I'm Heart, your friendly neighbourhood admin!

[Kelly]

Saturn's moons Pan:

 

 

And Atlas:

 

 

Are shaped like Ravioli.

 

(or flying saucers)

 

[daveb]

Wow, those are cool! I never saw pictures of them before. I wonder what the processes are that cause that sort of shape? Something to do with Saturn's gravity pulling on it? Or more to do with the moons themselves?

[Kelly]

Pan is in the Ecke Gap of the rings; its equatorial bulge may be ring material that it picked up.  Atlas is near the A ring, so it's shape likely happened the same way.  

[Heart]

Pan looks like a flower with eyes looking up (the two dents), and Atlas looks kinda like someone squished a macaroon

 

Speaking of, this song made me feel like I want to switch career tracks from antimatter to astronomy:

 

 

[daveb]

5 hours ago, Kelly said:

Pan is in the Ecke Gap of the rings; its equatorial bulge may be ring material that it picked up.  Atlas is near the A ring, so it's shape likely happened the same way.  

Fascinating

So, are the bulges are in the same plane as the ring? (that's what I would think would be the case, and it appears that Atlas is at least lined up with the linear pattern of the clouds on Saturn)

 

1 hour ago, Heart said:

Pan looks like a flower with eyes looking up (the two dents), and Atlas looks kinda like someone squished a macaroon

I see what you mean

[Kelly]

9 minutes ago, daveb said:

Fascinating

So, are the bulges are in the same plane as the ring? (that's what I would think would be the case, and it appears that Atlas is at least lined up with the linear pattern of the clouds on Saturn)

Yes.  Here are pics of Pan in the ring gap:

 

[Heart]

Holy high resolution! Where did these amazing pictures come from again?

[Kelly]

Very recent pics from Cassini.  It is exploring the rings at present, and should dive through them a few times.

[Heart]

So cool! Thanks for mentioning this Kelly, I've been so obsessed with the small things I forgot to keep tabs on the very big things in life!

[Dodecahedron314]

On 3/17/2017 at 5:58 PM, Heart said:

Pan looks like a flower with eyes looking up (the two dents), and Atlas looks kinda like someone squished a macaroon

 

Speaking of, this song made me feel like I want to switch career tracks from antimatter to astronomy:

 

 

Fun fact: posting an A Cappella Science video is a guaranteed way to summon me to a thread. And don't worry, Heart, most of this channel's songs have a lot more to do with your area of study than mine--have I subjected anyone here to Bohemian Gravity or Massless yet? If not, look those up--those have been some of my favorites since back when I thought I was going to be a physicist.  

[Heart]

Oh believe me. I think I've seen them all. Many times. Sometimes on repeat.

[Kelly]

An infinite number of mathematicians walk into a bar...

 

This one is likely easier than some others, and some may already know it.

 

Rodd had just completed his mathematics degree and headed to the Hilbert Hotel bar to celebrate.  An infinite number of mathematicians followed.  Since it was Rodd's day off, he poured himself a beer and drank it.  Steven was late, so that is why he poured it himself.

 

Steven showed up, and asked Rodd to help tend the bar, because tonight would be busy.

 

So, the two started pouring.  Since there were two taps and two bartenders, the mathematicians queued up according to whether their number was a multiple of two (that is, even and odd).

 

Since Rod had already been served and was the first patron, Steven started first with the even-numbered math geeks.  Rodd would serve the odd ones.

 

They used the Riemann Zeta function to determine the amount to pour.  They could do the zeta function of their number, for instance the one with number 2 would have Zeta(2) beers, which is about 1.6449 beers.  But that is too much to fit into one mug, so they served each number theorist Zeta(s)-1 beers, where "s" is their number.  So, the second one (first to be served after Rodd had his beer) would get about 0.6449 beers.

 

So, Steven will serve Zeta(s)-1 beers starting at number 2 (s = 2), and serve even mathematicians, and Rodd will do the same, starting at s = 3, for the odd ones.

 

 

How much beer do they serve the patrons once all have been served?

