Integral of (sin^n(x)+cos^n(x))^(1/n) as n goes to infinity

[@andrewhaar2815]

For those curious, to justify changing the limit with the integral we need to prove uniform convergence of the inside function to the piecewise defined sin(x) and cos(x) function. This is actually not all that hard. Call the inside function f_n and the limit function f. In the video pointwise convergence is shown, but we will show uniform convergence. First note that the uniform norm (the L^infinity norm) of f_n - f will clearly just be equal to f_n(pi/4) - f(pi/4). We can show this by taking derivatives and maximizing this function if we want to be rigorous, but I think it's decently obvious. Now we just need to show that the limit as n goes to infinity of f_n(pi/4) - f(pi/4) is 0. This is a fairly routine limit. It is limit as n to infinity of ((1/sqrt2)^n + (1/sqrt2)^n)^(1/n) - 1/sqrt2 = limit... 2^(1/n - 1/2) - 1/sqrt2 = 1/sqrt2 - 1/sqrt2 = 0 Therefore we have uniform convergence and we can switch the integral with the limit.

[@blackpenredpen]

Thanks!! This is great!

[@leefisher4418]

Dominated convergence theorem applies (sin^n(x) + cos^n(x))^n < 3 which is L^1 on [0,1]

[@paulluap3383]

Seems strange to me to go with uniform convergence in this case. Lebesgue's dominated convergence theorem will be more suitable and way more easier to use. You also can use Monotone convergence theorem in such cases.

[@paulluap3383]

Oh well, I thought about uniform convergence of an integral but then read your comment again. Uniform convergence of the function is great. The fact that you can swap integral and limit for uniform convergent sequence of functions is actually can be proved by using Lebesgue's dominated convergence theorem

[@jewlez8915]

@@paulluap3383 that would be a huge overkill to prove it like that. its literally a one line prove for the Riemann integral, which bprp is using here. No need for Lebesgue theory at all

[@SlimThrull]

Blackpenredpen: "As always pause the video and try this for yourself." Me: "Oh hell, no!"

[@markobavdek9450]

exactly

[@vivada2667]

what am i looking at 😐

[@youkaihenge5892]

Limit + Integral = Lintegral!

[@damianmatma708]

Limtegral ;)

[@itachi6336]

.

[@jevinliu4658]

Line + Integral = Linegral

[@Prxwler]

L'Hintegral

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@integralboi2900]

‘Please pause the video and try this first’ Do you expect me to be able to do this?

[@angelmendez-rivera351]

You should at least try it, it should not be too difficult if you know a few trig identities or some complex number algebra.

@@angelmendez-rivera351 That was the first idea I had to tackle the problem but ended up doing the same as done in the video. Could you give some advice on using complex numbers to solve that?

[@joaopaulorangel1010]

Strange, i found sqrt(2)/2

[@itachi6336]

I haven't been taught this

[@blackpenredpen]

Do I look half an year younger?

[@blackpenredpen]

Yes, bc this was filmed in December 2019

[@arjunswayamkumar2507]

@@blackpenredpen WOAhhh

[@akshatprakash5747]

Then why didn't you upload it then?

[@blackpenredpen]

Bc I took a break from YT and didn’t upload anything for 2.5 months.

[@eliardosoarescoelho4333]

@@blackpenredpen note 100000000000000.... integration is my live kkkk

[@thomaskim5394]

You have to show that you can interchange the order of limit and integral. If the function convergence uniformly, then you can interchange the order of the finite integral and limit.

[@steve2817]

Next Video Proof of lim n->inf (a^n+b^n)^(1/n) = max(a,b) for positive a and b... maybe

[@julespirony8748]

The reasoning misses a big thing : the proof of uniform convergence !!!! (without it you can't justify that the limit of the integral equals integral of the limit [integral of cosine + integral of sine]. Moreover the justification for value of the limit is dubious (for the first case if you reason like this then the limit of (1+1/n)^n is equal to 1 for exemple (while the right value is e)). You have to use the exponential form of the exponent 1/n (for both actually).

[@aashsyed1277]

Yes but it is okay in this case 👌

[@piyushabhay]

This is same as saying is your signature statement...I love the way u connect the dots and explains every dimension of specific topic...I salute u from the bottom of my heart

[@MacheTheFerret]

Back to the big board, I see. Stay safe, my guy.

