The geometric interpretation of sin x = x - x³/3! + x⁵/5! -...

[@mathemaniac]

Update: proof of donation: kzbin.infoUgkxMsm-mzZc9a-gD8cmiygHPBC62vcf1pdM I might change the title to "Without calculus..." because... how else do you combine "differentiation and integration" so that the title is snappier? The only thing that held me back from making this video a fundraiser is the fear of political comments, or any comment that seeks to create conflict, so please do me a favour by not making those comments. Anyway, donate if you can; if not, like, comment and share this video so that this gets more people’s attention. This video is a lot slower than my usual videos, but it might be better for understanding. If somehow you think this is too slow, you can always speed it up. The hope is that you don’t need any university-level maths to understand this video. NOTE: In the last part, k is supposed to be fixed, so each bracket do go to 1 when n tends to infinity.

[@mastershooter64]

[insert extremely polarized political comment here]

[@benYaakov]

Hello at 9:43 , how these lengths are L1 , L2 etc ? Shouldn't these lengths be equal because each angle is equal ( to x/5 ) . Sorry if this is a very dumb question.

[@maxwellsequation4887]

Woohohooo i support that side in the war I'm so bad

[@Icenri]

Thanks for your work. I really liked the video even though it can feel slow, I think it helps a lot that you don't take anything for granted and are very clear in every step. The clearer the better. Also, thanks for supporting children in need.

[@debblez]

@@benYaakov they are equal, he just decided to give them distinct names, which made sense if you watched the whole video

[@anwyl42]

This feels like tricking kids into eating vegetables. "Oh, taylor series is too complicated? Okay we'll just add up a bunch of lines tangent to curves!" I love it.

[@Tonizamikula]

Jjejw

[@stupidteous]

worked on me im in grade 11 lmao

[@stupidteous]

@Optaunix bro what

[@nightytime]

@Optaunix ?

[@parlor3115]

@@nightytime I think he's talking about how it's a bad idea to stuff our kids with veggies

[@blackpenredpen]

2:59 😮 thanks for an amazing video!!!

[@mathemaniac]

Wow! Thank you for stopping by!

[@henrymarkson3758]

Your content rivals that of 3b1b and there is no greater praise in the world of KZbin maths.

[@mathemaniac]

Wow thanks!

[@thedevansaini7634]

True my first thought was this seems like a channel of 3blue 1 browns caliber

[@nathanwestfall6950]

This is what I needed 20 years ago! A very satisfying approach that gives meaning to an otherwise non-intuitive result. Keep it up! :-)

[@mathemaniac]

Thank you!

[@3moirai]

Totally agree! I love these insights to help understand how beautiful mathematics can be.

[@benYaakov]

@@mathemaniac hello , at 9:43 , why these lengths are L1 L2 L3 etc ? Shouldn't they be same as each subtend equal angle ? Sorry if this is a very dumb question

[@Icenri]

@@benYaakov They are the same only at the beginning (in the 0th iteration) but you'll see that every time the process creates segments with different lengths.

[@benYaakov]

@@Icenri ok if it is same then why is there L1 , L2 , L3 in zero iteration

[@glarynth]

Great illustration! Also, e^x is hidden in there, as the total length of the spiral (including the horizontal segment from the origin to (1, 0)). This shows an intimate connection between the exponential and the sine/cosine, which typically isn't apparent without invoking the complex numbers.

[@spaz1810]

Can you elaborate please? Why is the length of the spiral e^x? I mean independently of the identity, as a means of actually demonstrating the identity

[@kallewirsch2263]

@@spaz1810 I don't know what Robert had in mind. I would look at the Taylor series for e^x, which is e(x) = 1 + x^1/1! + x^2/2! + x^3/3! + x^4/4! ...... I am sure you will find the individual terms by yourself in the graphics. So e^x just means: Sum up all the involutes

[@spaz1810]

Thank you @@kallewirsch2263. This doesn't really point to why the length of the spiral should, geometrically, be equal to e^x though. I'd love to see this construction used as a demonstration of Euler's identity rather than the other way round.

