An exponential trigonometric integral.

[@rbdgr8370]

keep the more saturated colour scheme more often. Adds quality to your production

[@MichaelPennMath]

Wait a minute, which is the better image?? I can't tell which has "better" quality, they just seem "different".

[@MichaelGrantPhD]

@@MichaelPennMath I prefer less color saturation, but with those settings you seem to have better contrast and sharpness, so overall I prefer it.

[@rbdgr8370]

@@MichaelPennMath So you use Nikon D610 as your main camera and for this video you used iphone XR too. Nikon's footage is a bit flat but skin tones are good where as XR's computational algorithms kicks more saturated colours which makes your chalkboard colours pop which is a good thing but your skin tones are over saturaed. So i guess you need to tweek a little bit of your Nikon's footage to increase saturation and thus increasing quality

[@tangpiseth8416]

@@rbdgr8370 Tommy🖤🖤

[@cphVlwYa]

This is also a beautiful integral to solve with type one contour integration. First realize that due to the symmetry of the cosine function the integral will also be equal to 1/2 the same integral from 0 to 2π. Then let z=e^(θi) then dθ = dz/(zi) and 2cos(θ) = z+1/z. Therefore we have the complex integral around the unit circle of 1/(2πi z)e^(z+1/z)dz. Now all we need to do is find the residue of the essential singularity at the origin. So start with the laurent series of e^z and (1/z)e^(1/z) centered at 0: e^z = 1+z+z^2/2! + z^3/3! + z^4/4! ... (1/z)e^(1/z)) = 1/z + 1/z^2 + 1/z^3*2! + 1/z^4*3! + 1/z^5*4! ... Now, all we care about is the coefficient of the 1/z term of the product of these two series, which we can see has to be the sum of all the coefficients of the z^n terms in the first series times all of the z^-(n+1) terms in the bottom series. And both of these coefficients are n! and so therefore we're left with the sum from 0 to infinity of (1/n!)^2 for the residue at 0, which will be equal to the value of the integral because the 2πi's cancel.

[@demenion3521]

this unsatisfying feeling of having a sum as the result of an integral... ^^

[@a_llama]

see sophomore's dream

[@thefranklin6463]

integrals are sums in disguise

[@demenion3521]

@@angelmendez-rivera351 i also found this solution. i don't know if this would be better. but exactly as you say, there is no clear line between standard functions and non-standard function, so one could also use the bessel functions for the result

[@tylershepard4269]

Demenion Normal CDF’s can only be written as infinite series as well. Q(x), Erf(x) are all expanded as Taylor series. Unfortunately we have to truncate them so if the series are expanded around zero, you can only use it “close” to zero with little error. There are other series expansions that distribute the error more evenly about the interval of interest.

[@WielkiKaleson]

@@angelmendez-rivera351 I hate Bessel functions. What I do like, however, is that the sum converges very quickly.

[@goodplacetostop2973]

12:55 8:52 That’s a good POV to stop

[@goodplacetostart9099]

Good place to start at 0:00 And Michael has his favourite equation T-shirt in this video

[@macfrankist]

Im a physicist and love watching the problems you work on.

[@soumadipsen8097]

I wonder why, This channel is so good, the topics, the questions, which are not there even in the paid platforms on internet, is available here, still the views and subs are so less....:(((((

[@erin-williams-vj8]

When we're swapping the order of the infinite sum and the integral, wouldn't we need to show the series is uniformly convergent? I'm sure it works out, but that would be another detail in the proof, right?

[@JoseTorresMates]

I think you're absolutely right. I've seen Michael doing that "trick" quite often without mentioning why he's allowed to do so

[@Arbmosal]

The easiest criterion I know is to use Fubini's Theorem (or Tonelli's Theorem). This needs some measure theory, so I would not consider it standard, but the gist is: Infinite sums can be viewed as integrals over the counting measure on the natural numbers. This means that if \int \sum |f_n(x)| dx is finite, then we can swap the stuff. In this case it is easy to see that we can just use |exp(2cos(t))|

[@amandeep9930]

Its very easy to prove. The absolute value nth term of the series of less than or equal to 2^n/n!. Since sum(2^n/n!)) converge to e^2 hence by Weistrass M Test the series is uniformly convergent.

[@JoseTorresMates]

Yes, Aman, it's not difficult to see. The thing is that "something" should be said, or written... For.the sake of completeness

[@amandeep9930]

@@JoseTorresMates yes, you are right. If a beginner watches the video he or she would think that its ok to do it in every situation.

