an infinite tangent product.

[@lukaboljevicboljevic]

21:38 you made a mistake, 8^3 = 2^9, not 2^6 Edit: As others pointed out, there are N-2 8s, rather than N-3, so even though the number of 8s is incorrect, 2/2^6 or 1/32 is correct. Great video anyways!

[@amebhattacherjee5848]

This...guy is amazing

[@איתןגרינזייד]

Your small mistakes of N-2 instead of N-3 and 2^6 instead of 2^9 actually cancelled them self's out and you surprisingly got the right answer

[@stephenbeck7222]

As someone else said, he was probably looking at notes sporadically as he went so he copied the 2^6 part from his notes and it sounded reasonable enough from his previous work on the board and he ended with the answer he wanted so he had no reason to go fix anything.

[@jkid1134]

10:00 this limit is usually required to derive l'hopital's rule as far as I can remember. I believe I have been specifically warned against this line of reasoning lol

[@Debg91]

You need to use the limit to obtain the derivative of the sine function, so using L'Hôpital's rule before proving this limit would be a circular reasoning. But once you've proven both things, I guess both ways are correct, but yeah, by calculating the derivative of the sine you already need the limit, so it would be kind of redundant (?)

[@beatoriche7301]

It all depends on how you define the sine function. What you mentioned is something they generally say in introductory calculus courses because there, trig functions are defined using the unit circle. In this context, the addition formulae for sine and cosine are derived using geometry, and using them to evaluate the limit associated with the derivative begs the question of the value of the limit of sin x / x as x tends to zero. However, once you get beyond calculus courses - into analysis, for example -, the sine function will generally not be defined in this way because it takes quite a bit of work to make this definition rigorous. Just one example of why this definition is problematic, what does it even mean to have an angle - in other words, the input of our sine function? If your answer involves the unit circle, keep in mind that, if you're in an introductory analysis course, you haven't actually defined what the length of a circular arc even means - and that's absolutely not a trivial question. Basically, if you define sine in this way, you have to do quite a bit of work to even precisely define what the trig functions mean, only to then derive the power series definition you'll always be using in proofs involving sine anyway because it's just way easier to work with. For this exact reason, basically any analysis course out there will define sine either via its power series or via the complex exponential in the first place. And if this is your definition of the trig functions, using L'Hopital's rule to evaluate this limit is absolutely a valid line of reasoning.

[@MA-bm9jz]

@@angelmendez-rivera351 the sinx/x limit can be easily done without the need of calculus,using l'h is circular(cause you need to know its differentiable everywhere except 0,and when you check the differentiability you need that limit,every respectable calculus course proves those limits(sinx/x;(e^x-1)/x and so on)without l'h

[@beatoriche7301]

​@@angelmendez-rivera351 Have you ever studied analysis before? No analysis course I’ve seen assumes knowledge of calculus. Most analysis courses start by introducing the real numbers, either using an axiomatic approach or constructing them explicitly from the axioms of set theory. And everything you learn in an analysis course is proven rigorously from these first principles. I’ve certainly never read an analysis textbook that says something along the lines of, “You’ve seen this before in an introductory calculus course, so we won’t prove it here.” I’ve checked a few analysis books I happened to have by my side right now, and it turns out all of them actually prove this limit using the power series expansion, which is what’s used to define the sine function. You can evaluate the limit using differentiation, or you can prove it directly from the power series definition of the sine function - most analysis textbooks introduce the sine function before introducing differentiation, though, so the latter way is probably more common. Still, you don’t need the sine function to introduce differentiation or prove L’Hopital’s rule, so using differentiation is valid, albeit a bit nonstandard. There is absolutely no reason you would have to know calculus before studying analysis, and in fact, in my country, we don’t actually have calculus courses - first-year math majors immediately learn analysis. Yes, most calculus courses do introduce a notion of arc length, and calculus students learn to compute arc lengths in various different ways, but how do you know this measure (which is the general mathematical term for length, area, volume etc.) is actually a useful construct? You’d first have to check that it is actually a measure function (which is pretty easy), and then you’d have to show that it is unique, that is, that there is a single extended real number you may call _the_ arc length of a curve on an interval. This is not actually obvious - it relies on a famous result in elementary measure theory known as the uniqueness of measures theorem. Furthermore, you’d have to show that it actually behaves in the same way as the familiar length, most notably, that it is, in fact, invariant under the familiar motions of Euclidean geometry. These are all things you discuss in-depth in introductory measure theory courses, and it usually takes quite some time until you’ve even proven that the Lebesgue measure exists. (Note that, in a measure theory course, you wouldn’t bother to define lengths using integrals because the Riemann integral has some inherent limitations - for instance, you’ll never be able to measure the length of all rational points on the real number line using the familiar integral from calculus -, and so you’ll first prove that such a thing as length exists, and then you’d define integrals using measures by introducing what is called the Lebesgue integral.) Just because something seems obvious doesn’t mean it’s true - a striking example of this in the field of measure theory is Vitali’s theorem, which states that there is no nontrivial measure function that can possibly assign a value to every subset of R^n. In other words, there are subsets of the real number line that have no length (not in the sense of zero length, but whose length is undefined) - so-called Vitali sets. So the intuitive notion that every set of points can be assigned a length is simply wrong - which is why, for instance, Lebesgue measure (the measure you’ll mostly be working with in analysis) is defined as only assigning values to what is known as the Borel σ-algebra - simply put, anything that can be reasonably approximated by rectangles. Maybe I worded my point a bit poorly - what’s tricky is not necessarily defining arc length, but showing that it is actually a useful construct.

