Solving Integrals is the First sin(sin(sin(sin(sin(...))))) of Madness

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Lambert W: • ODE Exercise #4 - Embr...
Infinitely Nested Integrals: • The Integral 1/(x^4+1)...
Today we evaluate infinitely nested, recursively defined integrals of sine, cosine, tan, exponential function and the natural logarithm :) The Dottie Number and Lambert W function make an appearance and we find out some curious facts about the convergence of these functions to their respective fixed point equations. Enjoy! =D
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Want to know more about me? Watch my QnA! =D • Question and Answer Ti...

[@PapaFlammy69]

*_Hey dick cheese, thanks for watching

[@anshulrai677]

Noice

[@anshulrai677]

🙃🙃🙃🙃

[@oni8337]

:d

[@FurryCombatWombat]

I was actually surprised that I got that nested sin one right. These were my steps I=int{sin[inf](x) dx} dI/dx=sin[inf](x) dI/dx=sin(dI/dx) dI/dx=0 dI=0dx I=int{0 dx} I=C Im not sure if I did anything wrong, but still got the right answer, the 3rd to last step felt sus to me

[@cicik57]

is it ok, if tan(x)-x = 0 has multiple solutions and exp(exp) diverges? with logs, ln(t) = t gives the same result as exp(t) = t

[@joirnpettersen]

The fundamental theorem of engineering will be the next axiom of choice. I see no reason why we shouldn't use it.

[@marcoponts8942]

Because it is not a real theorem! This is just some kind of weird joke that this channel is making! I agree for people not comfortable with math it is very hard to distinguish which videos are real math and which not on this channel. This is very dangerous!

[@loecg4588]

@@marcoponts8942 no way, I learned it in high school

[@NateROCKS112]

@@marcoponts8942 the axiom of choice isn't a real theorem either. Checkmate, liberal.

[@lienardgael808]

@@marcoponts8942 and you should be able to distinguish if this person is serious or not, shouldn't you?🤦

[@carmangreenway]

@@marcoponts8942 I know a few advanced math memes that I can't tell are joking but hearing the jokes makes me want to look more deeply into an idea, so fair trade

[@akselai]

We can see that sin(sin(sin(…))) = 0, due to the fact that the graphs of sin x and x have one intersection at the origin, so sin has one fixed point, and it is at 0.

[@abidhossain8074]

gay

[@akselai]

@@abidhossain8074 h🅾️mies 🅱️ like

[@VaradMahashabde]

Another argument is that max of sin is 1, sin is monotone increasing in it's own output interval of [-1,1], and sin(1) < 1, so the maxima of sin_n(x) follows a geometric series down to 0, with ratio r=sin(1). Same for the minima.

[@aug3842]

@@abidhossain8074 you called?

[@sharpedged7830]

Isn't the infinite natural log the same as the infinite exponential because: ln(t) = t => t = e^t

[@PapaFlammy69]

yas, should be :)

[@liamscully932]

If you graph them both functions would diverge: e^t keeps increasing with every iteration, and the domain of ln(t) keeps shifting further and further in the positive t direction. I also checked it for a few other pairs of functions that are also inverses of each other, and in general it didn't seem to work, so I'm not quite sure what's up with that.

[@decare696]

As you said - the result obtained in the video is false. The sequences of functions just don't converge, so it makes no sense to manipulate them as if they did (see also the -1/12 discussion). The value of -W(-1) is actually not a real number. If it were, we would have ln(x) = x = e^x and since the two functions don't meet, that is a contradiction.

[@shirou9790]

yes and since e^x = x has no solution (and neither does ln(x) = x), this shows that both are divergent

[@jadonjones4590]

@@shirou9790 e^x=x does have a solution, just not a real solution. The Lambert w function is defined for complex values as well

[@nikhilnagaria2672]

Well, by the fundamental theorem of engineering, the given function is just x. Now, notice already that x(x-x)=x^2-x^2=(x+x)(x-x) implying overall x being nothing but 2x. So our integarahl is I = int x dx = int 2x dx, subtracting gives 0 = int x dx immediately implying I = 0.

[@ay5960]

I think before trying to integrate the infinitely nested sins, you need to prove it is a well defined function. You assume that if the value of infinitely nested sins is t, then sin(t)=t and so t is zero. But here you are already assuming that a value t can be assigned to sin(sin(...(x))). Therefore you will first need to prove that for any real number x, the infinite nested sins converges. Now since it converges then it has to be zero. This can proved by using the fact |sin(x)|

[@tobybartels8426]

And unfortunately, only the sine and cosine actually converge.