[@minamagdy4126]

Don't know about a geometric intuition on why e^x, for real x, appears here (I just let e^x be defined by its Taylor polynomial). One nice intuition that this diagram proof lends itself to is Euler's identity. As long as you're comfortable with the Taylor polynomial definitions, and the idea of a counterclockwise 90 degree turn being equivalent geometrically to a multiplication by i, then transposing this diagram onto the complex plane proves Euler's identity.

[@woody442]

@@spaz1810 The series representation of e^x is the infinite sum of x^n/n! where n starts at 1 and goes all the way up to infinity. If you add the absolute values (lengths) of all segments in the spiral, you get exactly this series, converging to e^x.

[@yakov9ify]

While the Pascal part is quite beautiful I think it abstracts away a little too much from the geometrical intuition. We can simply note that the 'rate' of unwrapping of involute n+1 is proportional to the distance from the point along the involute n. Thus suggests that the distance of involute n+1 at the unwrap point is the integral of the previous involutes distance from the point of interest. Thus a recursive arguments makes the nth involutes have a distance function x^n/n!

[@mathemaniac]

Wow this is a much, much more natural explanation of x^n/n! :) The reason I went with the Pascal's triangle is that the paper I'm following, and the aim of the video, is to present this argument with no differentiation / integration involved.

[@yakov9ify]

@@mathemaniac And I think it achieves that purpose wonderfully, I just always feel like with videos like yours and 3blue1brown's, the individual details are very enlightening but the overall picture gets a lil bit too muddied because of the amount of steps involved, so I prefer shorter proofs personally. That being said I do think there is a value to reaching these facts and theorems without the prerequisites of calculus and so I still am very happy these videos exist.

[@bobtivnan]

I'm having trouble visualizing rate of unwrapping here. Rate w.r.t. to what quantity? Are you saying that the length of the (n+1)th involute section can be viewed as a dx and the distance from the nth involute section would k•dx? If so, what is the significance of k in the integration scheme? How would the recursion argument work? So many questions.

[@yakov9ify]

@@bobtivnan take a point Q_1 on the 1st involute (i.e the circle segment) and parametrize by the angle from the point of interest P. As you change that angle consider the point Q_2 on the second involute which is also on the tangent line to the circle at point Q_1. Now consider how the length of the curve between Q_2 and P changes as you vary theta. At any specific theta the rate of travel of Q_2 along this curve is proportional to the distance along the circle between P and Q_1. Thus the total length of the second involute is equal to the integral of the distance along the curve from P to Q_1. But that is precisely the integral of theta from 0 to x which is x^2/2.

[@bobtivnan]

@@yakov9ify Thanks! I think I got it. The bounded region between the tangent segment, the nth and (n+1)th involutes is essentially a sector in the limit as theta->inf. Label the arc as the differential ds and use the arc length formula ds=r* d(theta). Integrate from 0 to x to get the arc length of the 2nd involute. I had to convince myself that the sum of all d(theta) = x (the central angle and arc length on the circle), but it must since from geometry the sum of all exterior angles (which are the d(thetas)) sum to x. I think it's wonderful that we can analyze this with and without calculus.

[@nomanbinmorshed5584]

Another beauty of these videos are that: They make u feel jealous in a very positive way and u start thinking as to why I couldn't think of this masterpiece!!! Love for yet another beautiful channel

[@physicsboy1234]

This is an interesting video and the showing of the proof is simple enough that even I can understand you explained it well

[@mathemaniac]

Glad you like it!

[@yashrawat9409]

Woah the proof and its details are too good ( the entry of Pascal's Triangle and Binomial Coeffiecients too )

[@mathemaniac]

Yes! This is why I have to share this - it is too underrated!

[@yash1152]

yeah, the pascal's triangle entry was real cool 14:50 i was thinking it seems somewhat familiar, but got lost in summation series

[@coffeecup1196]

More of a "How to sneak calculus into a 'without calculus' explanation" video I love it, keep it up. There are a lot of people who will have a better intuition for calculus when they get to their intro class if they watch this stuff.

[@timermens350]

Very nice. Sice 3Blue1Brown and Mathologer their "moving average probability" of uploading content seems to be in a downwards trend, this might be the next channel showing the beauty of math. Pls keep doing these kinds of videos!

[@mathemaniac]

Haha thanks!

[@o_sch]

Its in an upward trend. Its all opinion.

