Showing this series converges is surprisingly difficult.

[@MichaelPennMath]

Head to squarespace.com/michaelpenn to save 10% off your first purchase of a website or domain using code michaelpenn

[@mstarsup]

at 1:20 : the definition of the measurement of irrationality is wrong: A (of the infimum mu(x) that comes from A) should be declared only when there are (at most) a finite number of natural numbers p and q satisfying the condition in A (it should not be defined when being satisfied "for all rational numbers p and q" as said at 1:45, otherwise the set A is empty (for any mu, and for any given q, there always exist a p large enough such that the condition in A doesn't hold)). I'm guessing a correct definition would be: For x in R and mu>0, define A(mu)={(p,q) in ZxN* : |x-p/q|0 : A(mu) is finite}

[@sbares]

Indeed. It seems to be a typo on the blackboard, since the equivalent definition below is in fact equivalent to the correct definition.

[@bonsairobo]

Thank you. I was super confused until I went to read the definition on Wolfram.

[@Anonymous-zp4hb]

Even if we say "for some integers p,q" rather than "for all rational p,q"... I still don't get how we get 𝜇(rational) = 1. Surely, for any integer x, there are no 𝜇 >= 1 such that 0 < | x - p/q | < 1/q^𝜇 So, if anything, 1 is the _supremum_ of A... e.g. 0 < | 3 - 13/4 | = 1/4 > 1/16 > 1/64 > ... edit: just realized I misread your comment. Yes, Given my example, there'd be no p,q satisfying for any 𝜇 >= 1, so since 0 is finite, they're in. But since there are infinitely many p,q that satisfy the condition when 𝜇 < 1, they're all out. makes sense now, thanks.

[@StratosFair]

Thank you. That part got me very confused

[@cmilkau]

Penn's "A" is Wolfram's "R". It's the set of all μ such that the 0 < ... < q^μ equation has at most finitely many solutions for integers p,q. [Gotta be careful not to repeat myself too much else yt is gonna yank the comment for spam.]

[@jacksonstarky8288]

The irrationality measurement of e being exactly 2 when e is proven to be transcendental (making the irrationality value the smallest possible) is worth a video (or series of videos), regardless of its complexity. Measurement of degree of irrationality has fascinated me for as long as I've known about the concept.

[@edcoad4930]

Isn't it because it can be expressed as a polynomial?

[@atreidesson]

that "I am interested, thus a video shall be made" sounds a bit selfish, you know

[@DeJay7]

@@atreidesson Not what they said though, you twisted their words quite a bit. "is worth a video regardless of its complexity" is an opinion, and personally a good one.

[@atreidesson]

@@DeJay7 That's it. You feel that Michael should do that video, though that takes infinite amount of effort from him, and infinitely small fraction of viewers will understand that. Which means, you don't value the time of viewers, nor author's time. Instead, you picture that as a "humanity's great accomplishment", which "deserves" our time. That's not selfish at the core, but sounds selfish because there are not many people wanting that content to be created. That's like a guy saying "I discovered an ancient god exists, so we should build a temple and sacrifice our young." It could make sense, but at first you can't easily respect such claims.

[@MrMctastics]

@@atreidessonYou're weird

[@kkanden]

i always thought of irrationality as somewhat an elementary idea since we learn it so early in school and are given an infinite amount of irrational numbers to work with (thanks to pythagoras), however upon learning that zeta function of 3 was proven to be irrational only in 1978 and is really the only thing we know about the zeta function at odd integer values really changed the way I view irrationality and famous irrational numbers. It went from something I would simply shrug off as not that interesting to something truly admirable. Great video as always!

[@darkshoxx]

Whoa that's interesting. About 10 years ago when I was in first year uni, I, and many others had the question "what's a series where we don't know if it converges or not" because as stupid freshmen we thought, we have so many tools, clearly this is a solved problem. We came across a Mathexchange article that gave that Flint-Hill series as an example. It's great to finally see the context to that problem after 10 years 😄

[@douglaszare1215]

There is a series that has radius of convergence 1 precisely if the Riemann Hypothesis is true: Consider 1/zeta(1/z) around z=1.

