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Quarks come in colours. There are six colours, three "colours" and three "anticolours". So the three colours are red, green and blue. The three anti-colours are yellow (or anti-blue), magenta (or anti-green) and cyan (or anti-red).

Notice a pattern? Well, green, red and blue form a basis of primary colours (for light, not paint ;) ). And so do cyan, magenta and yellow! If you have trouble believing me on the second set, check your computer printer. If it's a colour printer, it'll have cyan, magenta and yellow as colours, and it can make all the colours you print from those three. While these are only two examples of sets of primary colours, they are the ones we will use.

So, if we all had coloured lights around, we can experiment with mixing them. If we mix red and blue, we get a magenta colour. If we mix green and blue, we get cyan. Again, you have to think in terms of the way light mixes, not paint, because paint is a whole different kettle of fish. Mix all the colours of paint together and you'll get black, but mix all the colours of light together and you get white ;) But if there are any artsy people around, you'll know how all this works :D

So, on with the story. There's a fundamental rule saying that all quarks have to be in combinations that make the whole thing white. No ifs and/or buts. So, for example, a proton is made of three quarks. One blue, one red and one green; add blue, red and green together, and you get white! An anti-proton is made of the three antiquark colours; cyan + magenta + yellow = white.

There are other types of particles too. There are mesons, which have only two colours! They can be made with one quark and one anti-quark. For example, if I have a blue quark, I could pair it up with an anti-blue quark. If we remember from just above, the anti-blue is yellow. So yellow + blue = white. If you don't believe me, there are many good simulations out there for mixing light colours, just a google away :)

The big new thing now though, is this pentaquark. Can anyone think of a way of combining five quarks or antiquarks, or some combination thereof, to make white?

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Kelly

Fascinating. :)

Will these work?

I will use different terminologies in the same equations because it is easier:

cyan + magenta + yellow + blue + anti-blue

cyan + magenta + yellow + green + anti-green

cyan + magenta + yellow + red + anti-red

blue + green + red + cyan + anti-cyan

blue + green + red + magenta + anti-magenta

blue + green + red + yellow + anti-yellow

cyan + magenta + yellow + cyan + anti-cyan

cyan + magenta + yellow + magenta + anti-magenta

cyan + magenta + yellow + yellow + anti-yellow

blue + green + red + blue + anti-blue

blue + green + red + green + anti-green

blue + green + red + red + anti-red

Maybe?

I remember when I did my senior thesis on the colour of wine, as judged without opening the bottle. I needed to determine the transmission colour spectrum of the bottle first, and found that although it looked green, it was actually yellow and blue.

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Heart

Well done Kelly! I fear such a puzzle was too easy for you ;) I'm glad you put the answer in spoilers so others can still have a try!

Addition of light is an interesting thing. But fortunately, light addition is only an analogy for quark colours; they do not in fact "look" a colour. It would be kind of an odd world if anything made of quarks was white, eh? I wonder what colour electrons would be in such a world... (electrons are not made of quarks like protons and neutrons are, but they are a new type of particle called a lepton, and have no colour).

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Kelly

Sixty Symbols did a neat little blurb on the pentaquarks discovered by Heart and colleagues:

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littlepersonparadox

Ahhhhh - relish the math

Well done LHC. You have only been re-open for a short amount of time but giving us very nice work already :).

As for the simpson fermat trick i head David s. Coine used a computer program to calculate that equation to bugger the math nerds. :P Also if you look at Homer Treetop House of Horrors Fermat's last theorem comes back in the higher dimension.This time in slightly different equation. Along with another famous equation hanging over homers head: P= NP? A rather crude mark on homers intelligence as well as all us computer science geeks. :P

Alternatively interestingly enough another show written by one of the co-writers of the simpsions suggests in a episode of Futurama that P is NOT equal to NP. As shown by a bookshelf behind Fry that has two separate books one labeled P and another labeled NP.