 

How does this value compare to what Rodd drank?

 

And, of course, what is the difference in beers served between Rodd and Steven?

 

No calculus required for either the answers or proof.

 

[daveb]

47 minutes ago, Kelly said:

infinite number

Is that not the answer?

[Kelly]

That is the number of patrons, not the amount of beer.  

[daveb]

7 hours ago, Kelly said:

That is the number of patrons, not the amount of beer.  

But if you start with infinity doesn't the math work out to infinity? I mean, if you multiply or divide or add or subtract by or to or from infinity don't you end up with infinity? I'm obviously NOT a math wiz - just trying to look at it logically.  (I suppose/think there are maths that deal with these sorts of problems and that give non-intuitive or counter-intuitive answers - like infinity minus x = some value other than infinity? Beyond my ken.  )

[Kelly]

And yet:

 

On 5/27/2015 at 9:59 PM, Calligraphette_Coe said:

Infinitely many mathematicians walk into a bar. The first says, "I'll have a beer." The second says, "I'll have half a beer." The third says, "I'll have a quarter of a beer." The bartender pulls out just two beers. The mathematicians are all like, "That's all you're giving us?! How drunk do you expect us to get on that?" The bartender says, "Come on guys. Know your limits."

It is an infinite series puzzle.

[Kelly]

The answer is kind of cool, I think:

 

Spoiler

The Zeta function itself is quite interesting and difficult to evaluate without more advanced formulas.  For instance, Zeta(2) = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ...

 

What does that sum to?  That was a difficult question and was known as the Basel problem, proposed in 1644 and finally solved by the ingenious Leonhard Euler 90 years later.  It is, as mentioned, pi squared, over six (about 1.6449).

 

On to our questions.  When I first learned of the result, I was fascinated, and wondered why it could be so.  But then, while having lunch at work one day and thinking about it, it suddenly became obvious.

 

Since the Zeta function is the sum of an infinite series, and we need to sum an infinite number of such series, what we want to know now is the sum of an infinite series of infinite series.  Specifically:

 

 

Normally, the Zeta function starts at n=1, but I removed a redundant step.  For all of these Zetas, the first term would be 1/1^n, which is always 1 here.  Then we subtract 1, which removes the first term, so starting at n=2 completes those steps.

 

Now, this series of series actually makes the puzzle much easier to solve, if we think laterally.  Below are the first ten terms of each series:

 

 

The integers 2 through 11 on the top row are the s's of the series, and those in the second column to the left are the n's.  The zeta functions are thus the columns.  The first Zeta, with s = 2, we have:

 

 

At infinity, this will give pi squared over six, minus one.

 

The bottom row shows the partial sums for Zeta of 2 through 11.

 

The next, with s = 3, will give you Apéry's constant (minus 1), named after Roger Apéry, who proved that it is irrational relatively recently, in 1978 (not so trivial!).  The next, with s = 4, will give you a value that you can calculate, since a closed form for even positive integers does exist, but not one for odd positive integers.

 

Then you would need to sum up an infinite number of these.

 

However, if you sum up the rows, it is easier.  And it will give the same answer.  Partial sums of these are shown on the leftmost column.  The sum of the partial sums (both from adding columns or adding rows; the same thing) in on the bottom left cell.  For the first, with n = 2, we have:

 

 

And we know what that is.  Since we know that:

 

 

For n = 2, we get 1 (and for n = 3, we get 1/2; for n = 4, we get 1/3, and so forth).  For our series in question, we miss the first term, when n is to the power of 1.  So, we can subtract that.  For n = 2, we get 1 - 1/2 = 1/2.  For n = 3, we get 1/2 - 1/3 = 1/6.  For n = 4, we get 1/3 - 1/4 = 1/12. And so forth. 

 

And 2 = 2*1; 6 = 3*2; 12 = 4*3; and so on, meaning the sums are 1/(n*(n-1)).

 

That is, the rows in our series will be:

 

 

So, our rows will sum to 1/2, 1/6, 1/12, 1/20, 1/30, and so on.

