First, I would like to express my genuine admiration for your solution to this seeming monstrous solution. However, I do have one question: When you were applying L' Hopital's Rule on 2^(-k), it seems like you just applied the power rule directly onto the function. However, 2^(-k) is an obvious transcendental function, and the power rule should not apply here. The derivative of 2^(-k) is: -ln(2)/(2^k) Is your method some eerie technique I have never heard of before, or is your result equivalent to my derivative? Nonetheless, I do realize that even if my derivative is used, the numerator and denominator would still cancel out and produce the same result. Anyways, great video.
That is an excellent point! You are absolutely right: 2^(-k) should be treated as exponential function, not a power function. However, as you aptly mentioned, the mistake does not lead to an erroneous answer because two derivatives of 2^(-k) -- no matter what it is -- are destined to cancel out and be obsolete. I apologize for making such an elementary mistake. Thank you so much for pointing it out! =)
actually the chain rule saved him -- off the top of my head, any monotonic function would've had the same result as long as it was applied to k identically in the numerator and within sin in the denominator.
with sigmas and capital pi's, the answer is a related identity, or in this case, expand and look for a pattern (but with a trigonometric identity applied first)
At first I was like "Well, it's still sin(x)/x, it's not exactly pleasant" then he zooms back to the original and I see that random little x hanging out in front. I then throw up my hands and just about cheered in glee at all that ridiculousness ending up condensing into sin(x). EDIT: I think the final result would have been more amusing if it was just indefinite integral, or if they chose 2pi as the upper limit so the final answer was just 0. Hahaha.
And then you try to integrate sin(x)/x (because someone hasn't been so kind as to put a nice x * ...) and you realise that the monstrosity of the expression actually has a reason...
Absolutely elegant walkthrough and solution! I was just scrolling through videos and after noticing this was 17 minutes long, I was sure i'd leave early. But, your explanation and simplification of the expression as you went through was surprisingly satisfying & beautiful. Excellent Job!
12:22 you should not use L-Hopital rule. It's an elementary limit. Sin(y) = y when y tends to zero. Thus sin(x/2^k) can be replaced with x/2^k. Then 2^k will cancel off and you left with 1/x as solution for the limit.
The way I did the product was a little different. I saw that if you plugged, say, t into the value of x, the product would be cos(t/2)cos(t/4)cos(t/8)cos(t/16)..., but if you plugged 2t into the value of x, you’d get cos(t)cos(t/2)cos(t/4)cos(t/8)cos(t/16)..., which is basically the same thing as the original product but multiplied by cos(t). I couldn’t think of a function off the top of my head where f(2t)=f(t)cos(t). However, after you brought up the double angle formula later on, I realized sin(t) would (almost) for the description, the only difference is it’d return double the answer. After foolishly testing what would happen if I divided exponents, I realized if I had f(t)=sin(t)/t, f(2t) would be 2cos(t)sin(t)/(2t), or f(t)cos(t). So I kinda figured it out by way of continue the pattern from initial conditions, but I like how you did it too.
The derivatives do get cancelled out, but you should know that the derivative of 2^k with respect to k is actually ln(2)*2^k.. The power rule works in the opposite case - when your variable is the base, and the constant is the exponent.
I got click baited by that thumbnail. But dude, that was brilliant and beautiful! If we make that integral's upper limit to 2pi instead of pi/4, the final answer would have been a spectacular ZERO! Lol.
So you're differentiating the exponent function using the power rule. Not very nice, but you got lucky having it not matter since the wrongfully found derivatives cancelled each other out. Apart from that, cool vid!
SpaghettiToaster I do not think it matters. Consider the general expression f(x)/g[f(x)]. The derivative of the numerator is f’(x), and the derivative of the denominator is g’[f(x)]·f’(x). Then the quotient of the two is simply 1/g’[f(x)]. No conditions about f(x) or g(x) were assumed.
I think the limit of 2^k * sin(x / 2^k) as k goes to infinity is much more easily found by recognizing that as k becomes large the value inside the sin approaches zero. Then you can use the fact that sin(x) approaches x as x approaches zero giving 2^k * x / 2^k which is simply x.
I'm not where the sinx/x comes from in regards to the comment I responded to. Sorry if I lost track of something I've been watching alot of math videos over the last couple days.
Matisan Andrei x is not approaching 0, x/2^k is approaching 0 as k approaches infinity. That allows the small angle approximation, which makes this fomula valid. Note we don't let k get all the way to 0, that's impossible, we just examine what happens as it gets infinitessimally close, in which case the small angle approximation gets infinitessimally close to precise and the answer comes out correct without having to differentiate. It is a simpler way to do it. I think what he did here is actually just proving the small angle approximation, which is not necessary.
to be honest, once you do cosx = sin2x/2sinx, the question is standard for any math student. however having an eye to do that substitution takes immense lateral thinking
I solved the limit at 11:30 a different way... As k approaches infinity, sin(x/2^k) approaches x/2^k (sinx = x approximation at small angles) for finite values of x. This in turn cancels out the 2^k's and you're left with sin(x)/x a lot earlier
I'm not sure if I'm remembering it right, but that infinite product is very reminiscent of the derivation of the Viète series for π. By recognition, I got it in the first few seconds that the integrand is equal to sin x, which makes the evaluated integral 1-√2/2.