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Kelly

Awesome. :)

In a recent episode of The Simpsons, P vs NP appeared to be combined with the Traveling Salesman problem, or TSP. It had other equations, too. I recognized them (nerd moment) and took a screen dump:

http://www.asexuality.org/en/topic/35758-the-to-be-to-be-longest-thread-a-reprise/page-1549#entry1060972518

I also recently watched a fictional 2012 movie where a group of mathematicians solved P vs NP; the movie was called Travelling Salesman.

The Simpson peeps are as much of a group of math geeks as we are. :P

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littlepersonparadox

Probably because the head writers David s. Coine and David X. Coine (he changed his middle initial to X after joining the show) both have math degrees of there own. And not low level either. Because of that we get lots of math in both futureama and Simpsons.

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Heart

I love when I recognise little things like that on my own! I admit that it's not often, as I am usually watching things like that after a long day of work and my brain is no longer functioning, but when I do, I get all excited and stuff :D

Nerds are everywhere. We will rule the world one day ;)

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Kelly

So the mathematicians go to the bar across the street. It has two bartenders. The math peeps then queue into two lines, and the pouring of beers starts. Now we have two infinite lines of mathematicians (yet both lines are themselves infinite, so we have twice as many patrons as before, no?). :P

The first line has orders for one beer, 1/3, 1/5, 1/7, ... or (1/2n-1).

The second line has orders for 1/2 of a beer, 1/4, 1/6, 1/8, ... or (1/2n).

We still have the same divergent series as before, just split up into two. As expected, both bartenders will need to pour an infinite amount of beer each.

If both bartenders pour an infinite amount of beer, what is the difference in beer poured between them?

The bar at the Hilbert Hotel now has a third bartender; so, now there can be three infinite lines of mathematicians.

Rather than splitting up into lines where every third one is in a different line (with Rodd serving mathematicians 1, 4, 7, 10..., Pasquare serving 2, 5, 8, 11..., and Steven serving 3, 6, 9, 12...), each mathematician enters the door, takes a number, and then queues in the line according to how many factors the number has. For instance, primes have only one factor, themselves. Nonprimes may have two or three or more. For instance, 30 has three (30 = 2*3*5) so it has an odd number of factors. Six has two (6 = 2*3) so it has an even number. But what about, say, Twelve? 12 = 2*2*3. Does it have two factors (2 and 3) or three (2, 2, and 3)?

In these cases where there is the same prime factor more than once (a number divisible by a square), this can be the third category. We now have Rodd serving the math patrons with a number that has an odd number of factors, Steven serving those with a number with an even number of factors, and, in the middle line, Pasquare serving those with a number divisible by a square.

I watched the first million patrons being served and there were 300+ thousand served in each line, so the lines are equal enough to work.

As before, each will be served a glass with the amount of beer being one pint divided by their number.

To make this a little fairer, each mathematician will be served in the order of their number, N. This means that Rodd may serve two in a row (since 2 and 3 are both primes and thus both have an odd number of factors), but Pasquare will serve patrons 8 and 9 in a row because both are divisible by a square, and Steven will serve patrons 14 and 15 in a row because they both have two factors (an even number of factors).

But most of the time is taken in grabbing a glass and handing it to the mathematician patron. In all, there is a fairly constant movement of all three lines. And the bartenders do not need to think about when to serve the next patron, they just pour from their beer gun into a glass and pass the glass onto the patron when the beer gun stops pouring beer and starts to flow CO2 gas. The bartender then grabs another glass and waits until the beer flow starts and ends again, and repeats.

This is done by a silly but interesting system that I thought of while taking a break from AVEN and pouring a beer.

Since there is only one keg, the system needs to decide to which bartender to send the next beer. It does so by taking the number of the patron and setting up a series of 1V voltages in a circle in positions that correspond to:

VSum.jpg

Where k _|_ N means that the integers k are the coprimes of N. This will create an array of vectors that will sum to one Volt to the left (to Rodd's queue) and thus pull it over in that direction if the number N has an odd number of factors, pulled to the middle (Pasquare's) if the number is divisible by a square, and to the right (Steven's) if the number has an even number of factors.