[@kbin7042]

This was incredible. I came here with no expectations of understanding more than half of the proof (happens to me very often with these kind of videos), since I have no superior education yet, but I was amazed by how clear everything was, only in the last part I had to pause it for a while to understand Thanks for making your videos so accesible man, greetings from Perú

[@kshitijthakkar8074]

This just absolutely blew my mind mate, Keep up this great work. This video is just an exquisite example of how beautiful, connected and simple pure math can be. Congratulations for pulling this one. All the best for your future ventures ❤️❤️❤️👍.

[@mathemaniac]

Glad you enjoyed it!

[@codewordbw3340]

When you see videos like these and you understand mathematics to a certain degree, it's always awesome whenever you see a proof that is going in a direction that you've never seen before. But the best part is when you're partially through the proof and you can use your past knowledge to see where the proof is going and it finally clicks as to why the proof works. It's so cool, every time

[@codewordbw3340]

@Tech Keep watching these types of videos and keep learning my friend. Math is beautiful and there's actually so many different fields of math to learn about

[@RichieDevine8]

So true !

[@titan1235813]

This is the very first video of yours that I've watched, and I've already subscribed to your channel because this one is, by far, the best explanation I've ever come across with why sinx is expressed the way it is. You are an excellent teacher. Thank you!

[@sokunwu]

I like your explanation with using geometrical opinion to interpret Sinx. So amazing!

[@MichaelRothwell1]

Congratulations on bringing a geometric perspective to something so unexpected as a couple of Maclaurin series - and throwing in a favourite result of mine about the sums of columns of Pascal's triangles as a bonus! The whole approach was very clear, so I could see fairly early how this was going. The whole thing was a pleasure to watch, which is quite an achievement considering how much was going on here mathematically.

[@bobtivnan]

So, can we generalize this for other functions? Instead of of sin(x) and cos(x) as parameterizations that happen to form a circle could we use involutes to describe other parametric curves like say ? By the way, Euler's formula is beautifully hidden in here if we use the complex plane (follow up video?).

[@angeldude101]

Not just the complex plane. With just a small tweak, making all the involutes go outwards rather than spiraling inwards yields the unit hyperbola, and if you made all of them point in the same direction and plotted the total length against the starting angle, you'd get the exponential function. The only thing that an imaginary factor in the exponent does is dictate what direction you should fold the involutes. There's even a number system that causes every involute after the first one (the one with the lambda^1 factor, which the video called the zeroth) to self-destruct. Since this did seem like a geometric representation of the power series, you could probably do it with any function that has one.

[@orstorzsok6708]

I was thinking on the same but I came to the point that if you would like to do it with some other function (let it be e.g.: 'e ad x' or 1/x) then how would any involute generated...? From what... or of what...

[@orstorzsok6708]

@@angeldude101 Could you please send me a graphical illustration for that...? You can chose your preferred function...

[@bobtivnan]

My thinking is that we can represent a Maclaurin series for any n-times differentiable function in a similar way. e^x is the most obvious case since the arc lenths of each involute shown here is exactly the value of each term in its series. As someone pointed out, we would need to rotate each involute by 90⁰ clockwise on each iteration so they stack up vertically. But this idea can be extended to other functions as well...

[@bobtivnan]

The missing pieces are the coefficients. Actually, the factorial parts are already there, so we just need to scale each involute by the nth derivative at x=0. Then stack them up as described before. Not sure what purpose it serves other than a new way to view Taylor series.

[@rms_txrx]

This is precious, a gem of mathematical insight. You explained everything very carefully, so much that by 1/3 or so of the video I realized where you were going - and yet I chose to keep watching, because… DAMN! You’re good! Keep it up and you might be making videos like 3b1b in no time! I’ll wait for that!

[@mathemaniac]

Wow thank you!

[@anhthiensaigon]

dude this is so perfectly well done that it seems like a weapon by itself already.

[@1ab1]

Your videos are always a pleasure to listen to, even the slower ones! Well done for bringing a geometric POV to an algebraic phenomenon.

[@mathemaniac]

Glad you like them! I was surprised that nobody talked about this amazing link between the geometry and the algebra of sin series before, and so I decided to bring that up myself!