 

We just need to add all of these up to get our answer.  Well, 1/2 + 1/6 = 2/3.  Adding the next term (1/12) we get 3/4.  Adding the 1/20 to this gives us 4/5, and our next term will give 5/6.  Indeed, adding up to the nth term, we get (n - 1)/n.  At infinity, this will be:

 

 

And we have it.  The sum of all of the beer that Rodd and Steven served to the mathematicians was one beer, the same amount that Rodd had consumed before he and Steven started serving an infinite number of other people.

 

Easy!

 

OK, so what about the difference between what Rodd and Steven served?

 

We can get these from the rows, too.  Steven will serve Zeta(s) - 1 for the even numbered "s" mathematicians.  We will thus need to sum the even-numbered columns.  The first, when n = 2 is:

 

 

The next is:

 

 

The next sum will come to 1/15.  Now, 1/3 = 1/(1*3); 1/8 = 1/(2*4); 1/15 = 1/(3*5), and so on.  For each row N, the sum is 1/(N*(N+2))  And we need to add all of these up.  That sum is:

 

 

And we are actually done.  If Steven served 3/4 of a beer, and together with Rodd, served one beer, so Rodd served 1 - 3/4 = 1/4 of a beer.  But we can do the same logic for the odd "s" customers.  Their sum is:

 

 

So, the difference between what Rodd and Steven served was 1/2 of a beer.

 

So, when I first learned that:

 

 

I thought that it was one of the mysteries of the universe, but now it seems obvious.

 

But don't you hate it when a professor is giving a lecture, and then comes to a major part of it near the end, and writes an expression out of nowhere and says, "It should be obvious that this equals that."?  What?!  No, it isn't obvious!  Likely they do not know how to get there themself, and ducked out of the derivation by feigning "obvious."

 

[daveb]

I still don't understand any of it, or the significance of it. It's so far over my head "easy" or "obvious" are not even in the picture! 

 

And it still makes a lot more sense than some other recent threads around here.

[Heart]

 

I think some mathematicians have a bad habit of using the word "obvious" to mean that if you sat down with paper and pen and years of training and hours of time, you could figure it out without any further information

 

Which is a completely different definition than the general public uses

 

@Kelly, thank you! It took me a while to have time to sit down and read through this, but it was a treat. I really should figure out how Zeta functions work more one day.

[Emery.]

 

On 3/24/2017 at 4:25 AM, daveb said:

I still don't understand any of it, or the significance of it. It's so far over my head "easy" or "obvious" are not even in the picture! 

Mathematicians solve only trivial problems, as RichardFeynmann said  

 

9 hours ago, Heart said:

 I really should figure out how Zeta functions work more one day.

Me too. *adds to the "do it when you have time" list*

 

Heart, you're an admin! Congrats

[Guest]

On 24-3-2017 at 4:25 AM, daveb said:

I still don't understand any of it, or the significance of it. It's so far over my head "easy" or "obvious" are not even in the picture! 

 

And it still makes a lot more sense than some other recent threads around here.

At least you understand some, I don't understand anything at all, let be the Greek Kelly posted, or isn't that Greek? It definetly looks like it  

*Fans poor brain*

[Kelly]

It was puzzling to me, too, when I first looked at the Riemann Zeta function, daveb and Jayce.  And indeed, when I looked at a college math textbook before I started at the uni, I was freaked.

 

13 hours ago, Heart said:

 

@Kelly, thank you! It took me a while to have time to sit down and read through this, but it was a treat. I really should figure out how Zeta functions work more one day.

I am glad that you enjoyed it.  As mentioned, I found the result fascinating and perhaps unsolvable until I was nomming on pizza one day, and then, an epiphany.  It would be neat if the Riemann Hypothesis came that way.

 

4 hours ago, Emery. said:

Me too. *adds to the "do it when you have time" list*

 

It does take some math background, and is generally only enjoyed by math and physics geeks, graduate students, and professionals.  But, a good start would be Prime Obsession, by John Derbyshire.  He starts slowly, and only gets to the hypothesis half-way through.  A science or math major who has made in into their Sophomore year should be able to follow this enjoyable book.