For instance, 5 is a prime and has only one factor. It has coprimes of 1, 2, 3, and 4. Solving the sum above gives the vectors radiating from zero to:

cos[2π(1/5)] + sin[2π(1/5)]i

cos[2π(2/5)] + sin[2π(2/5)]i

cos[2π(3/5)] + sin[2π(3/5)]i

cos[2π(4/5)] + sin[2π(4/5)]i

Plotting looks like this:

CP5b.jpg

Imagine if we added the vector from (0,0) to (1,0). All vectors would sum to (0,0). But here, that vector is missing, so the sum of all of these would be (0,0) - (1,0), or, (-1,0). The sum of the vectors shown sums to (-1,0) and thus we have a potential to the left and it will pull the tap to the left. All integers N that have an odd number of factors will give vectors that sum to (-1,0) when the sum is evaluated. For instance, here is 30 (2*3*5). Its coprimes are 1, 7, 11, 13, 17, 19, 23, and 29. The vectors are:

CP30b.jpg

Here is 12. It is 2*2*3, so it is divisible by a square. The coprimes of 12 are 1, 5, 7, and 11. The vectors are:

CP12b.jpg

These sum to zero because each vector has a partner that is equal and opposite. All integers that are divisible by a square will give vectors that sum to (0,0).

Let us look at 6 (2*3). It has an even number of prime factors. Its coprimes are 1 and 5. Its vectors are:

CP6b.jpg

As always, the imaginary parts (y axis) always sum to zero since there are always a complex conjugate for every vector. And the positive parts for these two vectors are 0.5, so they sum to 1, and the vectors sum to (1,0). All integers N that have an even number of factors will give vectors that sum to (1,0) when the sum is evaluated, and the tap will be sent to Steven's system.

Between each serving, it will blow air and set up the next array of voltages. During the serving, it will send beer to the system of the intended bartender and send it for the time equal to one pintsecond (the amount of time to flow one pint) divided by N.

At the tenth round, we have the following pours and sums:

pours.jpg

Once again, I am asking what is the difference in the beer served between Rodd and Steven as we approach infinity.

(note: none of these sums converge; they all go to infinity)

For instance, let us look at Rodd's pours. He will pour an amount equal to the reciprocals of all numbers that have an odd number of factors. This includes all primes plus all integers that have three, five, etc. factors.

We already know that the sum of the reciprocals of primes diverges (proven by Euler nearly 300 years ago). This sum will have all of that, plus 1/30, 1/42, 1/70 and others, so it will sum to an even greater infinity than the sum of reciprocals of primes.

As mentioned, I watched the first million of the patrons, and I tallied the amount of beer poured by each bartender. Here is the one from Rodd:

Rodd1.jpg

When we take the log of the N (x) axis, we see:

Rodd2.jpg

This is clearly diverging. The first part is slightly choppy, but when we curve fit for the sum from 1,000 to 1,000,000, we see:

Rodd3.jpg

From the rather good curve fit equation, we can see that it will continue to climb as N climbs, and can even predict that the sum at one billion (1,000,000,000) should be about 6.821 and at one trillion (1,000,000,000,000), it should be about 8.920.

The other two bartenders will also pour an infinite amount of beer, too. At N equals one million, we see:

Others.jpg

And it is interesting to see that the amount of beer poured by Pasquare will catch up and exceed that of Steven at around N = 75,000.

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Heart

I think my brain just exploded o.O

Kelly, you're once again awesome.

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littlepersonparadox

I think my brain just exploded o.O

Kelly, you're once again awesome.

Same here. I want to take more math classes. II'd love to be able to solve stuff like that but it just was ... Huh?

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Kelly

Eeps; sorry.