[@beatoriche7301]

@@MA-bm9jz What do you mean by “easily done”? It’s easily done using power series, but then the derivative of sine falls out as well because the power series for sine converges everywhere in the reals, which allows us to differentiate it term by term. The problem with the approach in elementary calculus is that terms like “length” and “area” are not as clear-cut as they may seem.

[@benjaminbrat3922]

Very nice and elegant as always! At 19:06, there seems to be a small oversight, you say there is N-3 numbers between 3 and N but there is N-2, right? I guess that makes the final result pi^3/4?

[@bharatsethia9243]

It should be N-2 but at 21:50 he wrote 8³ = 2^6 hence cancelling off the mistakes.

[@ilonachan]

Yeah, that's right But his result is actually correct, because although the third to last equation says "1/8^3" his simplification below is only correct if that was "1/8^2" - which, as you pointed out, would have been the right factor anyway.

[@srenlaichinger3687]

no, the answer is correct, because than he can put an 1/8^2 out front instead of the 1/8^3, which makes the answer in fact 1/32*pi^3, because then the 1/8^2 is equal to 1/2^6.

[@juanixzx]

@@bharatsethia9243 saved by double errors.

[@juanixzx]

I cannot believe two errors still makes the result correct.

[@xaxuser5033]

that s obvious why you can't believe it

[@sergiokorochinsky49]

19:06 + 21:37 = error cancellation

[@xaxuser5033]

@@sergiokorochinsky49 exactly matter of compensation

[@hybmnzz2658]

@@xaxuser5033 what is your point

[@xaxuser5033]

@@hybmnzz2658 the second error fixed what was going on after the first mistake

[@goodplacetostop2973]

23:18

[@sumukhhegde6677]

Awesome:)

[@satyapalsingh4429]

Very good method of teaching . The best professor , I watched . Answer is correct .

[@nickruffmath]

L'hopital's rule can't be used to prove the sinx / x limit from first principles because the derivative of sine depends on that limit.

[@nikitabrilliantov6996]

It's a nice exercise proving that sec(π/4)sec(π/8)sec(π/16)...=π/2 And you can prove it geometrically. Converting 1-tan⁴(π/2ⁿ) to secants, you get the result almost effortlessly

[@tretyakov3112]

This solution looks like overkill. The problem can be figured out in 5 minutes. Without complex, just trigonometry.

[@mihaipuiu6231]

Prof. Michael Penn, you are the best. Fantastic!!!

[@factorization4845]

I converted the question into all cosines, then I multiple by a certain sine and 2^x to shrink the chains of products into a simple one by double angle formula for sine

[@geometrydashmega238]

I did the same. Still same effort I guess

[@zivssps]

Nice video but it's very hard for me to see you use L'hopital's rule to prove the sinx/x limit since it's a circular prove - You have to know that limit to know the derivative of sine, therefore you can't use L'hopital's rule to prove it.

[@zivssps]

Btw, you got a mistake on top a mistake which gave you the right answer. First, the number of appearances of the number 8 is N-2 (N-3+1), not N-3. Second, 8^3 is 2^9 ((2^3)^3 = 2^(3*3)=2^9), not 2^6. Those two mistakes together cancels each other so you got the right answer, but nevertheless they are mistakes.

[@coc235]

You can just say the derivative of sine is well-known, this video isn't supposed to build the entire math from scratch😅

[@mrminer071166]

10:00 My favorite way is to write out the power series for sin (x) , from which, being odd, it's easy to divide out the x and get 1 + a bunch of other powers that boil off.

