sin(314159...)

[@Epyxoid]

The result is quite interesting to me. First I thought it's 0. Then I saw you were starting to use series and limits, so I thought it's converges to 0, but it's not? So weird, because I'd thought as the series goes, it's getting closer to the multiple of π and the error is getting smaller. But since the number is getting bigger, I think the error level remains the same, thus there's no limit of the series. Quite interesting! And of course the series is always negative, since the argument of sine is always an even multiple of π, but a little less.

[@DrBarker]

This was basically my thought process when I first thought about it! I think it's tempting to think it's getting closer and closer to being a multiple of pi, but that's not actually the case, because the size of the error isn't getting smaller and smaller.

[@sp0_od597]

@@DrBarker Why cant you create a different sequence that is just f_n=sin(10^n*pi) without the floor function. In a sense, this sequence basically starts with pi and moves the decimal place n places. To me, I think that taking the limit as my sequence goes to infinity that you would find that it converges to 0. I don't understand why you need the floor function. Your sequence may diverge but it is not the only way to express sin(314159...) in sequence form

[@btd6vids]

@@sp0_od597 The sequence you mention just has every term equal to 0, that’s why

[@sp0_od597]

@@btd6vids That still does not answer my question of why you need the floor function.

[@Bl00drav3nz]

@@sp0_od597 Because sin(10^n*pi) is a completely different sequence than the one discussed in the video. 10^n is an even number for n>=1, and sin(2*m*pi) = 0 for all integer m. The limit of sin(10^n*pi) for n->inf is of course 0 (again, because 10^n*pi is an even multiple of pi for n>=1), but that's not the sequence we are interested in here. Hope this helps~

[@petersievert6830]

The collorary makes perfect sense when you look at it this way: With sin(31), sin(314), sin(3141),... you are just a bit before the sine of an even integer times Pi. So you are nearing the periode-starting 0 from the left, thus from below. (Ofc this basically is expressed by your formula, just the visual representation in the graph shows nicely what is going on.)

[@brandonklein1]

Shortly started, the error of sin in this sequence is just as random as the digits of π and thus does not converge as n tends to infinity.

[@cmilkau]

(10^n)π - floor((10^n)π) is basically a random sequence in [0,1], so this sine should basically look like sine(random) and not converge anywhere. (note that sin(x) = sin(x-(10^n)π) for n=1,2,3,...)

[@fiona1204]

i recommend looking at a plot of the counterpart to this series with a continuous variable: f(x)=sin(floor(10^x * π)). its hard to see a lot of it at once because its oscillations grow so quickly, but zooming in on any part in desmos there is a neat grid pattern.

[@renderize69]

This video is going to blow up! Great quality mate!

[@juanma4978]

amazing video and proof, really like your channel, keep it up! you've got a new subscriber

[@RH-qt2vk]

Instead of plugging into the expression for sin(A-B), you could say that 10^n*pi is an even multiple of pi, for n >= 1, and so subtracting it does not change the value of sine: sin(floor(10^n*pi)) = sin(floor(10^n*pi) - 10^n*pi) = -sin(10^n*pi - floor(10^n*pi)).

[@Qermaq]

8:30 127/110 = 1.1545454545... this has 2 digits that repeat infinitely. So you're saying that this does not identify an irrational number, but it is just a necessary property of irrational numbers, is that right? Also, just fun - my calculator claims that sin(floor(pi*10^19)) is about 3/4. I'm assuming that's where it does not know pi closely enough.

[@DrBarker]

Yes, it is a necessary, but not sufficient, condition for being an irrational number. I think this shows that your calculator only knows 19 or fewer digits of pi (or maybe 20 max), which is probably enough for most purposes! But the answer should definitely be negative!

[@angelmendez-rivera351]

The decimal digit string representation of an irrational real number satisfies two properties: -It does not terminate. -It is not periodic. These two properties uniquely identify the irrational numbers, since the decimal digit string representations of rational numbers do not have these properties: they either terminate, or do not terminate but are periodic. This can be proven, though it is somewhat tedious to prove this.

[@joseluisvazquez4221]

At 10:18 I would say that other digits less than a may appear in an irrational number even if they don't repeat infinitely often, so bn would we extrictly greater than 0.b. This does not change the outcome of the demonstration as b>a. Good video!!!

[@Reuleaux99]

Your videos are awesome. Very interesting content

[@Lodekac]

Amazingly explained!

