This is something that still confuses me. At 11:00, why can we arbitrarily decide that the upper limit of the integral increases at the same rate (1-x/n)^n converges to e^-x? Why dont we have to introduce a new variable and limit?
There are references below to the DCT being "required" to understand what is going on. In fact this is a legitimate question that arose in classical analysis long before Lebesgue. In his book " A Course of Pure Mathematics" Hardy deals with relative rates of convergence (see problem 16 pages 167-8) which are at the heart of Tauberian theory. Littlewood's 3 principles ( Every (measurable) set is nearly a finite sum of intervals; every function (of class Lp) is nearly continuous; every convergent sequence of functions is nearly uniformly convergent) ensure that most of the time you can get away with the "obvious" (assuming of course that you have checked that the hypotheses are satisfied).
23:05 The constant is defined without the 1/(n+1) bit. Doesn't change the result though, if you multiply it out, the limit of the remaining n / (n+1)^2 summand is 0.
a_n is a decreasing sequence, you can show this by using the same method used to show b_n is increasing but using a_n and replacing the integral of 1/x with the integral of 1/(n+1).
@ 6:24 Prof. Penn claims that a_n = 1 + 1/2 + .... + 1/n - ln(n) is an increasing sequence. We'll show that a_n is a strictly decreasing sequence. Let f(x) =: 1/(x+1) - ln((x+1)/x), for x > 0. Then, f'(x) = 1/(x(x+1)^2) > 0. So, f is strictly increasing. Since Lim(x -> infinity)f(x) = 0, it follows that f(x) < 0. Now, a_(n+1) - a_n = 1/(n+1) - ln((n+1)/n) = f(n) < 0. This also shows that a_n
Cool Video, thank you, Michael. If you mention the Euler-Masceroni constant, you also have to mention the di-gamma function (the log-derivation of the gamma function)
6:21 it's definitely not an increasing sequence, it's actually decreasing. Can anyone either explain why we know its bounded below by 0 (or always positive) or why it converges anyway?
Here's how I approached it. Start with the sequence defined by a_n = 1 + 1/2 + 1/3 + ... + 1/n - ln(n) Then let f(n) = a_(n+1) - a_n = 1/(n+1) - ln((n+1)/n) f(n) ... ...starts out negative (n=1): e < 4 implies ln(2) > 1/2 implies f(1) < 0 ... has positive gradient (n>0): f'(n) = 1 / n(n+1)(n+1) ...approaches zero in the limit as n increases: 1/(n+1) can approach ln((n+1)/n) only if ( (n+1)/n )^(n+1) approaches e which it does, by definition lol That tells us that f(n) < 0 for n>=1 and the fact that 1-ln(1) = 1 tells us that a_n = 1 and so gamma too must not exceed 1.
Since the mathematician from which the "Mascheroni" in "Euler-Mascheroni constant" comes, was Italian, it would correctly be pronounced "Maskeroni" and not "Masheroni".
don't see this one very often.... some thoughts: 1. are there other interesting constants coming from the difference between a sum an an integral? 2. What are the cases of γ appearing in unexpected places (like π often does)? even in physics? 3. using the methods in the video, can we approximate γ?
1. Other constants include the Mertens constant, closely related to the Euler Mascheroni constant. 2. Both of these numbers often appear in number theory, specifically when dealing with prime numbers. 3. There are like many other ways of approximating gamma by manipulating the original series or limit but yes.
It appears everytime in Quantum Field Theory. When you have to compute Feynman Diagrams with 1 or more loops you need to approximate the Gamma function. Since the derivative of Gamma(z) at z=1 is -γ you see it often
HW: 1. ln(n) = integral from x=1 to n of 1/x = sum from k=1 to n-1 of integral from x=k to k+1 of 1/x < sum from k=1 to n-1 of 1/(k+1) because 1/x is decreasing = 1/2 + 1/3 + ... +1/n < 1 + 1/2 +... + 1/n 2. Given a_n converges: lim n->infinity b_n = lim n->infinity a_n - 1/n = lim n->infinity a_n - lim n->infinity 1/n by a_n's convergence = lim n->infinity a_n which is given to converge Given b_n converges lim n->infinity a_n = lim n->infinity b_n + 1/n = lim n->infinity b_n + lim n->infinity of 1/n by b_n's convergence = lim n->infinity b_n which is given to converge
Euler was a religious person throughout his life.[20] Much of what is known of Euler's religious beliefs can be deduced from his Letters to a German Princess and an earlier work, Rettung der Göttlichen Offenbahrung gegen die Einwürfe der Freygeister (Defense of the Divine Revelation against the Objections of the Freethinkers). These works show that Euler was a devout Christian who believed the Bible to be inspired; the Rettung was primarily an argument for the divine inspiration of scripture.
