these numbers are "best friends"

[@okpgamingdk1093]

I think the method used to evaluate the first integral with tan^(n-2)(x)sec^2(x) was a bit overkill. The substitution u=tan(x) yields the same result almost immediately.

[@mahmoudalbahar1641]

I completely agree with you because I have done it in mind and I intended to comment but you already commented about it.

[@martinrosol7719]

​@@mahmoudalbahar1641junko

[@bigbadbith8422]

It’s also the way that I can get the right answer - but that’s for me to sort out😊

[@terryendicott2939]

I too was going to comment on this. Actually I kept yelling at my monitor as he was slogging the poor integral apart. I don't think he heard me. Maybe I should yell louder?

[@GiornoYoshikage]

​@@terryendicott2939this could be a demonstration of integration by parts. Videos are not about just problem solving, also they're about featuring different techniques, as somebody once said

[@gael8828]

Wolfgang gullich! Glad to see this reference

[@ethandasilva8243]

"And they were roommates..."

[@trueriver1950]

to me ln sqrt 2 seems to need simplification to (ln 2)/2. Yes it's the same thing, but back in the day when we'd need to look these things up in tables, the simplified version is less mental arithmetic. In fairness, the way Michael wrote it makes it look closer to the way the number turned up in this video.

[@aadfg0]

No wonder I see these numbers together in the answers to some flashy integrals.

[@edmundwoolliams1240]

Couldn't have you found the last result just using the polar representation of 1-i, then the result follows immediately?

[@TheBillBomb]

== root(2) exp(i pi/4) So ln(1-i) = ln(root(2)) - i pi/4

[@holyshit922]

4:05 Integration by parts ? Not t=tan(x) substitution

[@DOROnoDORO]

In Portuguese, the words for "prime" and "cousin" are the same, so that first bullet point isn't easily translatable

[@aniruddhvasishta8334]

To prove the second bullet point you could've used the first one. If you assume that the limit \lim_{n\to\infty} \int_0^{pi/4} \tan^n(x)dx exists and is equal to some value L, then we get L = lim_{n\to\infty} 1/(n-1) - lim_{n\to\infty} \int_0^{pi/4} \tan^{n-2}(x)dx. The first limit is 0, and the second limit is -L, so we get L=-L, giving L=0. Proving that the limit exists and doesn't diverge is a separate issue I guess, but before watching the proof I thought you were going to do it that way. Actually after thinking about it for a few seconds, showing that the limit exists should be trivial because \tan^n(x) is bounded by 0 and 1 on the interval [0,\pi/4] so for any n, the integral is at most \int_0^{\pi/4} 1 dx = pi/4 and is at least 0. Therefore, it doesn't diverge so L exists and is equal to 0. Edit: This proof is flawed because divergence isn't the only way for the limit to not exist. You also have to show that the integral is decreasing in n, which is actually a bit of work. Edit: Also at the very end you used the formula \sum_{n=0}^\infty z^n = 1/(1-z), but this only seems to be true for z where |z|

[@evanoneale9604]

For the last part: since we are integrating z between 0 and i, can't we say that 0

[@manstuckinabox3679]

24:55 or you could just memorize the formula to the *Principle* branch of the logarithm, you should've mentioned that this is the principle branch of the logarithm. Awesome vid as always.

[@miraj2264]

At 23:10, we can also use the complex log formula. log(z) = ln(|z|) + i*arg(z). First factor -1 out to get a similar form to the video: -ln(sqrt(2)) + i*pi/4 = - [ ln(sqrt(2) + i*(-pi/4) ]. In other words, we want to find a -log(z) where z has magnitude sqrt(2) and has an angle of -pi/4. From the unit circle, we know cos(-pi/4) + i*sin(-pi/4) = sqrt(2)/2 - i*sqrt(2)/2. But this obviously has magnitude 1 since it's from the unit circle. So multiplying by sqrt(2) to get the needed magnitude, we get 1 - i. Plugging that value in for z, we get -log(1-i). Neat trick at 6:00 for that integral. I've only ever seen it done as a u-sub. U-sub clearly easier in this particular instance, but nevertheless a trick I'll try to remember should it ever end up helping with a different integral :)

[@alexfekken7599]

I would start with ln(i+i) to have both of them pop up straight away....