I think it is simple if you use imaginary exponentials. The 1/2^k pops out immediately and the multiplication of the cosines is just the addition and subtraction of exponentials. Once the negative terms get factored the term will show you that you will get the 1/2^k * sin(x/2^(k-1))/2sin(x/2^k). Then you can proceed from there.
I did a different trig transformation on the infinite product that gave me an infinite sum, which happened to be a Riemann sum. From there an integration yielded the same function sin(x)/x.
@11:09 once you have lim sin(x)/(2^ksin(x/2^k)) you could have brought x in the integral in to get sin(x) lim (x/2^k)/(sin(x/2^k)), and that lim = 1 ( since siny/y --> 0 as y --> 0). Then your original integral becomes integral sin x.
So ~0.3, if you really insist. But then again any definite integral reduces to a "simple" number, including those that are impossible to calculate except through numerical methods.
...i don't think we need to use l 'hopital as 2^k tends to infinity 1/2^k tends to zero. so on re-arranging the expression we get lim(k -> infinity) 1/[{x.sin(x/2^k)}/{x/2^k}] which gives us just x as lim( A -> 0 ) sinA/A = 1
The problem becomes simpler if right in the beginning you take the product from k = 1 to k = n and place a limit (n -> infinity) before the product. The finite product can be simplified just as you showed, without dealing with infinity. Then you'll have a term sin(x/2^n)/(x/2^n) which goes to 1 in the specified limit. You're only left with sin x inside the integral.
d/dx(2^(-x)) = -2^(-x) log(2), should use the d/dx{f(x)^g(x)} formula... Main factor is possibly to notice that product rule or its inverse for infinite product is too hard to be used and something must be canceled a lot before integration starts.
A great video. I feel if calculus is taught the way you do then any Bee in the world can be cracked. Absolutely fantastic solution to a tricky problem. Great work as always.
In the expression of the limit, lim sin(x)/(2^k sin(x/(2^k))), as k increases without bound sin(x/(2^k)) can be approximated as simply x/(2^k) so that the denominator in the limit is just x leaving us with sin(x)/x, etc.
Great solution. In that last limit you can just replace 2^k with k (or substitute another letter if that's clearer). At that point it's just some quantity going to infinity; it doesn't matter that it's a power of 2.
Your solution is very convoluted: this integral easily follows by noting that the product telescopes, courtesy of sin(x)=2*cos(x/2)*sin(x/2)=2^2*cos(x/2)*cos(x/2^2)*sin(x/2^2)=...=2^n*[cos(x/2)*...*cos(x/2^n)]*sin(x/2^n), ergo product of cos(x/2^k) from k=1 to n is given by sin(x)/[2^n*sin(x/2^n)] Now, to find the limiting value of the product as n->infty, note that h:=x/2^n goes to 0 for any x; rewrite the earlier result as [sin(x)/x]*[(x/2^n)/sin(x/2^n)] and note that the second multiplicand goes to 1 as n->infty, courtesy of the standard limit sin(h)/h or equivalently h/sin(h) goes to 1 as h->0. This leaves us with sin(x)/x which when multiplied by x in the integrand makes it the integral of sin(x) from 0 to pi/4, which is straightforward. There is no need to use L'Hopital here. tl;dr: prnt.sc/nno2mq
Soon as i saw it, i knew how to simplify it because these type of expressions pop up in communication systems all the time for those studying electrical engineering
When calculating the limit, you could've used a first order taylor approximation of sin(x) around 0: sin(x/2^k) tends to x/2^k as x/2^k approaches 0, then the 2^k in the denominator would have cancelled out, and the limit would be sin(x)/x.
You can just multiply and divide both sides by x, without using L'Hôpital's rule. Since k goes to infinity, x/2^k goes to 0 and we know that sinw/w goes to 1 as w - > 0 and so does w/sinw. Therefore we obtain: sinx/x Lim ((x/2^k)/(sin(x/2^k)) = sinx/x * 1 = sinx/x
Comment about your limit (different from the differentiating part, since I know people seem to keep harping on you about that lol): Once you've gotten the expression at 13:00, it might be easier to replace 1/2^k -> u, and then take the limit as u -> 0. You'd be left with sinx * lim_{u -> 0} u/sin(ux) and from there L'Hopital's rule is very trivial to implement, or you may even recognize this limit immediately as being 1/x, which would give you sinx/x.
If one knows that sin(t)/t -> 1 as t -> 0 (which is easily checkable by taking sin definition with power series), it suffices to adjust the fraction a little bit as follows sin(x) / 2^N * sin(x/2^N) = (sin(x) / x) * 1 / [(sin(x/2^N)/(x/2^N)], indeed sin(x/2^N)/(x/2^N) -> 1 as N -> infinity (because x/2^N -> 0 as N -> infinity), and from this remark it follows that the whole limit converges to sin(x)/x.