The question is somewhat simple (as is the solution). We are looking for the difference in the sum of reciprocals of integers that have an odd number of prime factors and that of sum of reciprocals of integers that have an even number of prime factors (ignoring all integers that are divisible by a square number).

We have:

? = 1 - 1/2 - 1/3 - 1/5 + 1/6 - 1/7 + 1/10 - 1/11 - 1/13 + 1/14 + 1/15 - 1/17 ...

I tried to make it interesting by using the "an infinite number of mathematicians walk into a bar" scenario, and put most digressions into spoilers. But I also gave a big hint. ;)

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michaeld

I don't know if I'll give away too much but anyone interested in this puzzle should look up the Mobius function.

Actually I believe the question Kelly is asking is not an easy one, though it is not hard to guess the right answer. Proving it is non-trivial.

But to guess the right answer one just has to note that

1/zeta(s) = sum_n mobius(n)/n^s . (*)

This simply follows by inverting Euler's product for zeta and expanding it all out.

The sum Kelly wants is sum_n mobius(n)/n, which looks like it should be the limit of the right hand side above as s->1. We thus expect:

lim(s->1+) 1/zeta(s) = sum_n mobius(n)/n . (**)

This means that the answer to Kelly's problem (i.e. the right hand side of (**)) is ____ . (Fill in the blank! Just work out the limit on the left hand side.)

The above answer you get is correct but the reason proving its correctness is actually more difficult than this is that the step going (*) to (**) is not as easy as it looks. The basic problem is that one cannot swap the sum and the limit easily, because the final series one ends up with is absolutely divergent. (In other words if you forget the minus signs and just add the magnitudes up, you get infinity. This means that the classic theorems showing you when you can swap sums and limits won't help you.)

My recollection is that this particular summation was proved sometime in the 1800s but it is not trivial. It will probably be covered in a book on analytic number theory. In fact it is equivalent to the Riemann hypothesis in a special case - namely that the zeta function does not vanish on the line Re z = 1. It is also therefore equivalent to the Prime Number Theorem, as the first proofs of that first established the foregoing fact about zeta.

Basic point: this looks easy but beware, there are hidden traps and it's actually a much more subtle and difficult result than it seems at first.

(Oh and btw in the exponential sum defining the Mobius fun.. er I mean the voltages, the notation k _|_ N should mean that k is coprime to N and also that 1 <= k <= N. There are infinitely many integers coprime to N, which would make the sum diverge if there were no other restriction.)

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Heart

Eeps; sorry.

The question is somewhat simple (as is the solution). We are looking for the difference in the sum of reciprocals of integers that have an odd number of prime factors and that of sum of reciprocals of integers that have an even number of prime factors (ignoring all integers that are divisible by a square number).

We have:

? = 1 - 1/2 - 1/3 - 1/5 + 1/6 - 1/7 + 1/10 - 1/11 - 1/13 + 1/14 + 1/15 - 1/17 ...

I tried to make it interesting by using the "an infinite number of mathematicians walk into a bar" scenario, and put most digressions into spoilers. But I also gave a big hint. ;)

I don't know if I'll give away too much but anyone interested in this puzzle should look up the Mobius function.

Actually I believe the question Kelly is asking is not an easy one, though it is not hard to guess the right answer. Proving it is non-trivial.

But to guess the right answer one just has to note that

1/zeta(s) = sum_n mobius(n)/n^s . (*)

This simply follows by inverting Euler's product for zeta and expanding it all out.

The sum Kelly wants is sum_n mobius(n)/n, which looks like it should be the limit of the right hand side above as s->1. We thus expect:

lim(s->1+) 1/zeta(s) = sum_n mobius(n)/n . (**)

This means that the answer to Kelly's problem (i.e. the right hand side of (**)) is ____ . (Fill in the blank! Just work out the limit on the left hand side.)

The above answer you get is correct but the reason proving its correctness is actually more difficult than this is that the step going (*) to (**) is not as easy as it looks. The basic problem is that one cannot swap the sum and the limit easily, because the final series one ends up with is absolutely divergent. (In other words if you forget the minus signs and just add the magnitudes up, you get infinity. This means that the classic theorems showing you when you can swap sums and limits won't help you.)