[@ahmadkalaoun3473]

Here's an alternative proof for the second identity, which i think is also interesting; •Let a be an arbitrary number between 0 and pi/4 , born exclusive, break the integral into the sum of two pieces : - An integral from 0 to a ; this is smaller than a(tan a)ⁿ. -An integral from a to pi/4 ; this is less than pi/4 - a . • On taking the limit of infinite n , we see that the desired limit (which is of course positive) is less than pi/4 - a , which can be made arbitrarily close to zero , our limit is consequently null .

[@Egestus18]

Wouldn't it be enough for the second Integral to see that 0

[@andrewkarsten5268]

No, because other functions satisfy that which do not satisfy the limit, too such as y=1. What is enough, and I think you’re trying to point at, is that 0

[@diniaadil6154]

For both question you could have used a substition u = tanx. For the 2nd, it would give int u^n/(1+u²) du , which clearly goes to 0

[@deslomator]

That thumbnail got a like before watching

[@goodplacetostop2973]

25:21

[@mohamedbouloud7033]

BRO YOU HAVE NO LIFE

[@goodplacetostop2973]

@@mohamedbouloud7033 Yes 😎

[@mohamedbouloud7033]

@@goodplacetostop2973 AND THAT'S A GOOD PLACE TO REPLY

[@samueldeandrade8535]

​@@mohamedbouloud7033 man, don't be annoying.

[@Neodynium.the_permanent_magnet]

What is the point of marking the end of the video?

[@Archimedes_Notes]

These are close friends. They scare in the rxams

[@mrpsychodeliasmith]

Why not use Ln(4)/4 instead of Ln(Sqrt(2)) then the similarity to pi/4 looks stronger!

[@jamiepianist]

The thumbnail game is getting better 🔥

[@driksarkar6675]

For the second result, could you also use the first result? If the limit is L, then L=lim (n->inf) (1/(n-1))-L=0-L, so L=0.

[@kiranduggirala2786]

This idea is very interesting; some time back I remember thinking of some similar procedure that produces the same result with slightly easier integrals imo. They were integrals I_n = int_0^1 x^n/(1+x^2)dx. Just like in this case, you get the recursion I_n = 1/(n-1) - I_{n - 2} by a very simple integral trick (i.e add and subtract x^{n -2} in the numerator) and you get lim_{n \to \infty} I_n = 0 by DCT. So the rest of the procedure is the same. I assume the two procedures are the same up to some substitution, but this way was easier for me to think of.

[@nicholaselias9312]

A simple u=tan(x) substitution would have been much simpler.

[@hqTheToaster]

The (Pi^(1/2)/2) th order derivative of a smooth function at x = 1/e is ln(sqrt(2)) where y = 1 at x = 0 . Joking . I like this video.

[@bennyloodts5497]

I agree: they are cousins!

[@eiseks3410]

Brilliant

[@mohamedbouloud7033]

MICHAEL NEVER FAILS TO MAKE MATH EAS(IER)

[@bjornfeuerbacher5514]

Actually, as already noted by others, when he evaluated the first integral with tan^(n-2)(x)sec^2(x), he made the math much harder than necessary. ;)

[@mohamedbouloud7033]

@@bjornfeuerbacher5514 but he looks for content not the elegant answer

[@robertveith6383]

Stop yelling your post in all caps.

[@roberttelarket4934]

Are you Mike our best mathematician?

[@purplerpenguin]

Hmm. Not sure what deep meaning you see in this?

[@samueldeandrade8535]

Did you watch his video saying something like "Lucas and Fibonacci numbers are trigonometric functions?"?

[@benardolivier6624]

I don't get the whole part starting at 2:43... the derivative of tan^(n-1)x is (n-1)*tan^(n-2)x*sec²x*dx so you just need to evaluate tan^(n-1)x/(n-1) between 0 and pi/4 which just gives 1/(n-1).