@15：13 ,when video maker was showing how we can use L'Hôpital's rule to get the limit , it was wrong to derivative 2^(-k) as k*2(-k-1) ,it should be -ln2*2^(-k) . But due to the denominator also had 2^(-k) to derivative ,so actually ,even video maker was wrong here , it still can't affect the final result of this limit . Because it will cancel out eventually.
Using methods from your other video (@ you can (with some effort) write the product as 1/2^(m-1) * SUM cos(2i-1/2^m * x) for i = 1 to 2^(m-1) This is basically just the average of values cos(x) over the interval [0,x] Hence the the sum above becomes 1/x * INTEGRAL cos(t) dt for t = 0 to x = sin(x)/x
looks hard but it is actually easy multiply and divide with sinx and the limiting product equats to sinx/x . thus the integration reduce to sinxdx = 1-1/root2
Hiii! Great video, but I'd just like to point out that you made a small differentiating mistake then differentiating 2^k (though it won't affect the final result since the derivative gets cancelled out anyways)!!!!
Yes, you are absolutely correct! I treated the expression as a power function, when it is an exponential one; however, as you noted, the mistake does not need to an erroneous final answer due to the cancellation of the derivative, no matter what it is. Thank you for pointing it out! =)
11:54 correct me if I'm wrong we could have used Lim a->0 sin a/a =1. (Since k ->∞ so sin(x/2^k) -> 0) Put a as (x/2^k), multiply and divide by x/2^k. Which will give you 1. Please correct me (I'm a novice at calculus) thanks :)
Yes, you are right,Yash Mehan. Putting x/2^k =a, k tending to infinity, and multiplying and dividing by x/2^k, you get xsinx/x which is sinx. Using Sandwich Rule is easier than using L' Hopital Rule.
I did attempt to work out that question yesterday, but I was not able to obtain the answer provided (1/3). I may be wrong, but I believe the integral they gave us had no elementary solution or they might have provided the wrong solution. When you simplify both infinite geometric series, you obtain integral from 0 to pi/4 of (tanx)^2/(1+x^2). I attempted to make a substitution u=arctanx (which was somewhat natural given the presence of 1/(1+x^2)), but the resulting integral was still complex. Sure, there may be a very creative way of attacking the integral, but WolframAlpha provides the value of ~0.1565, which is obviously not 1/3. If you have figured out a way of evaluating it, I would really appreciate it if you could comment it! =)
@@LetsSolveMathProblems I don't know the question but tan(x)/(1+x²) cannot be simplified by neither u=tan(x) (A) nor x=tan(u) (B) . It would be (A:) u/(1+arctan(u)²) which to my knowledge can't be simplified or (B:) tan(tan(x))/sec(x)² which is bad
Mmmm.... The statement at 13:00 sounds invalid. The perplexities about applying de L'Hopital in this case gain further points. Looks like you have been kind of lucky this time :)
Okay I'm 6 years late but at 12:18 could you use small angle formulae? sin(x/2^k) approaches x/2^k because the expression becomes arbitrarily small as k tends to infinity. That gives you the same answer, as the limit equals 1/((2^infinity)*(x/2^infinity)) = 1/x, giving the whole product equal to sinx/x.
I’ve tried and tried and tried and tried to become good at math and have failed every time. I’m so angry and jealous of brilliant people like this, it just makes me want to cry. Why can’t I have a mathematical mind? It’s just not fair!
I am not so clever with trigonometric identities, so I came up with another solution: Writing cos(t) = [exp(-ix) + exp(ix)]/2, one can see that the N'th partial products look like N=1: [exp(-1/2 ix) + exp(1/2 ix)]/2, N=2: [exp(-3/4 ix) + exp(-1/4 ix) + exp(1/4 ix) + exp(3/4 ix)]/4, N=3: [exp(-7/8 ix) + exp(-5/8 ix) + exp(-3/8 ix) + ... + exp(7/8 ix)]/8 and so on. There are at least two ways to evaluate the limit: 1. One can see by the definition of the Riemann integral that the limit of the partial products is the integral of exp(itx)/2 from -1 to 1, which is easily evaluated to be sin(x)/x. 2. Each term is a geometric sum, so we can directly figure out with some effort that the N'th partial product is sin(x)/(2^N sin(x/2^N)), whose limit as N→∞ is sin(x)/x.
Wow! Such an elegant solution! However, the power rule shouldn’t be applied on an exponential. You have to use the rule for differentiating exponentials, which yields -ln(2)/2^k. Thankfully that didn’t matter because the false derivatives canceled each other out. Other than that, I loved this video! I never thought such a monstrosity of an integral would simplify so elegantly!
I think that there is a logical mistake in the end. I try to explain myself and I hope to be clear. The L'Hôpital's theorem can be used only with function that approaching to limit get 0/0 or inf/inf, as you said, but we need function and we have a succession. I know that there is the equivalence between successional limit and functional limit but I thing that it would be more genuine and correct using punctual convergence that lead us to a uniform convergence and finally using the asimptotic we get the result. I think that this solution is more elegant and correct. Hope to have your feedback! 😁
It is not an error. It can be proven that if a function (that is equal to the sequence at integral values) converges to L, the corresponding sequence converges to L as well. Of course, as you stated, you can try to solve this problem purely by analyzing the sequence itself. However, the solution presented in the video is probably the best one if you are pressed for time or wish for a more intuitive solution.