[@nathanisbored]

Thinking about it, I guess for math videos it always makes sense to have color contrast turned way up. In a normal KZbin video it would look odd and unnatural to do that cuz it messes with skin color and stuff, but in a math video you mainly care about the stuff on the board, and color contrast is just gonna make the colors more vibrant and easy to parse. It seems like it could be strictly better for this kind of presentation. One of the main reasons I like your videos is because of the very vibrant colors on the board, and I feel like upping color contrast would only make that effect stronger.

[@MohitRaj-1712]

Big fan of your videos. Keep making these videos. These videos help me in developing approach to a Maths 'Problem

[@knode1993]

I think this is the Only channel whicj discusse this kind of problem EVERY DAY.

[@bebarshossny5148]

Please do a video at one point on when and why we can switch the order of integration and summation

[@KhaledRadwan-ku2bh]

Another way: Consider the following contour integral which carries the same info of our integral: -i/2pi × closed int (e^z * e^1/z)/ z over a closed circle of radius 1 centered at z=0. This integral equal 2i pi × the residue at 0. Here the singularity is not a pole, but an essential singularity. You can expand the exponentials using Taylor series expansion to get the residue which is the coefficient of 1/z which will turn out to be sum 1/(n!)^2 from 0 to infinity, the same result. This is I_0, the modified Bessel function of the first kind.

[@ІгорСапунов]

The series representation of exp-function with trig power leads to series with integrals of (cos(x)^2n) from 0 to pi/2. Using the formula for such integrals we have same answer

[@FunctionalIntegral]

This would have been a good exercise for application of complex analysis!

[@ittoito4855]

These videos are just awesome

[@covfefe18225]

It's the modified Bessel Function I_0(2). I turned this sum into a hypergeometric function and figured out it was 0F1, so I immediately realized that it was a Bessel function.

[@tylershepard4269]

Enter the modified Bessel function. Similar integrals come up when studying class C power amplifiers!

[@Wurfenkopf]

I'm a little confused about what happened at 0:42. Maybe was it a good place to fade?

[@Kierio04]

Lol I’m just watching this and going “why can’t he just integrate it normally”

[@ConnorJScholten]

the antiderivative doesn't exist for this integrand.

[@petterituovinem8412]

I googled and the result is actually modified Bessel function of the first kind with n=0 I_{0}(2) evaluated at 2. The integral is one of it's representations

[@bulldawg4498]

Fantastic job!

[@polychromaa]

I solved this using Taylor series. I had already evaluated the integral of cos^n(x) from 0 to pi, so it was incredibly easy.

[@Notthatkindofdr]

Today I taught my calculus class (online) about Riemann sums and showed them I was wearing Michael's "My favorite equation" T-shirt that I got for my birthday this month. :-)

[@amandeep9930]

It can be proved that the integral is equal to the contour integral of the function exp(z+1/z)/z along the unit circle. Thus by Cauchy Residue Theorem the integral is equal to the residue of exp(z+1/z)/z at origin. But residue of exp(z+1/z)/z at origin is equal to the coefficient of 1/z in its Laurent Series at origin. The coefficient of 1/z in the Laurent Series expansion of this function is equal to the constant term in the Laurent Series expansion of exp(z+1/z) at origin. Which is sum(1/n!^2)

[@goodplacetostart4606]

Calc 3

[@CM63_France]

Hi, Could we calculate this infinite sum? May be starting from e=1 + 1/2! + 1/3! + ... About the camera : The camera is getting more and more down. This time we couldn't read at the very top of the board. PS : I am speaking about the A camera, not the Iphone.

[@tueur2squall973]

Where did go the 1/2Pi in front of the whole thing at the end ???

[@Manuel-pd9kf]

What's the approximation for the sum

[@mFix09]

Angel Mendez-Rivera 2.28 would be enough, bro...

[@rogerlie4176]

This sum converges quickly since (n!)^2 grows very fast. You would get a very good approximation by including terms up to n=7.

[@LordVoidFury]

Take a look at the Jacobi-Anger expansion!

[@MartinPerez-oz1nk]

PADRÍSIMO!!!, EXCELENTE!!!, GRACIAS!!, SALUDOS CORDIALES!!!

[@edwardjcoad]

Glorious!

[@lavneetjanagal]

There is another way. Using contour integration (but I think they are essentially same methods in disguise). convert integral from -pi to pi and multiply by 1/2. Do series expansion of Exp(2 cos(\theta)) and put cos(theta) = 1/2(z+1/z) where z = exp(i \theta). d z /(i z)= d theta . 2 pi i * residues is the answer. 2 pi i cancel, so the final answer is just the constant terms in the expansion 1 + (z+1/z)/1!+ (z+1/z)^2/2!+ ..., which is the desired answer.