[@simonwillover4175]

10:38, if a limit for b_n exists, then the end of pi (infonitely far out into its decimal expansion) is the same string of digits repeating a bunch of times over and over. The limit as n approaches infinity has to equal the limit as n approaches infinity+1, infinity+2, infinity+3, infinity+4, etc... Thus, if a limit for b_n exists, then pi is rational, which contradicts the fact that *PI IS IRRATIONAL.* Therefore... (read my reply...)

[@simonwillover4175]

Pi is not normal because all normal numbers must be rational.

[@galladeguy123]

@@simonwillover4175 Rationality is not a requirement of normal numbers. Champernowne's constant (0.1234567891011...) is an obvious counterexample, as it is irrational (transcendental even) and, by design, normal in base 10. In fact, it's conjectured that every irrational algebraic number is absolutely normal, meaning it's normal in every base. That's a property that no rational numbers have, since there will always be a base where any given rational number is a repeating decimal. As for pi, it's conjectured to be normal, although there's still no proof either way.

[@vinuthomas7193]

The part I liked most is: for non-negative integer n, 10^n is odd only when n=0 - I hadn't thought about that, before

[@JakubH]

nice, I had it solved in a few sedonds in my head, the argument is always a little off from a multple of PI, and how much off it is is (0,1)-random, so the sin of that is also random so no limit. But this video nicely shows how even the intuitive and simple things need to be proved rigorously, and that it is not that easy.

[@zhelyo_physics]

This is excellent!

[@ilonachan]

Couldn't you have just skipped the part at 3:04 and instead gone to 5:50 directly? sin(⌊ 10^n π ⌋) = -sin(-⌊ 10^n π ⌋) because sin is odd, and = -sin(10^n π - ⌊ 10^n π ⌋) for n≥1 because 10^n is even, alternatively = sin(10^0 π - ⌊ 10^0 π ⌋) for n=0. After that I think this would be an alternative solution, please point out if I've made a misstep: Because 10^n π is irrational, the fractional parts (input to the sine function) are dense in [0,1] (This is the statement you proved at 5:50 right?) Even if you skip the first n-1 terms, and then only take every nth term, you'll still densely cover [0,1]. And the same is true for the sine values dense in [0,sin(1)], again no matter how late in the sequence you start. In other words: no matter how far along the sine sequence you go, there will always be values just below 1 and just above 0. Formally: lim sup=1 and lim inf=0. For a converging sequence these two must coincide, so the sequence doesn't converge.

[@ImaginaryMdA]

I used the sin(x) ~ (x mod 2pi) approximation for small deviations from 2pi, to get some intuition about this problem, and worked it out using that intuition.

[@jarikosonen4079]

That looks good question... sin(x)=0 => x=n*π. Can it show that sin(n*π)=0 also when n->infinity? Or that sin(inf)=0?

[@DrBarker]

You can definitely say that sin(nπ) converges to 0, because it's essentially just the sequence 0, 0, 0, ... so its limit must also be 0. For sin(∞) = 0, it depends on how your sequence inside the sine function grows to infinity. For example, sin(2nπ + π/2) is always 1, so the sequence becomes 1, 1, 1, ... and its limit is 1.

[@pedrosso0]

​@@DrBarker Don't you have to strictly say that the limit is discrete? Otherwise n might as well be any non integer real number which leads to a non-zero result

[@scalenescott]

@@pedrosso0 Since he's finding the limit of a sequence (b_n, I think), then it is discrete as you're thinking of it. Essentially, he's evaluating the limit as n->infty of a function f(n)=b_n that takes positive integers to real numbers, as opposed to a function f(x) that takes real numbers to real numbers. :)

[@DrBarker]

@@pedrosso0 Yeah, strictly speaking you do need to specify that n is an integer, or like you say the limit wouldn't have to be 0. I'd usually use a different variable (perhaps x or t) if it was a continuous limit.

[@pedrosso0]

@@DrBarker Variables imply that sure, but best to specify anyway

[@evakuator8118]

Thank you for the fun video!

[@MyOneFiftiethOfADollar]

I liked the simple definition of Floor(x)= max { n member of Z : n less than or equal x} Some of the old timers in the US still use the gaudy and embarrassingly wordy Greatest Integer Function definition "greatest integer such that said integer is less than or equal to the argument" Generation after generation of students have suffered immeasurably laboring under that definition. Similar approach is used in producing a sequence of numbers that converge to any real number. Might be a Cauchy sequence?