[@rebellio619]

*yellow guy

[@washieman2445]

@@rebellio619 bruh

[@GrouchierThanThou]

Another way to think about that limit is that for any x, as n goes to infinity, the expression sin(x)^n + cos(x)^n will be dominated by whichever is the biggest of the two terms for that x. Therefore, if sin(x) != cos(x) the limit equals max(sin(x), cos(x)). At sin(x) = cos(x) you'll get a discontinuity though, as in that case the limit equals 2sin(x) instead. Of course, because of the integration, that discontinuity can just be ignored in this video.

[@zygoloid]

At sin x=cos x, you get (sin^n x + cos^n)^(1/n) = (2sin^n x)^1/n = 2^1/n sin x -> sin x, so there's no discontinuity.

[@GrouchierThanThou]

@@zygoloid Yes, you're quite right. Thanks.

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@felipelopes3171]

Why not do it directly without the tangent? The expression inside the root is dominated by the power whose base is larger, since it goes to 0 slower as you take higher powers of n.

[@turtlellamacow]

Probably for pedagogical reasons, since it reduces the problem to the basic limit laws. A similar example is (2n^2+n)/(n^2+n) -- we who are experienced can glance at this and see that the +n is irrelevant, but the "correct" way to show it is to rewrite it as (2 + 1/n) / (1 + 1/n) and then apply the limit laws.

[@buddythompson5284]

Yeah, I was thinking the same thing. Just show that for 0 to pi/4 the cos term dominates and for pi/4 to pi/2 the sin term dominates.

[@poolmorosanacona1930]

I just came to say thank you! I thought that I was never going to pass Calc II, this was going to be my third time on this course. But then I discovered your channel and it really helped me a lot, finally, I was understanding the topics and after a lot of your videos, I passed the course :)

[@blackpenredpen]

Glad to hear and BIG YAY on your success!

[@EnteiFire4]

(x^n + y^n)^(1/n) where n tends to infinity is the infinity norm, which is equal to max(|x|, |y|). So this could be written as integral from 0 to pi/2 of max(cos(x), sin(x)) dx (we can remove the absolute value because sin and cos are positive between 0 and pi/2). We can easily find that max(cos(x), sin(x)) = cos(x) for x between 0 and pi/4, and max(cos(x), sin(x)) = sin(x) for x between pi/4 and pi/2, which gives our final integral.

[@harendrayegr]

But what happens at pi/4? tan x =1.so under the root sign value becomes 2 which is not tan x raised to the power n which is 1.

[@wowZhenek]

Point pi/4 is forgotten? It does give zero in the limit though, but probably should be mentioned

[@lifeofphyraprun7601]

Oh,so Spiderman is your patron.😂

[@waitinblackout]

Quicker way is to simplify the integral by noticing cos(x)>sin(x) below π/4, thus the function can be replaced by (cos^n(x))^(1/n) = cos(x) when n goes to infinity. Same replacement with sin(x) above π/4.

[@BrendEcasa]

can you do it with a^2 in front of sin^2 and b^2 in front of cos^2? just the integral

[@craccocrai2778]

Are you ever going to start a series on calculus 3?

[@niyazkhan5789]

Please try JEE advaced integral question

[@nathanielhellerstein5871]

Lim(n->1/0) ((x^n)+(y^n))^(1/n) = max(x, y)

[@benjaminbrady2385]

I'm curious how bprp solves this, haven't watched yet. Let me give my solution though, I recognise this as the L_infinity distance to the point (sin(x), cos(x)). For those who don't know, this will just be the supremum of the two (or whichever one is bigger). The layman's way to write this could be max(sin(x), cos(x)). This means we can split up the integral into two by integrating whichever is bigger on the interval and we get: Integral from 0 to pi/4 cos(x) dx + integral from pi/4 to pi/2 sin(x) dx (these are strictly positive on these intervals so we don't need to worry about absolute values ruining our day). These are easy enough, they both evaluate to sqrt(2)/2 so the answer is sqrt(2). Did I get it right?

[@blackpenredpen]

Yes, you got it right!

[@alexismiller2349]

This was exactly my first thought!

[@hamiltonianpathondodecahed5236]

what do we mean by Linfinity distance sorry if it is a very stupid one

[@piyushabhay]

Cos decreases from 0 to pi/4 but still he took cos instead of sin while integrating from o to pi /4 Why??? Please explain Thanks in advance

[@TheOiseau]

@@piyushabhay It's not about whether cos decreases. It's about cos being _larger_ than sin on that interval. When you do the nth root of (a^n + b^n), whichever of a & b is largest will win (as n goes to infinity).