 

He also gives a very easy-to-follow proof of:

 

 

Here, we see that the Riemann Zeta function can be calculated using only prime numbers (p_n are primes).  After digesting the book, one may venture onto ways that the non-trivial zeros of the Riemann Zeta function are, in a sense, equivalent to the prime numbers; you get get one set from the other (I have given a few examples in previous posts).  Amazing.  And that is why the Riemann Hypothesis is so fascinating.

 

[Guest]

@Kelly That's nice to know! I can understand that it seems difficult..those study books always look scary in the beginning..let be if you have to study it as your main subject.I have done math in highschool but barely got a c for it because it never really made sense, despite the many hours i put in it trying to learn, this thread kinda makes it interesting to see how everything has evolved over the years, that batman reference is hilarious  

 

 

[Kimmie.]

The last posts here proves one thing: Kimmie is kind of dumb and not the brightest one in the bunch.

[Guest]

13 minutes ago, Kimmie. said:

The last posts here proves one thing: Kimmie is kind of dumb and not the brightest one in the bunch.

Awwww Kimmie! I only did highschool, never studied anything due to mental issues, do unskilled labor at a truckbuilder and i like posting in this thread because they have such cool stuff in here! No worries^^ i've seen people who went to uni who did dumb things *Remembers one who wanted to overthrow the local newsstation with a plastic gun and a fake story about being a member of a hacker group* Nobody is dumb here. *Hugs*

[Kimmie.]

Yeah math has never been my thing. Had to retake matematik 1 (the basic math course in gymnasiet our high school) to finnsh it. by the way the word High school has always confused me at first because if you translate it to swedish it is: Högskola which is our Uni.

 

But science is cool do.

[Guest]

51 minutes ago, Kimmie. said:

Yeah math has never been my thing. Had to retake matematik 1 (the basic math course in gymnasiet our high school) to finnsh it. by the way the word High school has always confused me at first because if you translate it to swedish it is: Högskola which is our Uni.

 

But science is cool do.

The dutch term for University is Universiteit en the dutch term for math is wiskunde, yeah words can be confusing indeed I like sience sometimes.

[Kimmie.]

8 minutes ago, Jayce said:

The dutch term for University is Universiteit en the dutch term for math is wiskunde, yeah words can be confusing indeed I like sience sometimes.

Yeah and ofcource universitet to I think they are the same thing

But universitet is just older. 

 

EDIT: Did some research and difference is that Universitys have the right to give out doctor exams. And things like that if understand it correctly. 

[Kelly]

This might be the last one for a little while.  Basically, all of the infinite mathematician puzzles posted here are what I am using in a recreational math book that I am writing, and each is used to introduce a new concept.  And I am pretty much caught up at this time.  This one brings up ideas that at first seem preposterous, but in reality bring up very important concepts that are key in the book.

 

An infinite number of mathematicians walk into a bar.

 

But the bartender (Steven is working this day) is in trouble.  His beer keg is empty.  Indeed, last night (Thursday), he drank away his troubles at another bar.  And he had just ordered his third beer before he discovered that he had just spent his last dime.  He told the other barkeep to stop pouring, but he already had one ounce (one-twelfth of a beer) in the mug by that time.  Rather than pour it out, he let Steven drink that one ounce, and owe the barkeep an ounce of beer, to be paid as soon as he can.

 

It is Friday, and the infinite number of mathematicians walk into the bar at the Hilbert Hotel, and each one is thirstier then the previous one.

 

Steven has no beer, and he is in debt; he cannot buy a new keg.  What shall he do?

[Kelly]

Steven has no beer, and he is in debt; he cannot buy a new keg.  What can he do?

 

 

We will let Bart and Lisa explain.  Bart came to Lisa's house for the weekly wine drinking and math discussion.  Bart arrived to see that the lights were out due to the recent storm.  They opted to take the wine to the treehouse and have their discussion there.