My recollection is that this particular summation was proved sometime in the 1800s but it is not trivial. It will probably be covered in a book on analytic number theory. In fact it is equivalent to the Riemann hypothesis in a special case - namely that the zeta function does not vanish on the line Re z = 1. It is also therefore equivalent to the Prime Number Theorem, as the first proofs of that first established the foregoing fact about zeta.

Basic point: this looks easy but beware, there are hidden traps and it's actually a much more subtle and difficult result than it seems at first.

(Oh and btw in the exponential sum defining the Mobius fun.. er I mean the voltages, the notation k _|_ N should mean that k is coprime to N and also that 1 <= k <= N. There are infinitely many integers coprime to N, which would make the sum diverge if there were no other restriction.)

Oh, don't you two worry about it. I've got this on my "to wonder at" list for my next day off. It'll happen and I'll be blown away, as per usual ;)

525122281_o.jpg

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michaeld

Here is how to work out those alternating zeta-like sums of odd degree.

Start with:

cot z = 1/z + sum_(n>=1) [1/(z-n*pi) + 1/(z+n*pi)]

Quick remark on this sum.

It follows from the better known Euler product formula for sine:

sin z = z product_(r>=1) (1 - z^2/(r^2pi^2))

To get to the cot summation from here, take logarithms(*) of both sides, converting the product into a sum. Then differentiate both sides.

(*) this needs a bit of care due to log's branch cuts on the complex plane, but it's not a real problem.

The Euler product can be heuristically justified by noting that sine has zeroes at r*pi for integer r. Therefore one can try to factorise it, z(1-z/pi)(1+z/pi)(1-z/(2pi))(1+z/(2pi))..., which gives the above. This is what Euler did, justifying the multiplicative constant being 1 by the fact sin z / z -> 1 as z->0. Then he read off zeta(2) by equating the coefficient of z^3 on each side.

That is not rigorous but it was how the value zeta(2) = pi^2/6 was originally arrived at by Euler.

How to more rigorously justify the sine product? Actually it's easier I think to justify the cot summation using a bit of complex analysis. Then if you want to go back to Euler's product, reverse the logarithmic differentiation step carefully.

To show that cot z = 1/z + sum_(r>=1) [1/(z-r*pi)+1/(z+r*pi)] one can use some complex analysis:

a) Show that the difference between each side is periodic with period pi. (Easy!)

b) Show that the difference is holomorphic on the entire plane. (This is not hard but you need to show that the poles cancel. Basically, the left hand side has a simple pole at r*pi for integer r with residue 1. So does the right hand side. Therefore the two cancel and the difference is holomorphic.)

c) Show that the difference is bounded in the strip 0 <= Re z <= pi. (Not hard.)

From the above we know the difference is holomorphic, bounded on the strip and hence bounded in the entire plane by periodicity. Therefore the difference is constant by Liouville's theorem. https://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29

We therefore know that

cot z = c + 1/z + sum_(r>=1) [1/(z-r*pi)+1/(z+r*pi)]

for all z where c is constant. But as the left hand side is an odd function, so must the right hand side be, which means that c = 0. We're done.

One final remark about this sum. It's tempting to write it in the form:

cot z = sum_(r integer) 1/(z-r*pi)

i.e. sum over the positive and negative integers and zero simutaneously. This is a good heuristic to keep in mind, but the problem is that each side of the sum diverges separately (they both act like the harmonic series in the limit) so you have to add them together in the right way to make it converge to the right thing. Therefore it's safer to write it in the form

cot z = 1/z + sum_(r>=1) [1/(z-r*pi)+1/(z+r*pi)]

OK now we can diifferentiate this:

cot'(z) = -sum_(integer r) 1/(z - r*pi)^2 (*)

Note that it IS now safe to write the sum in this more compact form instead of separating the two sides, because the entire thing is now absolutely convergent. This follows from the fact sum 1/r^2 is finite.