[@malawigw]

Nice book tip! It looks very good

[@petrie911]

The trig identity sin(x) = cos(pi/2 - x) shows that the integrand is symmetric about pi/4. So we can start by replacing the full integral by twice the integral from 0 to pi/4. This seems easier than trying to split the interval during the limit analysis.

[@comingshoon2717]

En este canal veo problemas más avanzados de todas las materias, por eso me gusta 💪👍 bien ahí amigo, aparte he aprendido bastante el inglés en matemáticas y cosas relacionadas 👍

[@StevenPhD4]

Mejor que unicoos o julio profe xd. Esos son un chiste comparado con estos problemas .

[@comingshoon2717]

Steven Chau see jaja

[@TheSenator007]

For any natural number k the expression 1/sqrt(k) is equal to 1/k*sqrt(k). That rule makes adding terms containing 1/sqrt(k) easier. For example in this video at the end we had: 1/sqrt(2)+1/sqrt(2) which based on this rule is equal to 1/2*sqrt(2)+1/2*sqrt(2) which is easy to compute to be sqrt(2) I found this rule very helpful, but only after I had proven it I started really using it. The proof I wrote down went as follows: Given: k is a natural number To be proven: 1/sqrt(k) = 1/k*sqrt(k) Proof: 1/sqrt(k) = (1/sqrt(k))*1 = (1/sqrt(k))*(sqrt(k)/sqrt(k)) = sqrt(k)/(sqrt(k)*sqrt(k)) = sqrt(k)/k = 1/k*sqrt(k) q.e.d.

[@sigmac30]

So happy to see this board back ! But I prefer when you do not spoil the result at the beginning, it is so satisfying when a simple answer comes out of nowhere at the end !

[@bollyfan1330]

pi/4 should be included in first interval and not second one, since tan(pi/4) = 1, raising it to nth power is still just 1. adding the other 1 inside the n th root, you get the value as n tends to infinity of the n th root of 2, which is also 1, just like the rest of the values in the interval.

[@walterpoelzing9412]

You are an incredible talent. I have a Masters Degree and you keep me at the top of my game everyday.

[@Kinghercules]

"Thats very nice!" 😄

[@alexismiller2349]

Can i just add that the integrand is basically the distance calulated in L_{infinity} norm of a point on the L_2 norm unite circle? It actually simplifies the problem because it just represents the max function :D

[@-ANaveenBagade]

So happy to see the big board after a long time.

[@iridium8562]

3.10 is cancelling the root and the power possible in every case? Because for some limits it might not be possible

[@ZackSussmanMusic]

This question was awesome!

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@forklift1712]

Your explanation that lim(x -> inf) ((tan x)^n + 1)^(1/n) = 1 when 0 inf)(1 + 1/x)^x and conclude that it is 1 and not e.

[@burk314]

The difference there is that lim(x->inf) (1+1/x)^x is 1^infty which is an indeterminate form while lim(n->inf)((tan x)^n+1)^(1/n) is 1^0=1 and not indeterminate. Now maybe the way he worded it is not the best, but there's not really a problem here.

[@winniethexiinwesttaiwan8578]

the inside part equals to max(cos(x),sin(x)), so the division boundary is very clearly pi/4

[@AhmedHan]

I can calculate it for n=2.

[@hamza_alsamraee]

Great explanation!

[@chessdominos]

I loved how you solved it. I reached the same conclusion by using the limit definition of e , exponential property and range of integration. I liked your better.

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@chessdominos]

@@georgepeterson2655 Whaaaat!? That is impressive. Thank you so much. It looks fantastic.

[@georgepeterson2655]

@@chessdominos welcome

[@thelightningwave]

I learned in my real analysis II course about norm and so our norm ||x||n= (sum from i =1 to m of (x sub i)^n)^1/n and so our usual norm is when n equals to two, which makes the infinity norm which is what black pen red pen went over is the basically equal to max{x sub 1, x sub 2..... x sub m}. (Please don't ask me to prove it, I really don't know how to prove that.) And he could have looked for the interval which cosine is the max which is [0, pi /4) and the interval which sine is the max which is (pi/4, pi/2] and then calculate the integral.

[@SylComplexDimensional]

tan(pi/2) is sin(pi/2)/cos(pi/2) & cos(pi/2) computationally is a floating point number 6.1232e-17 therefore tan(pi/2) is defined as 16331239353195370. 🎄

[@buzzfeedteen]

This channel makes me happier than it should

[@sherlockjunior8612]

This problem is a paid actor, explains how he solves everything!