[@zhuolovesmath7483]

GREAT video!! When I first subscribed you only had less than 50k subscribers, now you have 80k! Congratulations and please believe that your work is truly meaningful, inspirational, and we who love math will always be here to support you!!!

[@mathemaniac]

Wow, thank you!

[@AlexEMagnus]

Your best video yet. It was easy to understand, even without subtitles. Very good work.

[@mathemaniac]

Wow, thanks!

[@PetruRatiu]

A very satisfying proof, which I hadn't heard of before, thanks! The only small issue I see is that there's a skipped step to recursively prove the involute length formula (If I'm not mistaken, after the initial circle they're all cardioids). Somewhere around 11:00 there should be a note showing that the angles are preserved even if the lengths are not.

[@mathemaniac]

Yes, I deliberately glossed over this part because I don't want to make this longer than it should be - so essentially once you have proved that the angles between the different segments in the first involute is still x/n, or rather, pi - x/n, then you're good to go. (Like in the 0th involute, the angles between different segments are pi - x/n) This is because, as said in the video, we will return to exactly the same situation, just rotated, and the lengths of the little segments will change, so once you proved the angle for the first involute, then you have proved that for every involute.

[@itellyouforfree7238]

I knew someone had to spot this too!

[@othersidewanderer]

I loved this video! Never knew you could find the infinite series of sin(x) this way. Thanks such a clear and beautiful explanation!

[@mathemaniac]

Thanks!

[@oncedidactic]

I was so delighted to see this is also a fundraiser, instant donate lol. What an amazingly beautiful proof, thanks for putting up a video on it!

[@mathemaniac]

Thank you for the donation and your compliment!

[@The_King164]

كنت اكره الرياضيات ولكن بعدما شاهدت قناتك اصبحت احب الرياضيات جدا شكرا لك!

[@razikridzuan4662]

This is a different yet beautiful approach to McLaurin series. I love the way you explain one by one. We need more of this. I'm subscribing now.

[@mathemaniac]

Glad you enjoyed it!

[@notqueuebad4206]

Very interesting way to develop the Taylor series! I remember studying Chebyshev polynomials a few months ago, motivated primarily via generating function. Since this involved regular polygons (i.e., partitioning the circle into equal segments) and my approach was combinatorial, just like in the video, I noticed the coefficients in the polynomials of the second kind start to resemble Pascal's triangle. There's even a natural duality that appears to be demonstrated in the expressions: rewriting the polynomials as polynomials in 2x rather than x makes the binomial coefficients clearer, and in the video lambda has a coefficient of 2. I also happened to develop the Chebyshev polynomials out of stereographic projections rather than the standard (cos x, sin x) parametrization, which I find interesting for the fact that all of this can be done before you've written out the series for sine and cosine.

[@therealAQ]

Tangent lines to curves, inifinite processes and 'epsilons' going to zero, all that is calculus-ey enough for me. Complex analysis is quite geometric actually. 👌

[@adrianmisak07]

0:21… genius meme

[@mathemaniac]

Haha

[@enchantularity]

I request Mathemaniac to explain the geometric intuitions behind the following problem 1. Similarity transformation 2. Multi state Exclusive OR 3. Graph laplacian and Laplace Beltrami operators 4. Eigen vectors of graph laplacian and their geometric intuitions Thank you in advance.

[@rossjennings4755]

This sounds like a bit of a laundry list, but I think the graph Laplacian actually has a really neat "geometric" interpretation that might be worthy of a video from someone. Basically, to make head or tail of it, you have to think of it as a linear transformation, rather than as a matrix, which is how it's often presented. The vectors that it operates on are functions defined on the vertices of the graph. To get the value of the transformed function at a vertex, do this: for each adjacent vertex, subtract the value of the initial function at that vertex from its value at the vertex of interest, then add up all the differences. So if the value of the initial function was equal to the average of the values at adjacent vertices, you get zero. This corresponds really well to a bunch of systems, one of the simplest being a bunch of reservoirs of water (vertices) connected by pipes (edges). If the initial function gives the amount of water in each reservoir, then applying the graph Laplacian gives the rate at which water is flowing into or out of each reservoir (up to a constant factor). To get the amount of water in each reservoir at any later time, you can do a matrix exponential, and to do that, you need to think about the eigenvectors. The constant function is always an eigenvector with eigenvalue zero, and the others are "wave" patterns that decay away without changing shape, at a rate determined by the corresponding eigenvalue.

