a nice double sum.

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  • Опубліковано 18 гру 2024

КОМЕНТАРІ • 137

  • @seanm7445
    @seanm7445 4 роки тому +84

    The double sum of a triple integral isn’t how I expected to be spending my Saturday evening.
    Alas, here we are.

    • @nicholastessier8504
      @nicholastessier8504 3 роки тому +5

      A year later, this is how I spent my Saturday Morning!

    • @damyankorena
      @damyankorena Рік тому

      Another year lated I am spending my sunday morning

  • @LucaIlarioCarbonini
    @LucaIlarioCarbonini 4 роки тому +57

    In previous videos it has happened that I had difficulties in reading what you wrote on the blackboard, I thought it was because of the chalk colors or the illumination of the room. This video is perfectly clear. Please keep this same setting while filming, it works very well.

  • @camrouxbg
    @camrouxbg 4 роки тому +7

    Absolutely beautiful. Certainly nothing I would be comfortable trying on my own at my current level of math. But using so many different results and techniques... I just love it!

  • @kyintegralson9656
    @kyintegralson9656 4 роки тому +8

    Informative. It's straightforward to generalize from 2 infinite sums presented here to N such sums. E.g., N=3 would correspond to inf't sum over indices (m,n,k) of 1/[mnk(m+n+k)], & so on. The general sol'n is Gamma(N+1)xZeta(N+1), where "Gamma" denotes the familiar Gamma function, which is just the factorial over the natural numbers. Note that it works even for N=1, which is just the inf't sum over n of 1/n^2. Also, recall that the Zeta funct'n has a simple form over even natural numbers. E.g., for N=3 the triple sum is just (pi^4)/15, & for N=5 it's (8/63).(pi^6), etc.

  • @nombreusering7979
    @nombreusering7979 4 роки тому +34

    You can do it in another way with partial fractions, and get to the form Hn/n^2 which is 2zeta(3) (proved by another video of yours)

    • @zza7195
      @zza7195 4 роки тому

      Do you have the link of that video please ? I can't find it =(

    • @nombreusering7979
      @nombreusering7979 4 роки тому

      @@zza7195 ua-cam.com/video/5OPLW8wH_Po/v-deo.html

    • @zza7195
      @zza7195 4 роки тому +1

      @@nombreusering7979 thk

    • @gnochhuos645
      @gnochhuos645 4 роки тому +3

      Thank you obamasphere

    • @riadsouissi
      @riadsouissi 4 роки тому +2

      Thanks for pointing this out.
      This is exactly how I started it then got stuck on sum(Hn/n^2) and completely forgot that I solved it some time ago when Michael brought it up. So used the triple integral method which happens to be the same approach here, and honestly, it easier than going thru the harmonic series.

  • @BobY52944
    @BobY52944 4 роки тому +2

    Michael, I enjoy watching, learning about 10% of what you say, and learning how ignorant of high level math for the 90%. Fantastic!

  • @jpm3616
    @jpm3616 4 роки тому +48

    This is a great math channel - I don't know how I (or Google's algorithm) missed it for six months (!)

  • @nikitakipriyanov7260
    @nikitakipriyanov7260 4 роки тому

    13:20 these two integrals inside are exactly same thing. \int_0^1\frac{dy}{1-y z}=\int_0^1\frac{dx}{1-x z} , the only change is a private integration variable. No need to consider them separately.

  • @drpkmath12345
    @drpkmath12345 4 роки тому +15

    Third tool is especially useful. And I am also expecting some overkill series as well!

    • @RifatRahmanRimon
      @RifatRahmanRimon 3 роки тому

      Its a substituted version of one of the lemmas of Gamma function basically evaluated at 2

  • @DragonKidPlaysMC
    @DragonKidPlaysMC 4 роки тому +34

    Can you do a video about burnside’s lemma? And it’s applications to math olympiad it could be a part of your overkill series!

  • @azhakabad4229
    @azhakabad4229 4 роки тому +9

    Please start lectures on discrete mathematics!

  • @karolchojnacki3924
    @karolchojnacki3924 4 роки тому +9

    Such a nice sum!