If I hear one more person say, "It's all Greek to me!" when they see me watching these videos, I will start devising death traps for which the means of escape are solving these kinds of math problems.
This is a very good solution but has a slight flaw. I would like to point out that for differential of 2^(-k), direct power rule can't be applied. This however doesn't make any difference in the solution as d/dx(2^-k) cancels in the numerator and denominator.
My first instinct was to use an infinite sum that approximates cosine and see if things cancel out that way. That would probably have just made a mess.
L'Hôpital formula is used in the case of continuous functions. Here it is obviously a series (with respect to k). So you should use Stolz-Cesàro theorem instead of L'Hôpital.
Although it is true that we are only examining the values at discrete integers, it can be proven that if a function (that is equal to the sequence at integral values) converges to L, the corresponding sequence converges to L, as well. (As a side note, the converse is not true in general.) The manipulation in the video is justified. =)
the limit could have been solved using standard sint/t limit when t tends to 0, using sub. no need to differentiate!! great vid tho! im a jee aspirant, one of the toughest math exams in the world so yeah this one vid is great!
Everything was going pretty good until about 12:00 when he started calculating the limit of 2^k sin(x/2^k). The calculations in this video are at best amateurish. It's actually very simple and doesn't even require l'Hopital's rule. Just denote 2^-k by t, say, and your limit becomes sin(xt)/t as t->0 and that limit is easily seen to be x!!!
First, I would like to express my genuine admiration for your solution to this seeming monstrous solution. However, I do have one question:
When you were applying L' Hopital's Rule on 2^(-k), it seems like you just applied the power rule directly onto the function. However, 2^(-k) is an obvious transcendental function, and the power rule should not apply here. The derivative of 2^(-k) is:
-ln(2)/(2^k)
Is your method some eerie technique I have never heard of before, or is your result equivalent to my derivative?
Nonetheless, I do realize that even if my derivative is used, the numerator and denominator would still cancel out and produce the same result.
Anyways, great video.
That is an excellent point! You are absolutely right: 2^(-k) should be treated as exponential function, not a power function. However, as you aptly mentioned, the mistake does not lead to an erroneous answer because two derivatives of 2^(-k) -- no matter what it is -- are destined to cancel out and be obsolete. I apologize for making such an elementary mistake. Thank you so much for pointing it out! =)
I appreciate your response. This is still a great video nonetheless, and your content is quite appealing.
In overall that was great video. Thank you
actually the chain rule saved him -- off the top of my head, any monotonic function would've had the same result as long as it was applied to k identically in the numerator and within sin in the denominator.
I'm glad I found your comment as I was just about to say the same thing: d/dk(2^(-k)) = -ln(2)/(2^k).
"How do you attack this?" With nuclear weapons.
with sigmas and capital pi's, the answer is a related identity, or in this case, expand and look for a pattern (but with a trigonometric identity applied first)
IU ER r:whoooosh
@@iuer4643 r/woosh
From orbit.
lmao
I'm more impressed by the person who came up with the question
I think this infinite product was known to francois viete, a french mathematician, even before calculus is invented by newton and leibniz :)
@@monke9865 yea ur right
@@monke9865
The way Vieta found it is beautiful.
@@monke9865
Euler also found it.
Euler found basically everything
The fact that that monstrous expression reduced to sinx/x made me laugh
At first I was like "Well, it's still sin(x)/x, it's not exactly pleasant" then he zooms back to the original and I see that random little x hanging out in front. I then throw up my hands and just about cheered in glee at all that ridiculousness ending up condensing into sin(x).
EDIT: I think the final result would have been more amusing if it was just indefinite integral, or if they chose 2pi as the upper limit so the final answer was just 0. Hahaha.
And, when you realize that there is an x as factor in the integrand, you shit out.
from now on I'm writing sin(x) as x*(product of k=1 to infinity of cos[(x)/(2^k)]
How else would you write sin? I mean, x*(product of k=1 to infinity of cos[(x)/(2^k)]).
And then you try to integrate sin(x)/x (because someone hasn't been so kind as to put a nice x * ...) and you realise that the monstrosity of the expression actually has a reason...
Just passed my cal 2 class, and even I was able too follow long, I’m very impressed!!!!
Absolutely elegant walkthrough and solution! I was just scrolling through videos and after noticing this was 17 minutes long, I was sure i'd leave early. But, your explanation and simplification of the expression as you went through was surprisingly satisfying & beautiful. Excellent Job!
Thank you for your compliment. I'm glad you enjoyed the video. =)
12:22 you should not use L-Hopital rule. It's an elementary limit. Sin(y) = y when y tends to zero. Thus sin(x/2^k) can be replaced with x/2^k. Then 2^k will cancel off and you left with 1/x as solution for the limit.