[@changjeffreysinto3872]

5:16 nice magic you got there

[@erfanmohagheghian707]

shouldn't the exponent of 8 be N-2? from n=3 to n=N there are N-2 terms. You actually made two errors which cancelled out each other. The second error is that 8^3 is 2^9, not 2^6. Things worked out nicely though lol

[@sumukhhegde6677]

*python code* import math z=1 for n in range(3,100): z = z*(1-pow(math.tan(math.pi/pow(2,n)),4)) print(z)

[@112BALAGE112]

That's not a proof. You'd get zero marks for that in an exam.

[@pendawarrior]

@@112BALAGE112 jeff

[@sumukhhegde6677]

@@112BALAGE112 Ik that's, not a proof... but I love coding you know :)

[@sergiokorochinsky49]

Mathematica: Product[1-Tan[Pi/2^n]^4,{n,3, Infinity}]

[@sumukhhegde6677]

@@sergiokorochinsky49 woah

[@johanrichter2695]

Don't understand why L'Hôpital's rule is so popular in some places. It is a confusing rule with many prerequisities that students don't manage to learn, and even if they do learn it, it would be better if they learned other ways of computing limits that bring more understanding. In this specific application of the rule, there is great risk of circularity, depending on how the derivative of sine has been found.

[@oder4876]

The hopitals rule is not compatible to lim goes to 0 of sin(x) /x you need to approach it in 0+ and 0- because x is a lineare function but sin( x) is not

[@elliottmanley5182]

Have I missed something? What became of the yellow term @ kzbin.info/www/bejne/ooqchI2mpp6enq8 4tan^2(theta n)

[@factorization4845]

10:15 L'H cannot be used for sinx/x because the derivative of sinx requires the use of the limit of sinx/x as x->0, which becomes a loop. It has to be a geometric proof

[@azmath2059]

Of course you can use lhopital. What are talking about

[@beatoriche7301]

You don't, actually. If you define the sine function via its power series expansion, then the derivative of the sine function is fairly straightforward to evaluate directly using the power rule (since the power series of the sine function converges in the entire complex plane, which means it can also be differentiated term by term). It's all a question of how you define the sine function.

[@mathunt1130]

People should just know double angle formulae, so they don't need to be proven. as for the second, just work on the RHS to note that sin(4x)/(2sin(2x))=cos(2x)=cos^2x-sin^2x. Now what you do is you find cos^2x and sin^2x in terms of tan^2x, and put them together and they get the identity. For the third, that's the way I did it, but you could expand sin(x) as a power series to show that a_n=pi+O(2^-n) which clearly tends to pi as n tends to infinity.

[@nicolascamargo8339]

daba al final (pi)^3/256.

[@shanmugasundaram9688]

Product of two minuses is plus.Product of two mistakes is the correct answer.The problem is very interesting.Nicely presented.

[@nestorv7627]

Audio sounds bad at 10:26

[@112BALAGE112]

I find the idea of the red identity highly non-obvious. How would one come up with that?

[@insouciantFox]

112BALAGE112 Someone had too much time on their hands.

[@stephenbeck7222]

Obviously he had already worked the original problem and noticed something similar to the right side of the red identity popping up so he figured out a way to derive it ‘backwards’.

[@geometrydashmega238]

It's not that difficult after you notice the left hand side is actually cos(2x). Then you just multiply and divide by sin(2x) and use the double angle formula for sine (cos(2x)*sin(2x)= sin(4x)/2) and there you get it. The thing is, you can use the exact same reasoning to come up with the equality product of ( cos(x/2^k) , k=1..n ) = sin(x) / (2^n * sin(x/2^k)), just by expanding the product and multiplying and dividing by sin(x/2^n) and applying 2^n times the double angle formula for sine mentioned above. I used this one to solve the problem because I ended up with a slightly different expression.

[@112BALAGE112]

@@angelmendez-rivera351 I would have just simplified it to cos(2x) and got horribly stuck.

[@reshmikuntichandra4535]

Answer pi^3/32

[@НиколайШерстюк-ы7е]

Why there is this strange sound after 10:28?

[@mstarsup]

he prolly just opened the window. cicadas outside

[@rajballabhyadav5089]

An alternate approach which I used is as follows....... Write tan as sin/cos and then take LCM and further calculations which require a bit knowledge of trig formulas simplifies the expression as continued product of (cos@)/(cos©)^4 where @ and © are angles which happen to form a GP. Now actually this has a nice closed form(not of the whole expression but separately of numerator and denominator)which you can search on web(it's a pretty standard result, which we in India require to know for tackeling problems in competitive exams. There's a nice way to prove the result which I will explain at the end.) And finally when you carry out the Calcultions same limit formulas are required at the end to arrive at the ans. Proof: cos@ cos2@ cos 2^@ ..... cos2^(n-1). Multiply and divide by 2sin@ which converts the first term(cos@) to sin2@ and this further combines with cos2@ to form sin2^2@ but you have to multiply and divide by 2 again and this thing continues till the last term. This clever manipulation is applicable only when GP of angels has the common ratio as either 2 or 1/2 as we require double angle relation of sin. 😎😎

[@satyapalsingh4429]

Dear Raj Anubhav , your method is 100% correct and short . I also solve in that way . May God bless you !