[@pieterpost3606]

This video is a math-masterclass of a level not seen before. Just perfect. 👍 Thanks a lot.

[@ryantamburrino3289]

Beautifully presented! Well done. Love seeing something I haven't seen before, always good to have a new perspective on a common fact.

[@mathemaniac]

Glad you enjoyed it!

[@dylanparker130]

Utterly blew me away - beautiful video!

[@mathemaniac]

Glad you like it!

[@UberHummus]

This video has absolutely top tier visual animation and narration. Absolutely beautiful

[@mathemaniac]

Thank you!

[@G8tr1522]

0:52 i LOVE how you put this 'table of contents' of sorts! it makes it much easier to follow bc i will know what to pay attention to for the next step.

[@playerscience]

Absolutely beautiful explanation...!!!! I've never looked at Taylor series from this perspective...!!! Instantly subscribed 👍

[@mathemaniac]

Awesome, thank you!

[@winteringgoose]

I've never seen a Taylor series represented geometrically before! I'm no expert but I'm also no slouch in calculus, and still they always just seemed to come out of nowhere. But today I get sin explained to me with cos as a free bonus!? Best. Day. EVER.

[@mathemaniac]

Wow thank you!

[@idjles]

This was designed to show Sin and cos, but it also showed e^x as a side effect - e^x is the length of all the involutes as well as the length of the square spiral. Which beautifully shows that if you take an angle of 10 pi or more how fast that square spiral grows in length while sin and cos stay under 1. What i'd be interested to know is if from the graph of e^x we can show it's Taylor expansion and get sin and cos geometrically out of it.

[@enigmath0630]

what do you guys think about this proof? Is this the proof of the great mathematician Madhava of Sangamagrama? About the series/expansion of trigonometric functions....

[@johnchessant3012]

Wow, that was really unexpected and beautifully presented! The two most important ideas were right at the start, the diagram showing the involutes spiraling in and the similar triangles showing that each involute is related to sums of the previous involute. From there our intuition already sees the sums being roughly integrals! x -> x^2/2! -> x^3/3! -> ...

[@LittleCloveredElf]

One of the Most beautiful and elegant proofs I've seen

[@brucebehymer]

Truly beautiful and beautifully explained. I studied evolutes and involutes in both high school and college, but never did involutes of involutes, and was never aware of this. Just beautiful. Thank you!

[@theroyalgamer6560]

I just had an exam about working with Taylor polynomials, and looking at how sin can work with them to find a result for a larger series. This video was explained very well, and I enjoyed finding out the theory behind this concept that we didn’t explore in class

[@orstorzsok6708]

AMAZING! REVEALING! FANTASTIC! INGENIOUS! I am absolutely astonished to see this approach - I have not met this one yet, absolutely amazed! THANKS, many thanks!!!

[@mathemaniac]

Glad you enjoyed it!

[@Inspirator_AG112]

The "Finding lengths of involutes" section reminded me of an iteration formula that exists for repeating the sum of whole numbers (1 + 2 + 3 + ... + x). Πₖ₌₁ⁿ((x + k - 1) ÷ k) (With n being the number of iterations) *The highest degree term is xⁿ/n!.*

[@alganpokemon905]

Wow! What a fantastic approach to this problem, and delivered very well too. This video deserves a lot of credit.

[@mathemaniac]

Glad you liked it!

[@raulyazbeck7425]

Just amazing. You deserve way more views and subs than you currently have!

[@mathemaniac]

Wow, thank you!

[@sermarfe2584]

I have just discovered your channel on KZbin and I have to say that your content is amazing! Keep up with it!👏

[@mathemaniac]

Thank you so much!

[@luisfabricio6439]

This is insane!!! I wish I could show this to my maths teacher back in high school

[@jacksonstenger]

Coolest proof I've seen in ages, thank you for the video!

[@mathemaniac]

Glad you liked it!

[@Daviuliano]

Have never seen such a good explanation of sine approximation. Great job!

[@samuelkoski7694]

This is aweaome. How had i not heard of you before?

[@eric3813]

Such a magnificent Video!!! Never have i seen a geometric proof of the formula, thanks!