The way I did the product was a little different. I saw that if you plugged, say, t into the value of x, the product would be cos(t/2)cos(t/4)cos(t/8)cos(t/16)..., but if you plugged 2t into the value of x, you’d get cos(t)cos(t/2)cos(t/4)cos(t/8)cos(t/16)..., which is basically the same thing as the original product but multiplied by cos(t). I couldn’t think of a function off the top of my head where f(2t)=f(t)cos(t). However, after you brought up the double angle formula later on, I realized sin(t) would (almost) for the description, the only difference is it’d return double the answer. After foolishly testing what would happen if I divided exponents, I realized if I had f(t)=sin(t)/t, f(2t) would be 2cos(t)sin(t)/(2t), or f(t)cos(t). So I kinda figured it out by way of continue the pattern from initial conditions, but I like how you did it too.
The derivatives do get cancelled out, but you should know that the derivative of 2^k with respect to k is actually ln(2)*2^k.. The power rule works in the opposite case - when your variable is the base, and the constant is the exponent.
This comment should be pinned.
Ah, right, because we were deriving with respect to k. Good catch!
I like how his panicking about the problem lasts for the two minutes he explains what the Pi function is
The Pi function is actually something different. This Pi is an operator.
The pi function is the prime counting function, right?
@@agfd5659 i know it as a function similar to the gamma function that extends the definition of the factorial
@@trace8617 I think that is the gamma function
Agfd the gamma function is another function that operates similarly to the pi function. you know you’re one google search away
BRO, THIS IS ONE OF THE MOST EASIEST MATH PROBLEMS EVER IN MY PROFESSOR YEARS OF TRAINING.
It's taken me forever to finally stumble across an explanation of what capital pi means. Thanks.
Just send us to l'Hôpital XD
XD
:"v
I got click baited by that thumbnail. But dude, that was brilliant and beautiful!
If we make that integral's upper limit to 2pi instead of pi/4, the final answer would have been a spectacular ZERO! Lol.
d(2^(-k))/dk=-ln(2).2^(-k) not -k.2^(-k-1)
Thats what I thought! Was hoping someone else caught the mistake
My friend, you're saved by the virtue of cancellation hahaha
So you're differentiating the exponent function using the power rule. Not very nice, but you got lucky having it not matter since the wrongfully found derivatives cancelled each other out. Apart from that, cool vid!
Алексей Беляев Oops. I hadn't noticed that, but you're right.
Only if the function is monotonous, no?
SpaghettiToaster I do not think it matters. Consider the general expression f(x)/g[f(x)]. The derivative of the numerator is f’(x), and the derivative of the denominator is g’[f(x)]·f’(x). Then the quotient of the two is simply 1/g’[f(x)]. No conditions about f(x) or g(x) were assumed.
k is a constant
@@bigben6564
Rofl.
I am not even majoring math but I keep watching the video till the very end haha, you make hard things interesting, thank you! :)
I think the limit of 2^k * sin(x / 2^k) as k goes to infinity is much more easily found by recognizing that as k becomes large the value inside the sin approaches zero. Then you can use the fact that sin(x) approaches x as x approaches zero giving 2^k * x / 2^k which is simply x.
ib9rt exactly,plus you can’t use hospital when doing limits for sequences
Sin(x) approaches zero as x approaches zero though
Brandon Gaydusek it’s sinx/x as x tends to 0
I'm not where the sinx/x comes from in regards to the comment I responded to. Sorry if I lost track of something I've been watching alot of math videos over the last couple days.
Matisan Andrei x is not approaching 0, x/2^k is approaching 0 as k approaches infinity. That allows the small angle approximation, which makes this fomula valid. Note we don't let k get all the way to 0, that's impossible, we just examine what happens as it gets infinitessimally close, in which case the small angle approximation gets infinitessimally close to precise and the answer comes out correct without having to differentiate. It is a simpler way to do it.
I think what he did here is actually just proving the small angle approximation, which is not necessary.
to be honest, once you do cosx = sin2x/2sinx, the question is standard for any math student. however having an eye to do that substitution takes immense lateral thinking
Ajinkya Jumbad itachi is best bro
So, it’s easy, once you do the hard part.
once it was simplified to the integral of sinx I burst out laughing. the way it simplified so nicely into that was genuinely hilarious
I solved the limit at 11:30 a different way...
As k approaches infinity, sin(x/2^k) approaches x/2^k (sinx = x approximation at small angles) for finite values of x. This in turn cancels out the 2^k's and you're left with sin(x)/x a lot earlier
2:11 4:29 "cosiiiine"
I'm not sure if I'm remembering it right, but that infinite product is very reminiscent of the derivation of the Viète series for π. By recognition, I got it in the first few seconds that the integrand is equal to sin x, which makes the evaluated integral 1-√2/2.
I did it within 2 minutes. But your explaination overwhelmed me. Kudos!
You did? Because you recognized Eulers' expression or because you're some kind of genius?
I think it is simple if you use imaginary exponentials. The 1/2^k pops out immediately and the multiplication of the cosines is just the addition and subtraction of exponentials. Once the negative terms get factored the term will show you that you will get the 1/2^k * sin(x/2^(k-1))/2sin(x/2^k). Then you can proceed from there.
I did a different trig transformation on the infinite product that gave me an infinite sum, which happened to be a Riemann sum. From there an integration yielded the same function sin(x)/x.