[@rajballabhyadav5089]

@@satyapalsingh4429 just a correction sir, it's Raj Abhinav

[@pmj9925]

9:28 why is x approaching 0?

[@איתןגרינזייד]

x is approaching zero because it's equal to pi over two to the n and since the limit were talking about is when n goes to infinity, I thing you'll agree to that when n goes to infinity there, x goes to zero like, come on

[@krisbrandenberger544]

There has been a mistake. 8 raised to the 3rd power is 2 to the 9th power, not 2 to the 6th. Therefore, the final answer should be (pi^3)/256.

[@geometrydashmega238]

He also failed before that when writing 2^(n-3) as the product of two's, but there are n-2 two's. The final answer is correct

[@sergiokorochinsky49]

Obviously!, this is equal to the infinite alternating sum of reciprocals of odd numbers cubed. Duh!

[@sergiokorochinsky49]

1-1/27+1/125-1/7^3+1/9^3-1/11^3+1/13^3-1/15^3+1/17^3-1/19^3+...

[@ivoh2384]

To prove that lim sin x / x = 1 as x->0, you are not allowed to use l'Hopital's rule. Because to apply it, you need to differentiate sin (x), but the formula for d/dx sin(x) = cos(x) involves the fact that lim sin x / x = 1 as x->0. You need to use something else e.g. squeeze theorem or geometric proof.

[@azmath2059]

What the he'll are you on about?

[@beatoriche7301]

@Adam Romanov They're right, though, even if their comment wasn't very polite. There is no such thing as _the_ proof that the derivative of sine is cosine - depending on how you define the sine function, you may prove that in various different ways. I agree that, if you approach the question of the derivative of the sine function from its origins in trigonometry, the line of reasoning in the video is not valid - but it's not like there is only this one absolute way to define sine. (I hope you don't mind if I just copy and paste a comment I posted elsewhere.) It all depends on how you define the sine function. What you mentioned is something they generally say in introductory calculus courses because there, trig functions are defined using the unit circle. In this context, the addition formulae for sine and cosine are derived using geometry, and using them to evaluate the limit associated with the derivative begs the question of the value of the limit of sin x / x as x tends to zero. However, once you get beyond calculus courses - into analysis, for example -, the sine function will generally not be defined in this way because it takes quite a bit of work to make this definition rigorous. Just one example of why this definition is problematic, what does it even mean to have an angle - in other words, the input of our sine function? If your answer involves the unit circle, keep in mind that, if you're in an introductory analysis course, you haven't actually defined what the length of a circular arc even means - and that's absolutely not a trivial question. Basically, if you define sine in this way, you have to do quite a bit of work to even precisely define what the trig functions mean, only to then derive the power series definition you'll always be using in proofs involving sine anyway because it's just way easier to work with. For this exact reason, basically any analysis course out there will define sine either via its power series or via the complex exponential in the first place. And if this is your definition of the trig functions, using L'Hopital's rule to evaluate this limit is absolutely a valid line of reasoning. Let me show you an example of how to do this: For any complex number z, define exp(z) as the sum of z^n / n! from 0 to infinity. (In other words, we're using complex derivatives, but note that the complex derivative also captures the real derivative in the sense that, if the derivative of a complex function exists, it'll be the same as the real derivative if you plug in a real number.) Note that it is fairly easy to check using the ratio test that this series converges in the entirety of the complex plane. This lets us conclude that it is not only differentiable, but also differentiable term by term (that's usually one of the first theorems you encounter in a complex analysis course - I don't wanna prove it right now since it would be practically unreadable as a KZbin comment, but I can write down a proof in LaTeX if anyone's interested). Applying the power rule yields the result that, just like for real numbers, the complex exponential is also its own derivative. Now restrict exp (z) to the imaginary axis, which means the input z may be expressed as i*x, where x is a purely real number. We now define cos x = Re (e^ix) and sin x=Im (e^ix). Differentiating the sine function with respect to x yields, using the chain rule and comparing real and imaginary parts, d/dx (sin x) = d/dx (Im (e^ix)) = Im (d/dx (e^ix)) = Im (i e^ix) = cos x. This is a perfectly valid line of reasoning using some elementary complex analysis, and it doesn't require the limit in question at any point.