[@pierreabbat6157]

sin(sin(...(x)...)) converges very very slowly to 0 if x is real. If x=i/10^6, however, it diverges, starting very very slowly. I don't know if there's a complex fixed point where the absolute value of the derivative is strictly less than 1.

[@igxniisan6996]

0:19, That escalated from “sin of” to “son of a...” fast.

[@Jon-qh8dt]

* the first sin of Mathness

[@spurcalluth6300]

But tan(t)=t has infinitely more solutions than just t=0. What happened to those solutions?

[@shirou9790]

yeah and I'm actually quite sure tan(tan(...(x)...)) doesn't actually converge anyways

[@elael2]

@@shirou9790 exp(exp(...)) also doesn't seem to converge (at least not in R)

[@shirou9790]

@@elael2 yeah that one shouldn't converge, as it's pretty easy to see that e^x=x has no real solution. nor does log(x)=x for that matter

[@nikhilnagaria2672]

0:08 rare moment of papa not having strokes while doing the intro

[@PapaFlammy69]

:D

[@austinmitchell2652]

Love a classic integaral video!! This is the first papa video I've watched to completion (ohhh yeah) in quite a while cause your integrals were what really got me into the channel.

[@_..---]

infinitely nested integrals? mathematics induces madness

[@epicgamer4551]

It's approximately 0 because of the fundamental theorem of engineering smh

[@PapaFlammy69]

yeye

[@iispacedustii]

no jens constant = pure madness and a mathematicians nightmare

[@Kurtlane]

I first discovered the infinite cosine by pressing the cos button on my calculator. I saw that no matter what teh original number was, I would eventually approach the same number. Later I found the same problem in a textbook, where it was used to introduce students to Picard iteration.

[@Predaking4ever]

Papa Flammy: sin(t) = t Engineer: I knew it.

[@joannaford7137]

Enjoyed this Jens,,,,, keep up good work love it

[@senpaiii623]

IVE BEEN WAITING FOR THIS

[@saisanjaynarayanan8242]

Hmm... not really sure if I agree with integral tan(tan(.... (x).... )) = integral 0. Unlike for sin, the sequence of functions for tan doesn't uniformly converge to the zero function {in fact I don't think it even pointwise converges except at x = 0 . Correct me if I'm wrong though} Spicy video nonetheless :)

[@luggepytt]

I agree with you. If you plot y = tan(x) and y = x in the same diagram, you see that the two curves intersect in infinitely many points, meaning that there are infinitely many fixpoints. My intuition tells me that none of those is stable. The reason being that the derivative of tan(x) is always greater than the derivative of x, maybe? I am too lazy to prove it rigorously, but some experimenting with repeatedly hitting the TAN button on my calculator shows a rather chaotic behaviour that seems to never converge.

[@edskev7696]

I agree! Was going to comment this myself

[@edskev7696]

I'm also not sure about the exponential one

[@luggepytt]

@@edskev7696 Yeah... When I got to that part, I realized that either my calendar is out of sync and today is April Fools Day, or Papa has accidentally left his brain in OFF mode. So let's not speak about the exponential, OK? (Edit: Strange, this is the second of my favourite Math channels that posted something really bizarre today. I hope there isn't som new epidemic going on. "Infinite Recursion Syndrom" or something...)

[@saisanjaynarayanan8242]

@@luggepytt yup, that's exactly what I was thinking. Even if you restrict the domain of integration to, say (-π/2 , π/2) , the sequence of tan functions doesn't converge at any point except x = 0. Like you said, it blows up for any other point in the domain because the derivative is also strictly increasing without bounds (intuitively), so the integral doesn't make sense. Oh, and don't even get me started on the exponential function 😛

[@rollingsnowball9095]

Loved the video! The Dottie number for cosine can be approximated by an irrational number by truncating the Taylor series! You just end up solving a quadratic with (spoilers) t = -1 + sqrt(3), which is REALLY close to the actual value. This is especially interesting because the Dottie number is transcendental

[@PapaFlammy69]

nice!!!

[@myzensianquaternasius6848]

I ADORE THIS

[@nnotcircuit010]

tan(t) = t has infinitely many solutions, not just 0. So when tan converges it will not always be to 0. It also will not always converge. For example at approx 1.33 we can see that the tan(1.33) is about 4.07 but the tan(4.07) is back to 1.33. This would be a two-cycle and not converge.