  • @af9466
    @af9466 4 роки тому

    Very neat and clear. These multiple summation/integration tricks were really cool (did the transformations on my own up 'til that point, then was "watching & learning"). Like videos like that

  • @mtaur4113
    @mtaur4113 4 роки тому

    I think the change of order is more like Tonelli's theorem (it's like Fubini's Theorem but for positive functions, and it works with combinations of sums/integrals and integration w.r.t. sigma-finite measures, or whatever)
    Dominated Convergence Theorem helps you deal with +/- signs and functions that vary with the index but within a shared integrable absolute ceiling. I guess Monotone Convergence Theorem can also be applied, because adding positive terms to a sum will give you a pointwise-increasing total as n increases.

  • @lordfrogIII
    @lordfrogIII 4 роки тому

    Love your taps that cut the video so seemlessly

  • @shadali9045
    @shadali9045 4 роки тому +1

    elegant solution

  • @whiteblur13
    @whiteblur13 3 роки тому

    Still waiting for that "And that's a good place to stop"

  • @richardryan5826
    @richardryan5826 4 роки тому +1

    This video is fascinating. Forty years ago, the Dominated Convergence Theorem was not commonly taught at the undergraduate level because it is established using concepts from measure theory. Are measure theory and the Dominated Convergence Theorem currently taught to undergraduates?

  • @dankkush5678
    @dankkush5678 4 роки тому

    such a fucking fresh channel loving it man

  • @senhueichen3062
    @senhueichen3062 4 роки тому +9

    Indeed, very nice and very nice presentation. I wonder what happens if we have something similar, like triple sum.

  • @serkanbasatlk3322
    @serkanbasatlk3322 4 роки тому +1

    Crazy result, really unexpected

  • @elliottmanley5182
    @elliottmanley5182 4 роки тому +2

    Have you done something to improve the sound? This vid is much easier to listen to than previous ones. I'm glad because I love the quirky content.

  • @jimklm3560
    @jimklm3560 4 роки тому

    I think that was an incredible use of simple tools.

  • @tomoki-v6o
    @tomoki-v6o 4 роки тому +1

    A closer formula to this series for simply supported plate by the navier method

  • @adamluter
    @adamluter Рік тому

    "And that's a good place to stop".

  • @noahtaul
    @noahtaul 4 роки тому +4

    We can write 1/(m*n*(m+n)) as f(m,n)-f(m,m+n)-f(m+n,n) where f(x,y)=(x+y)/(x^2y^2). If we create a Stern-Brocot-like tree, where instead of fractions we put pairs, and (m,n) as a parent leads to (m,m+n) and (m+n,n), eventually we hit every pair of coprime numbers. If we add up 1/(m*n*(m+n)) level by level on this tree, we'll get f(1,1) minus the lowest-level leaves, so the sum is at most f(1,1)=2.
    It's exactly 2, however. If we go down any branch, there will eventually be a point where one of the two numbers will be greater than n. Let us add up f(a,b) where (a,b) is one of these points: that is, either a>n>=b and a-bn>=a and b-a

  • @davideventurini9699
    @davideventurini9699 4 роки тому +2

    Great video, simple and clear. Thank you!!

  • @aryangarg3200
    @aryangarg3200 4 роки тому +3

    Wow so early, you make excellent videos Michael

  • @nasim09021975
    @nasim09021975 4 роки тому

    Hi Michael,
    W.r.t. the double-sum that you solved in this video, here is a slight variation of it (I thought I would share with you):
    Find
    Sum(n=1,inf) Sum(m=1,inf) 1/(m*n*(m+n+2))
    The answer is 7/4
    Interesting how this one has a closed-form.

  • @slavinojunepri7648
    @slavinojunepri7648 Рік тому

    Great teaching style

  • @ranjeetsohanpal752
    @ranjeetsohanpal752 4 роки тому +2

    plz do summation of binomial coefficients also.

  • @AnishSarkarISIDelhi
    @AnishSarkarISIDelhi 4 роки тому +1

    You used twice the dominated convergence theorem, while I think in both cases it should have been easier by monotone convergence theorem. For using DCT, what would be the dominating function, in both case?

  • @satyapalsingh4429
    @satyapalsingh4429 4 роки тому +4

    Very good method of teaching . My heart is filled with joy .May God bless him .

  • @yanmich
    @yanmich 4 роки тому +1

    Very nice!!!

  • @StCharlos
    @StCharlos 4 роки тому +1

    That’s indeed awesome, amazing!