It doesn't seem like a correct move at 10:45 to replace e^-x in the integral this way. There actually should be two limits after this transform and two different numbers accounted for integral and for e^-x as a limit
Exactly my thought when watching this for the first time: seen from a standpoint of formal logic, it seems he's renaming a bound variable (the one in the limit for e^-x, which SHOULD be different from n) to a name that's (locally) free inside the integral - which COULD be an invalid rename!
Another viewer with more mathematical knowledge in measure theory than me commented that it is correct, but it requires the dominated convergence theorem and is thus non-trivial. If you wanna look for that comment. I looked up the theorem on wikipedia and although the theorem seems logical enough, I'm not sure how it explains that taking the limit this way is correct.
Sixty years ago l was taught that the exponential function was defined as the limit of (1+x/n)^n as n approaches infinity. Now MP says it is a result (at about 11.40). Funny old world!
Just separate the "s" from "ch": Mas-cheroni ("Mas-keroni") (English native speakers have their own way of pronunciation of foreign words / names: mostly false *lol*)
I hope other people will chime in with uses! I like Mathologer's video on the harmonic series, which also deals with the Euler-Mascheroni constant. Essentially, if you want to compute the nth partial sum of the harmonic series, this is quite tricky to do since there is no known nice formula for the nth partial sum. However, you can approximate it using the natural log function (which we can compute to arbitrary precision!). The error between this natural log approximation and the partial sums gets closer and closer to the Euler-Mascheroni constant. So if you want to approximate 1+1/2+1/3+...+1/n for some large positive integer n, you can compute ln(n+1)+γ, and this will be a very good approximation. But this is just one application, and it's not too hard to believe based on the limit Michael proved in this video. I would also love to hear what other people say!
One application of this constant, is the Laplace transform of natural log. A method that is commonly used for converting Calculus into Algebra, as a strategy for solving differential equations. I've tried to find an example problem where it would be practical to solve, that starts with natural log, and uses its Laplace transform to solve, but I can't seem to come up with one. If anyone can suggest one that works, please let me know.
First limite explanation could be summed up by the sentence: substracting an infinite series by its integral (turns out to be inferior or equal to one!!!)
The sum 1/n is more than the integral 1/x dx from 1 to n+1 (draw a box of height 1,1/2,...,1/n on graph y=1/x), and integral is ln(n+1) which is more than ln(n), for all n. Therefore the constant is strictly more than 0. The proof that it's less than 1 is in video 5:06 (it's also strict inequality)
@@eaglesquishy a_n is more than 0 for all n, and it is strictly increasing (also proven), therefore the limit is more than 0. And from the video the limit (not a_n) is proven to be less than 1 in the video
@@warmpianist a_n is actually decreasing (not proven or disproven in the video). Also, if you look closely, the limit was shown to be less than or equal to 1.
21:50 .. LEFT TOP................ gamma >0.......a countable number (rational orbit) calculated in a rational number 'Q' based formulae .............LEFT BOTTOM.........-gamma
@@Pablo360able ...kinda like asking equivalent geometric question about translating parallel lines (or the 2 lines endpoint extensions to infinity ...does a pair of parallel lines remain parallel(=) ...or.....notParallel(~=)....@ infinity
@@abdonecbishop gonna need to timestamp the relevant part because I don't see what singularities of pairs has to do with you thinking that an expression involving rational numbers can't have an irrational value
One application of this constant, is the Laplace transform of natural log. A method that is commonly used for converting Calculus into Algebra, as a strategy for solving differential equations.
@@carultch quite interesting, Math is about building tools to simplify analysis. Its always nice to know where do these tools are applied. Not just about learning about random symbolic facts. So thx for the application!
@@CppExpedition Indeed. Laplace Transforms are an awesome tool. I've tried to come up with an example where you could use the Laplace transform of natural log to solve a DiffEQ, but I haven't had any success thus far. Every example I try, seems to stump Wolfram Alpha. It's much easier to use the method of Laplace transforms when the DiffEq uses trig, exponentials, algebraic functions, and impulse/step/ramp functions, since they all have algebraic Laplace transforms that are practical to untangle.