[@beniborukhov9436]

I'm pretty confused by the first fact that's introduced around 9:40. Since sin is bounded between -1 and 1, isn't any x (absolutely) greater than 2 an obvious counter example? To take another, Sin(pi) = 0 which is clearly less than pi/2. What am I missing?

[@zikede]

He forgot to add the restriction that this holds for |x|

[@prismaphotonic2710]

The inequality holds as long as x is in [-pi/2, pi/2] (actually in a slightly larger interval), which suffices in this case, since it is being applied to a value that will always be less than or equal to pi/2 in absolute value. One way to see why that is the case is to realize that if n - q_n*pi were not in the interval [-pi/2, pi/2], then you could add or subtract an integer from q_n so that it falls in the interval (which would contradict the fact that q_n is the closest integer to n/pi).

[@veselindimov307]

I can't speak for other people, but I would LOVE to see a video about the measurement of the irrationality of some transcendental numbers. This topic is so poorly understood (I include myself in this group), that only Michael Penn (ft. Chalk) can clear things up!

[@DeJay7]

Only Brush can clear things up, you got Chalk messed up.

[@MarcoMate87]

If we substitute, in the initial series, n! with n^p, with the exact same machinery, at 16:25 we can conclude that, if p > 8 this series absolutely converges.

[@guerom00]

Hmmm i have a problem with |sin x| >= |x|/2 which is clearly not true for all x...

[@deinauge7894]

i suspect it's a running gag that every video has at least one basic mistake. The correct version is left as an exercise for the viewer😅

[@Jaeghead]

He forgot to add that we're only considering small x, for which it is true (plus maybe an explanation why the argument is small in the proof).

[@guerom00]

@@Jaeghead i forgot the exact argument inside the sine but there was n, which goes to infinity... But then, yeah there was this q_n pi... Probably that's what makes the argument inside the sine always small (?)

[@franzlyonheart4362]

Yeah, x=10 is an obvious counter-example.

[@franzlyonheart4362]

@@deinauge7894 Sounds like a lame excuse. Not really funny.

[@gcewing]

7:03 Aha! So the Ultimate Question of Life, the Universe and Everything is "What is the value of n such that pi is never closer to a rational number p/q than 1/q^n?" It all make so much sense now!

[@petergregory7199]

Michael, it doesn’t matter how drab your tee shirt is, your math is always dazzling.

[@CM63_France]

it's almost the easiest color to wash.

[@merrynoise]

1:47 "for all rational numbers p and q" makes no sense at all. It should be "for only finitely many Integers p and q".

[@lurkmoar3926]

5:22 The parenthesized note "i.e. Sqrt(2), phi" should say 👉e.g.,👈 meaning "for example", and not i.e., which means "that is".

[@markolson4660]

Disappointing! (which is unusual.) This looked very interesting, but there was too much magic involved -- like what the irrationality measure means and how it works. You sometimes go into detail on elementary algebra, but here skip over a decidedly less elementary point.

[@samueldeandrade8535]

Wow! This criticism looks interesting ...

[@bobernhardsson5345]

1.30 The set A is empty with this definition.

[@Reidemeistermoves]

More videos on diophantine approximation! It would be cool to see something on continued fractions. For instance, it is a bit difficult but not unreasonable to present a proof that every quadratic irrational is badly approximable (and thus irrationality measure 2).

[@minamagdy4126]

I kept staring at the video for a while in confusion before it finally clicked: it may be that mu(x) is actually the supremum of A rather than the infemum.

[@tracyh5751]

The entire set A is defined incorrectly. The set A should be the set of all mu so that the number of solutions to the equation 0 < |x-p/q|

[@Happy_Abe]

@@tracyh5751I was so confused the entire time That wasn’t explained at all!

[@cmilkau]

We could really have done this for any finite value of μ(π), so we really only needed to know that π is no Liouville number, right?

[@周品宏-o7w]

why μ(x) isn't -∞ for all real number x according to the first definition?