  • @kaushikmahanta5923
    @kaushikmahanta5923 4 роки тому +1

    Nice ,but we directly see Harmonics here
    S=Sigma from m=1 to infinity 1/(m^2) sigma from n=1 to infinity (1/n -1/(m+n))
    =sigma from m=1 to infinity H_m/m^2
    =2zeta(3)
    So easy

  • @112BALAGE112
    @112BALAGE112 4 роки тому +5

    What is the definition of a double infinite sum? What is the order of limits? Is it lim(sum(lim(sum(...))))? Or maybe lim(lim(sum(sum(...))? Perhaps the limit is multivariable? Is there even a difference between these?

    • @eta3323
      @eta3323 4 роки тому +4

      lim lim sum sum =?= lim sum lim sum
      There are a few theorems allowing you to interchange the middle lim and sum (in other words lim sum = sum lim). I'm pretty sure dominated convergence is one of them.
      So if the sum is absolutely convergent (sum |f| exists) the above statement is correct.
      Also sum_0^infty sum_0^infty is defined as lim sum lim sum.

  • @MrRod4000
    @MrRod4000 4 роки тому +4

    Wow
    I tried to learn this stuff 45 years ago, to no avail. I'm still confuzzled.

  • @thephysicistcuber175
    @thephysicistcuber175 4 роки тому +3

    Cool Apery sum.

  • @llawliet7177
    @llawliet7177 4 роки тому +1

    a nice math channel 👍

  • @panagiotisapostolidis6424
    @panagiotisapostolidis6424 4 роки тому +1

    i looove your videos

  • @tgx3529
    @tgx3529 4 роки тому

    Series integral....=integral series only when there is the uniform convergention. Series x^n doesn't convergent uniformly on (0,1),

  • @jiaming5269
    @jiaming5269 4 роки тому +1

    Wonderful!

  • @nasim09021975
    @nasim09021975 4 роки тому

    Hi Michael,
    Hope you get to see this comment.
    I can tell you have a passion for solving tough questions regarding limits, sums, and integrals (they are my favorite, too)
    Below I am providing one tough question that has a closed sum, thought it might interest you enough to create a video on it.
    Sorry I couldn't provide a picture of it (e.g. in LaTeX), I know that might have expressed the question better... anyway, here it is:
    Find
    Limit(n->inf) Integral(x=0,inf) [Sum(m=1,inf) (m+x)/((m^n+x^n)^n)] dx
    The answer is 3/2
    If you would like the steps to get to the answer, let me know, I can e-mail it to you.
    Thanks, and keep up the great work. Your videos are awesome.
    Cheers!

  • @Mr5nan
    @Mr5nan 4 роки тому +1

    Hi Michael, can you do a video about z-transformation application in determining series like fibonacci's?

  • @Patapom3
    @Patapom3 4 роки тому +1

    Amazing!

  • @coreymonsta7505
    @coreymonsta7505 4 роки тому

    When you used the geometric series it should (technically) be noted that the integrals were going from 0 to 1 right? @9:25

    • @Czeckie
      @Czeckie 4 роки тому

      you are right. There's small potential difficulty that the argument doesn't work when xz=1. Even though it doesn't work it doesn't matter, since whatever is the value of the function at a single point (or more generally on a null subset) - it can be even infinity, it doesn't change the integral.

  • @kaushikmahanta5923
    @kaushikmahanta5923 4 роки тому

    Basically the power of Harmonics is immense, it dominates logarithmic integrals and even such sums .We can escape those manipulations once we command harmonics

  • @bhumitbamel1748
    @bhumitbamel1748 4 роки тому +1

    Make more such videos

  • @mouadenoua6277
    @mouadenoua6277 4 роки тому +1

    Nice one!!

  • @JoonasD6
    @JoonasD6 4 роки тому

    12:20 Missed an edit there, did you? ;)

  • @liberalaccidental
    @liberalaccidental 4 роки тому

    Did the sum over three indices and got 3*2*zeta(4). I bet there is a general formula for any number of indices: t!Zeta(t+1) where t is the number of indices

  • @rishavgupta2117
    @rishavgupta2117 4 роки тому +2

    Counting this Sum other way gives
    sigma(H_n/n^2) from 1 to infinity where H_n is harmonic number from 1 to infininty
    implies zeta(3)=1/2(sigma(H_n/n^2))
    isnt it interesting🤔🤔🤔

    • @henriksensei
      @henriksensei 4 роки тому

      Yes, this is an example of a relation among so-called multiple zeta values. It corresponds to zeta(3) = zeta(2,1). See here for another proof: ua-cam.com/video/_HHwnVArDm4/v-deo.html

  • @williamhogrider4136
    @williamhogrider4136 2 роки тому +1

    This was a very nice double sum indeed 🍺🍺🍻.