So elegant... maths can be definitely better than porn
No
No
True
+
idk about that one chief
@11:09 once you have lim sin(x)/(2^ksin(x/2^k)) you could have brought x in the integral in to get sin(x) lim (x/2^k)/(sin(x/2^k)), and that lim = 1 ( since siny/y --> 0 as y --> 0). Then your original integral becomes integral sin x.
Just write that 1 / (2^k sin (x/2^k)) is equivalent to 1 / (2^k x/2^k) = 1/x since sin(x/k) is equivalent to x/k when k goes to infinity
So ~1.7
All that complex work, yet such a simple answer
So ~0.3, if you really insist. But then again any definite integral reduces to a "simple" number, including those that are impossible to calculate except through numerical methods.
Nah this problem only dealt with real numbers
...i don't think we need to use l 'hopital as 2^k tends to infinity 1/2^k tends to zero.
so on re-arranging the expression we get lim(k -> infinity) 1/[{x.sin(x/2^k)}/{x/2^k}] which gives us just x as lim( A -> 0 ) sinA/A = 1
Really nice
The problem becomes simpler if right in the beginning you take the product from k = 1 to k = n and place a limit (n -> infinity) before the product. The finite product can be simplified just as you showed, without dealing with infinity. Then you'll have a term sin(x/2^n)/(x/2^n) which goes to 1 in the specified limit. You're only left with sin x inside the integral.
Water Molecule thank you for your suggestion. I was wondering the same. Maybe I am not so “stupid” as I thought after all lol.
isn't derivative of 2^(-k) will be -ln(2)/2^(-k)?
Isn't it -ln(2)*2^(-k)?
You’re right since it is transcendental. But even though he got the derivation wrong the answer is still the same for a weird and lucky reason
@@drcommondrate12 Rather be lucky than good
right
d/dx(2^(-x)) = -2^(-x) log(2), should use the d/dx{f(x)^g(x)} formula...
Main factor is possibly to notice that product rule or its inverse for infinite product is too hard to be used and something must be canceled a lot before integration starts.
Imagine the x multiplier wasn’t there in the question, then you’d have to use Feynman’s technique after all the other stuff for the product
A great video. I feel if calculus is taught the way you do then any Bee in the world can be cracked. Absolutely fantastic solution to a tricky problem. Great work as always.
Incredible. The infinite product turns out to be a telescoping series. Without the x before the series, the integral will be even horrible to evaluate
Great video. A beautiful solution to a monstrous equation. Thanks! Keep doing these vids!
In the expression of the limit, lim sin(x)/(2^k sin(x/(2^k))), as k increases without bound sin(x/(2^k)) can be approximated as simply x/(2^k) so that the denominator in the limit is just x leaving us with sin(x)/x, etc.
Great solution. In that last limit you can just replace 2^k with k (or substitute another letter if that's clearer). At that point it's just some quantity going to infinity; it doesn't matter that it's a power of 2.
That use of L'Hopital to work that Infinity boi (á la flammable math) to 1 was SICK! What an elegant solution.
Never seen mathematics at this level before.
It is indescribable
beautiful intuition about the double angle formula!!
Your solution is very convoluted: this integral easily follows by noting that the product telescopes, courtesy of sin(x)=2*cos(x/2)*sin(x/2)=2^2*cos(x/2)*cos(x/2^2)*sin(x/2^2)=...=2^n*[cos(x/2)*...*cos(x/2^n)]*sin(x/2^n), ergo product of cos(x/2^k) from k=1 to n is given by sin(x)/[2^n*sin(x/2^n)]
Now, to find the limiting value of the product as n->infty, note that h:=x/2^n goes to 0 for any x; rewrite the earlier result as [sin(x)/x]*[(x/2^n)/sin(x/2^n)] and note that the second multiplicand goes to 1 as n->infty, courtesy of the standard limit sin(h)/h or equivalently h/sin(h) goes to 1 as h->0. This leaves us with sin(x)/x which when multiplied by x in the integrand makes it the integral of sin(x) from 0 to pi/4, which is straightforward. There is no need to use L'Hopital here.
tl;dr: prnt.sc/nno2mq
I knew I was not the only one who did this ^^
Soon as i saw it, i knew how to simplify it because these type of expressions pop up in communication systems all the time for those studying electrical engineering
Besides educational, this video was very entertaining. Keep up the good work!
When calculating the limit, you could've used a first order taylor approximation of sin(x) around 0: sin(x/2^k) tends to x/2^k as x/2^k approaches 0, then the 2^k in the denominator would have cancelled out, and the limit would be sin(x)/x.
d/dk (1/a^k) = - [ln(a)]/[a^k] , with "a" for arbitrary Real numbers.
13:06
Just multiply the top and bottom by x and use the sinx/x identity since x/2^k approaches 0 as k approaches infinity anyway.
Brian Lamptey perfect.
@@FrancoJ-c7p Thank you.