[@camrouxbg]

I'm confused about your 0.abbbbb... argument, but I had a hard time hearing you even with volume cranked up. The content is really interesting, and I'd like to understand it all when I have a chance to sit down and work through it carefully with you. Of course that will take me 17 hours to watch a 17 min video, because I'm slow hahaha Just please work on the mumbling and find a good microphone that works well. When you face away from the camera, we can hardly hear you at all.

[@DrBarker]

I've got the audio working a lot better on more recent uploads - it's a shame about some of my earlier videos like this one. Maybe one day I'll re-upload or re-record some like this one. For the 0.abbb... and 0.baaa... argument, basically we find 2 subsequences, along one of which the values are all ≥ 0.baaa... and along the other of which the values are all ≤ 0.abbb... So if a limit were to exist for the original sequence, it would have to be the same as the limit along each subsequence, but this is impossible because one subsequence is always ≥ 0.baaa... and the other is always ≤ 0.abbb...

[@camrouxbg]

@@DrBarker glad to hear the audio has improved. I hate to critique on things like that (would rather be able to critique the math) but sometimes we have to. This was the first of your videos I've seen, but I'll keep watching for sure.

[@smanzoli]

When aproaching 2π, we´re going up from negativo to get to zero. So N*10*π for any even N would be zero, so any multiple of 10π would be zero. When we go 314, 3141, 31415, 314159, 3141592 we are just getting 10 times a rounded down π... so does not matter how big N is, it will be ALMOST N*10*π, coming from down, so it will be always negative, aproaching, but never gettin to zero, sice π is irrational.

[@TylerHNothing]

Dr Barker, does this limit have anything to do with the "Normality" of Pi, eg the uniform distribution of the digits of Pi?

[@Xboxiscrunchy]

No he was very careful to avoid assuming that pi was normal. That’s why he used a and b as the smallest and largest infinitely repeating digits rather than saying it was 1 and 9 as would be the case if it was normal.

[@TylerHNothing]

@@Xboxiscrunchy sure, I definitely agree that he avoided this assumption. I wonder if this could be used to show it

[@TylerHNothing]

@@Xboxiscrunchy also normal would mean 0 and 9 would not only be the smallest digits, but normal also specifies that their distribution would be uniform and that they'd occur with the same frequency as 1-8

[@LukePalmer]

I don't know about the claim bn

[@DrBarker]

It's true that b_n isn't "always" ≤ 0.abbb... , but all we need to show is that b_n ≤ 0.abbb... "infinitely often" (for infinitely many n as you go through the sequence b_n). And similarly we see that b_n ≥ 0.baaa... "infinitely often". So we can find 2 subsequences, along one of which b_n ≤ 0.abbb..., and along the other of which b_n ≥ 0.baaa..., which shows that the limit can not exist.

[@LukePalmer]

@@DrBarker Ah that makes sense, thank you

[@comic4relief]

Maybe the video needn't end suddenly, but with a bit of a buffer at the end. Just a suggestion.

[@kenzoohi]

What are the accumulation values?

[@Mathspro1]

Nice video

[@gasun1274]

why is it not rigorous, to simply state that the limit DNE? because the argument of sin would have to diverge to infnity in order to fit all the digits of pi

[@gasun1274]

oh wait nvm i realized that its a redundant question🤦‍♂️

[@comic4relief]

Clever technique. It least it seems so to me.

[@eeddeellwweeiiss]

i thought it would be 0

[@alicewyan]

I can barely hear the audio, even with the volume maxed out :-(

[@NonTwinBrothers]

Why do all math youtubers have their volume so god damn low???

[@BlackZapdos]

Isn't this just saying that sin(inf) is undefined? Why not put the limit inside the sine? I don't get why this video is 17 minutes long.

[@ericb.4385]

because for example sin(n*pi) does converge to 0 for n going towards infinity. And also the result that these values are all negative, is quite remarkable so I guess you just dont see the beauty

[@BlackZapdos]

@@ericb.4385 it is only the case if it is a multiple of pi. Taking the floor function of any number will not give you a multiple of pi so the reasoning is still the same. Love the silly remark at the end though haha makes u look that much more insufferable

[@dr.palsonp.h.d815]

@@BlackZapdos i agree, but it was worth learning the "infinitely often" type of reasoning, havn't seen that before

[@ga3680]

Very interesting and, for me, most unexpected. At first I assumed it must be zero since the "complete" decimal expansion of pi (ignoring the decimal point) would be an infinite whole number power of 10 times pi. Of course the flaw is that pi has no complete decimal expansion. For these sines all to be negative shows that the digits of pi never become "random".