[@adityaekbote8498]

Noice

[@andyiswonderful]

Interesting. Do you have a copy of Gradshteyn and Ryzhik's Table of Integrals? I find it fun to peruse, and see the surprising results of some weird integrals.

[@quiveror]

8:46 OK, so I think that's pretty good. 😂

[@franzlyonheart4362]

I had to laugh as well. It's horrible!

[@edmundwoolliams1240]

Why didn't you just use the MacLaurin series of arctan and ln to get those sum results much faster that you spent 20 minutes deriving (albeit very rigorously!)?

[@karlmarxsteingoldberg-kike4046]

4:05 how real men integrate tan^(n-2)x * sec^2x

[@twelvethirteenyo]

Bruh, But how far can you Dyno??

[@chessematics]

2:49 that's some real overkill

[@AriosJentu]

Was the IBP the easiest way to achieve this result? I think it will be much easier with moving sec^2 into differential component as tangent, and then integrate as power function

[@cdkw2]

7:15 nice transition

[@saijan2118]

Why are titles being changed? Not too long ago it was "cousins" not "best friends"

[@SaidVSMath]

Fantastic!!

[@scebsy6524]

adam ondra !!

[@samueldeandrade8535]

What "adam ondra" means?

[@scebsy6524]

@@samueldeandrade8535 He's the rock climber in the thumbnail

[@samueldeandrade8535]

@@scebsy6524 oh ok. It's a name. Adam Ondra. I thought your saying something in Latin or some other language.

[@eduardochappa4761]

This video should be removed for several reasons. Unfortunately the end of the video is not good. The unjustified exchange of the order of the sum and the integral when one is integrating in an unspecified path from 0 to i, where the limit of integration is at the boundary of the circle of convergence, and the use of a natural logarithm function, which does not exist in the complex plane is too much for me. There is an analytic logarithm function, Log(z), not ln(z), and one can compute Ln(1 - i) and get the answer Michael got, just by using the definition of logarithm, but this is not the way to do it. Michael, keep posting videos, I like them. This just was not a good one.

[@gregsouza7564]

1. When he put the evaluation bar, it was not term by term it was for the whole sum so there was in fact NO exchange of order of the sum and integral. 2. Paths are left unspecified all the time. What is the integral of f(x) from 0 to infinity? You would have no problem with that even though it could be following the curve t+isin(t)/(t+1) for t in [0,infinity) or literally anything else 3. Given that the geometric series at i does not grow without bound and the fact that any single number contributes 0% of the integral, someone with your amount of knowledge should be able to tell that it doesn't matter. 4. Oh oh, you're being elitist! Maybe it doesn't matter for a fifteen second clip at the end of a video that was not about this integral to capitalize the L on some random function. You know the zeta function? Mathematicians seem to have no problem writing that for both the sum and the analytical continuation of the sum so clearly it is posible. I know sometimes it is inducing to have a fit on the internet but please, saying that the video "should be removed" for minute technicall details (on a section that I will remind you, the video is not about) to a content creator that you seem to like, is not the nicest thing.

[@edcoad4930]

Can one just say that tan x < 1 [0, pi/4) therefore tan^nx -> 0 as n -> inf?

[@robertveith6383]

Tangent is a function. Put x inside of a grouping symbol: tan^n(x).

[@edcoad4930]

@@robertveith6383 I mean, as the function is always less than one, raising it to any power reduces it and raising it to inf -> 0. Not sure why the video was needlessly complicated.

[@vekyll]

@@edcoad4930 The video absolutely is needlessly complicated (after all, it's _just_ a polar form of (1-i)⁻¹), but your reasoning is not correct, since the function might be nonuniformly less than 1. For example, 1-1/n < 1 for all n, but raising it to n and taking the limit doesn't give you 0, but e⁻¹.

[@edcoad4930]

@@vekyll even though the function is > 0 for all n and therefore tending to zero for n = inf?