 

It was Lisa who proposed the puzzle for Bart to solve.

 

"Order more beer," he suggested.

 

"They only take cash, and Steven is broke."

 

"Serve them Vodka instead."

 

"They want beer.  And this is a math puzzle."

 

"Well, he can hardly serve any beer if he doesn't have any to sell.  I give up.  What does he do?"

 

"It was that negative one-twelfth of a beer that saved him," she said.  "What Steven does is serve the first patron one beer, the second patron two beers, then three, four, and so forth."

 

"But he doesn't have any!"

 

"He will be saved after all are served, because he does have enough.  He has negative one-twelfth of a beer, and if you sum all of the positive integers, it equals minus one twelfth!"

 

"What?!  No it doesn't."

 

"Actually, it does.  Recall the Riemann Zeta function."  Lisa wrote:

 

 

"And zeta of minus one is minus one over twelve."

 

To help show this, here is a plot of the real part of the Riemann Zeta function from 0 to -2.  Note that it is -0.5 at x = 0 and 0 and x = -2.0.  At x = -1.0, it appears to be a little less in magnitude than -0.1.  Below the plot is the value for the zeta function at -1.  It is indeed -1/12.

 

 

"How soon you forget, and you are trying to solve Riemann's Hypothesis? Let me refine your zeta formula." He modified what Lisa wrote earlier to:

 

 

"It is only valid for the variable if the real part is more than positive one. Yours is certainly less than that."

 

 

Lisa agreed. "But it still works. Let me prove it another way."

 

"Lisa, the running total, as we add the terms, is 1, then 3, then 6, then 10, then 15 and so on. Indeed, it forms points through the right side of the equation:

 

 

"And that is a parabola. As x increases, the sum increases, and faster and faster as x increases. It does not converge, and, especially to a small negative number. You have lost fair and square. Let us finish the wine and have a nice chat about other things, like our medical research."

 

Lisa was not to be outdone. She began with "Let us take the following function:

 

 

"The Maclaurin expansion gives us:

 

 

Bart agreed.

 

"So," continued Lisa, "let x equal one and we have this."

 

 

"This will be the second series, or S2:

 

 

"Our original series is:

 

 

"Now, let us multiply that by 4."

 

 

"And subtract that from the original, giving us minus 3S1:

 

 

"And we know that that series is, it is our S2. We have:

 

 

"Thus,

 

 

"See?"

 

Bart just about spat out his wine. "No, I don't see. Lisa, go home; you're drunk."

 

"I am home, Bart. You came to visit me.  And I did not make a math error here."

 

Bart took another sip of wine.  "I think that you did some sleight of hand with that Maclaurin expansion.  And if it does equal minus one over twelve, than it should do so using any method.  There are methods to obtain an analytic sum; try one.  If it shows that it diverges, than you belong in a lunatic asylum."

 

Spoiler

"lunatic asylum" is an anagram of "analytic sum"

 

Lisa thought about that, and said, "There is the Ramanujan Summation for tricky infinite series."

 

She wrote:

 

 

"I will let you do the honors, Bart."

 

"OK.  Our function, f of k, is just k.  The first term will be, when k equals zero, is zero.  Now for the sum.  For the first term, we have Bernoulli numbers.  The first one will be, since two k is two, we need the second Bernoulli number, which is..."

 

Bart looks it up on his smartphone, "It is one sixth.  OK, then we have a derivative.  The first will be the two minus one derivative, evaluated at zero.  That would be the first derivative, which is one.  And nothing to evaluate at zero.  Also, there will be no higher derivatives, since they would all be zero.  We have just one term.  Lastly, we have a factorial.  Since k is one, we have the factorial of two, which is two.  So, we have one term, which is one sixth divided by two, and, oh, a minus sign in front of the sum.  That comes to...minus one over twelve!"

 

Bart is astonished.  How could all of the integers sum to a small negative number.  Yet it does.

 

Bart and Lisa continued sipping wine and the conversation moved over to their medical research.

 

Most importantly...

 

Spoiler

Happy April Fool's Day.