[@chalkchalkson5639]

you forgot to show that sin(...sin(x)...) actually does converge :( you only showed that if it converges what it converges to. It's not too hard to fix though, since sin maps R to [-1;1] we only need to show that sin(..sin(x)..) converges on that interval. sin(x)0, so the squence (x,sin(x),...,sin(...sin(x)..)) is definitely monotone and decreasing for x > 0, and for x€(0;1] is also has a lower bound since sin(x) > 0. So we have a monotone sequence with a lower bound, meaning it must converge. For x

[@shyeyebee]

Papa Flammy, I don't understand why we can assume the nested tan converges.

[@loecg4588]

Tell me if I'm wrong but the reason was the same for the sinus function. The tan_inf function is also tan(tan_inf) therefore tan(t) = t. But, I'm not sure if I'm answering your question nor if my explanation is stricly rigourous in terms of convergence.

[@auranous4501]

@@loecg4588 By convergence we mean the convergence of the sequence tan(x), tan(tan(x)), tan(tan(tan(x))), and so on. If you presume the sequence converges to some t, then one can use the calculation you presented to demonstrate that tan(t)=t. But we needed the existence of t to derive tan(t)=t, which we may or may not actually have. One should probably think of these types of calculations as heuristics to figure out a candidate for where the sequence converges to, rather than an actual proof of convergence.

[@dannydorito2954]

Sin sin salabim.. I dont know if thats racismus now lol

[@PapaFlammy69]

lol

[@taatuu25]

simsalabim comes from latin so nobody will seriously say it's racismus

[@kathanshah8305]

Dottie number , nice of them to name a number after a basement dweller

[@neilgerace355]

Sines, sines, everywhere sines Blocking out the chalkboard, breaking my mind

[@sharpman5772]

Nice video! Simple problem with a nice solution!

[@Dakers11]

Flammy, I went to a Math Bar for a few Drinks. A Lovely lady asked me my Sin. We converged rather well.

[@PapaFlammy69]

:^)

[@vincentt.7146]

I love recursion !

[@WhattheHectogon]

Awesome video! I did a video that involved the Dottie Number last year, about when an eclipse is half over :D It's a neat little constant.

[@sergioh5515]

Beautiful integral papa. More integrals like this 🥰

[@yassinezaoui4555]

Good work ^^ I hope you can make some videos on sequence and series of functions ( how to prove every type of convergence efficiently, difference between type of convergence) ... that would be intresting, no?

[@ethanwinchester4585]

That seemed to rely on the functions always converging in the domain of the integral but that wasn’t established so I couldn’t tell if this was all a joke or if I missed something

[@samuelthecamel]

The infinite tangent example doesn't work because even though there is a solution at t=0, surrounding values do not converge to 0. You can easily verify this with a calculator.

[@lordnavjot5921]

you should write a book on the fundamental theorem of engineering.

[@juijani4445]

DEM boi....youre gettin sirius bout the uploads

[@gregoired.4660]

yo those integral are really cool ! I just have a lil question : it seems that you missed a lots of solution too the equation "tan(t)=t". 0 is one of them but arn't they other on the other branches ? i dont know much about it so is that just because we ignore them while working with intégrale or is there really a misstake ?

[@bernat8331]

There is only one between -pi and pi

[@gregoired.4660]

@@bernat8331 ok so we just don't care about other branches ? Won't be interesting to see if there is other solution ? As I mentioned earlier, I don't know much about those case

[@sharpedged7830]

@@gregoired.4660 no, I'm pretty sure he fucked up on that one and the exponential (also the natural log), because when you infinitely iterate those functions they diverge.

[@gregoired.4660]

@@sharpedged7830 I don't know but sounds true. He actually suppose that those fonction converge, that may be where the problem is

[@EvilGeniusIIpi]

wth i’ve been thinking about this problem for a week you sorcerer.

[@anushkrajbordia1873]

The title is dope

[@ronaldiosua]

"Let's step away from the fundamentals of engineering and move onto real maths"..🤣🤣🤣🤣🤣🤣🤣

[@duggydo]

The nested video was back in Papa’s Fappable days.