It is! I think he's made a video about it before. You can also verify it by seeing how each side of the equation arises as the sum of the entries in a multiplication table.
just rename mathematics to findings of Euler
Eulerology?
Euleristics?
Euleritics?
In today's finding of Euler's class, we will learn ...
or Cauchy, Riemann or Gauss
@@petterituovinem8412 i don’t think that will “cauch” on.
Euleritics or Eulerology?
Ah yes, the Oily Macaroni Constant.
"Flammable!"
10:43 Replacing e^-x like that would require the dominated convergence theorem, it's not trivial
👍👍
The video ends and my mind auto completed with "and that's a good place to stop".
This is something that still confuses me. At 11:00, why can we arbitrarily decide that the upper limit of the integral increases at the same rate (1-x/n)^n converges to e^-x? Why dont we have to introduce a new variable and limit?
It is true, but it requires the dominated convergence theorem, it's not trivial
@@giovanni1946 il faut appliquer le théorème de convergence dominée à la fonction fn(x)=0 si x>n et fn(x)=(1-x/n)^n si x
@@Raphael-wg7zi Exact
There are references below to the DCT being "required" to understand what is going on. In fact this is a legitimate question that arose in classical analysis long before Lebesgue. In his book " A Course of Pure Mathematics" Hardy deals with relative rates of convergence (see problem 16 pages 167-8) which are at the heart of Tauberian theory. Littlewood's 3 principles ( Every (measurable) set is nearly a finite sum of intervals; every function (of class Lp) is nearly continuous; every convergent sequence of functions is nearly uniformly convergent) ensure that most of the time you can get away with the "obvious" (assuming of course that you have checked that the hypotheses are satisfied).
f(x)=(1+x/n)^n, n tends to infinity is a formal definition of e^x, proven by taking derivative.
Brilliant. Thanks!
You make me feel like I'm back in college again, except you only teach fun stuff.
@ 21:39 The sign of the second term in the numerator should be a minus, not a plus.
23:05 The constant is defined without the 1/(n+1) bit. Doesn't change the result though, if you multiply it out, the limit of the remaining n / (n+1)^2 summand is 0.
Small error at 11:51 , x->n from below. Thanks for the great content!
Yes. That is correct.
10:46 the substitution is not correct the n form the first limit is not the same as the n of the definition for e^(-x)
👍
Miss you good place to Stop 😥
a_n is a decreasing sequence, you can show this by using the same method used to show b_n is increasing but using a_n and replacing the integral of 1/x with the integral of 1/(n+1).
@ 6:24 Prof. Penn claims that a_n = 1 + 1/2 + .... + 1/n - ln(n) is an increasing sequence. We'll show that a_n is a strictly decreasing sequence.
Let f(x) =: 1/(x+1) - ln((x+1)/x), for x > 0. Then, f'(x) = 1/(x(x+1)^2) > 0. So, f is
strictly increasing. Since Lim(x -> infinity)f(x) = 0, it follows that f(x) < 0.
Now, a_(n+1) - a_n = 1/(n+1) - ln((n+1)/n) = f(n) < 0. This also shows that
a_n
😢
There's no good place to Stop 😢
The oily macaroni constant
Cool Video, thank you, Michael. If you mention the Euler-Masceroni constant, you also have to mention the di-gamma function (the log-derivation of the gamma function)
6:21 it's definitely not an increasing sequence, it's actually decreasing. Can anyone either explain why we know its bounded below by 0 (or always positive) or why it converges anyway?
it's increasing as you add the terms, not each term separately
The last limit can be calculated with only two uses of L'Hospital. Let w = n+1 0.
By L'Hospital,
(A) Lim(t -> 1){[(1 - t^w)^2]/(t - 1)} = Lim(t -> 1) { [(2wt^(2w - 1) - 2wt^(w - 1)] / 1}=0.
Using L'Hospital and (A),
Lim(t -> 1^-) {(1 - t^w)ln( 1 - t)} = Lim(t -> 1^-) {ln(1 - t)/(1 - t^w)^(-1)} =
Lim(t -> 1^-) { [1/(t-1)]/[(wt^(w-1)(1 - t^w)^(-2)]} =
Lim(t -> 1^-){1/[wt^(w-1)]}X Lim(Lim(t -> 1^-){[(1 - t^w)^2]/(t - 1)} = (w)(0) = 0.