[@gws1223]

Sin(220/7)

[@omkarnath4703]

Just sin(10^nπ} we know sin(nπ)=0

[@DrBarker]

I set it up slightly differently, so it was sin(\floor{10^n π}) rather than just sin(10^n π), which is easier to compute!

[@Inspirator_AG112]

Observation: sin(-1) < sin(floor(10ⁿπ)) < 0

[@aashsyed1277]

What does it converge to?

[@DrBarker]

We find out whether or not it converges around 14 mins into the video!

[@SCOTLANDCOOL2354]

pi

[@Bruh-bk6yo]

sin(3141592...) 3141592...=10ⁿπ sin(10ⁿπ)=sin(π)=0

[@DrBarker]

To be fair this is also a reasonable way to interpret the informal expression "sin(314159...)".

[@angelmendez-rivera351]

@@DrBarker Yes, that is how I interpreted it. This is why I dislike informal expressions in mathematics. It is the bad kind of ambiguity that mathematicians worked for centuries to get rid of, and there is good reason for that. If I want someone to find the limit of a sequence, I just tell them what the sequence is. Nothing against your video, though, you did it an excellent analysis, and you did specify what sequence you wanted to find the limit of.

[@arseniix]

Nope, nononono. 3141592... by any means is not a number. And that limit is not defined.

[@dmitrygolubev9398]

Alright, this proof is kinda sus. In particular, the part from 10:47 and forward when you state that the limit, if it exists, has to be both lower (or equal) than each individual sequence element that starts with the lowest repeating number and higher (or equal) than each individual element that starts with highest repeating number. Why is that? I don't see a single thing preventing a limit to be an in between number. sin(n)/n approaches 0 while oscillating from one sign to another. I think you should've gone with the "For all... /Exists..." definition of limit. Or at least reference why the limit can't be an in between number.

[@failsmichael2542]

OK so there are two subsequences b_1, n = 0.baaaaa. If b_n converged then the two subsequences would also converge to the same limit. But we know that limit preserves inequality so…

[@dmitrygolubev9398]

@@failsmichael2542 sin(1/n) < 1/n < tg(1/n) for all natural n, but lim(sin(1/n))=lim(1/n)=lim(tg(1/n))=0 for n -> ∞. It sure doesn't look like limits preserve inequality to me.

[@failsmichael2542]

@@dmitrygolubev9398 Let’s say weak inequality then. You get my point.

[@DrBarker]

There are 2 subsequences, along one of which the values are all ≥ 0.baaa... and along the other of which the values are all ≤ 0.abbb... More formally, this is saying that the limsup of the original sequence is ≥ 0.baaa..., and the liminf is ≤ 0.abbb..., but you need the limsup to equal the liminf in order for the sequence to converge. So the original sequence doesn't converge.

[@dmitrygolubev9398]

@@DrBarker I like this one. I guess, it goes like this: Assume sequence converges, then any of its subsequences also converge. Pick the subsequences with a and b as their starting numbers (0.axxxx... and 0.bxxxxx... ) and we start from where all not infinitely reapeating numbers fizzle out. One has a supremum That's lower than 0.abbbb..., the other has an infinum that's higher than 0.baaaa.... That would mean that limsup of the first one is lesser or equal to 0.abbbb..., and liminf of the other one is greater or equal to 0.baaaa. Since we assume that they converge, then the limsup = lim and liminf = lim for respective subsequences, and the limit should be the same for both of them, but lim >= 0.baaaa.... > 0.abbbb... >= lim for all a and b (self evident with any a and b besides 0 and 9, and it's also obvious there since 0.9 > 0.1). Now that I rewatch this video, this is clearly what you were going for, but you skipped the whole infinum to liminf to lim thing, which while relatively obvious, should still be said clearly.

[@lazaremoanang3116]

More than 17 mn? Why is it too long?

[@spudhead169]

Yeah but.... base 10 bruh. No.

[@cicik57]

to put it short, it is sin(pi*n)=0

[@willie333b]

0

[@pizzamidhead2183]

sin(31415...)=sin(∞π)=0. QED