[@goodplacetostop2973]

17:35

[@libregisin9878]

Here is a nice question about convergence, even if it's not so difficult to answer. Take the harmonique sum (sum (1/n). It is easy to see that it diverges. But if you take the sum over all natural numbers which does not contain the digit 9 in their writing in base 10. Does this new sum converges ? The question can be generalized by replacing 9 by any other digit, or any integer. I found the answer very interesting. To be clear, here is the sum = 1 + 1/2 + ... + 1/8 + 1/10 + 1/11 + ... + 1/18 + 1/20 + ... + 1/88 + 1/100 + 1/108 + 1/110 ... + 1/888 + 1/1000 + ... Thank you for your videos, very well presented.

[@GiornoYoshikage]

Impressive. I had to deal with something like this only once, while proving that radius of convergence of power series (x^n/sin^n(n)) is 0. Dirichlet's approximation theorem was enough, but still I was shocked how simple and beautiful the very connection between analysis and number theory is

[@whozz]

I think the convergence of the sum of 1/(n³ sin(n)) is an open problem

[@SpeedcoreDancecore]

Yes it is

[@wesleydeng71]

Anything involves n! obviously converges, but n^k will depend on k.

[@yanntal954]

If the irrationality measure of pi is at most 2.5, then this sum should also converge. It is much harder for 1/(n^3*sin^2(n)) to converge than the mentioned sum

[@leif1075]

Is it open because it's hard or just not enough information is known to solve it?

[@whozz]

@@wesleydeng71 thanks! This is probably what I had in mind

[@davidvilla2909]

What is a Liouville number on this scheme? Is it a number for which mu(x) in infinite?

[@japanada11]

yes, that's right

[@aweebthatlovesmath4220]

I knew that there was a way to measure irrationality of pi with convergence of sum but didn't knew how it exactly work... thanks for the video!

[@e2DAiPIE]

Can someone please help me understand in the second definition of irrationality given near the start of the video, what does restricting both p and q to be greater than some N actually restrict in the inequality? I feel like I could just take any p and q and multiply both by N and basically ignore the restriction.

[@e2DAiPIE]

Dumb question. I just realized that would directly affect the the inequality as q increases. Please disregard this question.

[@martin.thogersen]

I always learned that the golden ratio is the most irrational number, since its continued fraction is [1,1,1,1...] and converges as slow as possible. Conversely pi is well approximated by 22/7 and other higher fractions. I think these definitions seems to go against each other, and should have been discussed. Maybe an idea for another video, Penn?

[@japanada11]

Technically speaking, "most irrational" is a meaningless term; every number is either irrational or it's not. When we compare "how irrational" two numbers are, what we're really doing is choosing some property that rational numbers have, and seeing how well irrational numbers also have this property. And some of these properties actually work against each other! By analogy: is 1/2 or 999/1000 more like an integer? Well an integer n has the following two properties: (1) it is as close as possible to an integer (namely, it has distance 0 from itself), and (2) it is as far as possible from _other_ integers (namely, it has a distance of at least 1 from every integer that isn't equal to itself). If we measure "how integer" a number is by how well it achieves property (1), then 999/1000 is very much like an integer while 1/2 is _the most non-integer_ you can be (it's as far as possible from any integer). But if we instead measure "how integer" a number is by how well it satisfies property (2), then 1/2 is actually _the most integer_ a non-integer can be. The integers "repel" each other, so if you have to find a number that behaves like an integer, 1/2 is the best you can do: it keeps its distance from the integers around it, just like the integers do with each other. Meanwhile, 999/1000 is blatantly invading 1's personal space, which is something that integers just don't do. 1/2 does a much better job of pretending to be an integer than 999/1000 does. The same thing is happening here. A rational number x satisfies the following properties: (1) it can be approximated extremely well by rational numbers (namely, x is a perfect approximation for itself), and (2) it is extremely poorly approximated by _other_ rational numbers. The golden ratio is to rationals as 1/2 is to integers: it fits property (1) very poorly, but property (2) very well. Meanwhile, numbers that have extremely good rational approximations are more like 999/1000: they have a bunch of rational numbers that "stick" to them, which actually makes them very _unlike_ rational numbers (which like to keep their distance from each other). Irrationality measure is capturing property (2): the lower the irrationality measure of x, the better job x does at clearing out the rational numbers in its vicinity.

