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  • 4 months later...

It's been a while, but I assume this is still the AVEN maths & physics thread (disguised as a gender thread)?

 

Over the weekend I heard the sad news that the great mathematician John H Conway has died of COVID-19. Conway made important discoveries across a dizzying array of mathematical fields. He invented the surreal numbers, worked out the symmetry group of the Leech lattice, led the Atlas of Finite Groups, (jointly) formulated the Monstrous Moonshine conjectures as well as doing fundamental research on combinatorial game theory - among many other contributions. To the public he was probably best known (somewhat to his frustration!) for his Game of Life (and to a lesser extent, the game of Sprouts). Here is an amazing video on the former. Enjoy!

 

 

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  • 1 year later...

Just a brief interruption to let you all know I will be covering the Gender Discussion forum now that @Rynn has ascended to the heights of admin-ship. I want to test the waters and see if I can serve you well here. I may take over more long-term if it feels right. In the meantime, if you have any issues or questions for me feel free to post in this forum or to PM me. There are big shoes to fill here after mods like @Rynn and @Heart! :) 

 

We now return you to your regularly scheduled Math and Physics. :D 

 

09012AE7-AA8F-418E-AADCFC31D2C29EE2_sour

(image from https://www.scientificamerican.com/article/a-new-book-examines-the-relationship-between-math-and-physics/ )

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  • 2 weeks later...

And now it's official. I have stepped down from Hot Box and am now the official moderator of Gender Discussion. I hope I can serve you all well. Please report TOS violations I might miss, let me know if I get anything wrong, help new people, and look after each other. This is a great forum and I look forward to watching over it even more. :D 

 

thumb_maths-physics-memes-memes-are-life

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26 minutes ago, daveb said:

And now it's official. I have stepped down from Hot Box and am now the official moderator of Gender Discussion. I hope I can serve you all well. Please report TOS violations I might miss, let me know if I get anything wrong, help new people, and look after each other. This is a great forum and I look forward to watching over it even more. :D ...

Congrats! Moderating the gender discussion forum does make sense and seem like it'd be a good fit, for you, compared to Hot Box.

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There is a lot I don't know, so I'll be open to correction and input from all of you here. :D 

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4 hours ago, daveb said:

And now it's official. I have stepped down from Hot Box and am now the official moderator of Gender Discussion. I hope I can serve you all well. Please report TOS violations I might miss, let me know if I get anything wrong, help new people, and look after each other. This is a great forum and I look forward to watching over it even more. :D 

Welcome! :cake:❤️

 

On 9/28/2021 at 11:41 AM, daveb said:

We now return you to your regularly scheduled Math and Physics. :D 

 

 

OK. More math/physics.

 

An infinite number of mathematicians walk into a bar.

 

This time, we have a new bartender at the Hilbert Hotel Pub. His name is Faktoh Reyahl. And he is a bit too generous when it comes to serving beers, especially if you have a few mathematical patrons in front of you.

 

He serves the first mathematician one beer. He serves the second one beers. He serves the third four beers, and the fourth gets 18 beers. The fifth gets 96 beers!

 

The sum of beers served after the fifth mathematician is served is 120. This is also 5! (five factorial).

 

The number of beers served as well as the sum of beers served for the first ten mathematicians is thus:

 

beers.png

 

So, the sum of all beers served will be equal to the factorial of the patron number.

 

At the end of the night, after all mathematicians are served (an infinite number of mathematicians), how much beer in total has Faktoh Reyahl served?

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I remember taking a math class that included factorials at some point in my college career but I don't know the answer to "The Sum of All Beers".

5prce1.jpg

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Hmm this isn't far away from sum_{n>=0} (-1)^n n!, which is Borel summable to integral_0^{\infty} e^-x/(1+x)dx ~ 0.596.

 

However Kelly's version is actually 1+sum_{n>=1} [(n+1)! - n!], which is slightly different. I don't at the moment see a sensible regularisation approach for this one...

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One plausible approach towards regularisation is to note that, using the integral representation of the gamma function:

(n+1)!-n! = n*n! = integral_0^{\infty} nt^n e^-t dt.