  • @benjensen7251
    @benjensen7251 4 роки тому +3

    This is why I love math. That is a very nice sum!

  • @oiuhnp98hp33
    @oiuhnp98hp33 4 роки тому +1

    Nice touching the board and having the colour appear.

  • @RajamQED
    @RajamQED 4 роки тому +1

    Greetings from Chile. I wonder what's the thought process to change a fraction to an integral to make the problem solvable O_o

  • @doodelay
    @doodelay 4 роки тому +2

    Michael I'm glad you're teaching at a slower pace now, in the past I've had to slow you down to .85x in order to follow without proof by intimidation lol
    I do have a question, u never explained why we're evaluating this as an integral to begin with. And why are we using those specific tools? Some of this stuff I can gather on my own but it'd be nice to hear the logic explained!

    • @camrouxbg
      @camrouxbg 4 роки тому

      Starting at 6:40 he explains why we are using integration. Basically it is something a little easier to work with, oddly enough.

  • @a-manthegeneral
    @a-manthegeneral 4 роки тому

    That editing though

  • @andy_lamax
    @andy_lamax 4 роки тому +3

    I was hopping for PI or e to pop outa no where. Guess not this time then

    • @emanuellandeholm5657
      @emanuellandeholm5657 4 роки тому

      wikimedia.org/api/rest_v1/media/math/render/svg/b786732ab9ec65d7792e594026e43d1cb659188c

  • @ayushbanerjee1187
    @ayushbanerjee1187 4 роки тому +7

    Isn't the Riemann zeta function at 3 equal to apery's constant?

  • @asifiqbal.apurba.7
    @asifiqbal.apurba.7 3 роки тому

    Awesome! 😍🔥

  • @tomoki-v6o
    @tomoki-v6o 4 роки тому

    At the start first thing came up to mind iz the reiman zeta function

  • @rickshaw1641
    @rickshaw1641 4 роки тому +1

    Nice video. It's a toughest question.

  • @elliottmanley5182
    @elliottmanley5182 4 роки тому +2

    This is incredibly counter-intuitive to me. You start with a discrete function on a rational domain with a lower bound of 1, apply a bunch of contiinuous transformations with real domains and end up with an analytically continued function with a complex domain that excludes {1}.

    • @SugarBeetMC
      @SugarBeetMC 4 роки тому

      It doesn't need to be analytically continued to be defined at 3.

    • @elliottmanley5182
      @elliottmanley5182 4 роки тому

      @@SugarBeetMC true but picky! :)

    • @UnCavi
      @UnCavi 4 роки тому

      Why should it be a problem that the Riemann zeta function is not defined in 1? Zeta(3) doesn't care what happens at 1

  • @tioa.p.1058
    @tioa.p.1058 3 роки тому

    Is there any other solution without using integral? if there is, can someone explain it to me?

  • @zeravam
    @zeravam 4 роки тому

    Hello my friends. I come from the future and Michael is going to coin a catchy phrase in the end of his videos "This is a good place to stop"

  • @linggamusroji227
    @linggamusroji227 4 роки тому

    Let n be fixed, then
    1/mn(m+n)=1/n[1/m(m+n)]=1/n^2[1/m - 1/(m+n)]
    Sum 1/mn(m+n) for m=1 to inf
    =1/n^2 [Sum(1/m-1/(m+n)) for m=1 to inf]
    =1/n^2 [1/1-1/(n+1)+1/2-1/(n+2)+...+1/n-1/(n+n)+1/(n+1)-1/(n+n+1)+...], this is telescoping series
    =1/n^2[1/1+1/2+1/3+...+1/n]
    =Hn/n^2
    And it converges to 2Zeta(3)

  • @mtaur4113
    @mtaur4113 4 роки тому

    TL;DR - There is no closed form expression involving pi, exp, trig, inverse functions, roots, and arithmetic, but there is a much nicer infinite single-sum form.

  • @pikkutonttu2697
    @pikkutonttu2697 4 роки тому

    Is the sum equal to sum of H_n/n^2, where H_n is the nth harmonic number.

    • @pikkutonttu2697
      @pikkutonttu2697 4 роки тому

      I see that someone else noticed the same thing.