We can just use the form Lt{y->0} sin(y)/y = 1 and we will directly get cosx. In this case y = x/2^k
You can just multiply and divide both sides by x, without using L'Hôpital's rule. Since k goes to infinity, x/2^k goes to 0 and we know that sinw/w goes to 1 as w - > 0 and so does w/sinw. Therefore we obtain:
sinx/x Lim ((x/2^k)/(sin(x/2^k)) = sinx/x * 1 = sinx/x
Comment about your limit (different from the differentiating part, since I know people seem to keep harping on you about that lol):
Once you've gotten the expression at 13:00, it might be easier to replace 1/2^k -> u, and then take the limit as u -> 0. You'd be left with sinx * lim_{u -> 0} u/sin(ux) and from there L'Hopital's rule is very trivial to implement, or you may even recognize this limit immediately as being 1/x, which would give you sinx/x.
If one knows that sin(t)/t -> 1 as t -> 0 (which is easily checkable by taking sin definition with power series), it suffices to adjust the fraction a little bit as follows
sin(x) / 2^N * sin(x/2^N) = (sin(x) / x) * 1 / [(sin(x/2^N)/(x/2^N)],
indeed sin(x/2^N)/(x/2^N) -> 1 as N -> infinity (because x/2^N -> 0 as N -> infinity), and from this remark it follows that the whole limit converges to sin(x)/x.
If you start by graphing it, even on paper, it becomes obvious that the quantity to be integrated is just equivalent to sinx.
@15：13 ,when video maker was showing how we can use L'Hôpital's rule to get the limit , it was wrong to derivative 2^(-k) as k*2(-k-1) ,it should be -ln2*2^(-k) . But due to the denominator also had 2^(-k) to derivative ,so actually ,even video maker was wrong here , it still can't affect the final result of this limit . Because it will cancel out eventually.
The derivative of 2^-k w.r.t k is NOT -k*2^(k-1)
Using methods from your other video (@ you can (with some effort) write the product as
1/2^(m-1) * SUM cos(2i-1/2^m * x) for i = 1 to 2^(m-1)
This is basically just the average of values cos(x) over the interval [0,x]
Hence the the sum above becomes 1/x * INTEGRAL cos(t) dt for t = 0 to x
= sin(x)/x
looks hard but it is actually easy multiply and divide with sinx and the limiting product equats to sinx/x . thus the integration reduce to sinxdx = 1-1/root2
15:26 uhm... I know you get the right answer in the end but that differential is not correct
Hiii! Great video, but I'd just like to point out that you made a small differentiating mistake then differentiating 2^k (though it won't affect the final result since the derivative gets cancelled out anyways)!!!!
Yes, you are absolutely correct! I treated the expression as a power function, when it is an exponential one; however, as you noted, the mistake does not need to an erroneous final answer due to the cancellation of the derivative, no matter what it is. Thank you for pointing it out! =)
Brilliant! ⭐️ Thank you for making these videos.
Me: "I hope that's just a really weird looking Pi symbol"
you are a charismatic teacher
The derivative of 2^k can't be the same with 2^-k, one is k*2^k-1, and the other one is is -k*2^-k-1, isn't it?
I derived this product at one point on accident. Finally found it on yt.
11:54 correct me if I'm wrong we could have used Lim a->0 sin a/a =1. (Since k ->∞ so sin(x/2^k) -> 0)
Put a as (x/2^k), multiply and divide by x/2^k. Which will give you 1. Please correct me (I'm a novice at calculus) thanks :)
Yes, you are right,Yash Mehan. Putting x/2^k =a, k tending to infinity, and multiplying and dividing by x/2^k, you get xsinx/x which is sinx. Using Sandwich Rule is easier than using L' Hopital Rule.
You explain things very very well. I loved the video
School the screen more so that the proof is clean and easy to see
I actually figured this out while watching the 2015 bee on YT. It's a fun one.
Can you do the ratio between infinite sums from 2015?
I did attempt to work out that question yesterday, but I was not able to obtain the answer provided (1/3). I may be wrong, but I believe the integral they gave us had no elementary solution or they might have provided the wrong solution. When you simplify both infinite geometric series, you obtain integral from 0 to pi/4 of (tanx)^2/(1+x^2). I attempted to make a substitution u=arctanx (which was somewhat natural given the presence of 1/(1+x^2)), but the resulting integral was still complex. Sure, there may be a very creative way of attacking the integral, but WolframAlpha provides the value of ~0.1565, which is obviously not 1/3. If you have figured out a way of evaluating it, I would really appreciate it if you could comment it! =)
@@LetsSolveMathProblems I don't know the question but tan(x)/(1+x²) cannot be simplified by neither u=tan(x) (A) nor x=tan(u) (B) .
It would be (A:) u/(1+arctan(u)²) which to my knowledge can't be simplified or (B:) tan(tan(x))/sec(x)² which is bad
Mmmm.... The statement at 13:00 sounds invalid. The perplexities about applying de L'Hopital in this case gain further points. Looks like you have been kind of lucky this time :)
1/(2^k)/f(x)= 1/ (2^kf(x)) Seems valid. 1/infinity is not 0 rigorously speaking but it would lead me to apply L'Hopitals' rule
I don't even recognise that pi symbol thinking that perhaps it's a strain energy portal deflection problem.
Bro you are crazy, simply amazing
Beautiful.