[@mathemaniac]

Glad you liked it!

[@blue-cuboid]

Is there a similar geometric proof for the Taylor seires of tan(x) or other trigonometric functions?

[@mathemaniac]

Yes! At least for tangent and secant, there is a similar involute-y proof, but I haven't looked into those in detail. For the paper I am looking at, it is a bit more involved, because the coefficients of these series themselves are more involved. But, who knows, I might make another video addressing tangent and secant series if this video performs well.

[@mastershooter64]

every trig function can be represented in terms of sin(x) so yes!

[@blue-cuboid]

@@mastershooter64 That's true, but I meant to ask if there was a proof that directly uses tan(x) and does not have to use sin(x).

[@fattimiv]

Great vid. Never seen this proof before. I like that the series for cosine also comes out of this, almost for free.

[@mathemaniac]

Glad you enjoy it!

[@TruthOfZ0]

Now i understand e^ix even better!!

[@angeldude101]

I had 2 main thoughts: 1) I may be a programmer, but I would label the involute with coefficients of 1 as the first involute rather than the zeroth. Why? Because they're all lambda^1, which would make the second have lambda^2. There is also a hidden extra zeroth involute: the line segment going from the origin to the point on the circle where theta is 0. This also makes it so the nth involute always corresponds to the x^n term of the power series. 2) Here, the involutes all spiraled inwards, meaning that half of them were subtracted rather than added. If you were instead to only travel outwards, adding every involute, the result would not be a point on the unit circle, but rather the unit hyperbola.

[@alejrandom6592]

I love how they start by the visual part, It's kinda like having the trailer of a movie inside the movie.

[@2dboys230]

Why are you guys so awesome that for some reason these feels like at a documentary level and that also for free , like Khan academy but with different personalities and styles and for some reason you guys fulfill a desire to learn maths and see somehow it's beauty .And when I tell my classmates about you and hear there response , one word: nice

[@mathemaniac]

Aww thank you!

[@2dboys230]

Mate didn't expect you to reply so early

[@2dboys230]

And I can see that sparkle of happiness in your words you well deserved it my friend

[@2dboys230]

Btw for all my friend's. Who watch your channel can you pls give me a shoutout in the next vid it's just a appeal but would really appreciate it but yeh even if you don't I still appreciate the time you placed to read my comment

[@2dboys230]

And yeh i watched your complex analysis really loved it

[@БондаренкоАлександр-ц8и]

This is brilliant! Pure genius! So simple and elegant. I always wandered how those math guys invented Taylor series (if we assume that we don't know calculus) - they just tried to approximate curved lines with straight ones. And seems like this video could be generalised to functions other than sin (: Thank you, sir

[@sabarishssibi3382]

Falling in Love with maths by your teaching

[@mathemaniac]

Wow! Glad to know that!

[@egohicsum]

Thank you much! Thats a very interesting and beautiful proof! Great visualisation and narration too!

[@mathemaniac]

Thank you very much!

[@p0utan]

Finally,I understand “Euler’s identity” geometrically!

[@yinq5384]

Is there a geometric proof for e^x?

[@Adityarm.08]

wow, this was amazing. thank you for putting this together!

[@web2wl00p]

Truly beautiful. And yes, you are definitely on par with 3b1b and mathologer!

[@mathemaniac]

Oh thank you!

[@loggat3804]

I searched for this not long ago and didn't find it. I really wanted to know this. thanks

[@kuralpozhilann4355]

Sir, This is a beautiful proof. You have my gratitude for this. Kindly don't stop making videos like this. Big fan.

[@fowlerj111]

The involute does have n segments rather than n+1, to bring the taut line up to vertical - but that's just one more term in the series and it doesn't affect the limit as n→∞

[@mathemaniac]

However, then the final isosceles triangle will not have the angle being x/n anymore - it would instead be x/2n, so it will complicate the calculation. But, as you pointed out, in the limit, it doesn't really matter.