 

Now sum over n>=1 and use sum_n nt^n = t/(1-t)^2 for |t|<1, ignoring the lack of convergence:

sum_{n>=1}[(n+1)!-n!] = integral_0^{\infty} te^-t/(1-t)^2 dt.

 

The problem is that the integrand on the right has a double pole at t=1.  So it still diverges.

 

As it's a double pole, the Cauchy integral value doesn't exist. However you can remove an inteval (1-eps,1+eps) and then the result will be of the form A + B/eps. You can regularise by getting rid of the B/eps term leaving you just with the constant A.

 

Is that approximately what you had in mind, before I go ahead and work out the details?

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15 minutes ago, michaeld said:

One plausible approach towards regularisation is to note that, using the integral representation of the gamma function:

(n+1)!-n! = n*n! = integral_0^{\infty} nt^n e^-t dt.

 

Now sum over n>=1 and use sum_n nt^n = t/(1-t)^2 for |t|<1, ignoring the lack of convergence:

sum_{n>=1}[(n+1)!-n!] = integral_0^{\infty} te^-t/(1-t)^2 dt.

 

The problem is that the integrand on the right has a double pole at t=1.  So it still diverges.

 

As it's a double pole, the Cauchy integral value doesn't exist. However you can remove an inteval (1-eps,1+eps) and then the result will be of the form A + B/eps. You can regularise by getting rid of the B/eps term leaving you just with the constant A.

 

Is that approximately what you had in mind, before I go ahead and work out the details?

Excellent! :cake: But I took a different approach. I will post my solution later today.

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The last patron would have the number infinity, so Factoh Reyahl would serve a total of infinity factorial (!). "!" indeed!

 

So, how might we evaluate !?

 

We recall the Riemann zeta function. It can be plotted on the complex plane, but let's get real and look at it as a function of the real number line.

Plotting it from x = -19 to +10, we have:

 

zeta.png

 

Note that with the exception of the break at x = 1, the curve is continuous. That means that we can differentiate it. Differentiation gives us the slope of the curve at each value of x. The differential of the zeta function ("zeta prime") is:

 

diff.png

 

There is also a formula for zeta prime. It is:

 

f1.png

 

Where "s" is, here, the x-axis, and zeta prime returns the value of y for the function.

 

This formula is actually very useful here. And most importantly we will evaluate this for s = 0 (x = 0 in our plot). Since n0 = 1 for any n, we lose the denominator, giving us:

 

f2.png

 

Now, logarithms have the property that:

 

ln(a) + ln(b) + ln(c) = ln(a*b*c), we now have:

 

f3.png

 

In the big brackets, we have 1*2*3*…*, or, !

 

f4.png

 

Now we are getting somewhere!

 

So, what is z(0)? If we look at the plot of zeta prime, it looks like that value at zero is about negative one, but not quite. It is about -0.918938… It is actually an irrational number. But it can be written in closed form. It is:

 

f5.png

 

Thus:

 

f6.png

 

And:

 

f7.png

 

We are almost there. Let's get rid of the logarithms by taking the exponent:

 

fixed.png

 

Therefore:

 

f9b.png

 

So, at the end of the night, Faktoh Reyahl served a total of just over two and a half beers. Cheers!

 

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Nice, so this is actually zeta regularisation. See the Relation to Dirichlet Series section:

https://en.wikipedia.org/wiki/Zeta_function_regularization#Relation_to_Dirichlet_series

 

It would be interesting to see if the two methods give the same result. It's curious how, even though in theory regularised divergent series are infinitely underdetermined, very often natural looking approaches reach the same answer.

 

Another possibility here is to use Stirling's formula for log Gamma. This is well known to be the following (copying the formula from Wikipedia):

8d257681480aa1bed0237cb9239f1f722a1786e7

(except the notation is slightly misleading as it's no longer asymptotic, rather the error is o(1) as n->infty).

 

Now if you strip off the z log z, z and -1/2 log z divergent terms, and the o(1) terms, you are left with 1/2 log(2*pi), leading to sqrt(2*pi) again. This surely isn't a coincidence...

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26 minutes ago, michaeld said:

Now if you strip off the z log z, z and -1/2 log z divergent terms, and the o(1) terms, you are left with 1/2 log(2*pi), leading to sqrt(2*pi) again. This surely isn't a coincidence...