Additionally, if w = 0, then Lim(t -> 1^-) {(1 - t^w)ln( 1 - t)} = Lim(t -> 1^-) {0} = 0 .
I really like that you upload almost every single day, your videos are fun to watch and I just wait for them.
Based. Underrated constant
Totally! ^.^
Based replier🙏
Here's how I approached it.
Start with the sequence defined by
a_n = 1 + 1/2 + 1/3 + ... + 1/n - ln(n)
Then let f(n) = a_(n+1) - a_n
= 1/(n+1) - ln((n+1)/n)
f(n) ...
...starts out negative (n=1):
e < 4 implies ln(2) > 1/2 implies f(1) < 0
... has positive gradient (n>0):
f'(n) = 1 / n(n+1)(n+1)
...approaches zero in the limit as n increases:
1/(n+1) can approach ln((n+1)/n)
only if ( (n+1)/n )^(n+1) approaches e
which it does, by definition lol
That tells us that f(n) < 0 for n>=1
and the fact that 1-ln(1) = 1
tells us that a_n = 1
and so gamma too must not exceed 1.
11:52 it is as x approaches n, not 1
Fun fact: we don't even know whether gamma is rational or not.
Since the mathematician from which the "Mascheroni" in "Euler-Mascheroni constant" comes, was Italian, it would correctly be pronounced "Maskeroni" and not "Masheroni".
don't see this one very often....
some thoughts:
1. are there other interesting constants coming from the difference between a sum an an integral?
2. What are the cases of γ appearing in unexpected places (like π often does)? even in physics?
3. using the methods in the video, can we approximate γ?
3) plug in n=99
1. Other constants include the Mertens constant, closely related to the Euler Mascheroni constant.
2. Both of these numbers often appear in number theory, specifically when dealing with prime numbers.
3. There are like many other ways of approximating gamma by manipulating the original series or limit but yes.
It appears everytime in Quantum Field Theory. When you have to compute Feynman Diagrams with 1 or more loops you need to approximate the Gamma function. Since the derivative of Gamma(z) at z=1 is -γ you see it often
No place to stop!
I guess there's no good place to stop with this one 😮💨
HW:
1. ln(n) = integral from x=1 to n of 1/x
= sum from k=1 to n-1 of integral from x=k to k+1 of 1/x
< sum from k=1 to n-1 of 1/(k+1) because 1/x is decreasing
= 1/2 + 1/3 + ... +1/n
< 1 + 1/2 +... + 1/n
2. Given a_n converges:
lim n->infinity b_n
= lim n->infinity a_n - 1/n
= lim n->infinity a_n - lim n->infinity 1/n by a_n's convergence
= lim n->infinity a_n which is given to converge
Given b_n converges
lim n->infinity a_n
= lim n->infinity b_n + 1/n
= lim n->infinity b_n + lim n->infinity of 1/n by b_n's convergence
= lim n->infinity b_n which is given to converge
Oh I think I did at least some extra work
I think u mean 1/k not 1/(k+1), pls be careful about the +- sign of the decreasing function
Wow, incredible. Just today worked on the video's integral and derivatives of gamma of bigger order and there comes yours video on the subject 😼
Is there a good place to stop?
Euler was a religious person throughout his life.[20] Much of what is known of Euler's religious beliefs can be deduced from his Letters to a German Princess and an earlier work, Rettung der Göttlichen Offenbahrung gegen die Einwürfe der Freygeister (Defense of the Divine Revelation against the Objections of the Freethinkers). These works show that Euler was a devout Christian who believed the Bible to be inspired; the Rettung was primarily an argument for the divine inspiration of scripture.
Ok, thank you very much for the proof of the integral form. But any ideas what motivated the idea that it could be repesented that way?
That constant poped up in a proof in my dissertation regarding some prime densities. It came out of nowhere.
It doesn't seem like a correct move at 10:45 to replace e^-x in the integral this way. There actually should be two limits after this transform and two different numbers accounted for integral and for e^-x as a limit
Exactly my thought when watching this for the first time: seen from a standpoint of formal logic, it seems he's renaming a bound variable (the one in the limit for e^-x, which SHOULD be different from n) to a name that's (locally) free inside the integral - which COULD be an invalid rename!
