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

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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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[PapaFlammy]

*_Hey dick cheese, thanks for watching <3_* Make sure to check out some of deeze relevant links-desu to support the channel-chan

[joirnpettersen]

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

[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.

[sharpedged]

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

[ay]

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)|<|x|. This implies that the absolute value of sins are a bounded decreasing sequence so it converges. Now if |sin_infty(x)|=t then sin(t)=+/-t so t is zero. You will need to prove the convergence in your all examples.

[nikhilnagaria]

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.

[igxniisan]

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

[nikhilnagaria]

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

[spurcalluth]

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

[epicgamer]

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

[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.

[_..---]

infinitely nested integrals? mathematics induces madness

[rollingsnowball]

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

[Predakingever]

Papa Flammy: sin(t) = t

Engineer: I knew it.

[iispacedustii]

no jens constant = pure madness and a mathematicians nightmare

[nnotcircuit]

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.

[chalkchalkson]

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)<x is true for 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<0 the same works in reverse, monotone increase with upper bound.

[saisanjaynarayanan]

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 :)

[Dakers]

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

[angelmendez-rivera]

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.