I am sure that that is not a coincidence. It is the right answer, AFAIK. For instance, in another seemingly infinite series that would seem to get an infinite answer, I know of several different ways of summing all natural numbers (1 + 2 + 3 + ... to infinity) and the sum is always -1/12 and never anything different.

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There's an old but classic book that is entirely devoted to this subject: Divergent Series (1949) by the great G.H.Hardy.

 

*dusts off my copy from bookshelf, a Christmas present from my sister a few years ago*

 

AM-JKLU1QsGSoEpXei7oW5yPso2iujMltpEpyqSO

 

They cover this very example towards the end:

 

AM-JKLUstwJ5fbcVPXT-Da3Wdl_FB6FFYuRtub-5

(see equation 13.11.11)

 

(Can't seem to embed images at the moment, not sure why...) [edit:fixed]

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Way cool. Hardy did some amazing work.

 

I will seek out that book. I am sure that I will enjoy it immensely.

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Calligraphette_Coe
On 10/10/2021 at 7:21 AM, s.Kellyton said:

Way cool. Hardy did some amazing work.

 

I will seek out that book. I am sure that I will enjoy it immensely.

This part of the thread really made me smile, I just love when you and Michael expound on these problems. I doubt I'll ever be smart enough to decipher all the mathematical beauty in these, but after having spent a year in a crippling depression, paradoxially, I find that turning my mind loose on things like the Ponicare' Conjecture does, in fact, help me from the abyss. If only for the fact that Grigory Perelman turned down the million dollar prize and the Fields Medal for solving it. We desparately need more folks like him!

 

Does anyone recommend any particular books on the Golden Ratio? Lately, I've been thinking a lot about that one. That, and wish that our United States Congress would take on the Collatz Conjecture.

 

(BTW, if you don't mind my asking, how did your book do on Amazon?)

 

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22 hours ago, Calligraphette_Coe said:

Does anyone recommend any particular books on the Golden Ratio? Lately, I've been thinking a lot about that one. That, and wish that our United States Congress would take on the Collatz Conjecture.

 

(BTW, if you don't mind my asking, how did your book do on Amazon?)

The Golden Ratio is rather fascinating. I have a few books on it, but none that are really outstanding. Some focus on its use in art and architecture, others focus on Fibonacci and math history. I had thought about writing a book on it myself.

 

Regarding my math book, I never finished it. However, my novel got good reviews:

 

https://www.amazon.com/Doubloon-Cove-Secret-Ancient-Visitors/dp/0999035703/

 

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Unpinning this thread, but feel free to carry on with maths and physics stuff to your heart's content. :D 

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 This one should be easier. :)

 

An infinite number of mathematicians walk into a bar.

 

The bartender today is Pourzov Deaux. She plans on serving each patron two to the power of the number of the patron's position as they walk in, beginning with zero. Their numbers are written on their name tags. First, she samples the beer to make sure that it is OK. She serves herself one full beer. It tastes rather good.

 

When she is finished, the patrons start heading to the door. She sees that the one at the head of the queue has "1" on the name tag instead of "0". She should have known that. So, she names herself Patron Zero. Two to the power of zero is one, so she serves herself another beer. Again, it is refreshing.

 

She serves the first patron two beers (two to the power of one is two). Patron two gets four beers (2^2 = 4). The third gets eight beers (2^3 = 8), etc.

 

At the end of the night, after all infinity of mathematicians are served, how many beers in total did Pourzov Deaux serve?

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I'm experencing déjà vu ... I'm sure that was posted before!

 

If anyone wants a hint, look up geometric series. :) Then use the formula for the infinite series 1+x+x^2+... for x=2, forgetting that the series only converges for |x|<1.

 

I'm sure this example must also be covered in the early chapters of Hardy's Divergent Series as motivation for other more involved examples, but I am away from my copy this weekend...

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  • 2 weeks later...
On 10/17/2021 at 11:50 AM, michaeld said:

I'm sure this example must also be covered in the early chapters of Hardy's Divergent Series as motivation for other more involved examples, but I am away from my copy this weekend...

He did cover it early in the book, using a technique similar to what we will use here.