[@PapaFlammy69]

ayyyy :^)

[@mastershooter64]

papa flammy do a video on the fractional derivatives of the riemann zeta function and the dirchlet eta function

[@PierreSoubourou]

"theorem of engineering", i want that t-shirt! :'-D

[@PapaFlammy69]

Always available at my Teespring shop :p Link is in the description:3

[@nose766]

What if one of these recursive equations gives me two values like in a quadratic formula? Which one is it? D:

[@Simon-hy2fh]

Super Video! Der Schwabe hat sich darüber gefreut!

[@PapaFlammy69]

:)

[@luggepytt]

Surely, you're trolling us, Mr Flammy!

[@stapler942]

I can never tell if Flammable Maths is the ultimate math shitposter or if these techniques are actually legitimate.

[@p0gr]

usually he's right, but here things are wrong. gcd for real numbers?

[@wildalfred]

An important argument to get this is the Banachscher Fixpunktsatz.

[@ajiwibowo8736]

Can you do this for all other trigonometric function???

[@annonyme8529]

And what about the integral of the infinite nested Lambert function ?

[@alberteinstein3612]

Forget the Multiverse, THIS is a true state of madness

[@daphenomenalz4100]

Exponential one was interesting, thnx :>

[@PapaFlammy69]

Np :3

[@skylardeslypere9909]

11:03 it's the same solution as the integral of \exp_{\infty}(x)dx since t = log(t) is equivalent to t=exp(t)

[@waitinblackout]

Diddl you notice the rabbit game graph on the top right corner? Worth a full episode!

[@tobybartels8426]

But you never prove that any of these infinite iterations of transcendental functions converge! All you prove is that _if_ sin_∞(x) exists, then it must be 0, and similarly for the others. To see that the iterated sine _does_ converge to 0, notice that |sin(t) - 0| < |t - 0|, so it is indeed getting closer to 0 as we iterate. This also works with the cosine; you don't even have to know the value of r to see that |cos(t) - r| < |t - r| if cos(r) = r, although you do need some trig identities. Unfortunately, this doesn't work for the tangent, and in fact you can make tan_∞(x) diverge to infinity or even be undefined. The same goes for exp_∞(x) and ln_∞(x). (Indeed, since the exponential of a real number is always real, there's no way that exp_∞(x) could be imaginary, did you prove that it has to be imaginary if it exists.) So these integrals don't make sense.

[@goodplacetostop2973]

13:57 Infinite square root integral ? It should be x+c, right?

[@boterham7791]

I love you so much I will never substitute u

[@PapaFlammy69]

[@daphenomenalz4100]

Bo Terham uses substitute, it was not very effective

[@LeviATallaksen]

This should only work if the fixed point acts as a stable equilibrium upon iterating the function, right? E.g. iterating x=0.5 under the tan-function gives: 0.5->0.55->0.61->0.70->0.84..., suggesting that it doesn't actually converge to the fixed point at 0. Or is it guaranteed to ever hit a multiple of pi?

[@aneeshsrinivas892]

now do this while running

[@angelmendez-rivera351]

Since sin : R -> R, the function sin°sin : R -> R is well-defined, and hence sin^n : R -> R is well-defined for any natural number n. It can be shown that limsin := lim sin^n (n -> ♾) : R -> R is well-defined, and it can be shown that limsin = 0, because sin : R -> [-1, +1] is a surjection, so sin^n has range [-(sin^[n - 1])(1), +(sin^[n - 1](1)], and the sequence n |-> (sin^[n - 1])(1) has infinum 0 and is monotonically decreasing, so it converges to 0, hence range(limsin) = {0}, so limsin = 0. Every constant function k : R -> R has derivative 0 : R -> R, hence every such k is an antiderivative of limsin. With cos, the same analysis as with sin applies, except that lim cos^n (n -> ♾) = r. Working with tan, this is significantly more complicated, since the dom(tan) is not R, but R\Π, where Π := {π/2 + n·π in R : n in Z}. Therefore, tan°tan is well-defined, but not function. What we need is to restrict dom(tan) by looking at the preimage of R\Π under tan, and make this its new domain. However, then tan^3 is not a function. In order for lim tan^n (n -> ♾) to have a chance to be a function, we have a sequence of sets D[n] given by D(0) = R/Π, and D(n + 1) = [tan^(-1)][D(n)], where [tan^(-1)][Y] is the preimage of Y under tan. We need lim D(n) (n -> ♾) =: D to be the domain restriction in order for tan^n to be well-defined for every natural n. It is unclear if D exists, or if D is an open set where a function g can be differentiable on. Therefore, it is unclear if lim tan^n (n -> ♾) can have well-defined antiderivatives. Also, it should be noted that the fixed point equation tan(t) = t has infinitely many solutions. It is thus unclear whether tan^n converges. With exp, we have that exp : R -> (0, +♾) is a surjection, hence exp°exp : R -> (+1, +♾), hence exp^3 : R -> (+e, +♾), and in general, for natural n > 0, exp^n : R -> (+e^^[n - 2], +♾). Since lim e^^[n - 2] (n -> ♾) = +♾, the sequence exp^n does not converge to a function, hence the function exp♾ defined in the video does not actually exist, so the its antiderivatives are obviously undefined. The same would be true of the hyperbolic function cosh.