[@stupidteous]

i only understood a fraction of this vid bc im in grade 11, so idk the pascals triangle notation and how it turned into what it did turned into what it did, or the math you did afterwards, but for everything else i understood it bc you explained everything so clearly and amazingly, and at an amazing pace, so when i slowed down, i could think about it and understand it. thank you so much for this vid, this is amazing. this is why im continuing math in uni, these things are amazing

[@janemillervideos]

If you want to choose r objects out of n objects, then you can do the following: 1. You choose the first object and then choose r - 1 objects out of the remaining n - 1 objects. 2. Or you don't choose the first object and then choose r objects out of the remaining n - 1 objects. Defining nCr as the number of ways to choose r objects out of n objects and following the above logic, you get: nCr = (n-1)C(r-1) + (n-1)Cr Now, look at the Pascal's triangle in Wikipedia, and you would see you are adding two numbers to get a number in the next row. You're basically following the above formula.

[@MatthewWroten]

This is the most beautiful geometric proof I’ve seen this year :)

[@miki2525]

Wow! This video is amazing! There was a Chinese video porter who added Chinese subtitles to this video and posted it on “Bilibili”. After I watched I can’t wait to find this excellent KZbinr and subscribe!

[@petebunch5800]

This is fabulous. Wish I'd been shown this 20 years ago. Thank you!

[@Sheikxlove123]

Omg! This is an amazing video! I've never heard of involutes before!

[@mathemaniac]

Glad you enjoyed!

[@bayesian0.0]

This is fascinating! Great video :p

[@mathemaniac]

Thank you!

[@qpzmwoxneicbv]

I absolutely love this channel

[@anthonyheak3479]

Great explanation. First that I have seen it explained geometrically!

[@mathemaniac]

Glad you liked it!

[@m.venkadesen9037]

Im always using this expansion series to solve lot of limit problem i love it

[@PollyMath]

Excellent video. So nice to have an intuition for the expansion. Thank you!

[@eliyasne9695]

In effect, the involution process is a geometric interpretation of integration employing only lengths, without any areas!

[@guntherroll3106]

Truely a gem. Great topic. Great explanation. Great visualization.

[@player6769]

Coming back to say this is perhaps the best math video I have seen (and I’ve seen a lot)

[@josephyoung6749]

I've seen this image of the involutes somewhere, but never knew the meaning. Very enjoyable video for laypersons.

[@phaniramadevu2449]

Such a beautiful way to derive the sine!

[@mathemaniac]

Which is why I have to share this!

[@chinskiszpieg984]

A beautiful proof! People passionate about about math live for such elegant constructs!

[@matheusaugustodasilvasanto3171]

Between mathologer, 3b1b, and you, the math nerd in me is being kept very happy haha. Any other math channels you recommend?

[@hqs9585]

Just GREAT! A woderful explanation that many many mathematicians probably never envisioned! Loved it

[@Omar-hm6pu]

This guy is very underrated

[@mathemaniac]

Thank you for the compliment!

[@WorldMover]

This is such an important video. I waited 35 years for this

[@davidbrisbane7206]

Truly amazing demonstration.

[@supriya1729]

Mind blown brother,just amazing 🎉🎉🎉

[@mathemaniac]

Glad you liked it!

[@masonfrias8292]

I just found your youtube channel and am so interested!

[@mathemaniac]

Thank you!

[@TheQEDRoom]

i was just thinkibg about this the other day. not the details of the proof but the existence of such proof.

[@franciscoxaviergonzalezrom4648]

marvelous video! awesome! I almost cried! Thank you!!!!

[@mathemaniac]

Wow, glad you like it!

[@robbiechen3707]

Great video! Maybe it's better to extend one more segment when constructing the next segmented involute, so the number of segments stays the same rather than decreasing, and prove that 1. the last segment goes to 0 as n goes to infinity and 2. the true length lies btw n segments and (n-1) segments, then with the sandwich theorem we get the length of any k-th order of the involute when n goes to infinity.

[@brawldude2656]

I can't believe I understood all that. That was one of the best lessons I've ever had!

[@ChronusZed]

Very cool way of visualizing this series! I think for the part at the end (going from the table with the lambdas to the formula for the lengths), it would have been clearer to leave the yellow part as just lambda, since you didn't actually use anything about sin, just the fact that n*lambda tends to x.

[@mathemaniac]

Thank you for the comment - yes indeed that will be a bit better.

[@venkybabu8140]

A geometric series. Can be represented as fractional distillation. A geometric series is somewhat like stirring and fractional distillation the waves.