[@azmath2059]

No need for first principles here. That's just pedantic. For the purposes of solving the main problem it's best to keep it simple and use good old lhopital.

[@azmath2059]

This problem is not about proving lim of sinx/x. If you wanna do that it's a separate problem and you don't need geometry or the squeeze theorem to prove it. You can find the derivative of sinx without going to first principles but by using the definition that d/dx ( e^x) = e^x and then go to lhopital.

[@beatoriche7301]

@@angelmendez-rivera351 That is something I take issue with - words mean whatever we want them to mean, and just because the sine function has its origins in trigonometry doesn't mean it has to be defined in this way. Since these definitions are all equivalent anyway, we may as well choose any particular one that is convenient for us. There's a visual definition using the unit circle, which takes quite a bit of work to make rigorous but still serves as a nice introduction for beginners, and a power series definition that is generally much easier to work with. If you first define something, you get to choose the conventions yourself - think of division by zero as a classical example. In algebra, division by zero is generally undefined because it is inconsistent with the field structure, but the extended complex plane - which is extremely useful for studying Möbius transformations and, by extension, non-Euclidean geometries, among many other things - is obtained by defining 1/0 = ∞ and 1/∞ = 0. The extended complex numbers are not a field anymore, which makes them somewhat inconvenient to work with in some contexts, but in other contexts, they're still useful. Whether we define division by zero is an arbitrary decision we make depending on what's convenient for us. To use another example - in calculus and analysis, 0 * ∞ is an indeterminate form in the context of limits, but in measure theory, the general convention is 0 * ∞ = 0. There's no need to legitimize this convention in some way - it's just what's convenient. I hate to be pedantic, but this definition doesn't work. First off, what is the number π? Of course, you may define π as being four times the limit of the Leibniz series or something along those lines, but that begs a much more fundamental question: What is a rotation by an angle? I certainly see no way to define a rotation without first either defining length or using the trigonometric functions - it's easy to define rotation by simple angles like π/2, but all the conventional ways to define rotation by an arbitrary angle I can think of presuppose the concept of an angle in some way, be it the complex exponential, rotation matrices, or the two reflections theorem. Of course, that doesn't mean there is no way, but if you know of one, I'd definitely be interested. I've already addressed this point in another thread - calculus courses do introduce arc length, but they don't actually prove that it is a sensible notions with the properties you might expect, such that the existence of one unique concept of arc length; this is what you learn when studying measure theory, which I guess you can do before studying analysis, though you'll definitely need some basic concepts from analysis such as the real numbers. Well, this is really a case of anecdote against anecdote. All I can say is that no analysis textbook I've ever read defines the trig functions using the unit circle, and maybe you have read one that does so. I can tell you the names of the textbooks I've read, though most of them aren't in English because English isn't my native language. Well, I guess you're right that complex derivatives are not introduced in real analysis courses, but they're defined completely analogously, and the theorem that any complex power series can be differentiated term by term within its open disk of convergence is one of the most basic theorems in complex analysis - it's literally the second page of the complex analysis textbook I happen to have by my side right now. It's not like it requires a large theoretical apparatus that is not available to a real analysis student. Since the complex numbers also form a valued field, the basic definitions and theorems relating to the convergence of infinite series like absolute convergence being a stricter property than regular convergence and the ratio test can be immediately generalized.

[@jesusalej1]

I forgot give like. Awesome, no more to tell.

[@tonk6812]

Yes it should be 8^ N-2

[@user-cg9zb3uc8b]

awesome!

[@nitayderei]

You shouldn't use Lhospital rule on this specific limit.

[@azmath2059]

Of course you should and he did

[@LucaIlarioCarbonini]

8^3 = 2^9

[@geometrydashmega238]

He also failed before that when writing 2^(n-3), it was 2^(n-2). Errors cancel :D

[@LucaIlarioCarbonini]

Thanks @@geometrydashmega238 I missed that

[@함함-b9t]

nice pro.

[@catzgaming7197]

Can you help me solve this? (2^x)3x +3x =1

[@sergiokorochinsky49]

Read about Lambert W function.

[@sergiokorochinsky49]

@@angelmendez-rivera351 , thanks, I didn't know that. I just assumed that mixing linear functions with exponentials can always be untangled with W, but obviously not.

[@SlidellRobotics]

Really? Using Euler's formula to prove a double angle formula? Those can be shown with isosceles triangles and alternative ways of calculating area. You're killing a wolf with a nuke here.

[@FaranAiki]

Yes, it is much simpler.