  • @Manuel-pd9kf
    @Manuel-pd9kf 4 роки тому

    everyone gangsta till he pulls out the triple integral 😳😳😳😳

  • @void7366
    @void7366 4 роки тому +1

    Hey @Michael Penn , i'm a big fan of your channel , i was wondering if you could solve a problem from a math contest in my school , the problem reads : a , b , c are three positive real numbers , prove that
    2a/(3a²+b²+2ac) + 2b/(3b²+c²+2ab) 2c/(3c²+a²+2bc) ≤3(a+b+c)

    • @thiantromp6607
      @thiantromp6607 4 роки тому +1

      v o i d the inequality doesn't appear to hold. For a=b=c the left hand side becomes 1/a, and the right hand side becomes 9a. That doesn't hold for all real positive a. Just pick a small enough number.

    • @looming314
      @looming314 4 роки тому +1

      do you actually mean 2a/(3a²+b²+2ac) + 2b/(3b²+c²+2ab) 2c/(3c²+a²+2bc) ≤3/(a+b+c)? if so, you just need to notice that 2a/(3a²+b²+2ac)≤1/(a+b+c) by plain calculation and that's the end of story

    • @prasannviswanathan6294
      @prasannviswanathan6294 4 роки тому

      @@looming314 Yeah she mostly meant that.

    • @thiantromp6607
      @thiantromp6607 4 роки тому +1

      For completeness, if you meant to divide by (a+b+c), the steps would be as follows. We know squares are nonnegative, so
      (a-b)²≥0,
      a²-2ab+b²≥0,
      a²+b²≥2ab,
      3a²+b²+2ac≥2a²+2ab+2ac, (added 2a²+2ac)
      3a²+b²+2ac≥2a(a+b+c),
      1/(a+b+c)≥2a/(3a²+b²+2ac), where we can divide because the variables are all positive.
      We can repeat the steps for the other two terms and add the inequalities to finish the proof.

    • @heejune319
      @heejune319 4 роки тому

      2a/(3a^2 +b^2+2ac) = 2/(a+b^2/a + 2a + 2c) =< 2/(2b +2a +2c) by the AM-GM inequality.

  • @GG-tw8rw
    @GG-tw8rw 4 роки тому +1

    Can anyone tell me the solution of this functional equation or its source please :)
    The problem:
    Let α,β ∈ Q* . Find all functions f: R ->R
    Such that:
    f([x+y]/α)=[f(x)+f(y)]/β

    • @lubosdostal8523
      @lubosdostal8523 4 роки тому

      Solution:
      If α≠β, then f(x)=0 only. [f(0)=0, f(1/α)=C, f(1)=C/β, f(2)=2C/β, f(2/α)=2C, ... by induction f(k)=kC/β and f(n/α)=nC. Use α=p/q for either C=0 or contradiction with α≠β. You can choose f(x/α)=C, f(x)=C/β for arbitrary x and repeat this argument. So f(x)=0. ]
      If α=β, then let B is a base of Q in R. I.e. B is a uncountable set, such that every real number can be written a FINITE linear combination r = q_1*b_1+...+q_n*b_n with rational coefficients. Then for every element b∈B choose arbitrary coefficients c. This give you all possible solutions f(r)=f(q_1*b_1+...+q_n*b_n)=q_1*c_1+...+q_n*c_n.
      Example choice b_1=1, c_1=1, and for all other b∈B/{1} choice c=0:
      f(x)=x for rational and f(x)=0 for irrational.

    • @GG-tw8rw
      @GG-tw8rw 4 роки тому

      @@lubosdostal8523 thanks
      I'm sorry I forgot to write that α≠β
      But I have a few questions
      I think when you plug x=y=0 we will get
      That βf(0)=2f(0)
      I think we should discuss two cases when f(0)=0 & β=2
      And one more question what is c and what did you plug to get f(1) & f(1/α)
      I'm so sorry for asking a lot of questions 😅