[@martin.thogersen]

@@japanada11 great take, thanks for writing this up.

[@Happy_Abe]

The set A is defined incorrectly here

[@radadadadee]

which is a shame, since it's a fundamental topic of this video

[@byronwatkins2565]

At 9:40, sin(pi)=0 and certainly is NOT >= pi/2...? Also, sin(5 pi/2)=1 and is NOT >= 5 pi/4...?

[@coc235]

Yes, that was a mistake. Michael should probably have said that this inequality holds for x in (-pi/2, pi/2), which is enough for the use case below, in fact this is the exact reason we need that "closest integer to n/pi" function - to make sure we use the inequality for an argument between -1 and 1.

[@byronwatkins2565]

@@coc235 As I recall, he used it for n>34...

[@coc235]

@@byronwatkins2565 I only recall him doing that at 10:40 for n-(q_n×pi), which has an absolute value less than or equal to 1 by construction

[@GatesOlive]

I'm curious about the use of | sin x | >= | x | / 2 as it is clearly false for x=10,000,000 or a high value like that

[@WackyAmoebatrons]

I can reduce that claim to being false for x>2. 🙂

[@MarcoMate87]

@@WackyAmoebatrons it's false even for x = 2.

[@romajimamulo]

... wait the paper you flashed on screen had the inequality in the other direction. Did you mess it up on the board?

[@quazzydiscman]

I like falling asleep to these videos. It makes me feel like my subconscious is getting smarter.

[@kozokosa9289]

What if I prove that there is no such N such that |Pi*n-N|< every epsilon greater than zero, thus showing there is no such "explosion" in the denominator?

[@StratosFair]

I don't understand the definition of irrationality measure, I'm especially confused by the fact that there are no quantifiers on the rational numbers p/q. Is the inequality meant to hold for ALL rationals ? Because in that case I'm pretty sure A would always be empty regardless of whether x is rational or not. The equivalence with the other definition is not clear at all either. I guess the original definition should be understood as "the set of mu such that there exists a rational p/q such that this inequality holds", but I'm not too sure.

[@Mystery_Biscuits]

Yesterday I was thinking about rational approximations of pi (as you do) and today this video gets uploaded. What timing!

[@pierreabbat6157]

The in "Zeilberger" is pronounced /ts/, as you can see from the Hebrew.

[@gutschke]

English frequently favors approximating English spelling/pronunciation rules for foreign names, instead of mimicking the "correct" native rules. This has been true for a long time, and it gives us funny ways of saying "van Gogh". It also explains why Chinese names tend to be almost unintelligible to native Chinese speakers, when pronounced in English. And it regularly confuses German speakers, when English speakers keep insisting that there is no such thing as umlauts. Mathematicians are actually bucking the overall trend a little bit, and I am surprised that names such as "Euler" or "Gauss" are pronounced closer to the way that a native speaker would say it as opposed to what you would expect if you applied English rules. Nonetheless, barring an established exception for a particular name, it is idiomatically correct to adjust names when used in English speech. Having said that, I do applaud your efforts to educate people on the native pronunciation, no matter how futile this effort is likely going to be.

[@tracyh5751]

@@gutschke This might have been true 50 years ago, but in modern English speaking mathematics, it's usually held that a speaker should use the closest approximates of the speaker's mother tongue to pronounce a foreign name (do the best you can). The sound 'ts' is closer to a German 'Z' and Michael can comfortably pronounce either, so Pierre is correct here.

[@KramRemin]

This is hilarious, it's like two fighting siblings in a get-along shirt. WHO WILL WIN? ONE WELL-BEHAVED BOY-- 1/n! "I WILL CONVERGE NO PROBLEM! ALL I WANT TO DO IS CONVERGE!" OR ONE IRRATIONAL BOY -- 1/sin(n), n being an integer, so essentially a random number generator, including numbers arbitrarily large. "I DO WHAT I WANT!"

[@demenion3521]

did it really matter for the proof how irrational pi is? we ended essentially with the convergence of n^7/n! with the 7 being the floor of mu(pi). but replacing the 7 with a 1 or a 100 doesn't change anything at all

[@General12th]

Hi Dr. Penn! Very cool!