Another viewer with more mathematical knowledge in measure theory than me commented that it is correct, but it requires the dominated convergence theorem and is thus non-trivial. If you wanna look for that comment. I looked up the theorem on wikipedia and although the theorem seems logical enough, I'm not sure how it explains that taking the limit this way is correct.
Ah yes, the Euler-Macaroni constant, my favorite!
∞ - ∞ ~ ½, obviously. ;)
It's interesting how often this constant appears in Physics.
12:25 is where he goes full speed ahead until the end. 😮
which is about the point I fell off the oxcart, and from the edge of the road watched it recede into the distance
Euler's Macaroni
Some mathematical constants :
Ω = 0.5671432904… (Omega constant)
γ ≈ 0.5772156649… (Euler-Mascheroni constant)
δ ≈ 0.5963473623… (Euler-Gompertz constant)
G ≈ 0.9159655942… (Catalan's constant)
ζ(3) ≈ 1.2020569032… (Apéry's constant)
ρ ≈ 1.3247179572… (Plastic ratio)
√2 ≈ 1.4142135624… (Pythagoras' constant)
μ ≈ 1.4513692349… (Ramanujan-Soldner constant)
φ ≈ 1.6180339887… (Golden ratio)
√3 ≈ 1.7320508075… (Theodorus' constant)
P₂ ≈ 2.29558 71494… (Universal parabolic constant)
e ≈ 2.71828 18284… (Euler's number)
π ≈ 3.14159 26536… (Archimedes' constant)
δ ≈ 4.6692016091… α ≈ 2.5029078751… (Feigenbaum constants)
Sixty years ago l was taught that the exponential function was defined as the limit of (1+x/n)^n as n approaches infinity. Now MP says it is a result (at about 11.40). Funny old world!
Hi! love your videos. btw "Mascheroni" is pronounced with a "K" sound, not "C", like "Mask". Hope this helps!
I pronunce it with a Spanish "sh" also know as German "Sch"
@@friedrichhayek4862 it's a common mistake, but that's italian
@@friedrichhayek4862 Complete nonsense! It´s neither German nor Spanish, but Italian!
@@friedrichhayek4862 Your pronunciation is wrong. It is a Italian name. Pizzamid Head is right with his commentary.
Just separate the "s" from "ch": Mas-cheroni ("Mas-keroni")
(English native speakers have their own way of pronunciation of foreign words / names: mostly false *lol*)
3:05 integral from 1 to 2 plus integral from 2 to 3 plus ... plus integral from n-1 to n, for a total of n-1 integrals?
Correct, think about it. If n=2, you'd only have 1 integral, from 1-2. If n=3, you'd only have 2: 1-2 and 2-3. For arbitrary n, n-1 integrals
And that's a good place to stop.
He's right bro
Is this the debut of eraser sleeves? Well, I suppose a high-level climber has a lot of experience getting chalk out of his clothing!
i don't know which is the good place to stop for today :(
Thank you for the video as always! Did anyone else hear a little interference with the microphone today?
Pivot between a sequence, and a function.
OK, where to get that t-shirt/hoody? sorry, couldn't find it on the internetz. ... oh whait: found it! (under Merch!)
Did I miss "That's a good place to stop"?
A nice "mathematical details" follow-up to Numberphile's video. 🙂
Which is the use of this constant?
Just like pi, e, zeta(3) and other constants, it often appears as a part of answers for weird/non-elementary integrals, sums and other problems.
The random wikipedia equations which mathematicians cook up for no reason
I hope other people will chime in with uses!
I like Mathologer's video on the harmonic series, which also deals with the Euler-Mascheroni constant. Essentially, if you want to compute the nth partial sum of the harmonic series, this is quite tricky to do since there is no known nice formula for the nth partial sum. However, you can approximate it using the natural log function (which we can compute to arbitrary precision!). The error between this natural log approximation and the partial sums gets closer and closer to the Euler-Mascheroni constant. So if you want to approximate 1+1/2+1/3+...+1/n for some large positive integer n, you can compute ln(n+1)+γ, and this will be a very good approximation.
But this is just one application, and it's not too hard to believe based on the limit Michael proved in this video. I would also love to hear what other people say!
One application of this constant, is the Laplace transform of natural log. A method that is commonly used for converting Calculus into Algebra, as a strategy for solving differential equations.