[@matejcataric2259]

Cant wait to see BPRP comment :)

[@runyujiang8225]

the tangent one is wrong: tan(x)=x has more than one solutions other than x=0; in fact, in different interval tan_infinite(x) converges to different constants, so it is a step function, which looks like a ladder.

[@ja-vishaara]

I swear mathematicians just do a line of coke and then come up with this stuff to have something to do while real scientists do actual stuff

[@aniketeuler6443]

I like virtual math because it consists of fundamental theorem of engineering 😂😂😂

[@normalstickman1160]

i dont know how to do calculus but yes

[@pedrocusinato1743]

I wanna cry about the tan

[@ezras7997]

My favorite arbitrary constant was zero though, amazing.

[@spiroxylo2247]

"- can you tell me the value of this integral ? - yes, the value of this integral equals : some number - wow thanks math man"

[@dave_xc]

What about doing a square root without using square roots?

[@TI5040]

Nice infinite boi

[@jorex6816]

plot twist: the sin(sin(sin(...))) is with respect to t

[@jamesyeung3286]

i was just researching Dottie's number :D

[@tsawy6]

Doesn't exp_infinity(X) diverge for many starting values of X? E.g starting at X=1 we just get an infinite tower of e's

[@erik9671]

Smhmh, according to the 4th fundamental Axiom of Engineering, any infinite recursion can be approximated with typing in like, 10 of the recurring elements into your calculator, and the magical mushroom gnomes inside of it will find a proper solution. Also, genuine question, shouldn't/doesn't sin(t) have infinetly many solutions for t = 2PI*n, with n a Natural Number (or 0)? I know this stuff is sometimes relevant in complex numbers, but is it relevant here too? That would make the solution 2PI*n*x+C if I am not mistaken?

[@shaiavraham2910]

Wait a minute... there are mathematicians AMONG US. That's kinda sus, not gonna lie. VOTE EM!!

[@fuji_films]

C R I N G E.

[@fynnkessels2488]

I thought it would be necessary to prove that the function actually converges not only "if it converges it converges to..." (since we're doing math not physics.)

[@flopsiturtle1333]

doesnt tan(t)=t have infinetly many solutions? if you plot the function f(x)=tan(x) and g(x)=x and look for the intersections you will find infinetly many no?

[@fedorlozben6344]

Why you told t/1=sin(x)/cos(x) is hold only if t=0 I know it is true,but you proved it some strange... I rewrote cos as sqrt(1-sin2) and after got sin(x)=x/sqrt(1-x2) But rhs is Bigger than x (because denominator min is 1) when x>0 and less than -x for x

[@thomaslemeunier6533]

Can someone solve the integral of the Lambert-W-function_infinity(x) with respekt to x ? 😇😇

[@skylardeslypere9909]

So solve the equation W(t)=t Now, think graphically; if a function intersects the line y=t, then its inverse function also intersects the line y=t So W(t)=t => t exp(t)=t which yields only 1 solution, t=0

[@derfret1365]

isn't sin(sin(...sin(x)...)) just x?

[@PapaFlammy69]

ye, by FToE

[@boterham7791]

I will sin for u

[@the_nuwarrior]

Banach fixed-point theorem

[@pardeepgarg2640]

pi × e = pie = ⭕ has area pi r^2 So e = r^2 And hence no matter how much your pie cost it always has same radius i.e sqrt(e) But pi × e = pie = ⭕ has circumference 2πr So r= e/2 Taking Sum of both r i.e (e/2 + sqrt(e)) = 3 But π=3 So *Fundamental Theorem of Engineering* is proved QED

[@gauravbhardwaj4010]

Lamda w when i quit this video

[@ScrotN]

At this point I’m curious about who is your waifu than the Mathematics

[@PapaFlammy69]

My wife Lisa :p

[@boterham7791]

I love you

[@Convergant]

If finding the value of r is so slow why not just use Newton Raphson method? Im pretty sure that converges quadratically

[@telotawa]

when you say t i always thing of it spelled as "tieh"

[@kasuha]

tan(t) = t has infinite number of solutions. Besides 0 it they are values near 4.5, 7.9, 10.9 and so on. But more importantly, chaining tan() function does not converge, therefore tan(tan(tan...(x)...)) does not have value anywhere besides the solutions and the integral does not make sense.