I haven’t even started calculus why tf am I here
Do I get a full ride if I solve this
I recommend asking the admission officers at MIT for a definitive response. =)
This is child's play. My professors say unless you master Calculus in elementary school don't join the Physics program - _-
Okay I'm 6 years late but at 12:18 could you use small angle formulae? sin(x/2^k) approaches x/2^k because the expression becomes arbitrarily small as k tends to infinity. That gives you the same answer, as the limit equals 1/((2^infinity)*(x/2^infinity)) = 1/x, giving the whole product equal to sinx/x.
Instead of applying l'Hoptilal's rule, it might be easier to use equivalence like sin(x)=x+o(x^3) provided x is vanishing
Other than using LHopital, I have another thought.
Since sin(x)=x as x approach 0, sin(x/2^k) = x/2^k as k approach infinity.
I’ve tried and tried and tried and tried to become good at math and have failed every time. I’m so angry and jealous of brilliant people like this, it just makes me want to cry. Why can’t I have a mathematical mind? It’s just not fair!
Bowtoyoursensi!!238 keep trying
I am not so clever with trigonometric identities, so I came up with another solution: Writing cos(t) = [exp(-ix) + exp(ix)]/2, one can see that the N'th partial products look like
N=1: [exp(-1/2 ix) + exp(1/2 ix)]/2,
N=2: [exp(-3/4 ix) + exp(-1/4 ix) + exp(1/4 ix) + exp(3/4 ix)]/4,
N=3: [exp(-7/8 ix) + exp(-5/8 ix) + exp(-3/8 ix) + ... + exp(7/8 ix)]/8
and so on. There are at least two ways to evaluate the limit:
1. One can see by the definition of the Riemann integral that the limit of the partial products is the integral of exp(itx)/2 from -1 to 1, which is easily evaluated to be sin(x)/x.
2. Each term is a geometric sum, so we can directly figure out with some effort that the N'th partial product is sin(x)/(2^N sin(x/2^N)), whose limit as N→∞ is sin(x)/x.
This is mind blowing! Love this!
Isnt this wrong? 15:00 you can only apply power rule to constants not functions it cancels anyway so the answer is still not wrong.
Wow! Such an elegant solution! However, the power rule shouldn’t be applied on an exponential. You have to use the rule for differentiating exponentials, which yields -ln(2)/2^k. Thankfully that didn’t matter because the false derivatives canceled each other out. Other than that, I loved this video! I never thought such a monstrosity of an integral would simplify so elegantly!
This is what monster from hell looks like.
I think that there is a logical mistake in the end. I try to explain myself and I hope to be clear. The L'Hôpital's theorem can be used only with function that approaching to limit get 0/0 or inf/inf, as you said, but we need function and we have a succession. I know that there is the equivalence between successional limit and functional limit but I thing that it would be more genuine and correct using punctual convergence that lead us to a uniform convergence and finally using the asimptotic we get the result. I think that this solution is more elegant and correct. Hope to have your feedback! 😁
It is not an error. It can be proven that if a function (that is equal to the sequence at integral values) converges to L, the corresponding sequence converges to L as well. Of course, as you stated, you can try to solve this problem purely by analyzing the sequence itself. However, the solution presented in the video is probably the best one if you are pressed for time or wish for a more intuitive solution.
If I hear one more person say, "It's all Greek to me!" when they see me watching these videos, I will start devising death traps for which the means of escape are solving these kinds of math problems.
just the sinc function, beautiful
Well explained, and great comments as well. Thanks!
I think using L'Hopital's rule is a little clunky. I think it's cleaner to use the Taylor expansion for sin(x/2^k).
I was given this in a test yesterday.
Goddamnit way to go demotivating high school students.
It's easier to apply Taylor's theorem on sin in the denominator. This way you avoid L'Hopital's rule entirely.
This is a very good solution but has a slight flaw.
I would like to point out that for differential of 2^(-k), direct power rule can't be applied.
This however doesn't make any difference in the solution as d/dx(2^-k) cancels in the
numerator and denominator.
My first instinct was to use an infinite sum that approximates cosine and see if things cancel out that way. That would probably have just made a mess.
L'Hôpital formula is used in the case of continuous functions. Here it is obviously a series (with respect to k). So you should use Stolz-Cesàro theorem instead of L'Hôpital.
Although it is true that we are only examining the values at discrete integers, it can be proven that if a function (that is equal to the sequence at integral values) converges to L, the corresponding sequence converges to L, as well. (As a side note, the converse is not true in general.) The manipulation in the video is justified. =)
This is a repeated questions of 2004 or 2005 mit integration bee , the winner was decided from this question
This product came up in STEP 2 this year!!!
Beautiful!! Cheers to your skills!!
the limit could have been solved using standard sint/t limit when t tends to 0, using sub. no need to differentiate!! great vid tho! im a jee aspirant, one of the toughest math exams in the world so yeah this one vid is great!
It is just easy to realise because it is the same like root canecelling out
Everything was going pretty good until about 12:00 when he started calculating the limit of 2^k sin(x/2^k). The calculations in this video are at best amateurish. It's actually very simple and doesn't even require l'Hopital's rule. Just denote 2^-k by t, say, and your limit becomes sin(xt)/t as t->0 and that limit is easily seen to be x!!!