    • @lubosdostal8523
      @lubosdostal8523 4 роки тому

      ​@@GG-tw8rw Ok, you are right, the case β=2 has to be done separately... Its similar to the case α=β.
      The case β≠2: Sorry for previous typoes. Set f(α)=C and try to insert possible x,y:
      1 = (α + 0) / α => f(1) = (f(α) + 0)/β = C/β
      2 = (α + α) / α = (2α + 0) / α => f(2) = (f(α) + f(α))/β = 2C/β = (f(2α) + f(0))/β = f(2α)/β => f(2α) = 2C
      3 = (2α + α) / α = (3α + 0) / α => ...
      ...
      0 = (x + (-x)) / α => ...
      ...
      For α=p/q then: ? = f(p) = f(qα) = ? ... => C=0 and f(x)=0 for x ∈ Q
      ...
      Start again with arbitrary irrational r ... f(r)=D to show that D must also be 0.
      And finish it by yourself, you can do it 😅

    • @GG-tw8rw
      @GG-tw8rw 4 роки тому

      @@lubosdostal8523 thank you so much 😍❤

  • @balazsgyekiczki1140
    @balazsgyekiczki1140 4 роки тому

    Sick

  • @ashutoshtiwari3129
    @ashutoshtiwari3129 2 роки тому

    2 × Apery's constant = sum( lnx/x^2), x=1 to infinity.. 😎😌

  • @FrozenArtStudio
    @FrozenArtStudio 4 роки тому +1

    Thank you for the video!
    By the way, right now I'm struggling with an infinite sum of square powers of x, |x| < 1 (1 + x + x^4 + x^9 + x^16 + ...)
    Do you think there is a way to represent it as some kind of integral or connect it to another sum?
    Any thoughts appreciated)

    • @riadsouissi
      @riadsouissi 4 роки тому +1

      Maybe Google Theta function or Jacobi Theta function

    • @smiley_1000
      @smiley_1000 4 роки тому +2

      @@angelmendez-rivera351 this is certainly an amazing first step, but I think it would be needed to continue to transform this sum into a closed form

  • @elaadt
    @elaadt 3 роки тому

    Still looking for a good place to stop.

  • @giuseppemalaguti435
    @giuseppemalaguti435 4 роки тому

    A me risulta.... Sommatoria m from 1 to infty di (1/m^2)×Hm...È corretto? Non riesco a semplificarlo ulteriormente. GRAZIE

  • @duckymomo7935
    @duckymomo7935 4 роки тому

    I remember this from StackExchange

  • @UnforsakenXII
    @UnforsakenXII 4 роки тому +4

    The sort of calculation you'd see in string theory. = )

  • @phat5340
    @phat5340 4 роки тому

    Nice one

  • @MathHammer
    @MathHammer 2 роки тому

    What!? Not a good place to stop? 🤣

  • @b43xoit
    @b43xoit 4 роки тому

    So about 2.404. My guess had been about 2.

  • @777MrSteve777
    @777MrSteve777 4 роки тому

    Thanks 4 u :) U does not speak as fast as usualy eng speakers speak, so it is easier 4 people from other coutries to understand😁 sry 4 my english)) Привет из России!😘
    Can you explain the proof of Stirling's approximation?

  • @savantshuia
    @savantshuia 2 роки тому +1

    2.388536

  • @math_infinity_
    @math_infinity_ 4 роки тому

    Super

  • @mathissupereasy
    @mathissupereasy 4 роки тому

    You missed the thing " and this is a good place to stop."

  • @johnloony68
    @johnloony68 4 роки тому +2

    He's driving me up the wall because he keeps on saying "zee" instead of Z

    • @camrouxbg
      @camrouxbg 4 роки тому

      It is hard on Canadians for sure.

    • @face2708
      @face2708 4 роки тому

      Canadians need to cope lmao

    • @face2708
      @face2708 4 роки тому

      I’m sorry that was mean

  • @kqp1998gyy
    @kqp1998gyy 4 роки тому +1

    Great as usual. One remark: when watched landscape on my phone the quality is poor.

  • @ВикторМиллер-ь7л
    @ВикторМиллер-ь7л 4 роки тому

    11 обезьян поставили дизлайк XD

  • @ashishchoudhary594
    @ashishchoudhary594 4 роки тому

    End is anticlimactic. I thought some π type thing will show up.

  • @plushbeery
    @plushbeery 4 роки тому +2

    1st

  • @giuseppemalaguti435
    @giuseppemalaguti435 4 роки тому

    A me sembra corretto e l'ho fatto in 3 passaggi... Complimemti

  • @Cyrcyr67
    @Cyrcyr67 4 роки тому

    Nice one

  • @mathissupereasy
    @mathissupereasy 4 роки тому

    You missed the thing " and this is a good place to stop."