I've tried to find an example problem where it would be practical to solve, that starts with natural log, and uses its Laplace transform to solve, but I can't seem to come up with one. If anyone can suggest one that works, please let me know.
First limite explanation could be summed up by the sentence: substracting an infinite series by its integral (turns out to be inferior or equal to one!!!)
Yes, I know of the Euler gamma and his big brother \Gamma(\epsilon)
Hey, it's the OILY MACARONI constant! (Sorry, schoolboy humor.)
can we use L'Hopital's rule for 0/(1/0) i.e 0/infinity ?
There's a nice geometric argument that the E-M constant has a value greater than 1/2.
10:18 everybody: the FACT.
idea for a future video, fractional harmonic numbers.
What ? There were no good place to stop today ?! 😮
Wonderful.
How was gamma proven to be strictly between 0 and 1?
The sum 1/n is more than the integral 1/x dx from 1 to n+1 (draw a box of height 1,1/2,...,1/n on graph y=1/x), and integral is ln(n+1) which is more than ln(n), for all n. Therefore the constant is strictly more than 0.
The proof that it's less than 1 is in video 5:06 (it's also strict inequality)
@@warmpianist What you just said are arguments to show that a_n is strictly between 0 and 1. The limit then could still be equal to 0 or 1.
@@eaglesquishy a_n is more than 0 for all n, and it is strictly increasing (also proven), therefore the limit is more than 0.
And from the video the limit (not a_n) is proven to be less than 1 in the video
@@warmpianist a_n is actually decreasing (not proven or disproven in the video).
Also, if you look closely, the limit was shown to be less than or equal to 1.
Michael's hair length suggests this one was recorded out of order :)
19:02, t^(n+1)+1 :/
euler-macaroni best constant. Actually, in Calculus 1, I had to prove that the limit existed during an exam.
rubbish
this math is getting more esoteric all the time ...
Г'(1)=-γ
According to Wikipedia it can also be defined by an integral involving the floor function. Segway to your next video? 😜
Segue (Italian for 'follows') - a Segway is an electric vehicle! 😉
I never knew how to pronounce it before
A sign of missing education.,
Michael gets it slightly wrong: the C is hard, so it starts like 'mask', not 'mash'.
@@davidgould9431 thx
Anyone else want to say it as the, “Euler Macaroni constant?”
You mean, "oily macaroni". ;)
Props to Papa Flammy for coining the alias "oily macaroni constant"
sketchy :))))
21:50 .. LEFT TOP................ gamma >0.......a countable number (rational orbit) calculated in a rational number 'Q' based formulae
.............LEFT BOTTOM.........-gamma
no, both are true
@@Pablo360able ...kinda like asking equivalent geometric question about translating parallel lines (or the 2 lines endpoint extensions to infinity ...does a pair of parallel lines remain parallel(=) ...or.....notParallel(~=)....@ infinity
@@abdonecbishop it's like exactly none of that. what you are saying is mathematical word salad.
@@Pablo360able you sound really confident...to bad you are wrong
.. th-cam.com/video/DOnlcsihEcM/w-d-xo.html
@@abdonecbishop gonna need to timestamp the relevant part because I don't see what singularities of pairs has to do with you thinking that an expression involving rational numbers can't have an irrational value
Euler-Mascarpone constant
256th like
but how that integral is important? i was expecting an application :P
One application of this constant, is the Laplace transform of natural log. A method that is commonly used for converting Calculus into Algebra, as a strategy for solving differential equations.
@@carultch quite interesting, Math is about building tools to simplify analysis. Its always nice to know where do these tools are applied. Not just about learning about random symbolic facts. So thx for the application!
@@CppExpedition Indeed. Laplace Transforms are an awesome tool.
I've tried to come up with an example where you could use the Laplace transform of natural log to solve a DiffEQ, but I haven't had any success thus far. Every example I try, seems to stump Wolfram Alpha.
It's much easier to use the method of Laplace transforms when the DiffEq uses trig, exponentials, algebraic functions, and impulse/step/ramp functions, since they all have algebraic Laplace transforms that are practical to untangle.
@@carultch don't worry, either way i would calculate any fourier transform through numeric FFT.
I'm curious of your sweatshirt! Is that 'identity' true?
It is! I think he's made a video about it before. You can also verify it by seeing how each side of the equation arises as the sum of the entries in a multiplication table.