[@kingbeauregard]

What I like about this problem is, I am incapable of coming up with any intuitions about the original integrand, but after factoring out the cosine term I can pretty easily see what I'm looking at. And then, MUCH later, I can look back at the original function and almost understand it intuitively: since we're talking about infinities, only the sine term or the cosine term will be significant, while the other will diminish to irrelevance. What I'm saying is, I'm a little smarter after seeing this.

[@gekkouga2828]

It's interesting and amazing 😍😍(shocked and bewildered😂😤😲) to see how such a difficult and complex (not complex numbers) problem transformed into such easy question !!!!!! You're kidding me! 😁 I mean, look at that transformation !!!!!!!💖 Truly stunning !!!😃 BlackPenRedPen, you are awesome!💞💟💕💖💓

[@mohammadfahrurrozy8082]

I know right Im a highschooler ,learned calculus and this looks so amazing

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@PureMathGuy]

I don't understand how you computed lim n->inf (tan^n(x) + 1)^1/n. If you take the limit inside there and compute it, then its similar to calculating e = lim n->inf (1 + 1/n)^n = 1^inf = 1 which is incorrect. So how do you justify this?

[@adamrjhughes]

what about +c ??

[@sinecurve9999]

This problem reminds me of the infinity-norm. Lim{p->infinity} (sum_i(x_i^p))^(1/p) = max{x_i}

[@prashantshukla6018]

You should try and make videos of prmo previous year paper u can download the pdf from Google please

[@eustacenjeru7225]

Excellent reasoning

[@divyt8446]

Welcome to another episode off every thing bounces of our brain...

[@yafqakhan3913]

What happens if you differentiate under the integral sign with respect to n?

[@Дмитрий-в2з1ч]

why u can switch the order of limits (limit and integral)? it needs proof

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@OlympicMaths]

It was interesting. Regards, Blackpenredpen!

[@criskity]

That's bizarre! I would have thought that an infinite power of a sine or cosine would look like a flat line of zero height with a blip of height 1 or -1 at every pi steps. So my intuition said the answer to this problem had to be zero.

[@stephenbeck7222]

That’s where the n-th root does a lot of work, raising all those little flat line parts up. Remember for 0

[@cable4751]

can you do the integral of the 5th root of tanx pls

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@tomkerruish2982]

Just use lim(a^n + b^n)^(1/n) = max{a,b} as n goes to positive infinity for nonnegative a,b.

[@Mustapha.Math_at_KUSTWUDIL]

In this case, What will be the max?

[@medmoufahim]

where the fountain is located please ? by the way you do a great job simpifying math (y)

[@AbhayKumar-hb7il]

Why x is not equal to pi/4 in first interval itself bcuz exact one raised to power infinity is 1 and 1 is finite Plz explain sir

[@edward_0909]

你没说到为什么 极限号 可以 “穿入” 积分号内😥

[@subhadeepsarkar5606]

hey bprp, today i watched your simple setup video.i just wanted to ask which hard drive you used for storing your videos??

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@subhadeepsarkar5606]

@@georgepeterson2655 u can use big white board. It will be very good

[@NovaWarrior77]

Good stuff!

[@gustavorc25]

Very impressive but you can resolve: ∫(tan(x)^(1/n))dx=?

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@eliardosoarescoelho4333]

Note 100000000ⁿ10100 kkkkk parabéns Blackpenredpen 👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏

[@ameerunbegum7525]

At last, you got the big board, yeah ! big is nice.

[@drpeyam]

Whoa!!!

[@blackpenredpen]

Dr Peyam thanks! Some steps apparently aren’t rigorous and unjustified but I tried!

[@xCwieCHRISx]

0:14 and 99% of people are just waiting 5sec

[@carlfarruggio3835]

Would you pleas be kind and explain to me where ever I’m going to use this... and what’s the point of it.

[@urmemlel8721]

Damn, i sounded elitst as fuck, imma delete this

[@purim_sakamoto]

やああ　理屈はわかったけど、 なんか見落としてる気がしちゃうんだけど、これで合ってるわけだねええ 無限乗のインテグラルってそうやって扱うのかあ

[@Dionisi0]

7:30 not justfied

[@angelmendez-rivera351]

The integral sequence converges uniformly, so it's fine. He didn't justify it in the video, but we know it's true. It's almost trivial to prove, so any person watching should be able to gather that much.