[@edcoad4930]

Why did the > change to < for the n/3 part? Seemed right initially?

[@RandomBoxOfWeird]

Does the irrationality measure have fixed points? Also very satisfying that e has mu(e) = 2.

[@jay_13875]

Is the exact value of mu(pi) even important? It looks like you could prove that this particular series converges for any finite upper bound on mu(pi), some of which may (?) be easier to derive.

[@reeeeeplease1178]

I don't think it does in this case. If we want to know the convergence of 1/(n^r * sin(n)), we know that it converges for all real r>8 just by changing the conclusion of the video. This exponent does depend on mu(x).

[@titusandronicus5975]

It's not true - in 2:44. Let's take p=N, q=2N and x=1/2. By definition these numbers should exist. Then the left side is |x - p/q| = |1/2 - N/2N| = 0. mu(x) for all rational x should be infinite.

[@japanada11]

He made a ton of mistakes in the definition. One of these is that p and q have to be restricted to coprime integers so that p/q is in lowest form.

[@Alan-zf2tt]

All I can contribute today is: I ain't 'fraid no scary monsters whether understood or not

[@StratosFair]

Not done watching the video but the sinx inequality is clearly not true outside of a neighborhood of zero. However I think it's pretty easy to show that with the way q_n is define, n-q_nπ will always be close enough to zero for the inequality to hold

[@Geenimetsuri]

Didn't know of a measure of irrationality! Great stuff! Thanks!

[@jaje9004]

I don’t understand why mu(x) is obviously 1 for rationals…. Can someone explain?

[@japanada11]

The definition in the video actually has a ton of mistakes in it. Here's a more accurate definition. We say mu is the irrationality measure of x if for all e>0: A. there exists N>0 such that for all relatively prime integers p,q with q>N, |x - p/q| > 1/q^(mu+e); B. for all N>0 there exist relatively prime integers p,q with q>N and |x - p/q| < 1/q^(mu-e). (Informally: there are only finitely many rational approximations within 1/q^d of x if d>mu, but infinitely many within 1/q^d of x if db^(1/e), we have q^(1+e) = q*q^e > qb, so 1/q^(1+e) < 1/qb. Combining these we see that if q>max(b,b^(1/e)), then we obtain |x-p/q| = |aq-bp|/|bq| >= 1/qb > 1/q^(1+e), proving A.)

[@StratosFair]

Thanks for the explanation, I finally got it ! It's a shame it wasn't explained (or even defined properly) in the video...

[@ANTONIOMARTINEZ-zz4sp]

Absolutely heavy stuff... Would it be possible to get the sum in this case? Thanks for sharing, Prof. Penn.

[@ldhung308]

Why cant use lim an+1/an

[@Coffeyhandle]

1 + 2[sum of sin(n)/n for n from 1 to infinity] = pi.

[@thechannelwithoutanyconten6364]

So, this converges absolutely, and how about plain convergence?

[@luissimaoaf]

Absolute convergence implies plain convergence

[@thechannelwithoutanyconten6364]

@@luissimaoaf OOO, yeah, I forgot xD

[@Mikey_AK_12]

There are some errors in this one which the other commentors have probably already pointed out. First, the definition of the irrationality measure (p is arbitrary, does not have to be bigger than N), which was confusing until I went to Mathworld to understand it. Second, |sin(x)| ≥ |x|/2 only holds for |x| sufficiently small, |x| ≤ 3/2 is good enough and this is the case when this inequality is implied. This is relevant especially since the proof is handwaived. These really should be corrected in the description or pinned comment, it would be very sad if the sponsor owns this space and doesn't allow it. Also simply using n/π + 1/2 ≤ n/3 for n > 33 seems way more straightforward than "adding and subtracting zero." I feel like you can do better, Michael.

[@robshaw2639]

Do we know the value it converges to?