[@sakanagakyoko]

From the fundamental theorem of engineering you know that the sin is a useless function, so you have that sin-infinity(x)=x so the solution is a cuadratic, well, aproximately

[@orphixigl1476]

x^2/2=0

[@CoffeeVector]

Doesn't the tan one have more solutions? Graphically, you can see that tan(t) = t has other solutions right?

[@gamingstars8956]

Yeah we know sinx

[@StepanSmith]

Iterated tan(x) converges only for x = n*pi. Iterated exp(x) or log(x) do not converge for any real number. So you can’t really talk about integrals of these functions. Getting an equation like tan(t)=t only tells you that if it were to converge, it would converge to one of the solutions of this equation; which for example tan(t)=t has infinitely many in real numbers. (Edit: every fixed point x_i of tan(x), which are all repulsive and there is countably many of them, gives rise to countably many points x_i + n*pi for which the iteration converges. Now every point x which converges again gives rise to countably many points arctan(x)+n*pi, which also converge. Iterating this process will give you more and more points that converge, but it will always just be a countable union of countable sets. So there is countably many points which converge, hence there is still uncountably many divergent points, so you can’t talk about the integral)

[@NateROCKS112]

n = 0 (and thus x = 0) is the only solution there, because tan(n pi) = 0 for all (edit: other) n. Besides 0, there are no closed-form solutions to tan(x) = x iirc. Also, there are only countably many solutions to tan(x) = x, so any definite integral will always be 0, regardless. (Edit:) Therefore, the indefinite integral must be a constant.

[@StepanSmith]

@@NateROCKS112 Not sure what you mean, but the equation tan(t)=t, the form of its solutions and the countability of them doesn’t really have anything to do with the integral. Infinitely iterating tan(x) does diverge for all x other than n*pi (Edit: and countably many more points, which converge; see edit in my first comment), and hence you can’t talk about the integral of such a function, when it isn’t even well defined. The countability would play a role if it were the case that only countably many x diverge, and hence you could just ignore them; but it’s the opposite in this case.

[@NateROCKS112]

@@StepanSmith tan(pi) = 0, so there's at least one counterexample to your statement. Not that it matters for your broader point necessarily; just that these solutions exist and that the neighboring points diverge, with which I agree, so yeah there can't be an integral.

[@StepanSmith]

@@NateROCKS112 I’m not saying x = pi is a fixed point, I’m saying iterating tan(x) converges for x = pi, since tan(pi)=0, and tan(0)=0 is a fixed point, hence x = pi converges to 0 by iterating tan(x). Similarly for every other fixed point x_i, the set {x_i + n*pi} will converge

[@mayankchandrakar8333]

Papa we are back

[@PapaFlammy69]

b a c c

[@mayankchandrakar8333]

Yup b a c c

[@lordofcastamere9376]

Did you just write sin(x) = x ? Papa does Engineering

[@mrbeancanman]

anyone have an intuitive proof for why 0 is the only solution to sin(t) = t ?

[@luggepytt]

Just draw the functions y = sin(x) and y = x, and you see directly that they only intersect at x = 0.

[@mineblox2313]

For sin(sin(...)) Sin(somethin) can have values between -1 and 1 So sin(sin(somethin)) is sin(1 or -1) If x is small sin (x) = x so sin(sin(..)) = sin(1 or -1) = -1 or 1 And 1-1 = 0 Ez

[@EminTuralic]

Flammy, let me make a confession. I'm an engineer. To make things worse, from Bosnia, and I know German. I love your content to death, your accent, and memey sense of humor. I watch 80% of your content drunk, alone, binging on roasted sunflower seeds. Thank you for keeping up with the KZbin pressure even though the algorithm is not favouring you. If I ever stop being a dogshit-nonEU citizen I'll support you as much as I can. Please keep it up. With love, future immigrant

[@neuralwarp]

I suppose you could also write (sin▪)^inf(x)