[@GaryTugan]

holy crap. and I love the end: "and that's a good place to stop". Mind Blown

[@violintegral]

By the way, this is the average value of e^(2cosx) on the interval [0,π]

[@emmepombar3328]

That camera switch. From a high end camera stolen from Peter Jackson's last set to a toaster, back and forth.

[@divyakumar8147]

thanks

[@damascus21]

Yo, what a crazy result. The integral of this function is an infinite sum?!

[@CaridorcTergilti]

Kind of, the terms go to zero INCREDIBLY fast. But technically infinite yes

[@topraramx9897]

I dont understand that how can we switch summation's position to front integral. Some one pls explain me

[@afonsobarra582]

Neither do I. But I was looking for it and then I found this: math.stackexchange.com/questions/1334907/reversing-the-order-of-integration-and-summation

[@rc4828]

1/π integral_0^π exp(2 cos(x)) dx= 2.27

[@coderdemo9169]

Can you please explain more about mit

[@cicik57]

yes, there must b the way to simplify last statement, do it pls!

[@LucaIlarioCarbonini]

I feel pretty lucky that I follow this channel from its birth or little later because "we have already proved" might be a small push to leave, might be something that someone could consider enough to be discouraged.

[@JoseTorresMates]

And, what's an approximate.value.of that series at the end?

[@otakurocklee]

Very nice!

[@Whoeveriam226]

How can I contact you to suggest some problems? I have a few problems on number and set theory and especially one geometry question that I think would be a great idea for a video

[@呂永志]

3:12 the integral seems wrong?

[@77Chester77]

Breaking the WALL at 4:32

[@holyshit922]

We can avoid complex numbers using integration by parts After integration by parts we will get easy recurrence relation

[@minh9545]

Did he said that m is an integer from the beginning or im so stupid to not remember it?

[@k3dr1]

Nice camera

[@user-A168]

Good

[@mbpl10]

What's up with the camera angles?

[@peremarquez6489]

so goood video!!!

[@javizaragoza1463]

3:00 hurt my eyes when he missed the parenthesis on the sin

[@jamesmonteroso824]

hahaha nindut ug agi

[@berzerksharma]

Maths is so beautiful

[@rbdgr8370]

can we make 8:52 into a meme/gif?

[@aliberro]

What Camera are you using?

[@MichaelPennMath]

My A-camera is a Nikon D600 and my B-Camera is my iPhone XR. Over the next several months I hope to upgrade a bit, maybe to the cannon R5/R6 or similar.

[@aliberro]

@@MichaelPennMath You're the best❤️

[@MichaelPennMath]

My pie in the sky goal is to have ridiculous MKBHD level production for simple chalk-board math videos...

[@aliberro]

@@MichaelPennMath you deserve the best, Mr. Michael, I was asking since you've inspired me to do similar thing in my college

[@reshmikuntichandra4535]

Please do 1988 IMO Q.6

[@Dalfi91]

12:55

[@a_llama]

bruh.

[@itamargolomb8530]

In academic terms, what level is the math in this video? (I'm 15, never took a calculus 1 class and understood everything)

[@MichaelPennMath]

You are right about Calculus 2, but in my experience this course never "gets" to the point where interesting things like this are calculated. I am a bit of an outsider (being an algebraist) but as far as I can tell stuff like this mostly has a weird internet following.

[@JB-ym4up]

How about a pointless math series like overkill. Do stuff like factoring 0 as the difference of 1 and i⁴ (1-i²)(1+i²), or a proof that for y=a1+a2...an X = kth root of y => x = y(1/k) with (an) in R and (n,k) in N that x= a1/(x^(k-1)) + a2/(x^(k-1)) ... + an/(x^(k-1)).

[@JB-ym4up]

@UCX_AAFYfjhEroxEeVkw5Ohw its defining a relationship between a kth root of the number Y which is a sum of other numbers a1 etc. Which is spilt into a number of fractions with the a1 a2 as the denominator. I have yet to find a use. a1 a2 a3 .... an are reals n is a integer Y is the sum of all a X is the kth root of Y, or Y to the (1/k) power. K is a integer. Where it says n and k are in N it means positive integers.

[@FirstLast-gl5jq]

@J B, @anyone Hmmm! - that 2nd point looks interesting, but the notation isn't clear to me. Could you ( or anyone else ) clarify it for me please.Thanks.

[@mathadventuress]

-10 points you forget a dtheta at 8:20 or so 😛

[@mathadventuress]

Lol oh I see you got it ten seconds later