[@Dionisi0]

@@angelmendez-rivera351 que no justifica los valores asignados en el momento de separar la integral en 2 partes, eso lo asume el, no signifca que sea correcto

[@angelmendez-rivera351]

Zamir C No, te equivocas, él lo justificó en la sección previa. No es una asunción.

[@egillandersson1780]

Wow ! Your whiteboard grew up again !

[@kallolmanna8664]

If we chan the limit of the integral the it is possible.. that is twice the integral..

[@rezamiau]

Great! Thanks

[@insouciantFox]

For once thinking about it geometrically helped.

[@Rahulsingh-lw8hk]

hi , can you please upload a video on visualising complex roots of a cubic equation graphically ?

[@nuctang]

Whiteboard is back!

[@blackpenredpen]

This was recorded back in December.. sorry

[@mathadventuress]

Black pen red pen.... Blue pen Too much to process 🤪🤪

[@Peter_1986]

He also used a purple pen in some other video from maybe a couple years ago.

[@mathadventuress]

@@Peter_1986 oh man...

[@elizabethparnell4724]

ooo yay big whiteboard!!

[@tarunverma3335]

Limit + integral = Limtegral

[@sasoribi1341]

nice potential😎

[@UltimateGaminggaurang]

Hey! Can we solve using LOGARITHM??? Those properties make the calculations much more easier!!!!

[@oussamalaaroussi5467]

Hello, I was thinking why we forgot the +1 in the second case where tanx >=1 , and thanks

[@dlevi67]

Because tan(x) to the nth power when tan(x) > 1 goes to infinity, and infinity + 1 is still infinity, so at the limit we can neglect the 1 and you end up with nth root of [tan(x)]^n, i.e. tan(x). There's a fair amount of hand-waving in this one...

[@joryjones6808]

Watching this instead of preparing for calc 2 test tomorrow.

[@georgepeterson2655]

Just watch this Math channel... Very impressive kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@joaopaulorangel1010]

I found sqrt(2)/2

[@joaopaulorangel1010]

Oh, i Just saw my mistake

[@miso-ge1gz]

what. This video is 4 years too advanced for me

[@bandamkaromi]

Wow! Big Big Big Board.

[@thedoublehelix5661]

unjustified interchange of limits smh

[@leefisher4418]

you could apply bounded convergence theorem to interchanged the limit and integral.

[@markuschen6729]

I just have the same question. The proof of the interchangability of limit and integral is essential for the solution.

[@ishrakhameem2313]

3rd comment, love from Bangladesh 🇧🇩❤

[@phythematics2188]

Me also

[@jochoa2009]

Those small whiteboards stressed me out.

[@mamadou3076]

Blackpenredpen I've something for you Prove that x^2-y^2=3 has no solution in Q world

[@willnewman9783]

But 2^2-1^2=3

[@mamadou3076]

@@willnewman9783 yes sorry I did a mistake it is not "minus" but "plus" So the question is : Prove that x^2 + y^2 = 3 has no solution (so show that it is impossible) in the Q world

[@8nagesh]

I am new to this and I am lost in jungle😀

[@HardyPinto]

I could not resist to make a small python code that numerically calculate the integral (Euler's method) ###--- CODE BEGINS HERE import numpy as np import matplotlib.pyplot as plt #uses 10000 steps! k=10000 h=(np.pi/2)/k #stepsize! x=np.linspace(0,np.pi/2,k) #sweeps n from 1 to 998 for n in range(1,1000,2): y=((np.cos(x))**n+np.sin(x)**n)**(1/n) plt.plot(x,y) plt.axis([0,np.pi/2,0,1.5]) #Euler Integration here yi=np.sum(y*h) #Integral ei=np.abs(yi-2*np.sqrt(2)) #Absolute error ep= ei/yi*100 #relative error (percentage) print (f'{n:02d} | {yi:1.5f} | {ei:1.5f} | {ep:>8.5f} %') plt.show() ###--- CODE ENDS HERE

[@coderdemo9169]

Ohh no ☹️ 😔😟

[@Charles_Reid]

Who thinks of these problems??!

[@albertusventer2896]

cos(0) = 1 not zero :)

[@bengtbengt3850]

I found it very annoying that you just interchanged the limits without even mentioning that it’s not a trivial thing.

[@sagnikbiswas3268]

My intuition is 0

[@pabloszuban3107]

Wow

[@52.yusrilihsanadinatanegar79]

yay big board