 

We have this many beers served:

 

1 + 2 + 4 + 8 + … on to infinity.

 

What we are looking for is the following sum:

 

1.png

 

How can we evaluate that?

 

There are several ways. First, we can use the Taylor/McLaurin series expansion of 1/(1-x). It is:

 

2.png

 

Let's try that with a few values of x. For x = ½, we have 1/(1 – ½) = 2

 

Summing up terms up to n = 50:

 

½0 + ½1 + ½2 + ½3 + … ½50 = 2.0000…

 

It is 2 to several decimal places.

 

For x = ¾, we have 1/(1 – ¾) = 4, and summing the first 50 terms is also 4.0000... to several decimal places.

 

We can't let 1 = 1, since we would be dividing by zero, but we can analytically continue the expansion above 1. So, let's let x = 2. 1/(1 – 2) = -1.

 

Here is the function:

 

plot.png

 

Thus:

 

3.png

 

If that does not seem to make sense, let's try that differently.

 

Take the original series:

 

S = 20 + 21 + 22 + 23 + 24 + …

S = 1 + 2 + 4 + 8 + 16 + …

 

We can rewrite that as:

 

S = 1 + 2(1 + 2 + 4 + 8 + …)

S = 1 + 2S

 

Solving for S, we have:

 

S = -1

 

So, we get the same answer. 20 + 21 + 22 + 23 + 24 + … = 1 + 2 + 4 + 8 + 16 + … = -1

 

So the bartender served minus one beers?!

 

Well, recall that before she started the series, with her being Patron Zero, she served herself one beer. We need to add that beer to the sum.

 

We have:

 

Total served = 1 + (20 + 21 + 22 + 23 + 24 + …) = 1 + (1 + 2 + 4 + 8 + 16 + …) = 1 + (-1) = 0

 

Therefore, she served a total of zero beers.

 

And a good time was had by all.

 

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Calligraphette_Coe
On 10/25/2021 at 2:17 PM, s.Kellyton said:

He did cover it early in the book, using a technique similar to what we will use here.

 

We have this many beers served:

 

1 + 2 + 4 + 8 + … on to infinity.

 

What we are looking for is the following sum:

 

1.png

 

How can we evaluate that?

 

There are several ways. First, we can use the Taylor/McLaurin series expansion of 1/(1-x). It is:

 

2.png

 

Let's try that with a few values of x. For x = ½, we have 1/(1 – ½) = 2

 

Summing up terms up to n = 50:

 

½0 + ½1 + ½2 + ½3 + … ½50 = 2.0000…

 

It is 2 to several decimal places.

 

For x = ¾, we have 1/(1 – ¾) = 4, and summing the first 50 terms is also 4.0000... to several decimal places.

 

We can't let 1 = 1, since we would be dividing by zero, but we can analytically continue the expansion above 1. So, let's let x = 2. 1/(1 – 2) = -1.

 

Here is the function:

 

plot.png

 

Thus:

 

3.png

 

If that does not seem to make sense, let's try that differently.

 

Take the original series:

 

S = 20 + 21 + 22 + 23 + 24 + …

S = 1 + 2 + 4 + 8 + 16 + …

 

We can rewrite that as:

 

S = 1 + 2(1 + 2 + 4 + 8 + …)

S = 1 + 2S

 

Solving for S, we have:

 

S = -1

 

So, we get the same answer. 20 + 21 + 22 + 23 + 24 + … = 1 + 2 + 4 + 8 + 16 + … = -1

 

So the bartender served minus one beers?!

 

Well, recall that before she started the series, with her being Patron Zero, she served herself one beer. We need to add that beer to the sum.

 

We have:

 

Total served = 1 + (20 + 21 + 22 + 23 + 24 + …) = 1 + (1 + 2 + 4 + 8 + 16 + …) = 1 + (-1) = 0

 

Therefore, she served a total of zero beers.

 

And a good time was had by all.

 

::::face palm::::  So, THIS is how the Accounting industry came up with the  "Mark to Market" accounting alchemy, allowing corporations to pay zero taxes?!

 

Which raises another conundrum.... could one use this same technique to explain The Feeding of the 5000 with 5 loaves and 2 Fish Miracle?

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