If one adds 1/z^2 to each side and takes the z->0 limit one quickly finds:

lim(z->0) (cot'(z) + 1/z^2) = -1/pi^2 sum_(r not 0) 1/r^2 = -2/pi^2 zeta(2)

the last equality following from the evenness of squaring, meaning the negative and positive terms double up.

As cot'(z) = -cosec^2(z) one can quickly evaluate the left hand and show it comes to -1/12, which gives us zeta(2) = pi^2/6. But we already knew that. Let's go further.

Differentiate again:

cot''(z) = 2 * sum_(integer r) 1/(z - r*pi)^3 (**)

What happens if one tries to do the same thing to evaluate zeta(3)? Add -2/z^3 to both sides and take the limit:

lim(z->0) cot''(z) + 2/z^3 = 2 * sum_(r!=0) 1/(-r*pi)^3

Alas here we see the difficulty. Above, the negative and positive terms double up. Here they cancel - because cubing is an odd function not an even function. Hence the right does not involve zeta at all - it just becomes 0. The limit on the left is also easily seen to be 0. So the equation just gives us 0 = 0, i.e. nothing interesting.

That's a shame. But if it were that easy, mathematicians would have discovered a closed form value of zeta(3), which they haven't. So we expected that.

But can we do anything else interesting? The z->0 limit is not interesting. What other values can we try? z->r*pi is the next most obvious option, but that's really no different to z->0 because both sides are periodic with period pi.

So how about z->pi/2? If you do that you end up with another boring 0 = 0. So what about z = pi/4....

Aha now this is more interesting. If one puts z = pi/4 into

cot''(z) = 2 * sum_(integer r) 1/(z - r*pi)^3

then we get

cot''(pi/4) = 2 * sum_(integer r) 1/(pi/4 - r*pi)^3

The right is just 128/pi^3 sum_(integer r) 1/(1-4*r)^3.

What does that come to? Well for r>=1 the sum gives us:

1/(-3)^3 + 1/(-7)^3 + 1/(-11)^3 + ...

and for r<=0 it gives us

1/1^3 + 1/5^3 + 1/9^3 + ...

Aha! We now have our alternating sum. As the series is absolutely convergent we can re-arrange at will and we get:

cot''(pi/4) = 128/pi^3 * (1/1^3 - 1/3^3 + 1/5^3 - 1/7^3 + ...)

Evaluating the left by calculus we get:

1/1^3 - 1/3^3 + 1/5^3 - .... = pi^3/32

What if we keep on going? Well the same pattern keeps on happening. After differentiating cot sum k-1 times one gets:

cot^(k-1)(z) = (-1)^(k-1)(k-1)! * sum_r integer 1/(z-r*pi)^k

When k=2*m is even the z->0 limit allows us to derive the value of zeta(2m) and z=pi/4 doesn't give anything interesting beyond re-deriving zeta(2m) (try it!) One finds that zeta(2m) is always a rational multiple of pi^(2m).

When k=2*m+1 is odd the z->0 limit gives us a tautology 0 = 0 but z=pi/4 gives us the value of the alternating summation

1/1^(2m+1) - 1/3^(2m+1) + 1/5^(2m+1) - ...

as a rational multiple of pi^(2m+1).

At the moment though the above answers are all in terms of successive derivatives of cot, which you can evaluate by basic calculus. But doing it gets tedious after a while. Isn't there a simpler way to express the final answer? In fact the zeta(2m)s are all expressible interms of Bernoulli numbers and one can derive this from the above by calculating the Taylor series of cot. One can also find a recurrence for the rational coefficients appearing in the alternating summations. But I'll leave that for now.... next time maybe.

Btw I answered the question of 1/1^3 - 1/3^3 + 1/5^3 - ... a slightly different (but related) way a long time ago on another message board...

http://math2.org/mmb/thread/13108

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