[@PhilippeLacoude]

Using 100 terms, I get 2.850914

[@PhilippeLacoude]

# Sum 1/(n! * sin(n)), n=1 to infinity # kzbin.info/www/bejne/hXe0amhvj7OblaM from math import factorial, sin # Compute the Sum of terms 1/(n! * sin(n)) from n=1 to terms, # where terms is an integer def InfiniteSum(terms): total = 0 for n in range(1, terms+1): total += 1/(factorial(n) * sin(n)) return total # Compute the sum using the first 100 terms num_terms = 100 # Increase for a better approximation result = InfiniteSum(num_terms) print(f"Approximate sum of the series using {num_terms} terms: {result:.6f}")

[@robshaw2639]

@@PhilippeLacoude doesn't seem to be any famous constant directly...

[@szkoclaw]

Can't ever square anything with here, completely irrational.

[@adb012]

sin x > x/2??? sin π = 0 < π/2

[@gp-ht7ug]

Complicated machinery for me but really interesting Thanks

[@jplikesmaths]

Pi is the most rational irrational number

[@prydin]

The only rational (no pun intended) answer to your question must be the Collatz conjecture.

[@erfanmohagheghian707]

Jesus! All this work to show it's absolutely convergent?!!! Are you kidding me? This series is absolutely convergent by ratio test. Done! Or am I mistaken?

[@Meni_Rosenfeld]

You are, indeed, mistaken. The ratio of subsequent terms is unbounded, so the ratio test does not apply. For example, the ratio between terms 22 and 21 is ~4.3. The ratio between terms 355 and 354 is ~78. As Michael explains in the first minute, the whole point is that sin(n) can get very small when n is close to a multiple of pi, and then 1/sin(n) is large and the whole term get bigger than the one before it. To know that the series convergence you must put a lower bound on just how small sin(n) can be, and that requires nontrivial considerations of pi's irrationality measure (doesn't matter what it is, just that it has one).

[@erfanmohagheghian707]

@@Meni_Rosenfeld I did some numerical experiment up to n=50 million. I don't trust machine precision, that's why I did not go higher. It does not seem unbounded and the max occurs at n=354 at the value you mentioned. But yeah the limit at infinity is not zero. Mathematica was not able to evaluate it. I actually watched the video while I was in bed and put this comment before I fully load after waking up lol. Thank you.

[@Meni_Rosenfeld]

@@erfanmohagheghian707 Interesting. Yeah, I was wrong to say it's unbounded. In fact, now I'm not even convinced that it doesn't converge to 0... I'll take a closer look later. But whatever it is, it's not trivial to prove. But to find terms for which the ratio is relatively high, you don't need to look at every n, just numerators of good rational approximations of pi - which themselves can be found by truncating its continued fraction expansion.

[@erfanmohagheghian707]

@@Meni_Rosenfeld I've had a look at the behavior. for large n, the value is small (around 10^(-7) for example) but it fluctuates and I don't see convergence to zero. Though I don't trust software fully but I find that the limit (I mean the limit of the ratio test) does not exist after trying both Maple and Mathematica. I think it should not be too hard to prove that the limit does not exist. I'm gonna watch the video again now that I'm fully awake lol, but I trust Michael; not only he's got a PhD in math (which I don't), he is really smart and definitely he has not invented this problem.

[@erfanmohagheghian707]

@@Meni_Rosenfeld Can you show that the series passes the divergence test? i.e., can you show that lim(1/(n!) sin(n)) n-> inf is zero? It's not a piece of cake or is it?

[@PawelS_77]

According to Wolfram Alpha, this sum diverges.

[@mantisbog]

How can we measure the irrationality of my ex-wife?

[@Anonymous-zp4hb]

Not surprised 𝜇(e) = 2. I'm curious, do we know if 𝜇(𝜋) > 2 ? It feels like it should be, given that almost (if not all) infinite sums for 𝜋 are just terrible, practically speaking And if not 𝜋, do we know of _any_ x for which 𝜇(x) > 2 has been proven?

[@tracyh5751]

Any Liouville number has mu(x)=infinity by definition. A bunch of these are known, the first one being Liouville's number, which is 0.a_1a_2a_3... where a_n is equal to 1 if n=k! for some integer k, and 0 otherwise. Or as a decimal approximation, the number is approximately 0.110001000000000000000001...

[@OrbitTheSun]

For x ∈ ℚ shouldn't µ(x) be minus infinity?