Can you solve these 3 calculus tiebreakers the fastest? (2023 Berkeley Math Tournament)

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  • เผยแพร่เมื่อ 4 ธ.ค. 2024

ความคิดเห็น • 99

  • @blackpenredpen
    @blackpenredpen  หลายเดือนก่อน +7

    If you are participating in the BMT and you want to honor your coach, then apply for the BMT x BPRP Student Mentor Appreciation Scholarship here: berkeley.mt/news/black-pen-red-pen-award/

  • @randomjin9392
    @randomjin9392 หลายเดือนก่อน +93

    The last sum can be cheesed neatly. For n = 0 we have n³/n! = 0, so we can start with n = 1 and then reduce: n³/n! = n²/(n-1)! therefore reducing the power by 1. We then go with n-1 = k which is legit since now n starts at 1. We then get ∑(k+1)²/k! starting with k = 0. This is same as ∑k²/k! + 2∑k/k! + ∑1/k!. The last one is just e, the one in the middle is also e: discard k = 0 term (it's 0) and reindex. As for the first: we just do the same trick with power reduction again, finding it's 2e. So ∑n³/n! = 2e + 2e + e = 5e

    • @deinauge7894
      @deinauge7894 หลายเดือนก่อน +3

      or in a different way: by differentiating x e^x three times we get Sum(n^3-n)x^n/n!

    • @gargisharma6697
      @gargisharma6697 หลายเดือนก่อน

      this question was in my notes copy, im literally in 11th grade

    • @mfol2374
      @mfol2374 หลายเดือนก่อน

      Yeah, also solved in this way. Much simplier )

    • @Osirion16
      @Osirion16 หลายเดือนก่อน

      I did it the same way !

    • @anastasissfyrides2919
      @anastasissfyrides2919 25 วันที่ผ่านมา

      i was looking for this comment, this approach looked obvious to me

  • @MyzticF
    @MyzticF หลายเดือนก่อน +11

    Managed to solve all of them with relative ease - super fun exercise

  • @kornelviktor6985
    @kornelviktor6985 หลายเดือนก่อน +21

    3:45 no way, I'm in the video. It makes me so happy😊. Very clever solution btw

  • @moazamabbasi1899
    @moazamabbasi1899 หลายเดือนก่อน +22

    Hi. I solved the integral problem myself. I am in University, I saw your notification, slide it down, and started solving on my Tablet. In a nutshell I found the answer(1-π/4), when I Rechecked, it matches your answer. Very Happy.🎉😊❤

    • @IamACrafter
      @IamACrafter หลายเดือนก่อน +1

      my unborn child could do this question without a tablet. in its sleep.

    • @mohammadfahrurrozy8082
      @mohammadfahrurrozy8082 หลายเดือนก่อน +1

      congrats bro! its good to see another person who got excited after solving an integral like me (for the original comment)
      nobody asked (for the reply above me)

    • @moazamabbasi1899
      @moazamabbasi1899 หลายเดือนก่อน

      @@mohammadfahrurrozy8082 yup.

    • @YoungPhysicistsClub1729
      @YoungPhysicistsClub1729 หลายเดือนก่อน

      @@IamACrafter fr, just rationalise it

  • @IamExeller
    @IamExeller หลายเดือนก่อน +9

    Please don't retire so soon, you're a rare and excellent internet teacher.

  • @PRABALBAISHYA-xi1fd
    @PRABALBAISHYA-xi1fd หลายเดือนก่อน +4

    limits talk about the limiting behaviour, for the 1st problem its important to note that arctan x ~x as x approaches 0.( similar to sinx) so for extremely small values of x the expression reduces to 1/2, cubing it we get 1/8. The knowledge of limiting behavior saves us from the hazarad of blindy applying the differentiation rule all the times.

  • @SlipperyTeeth
    @SlipperyTeeth หลายเดือนก่อน +5

    For the last one, you can just aim to cancel the largest term of the factorial and then add and subtract the same value to cancel the next largest term, etc. until you only have constants in the numerator.
    I'll skip the bookkeeping of the index of the sum, but it works out with the first term always being 0 - allowing you to shift things up as needed and sticking to the factorial of natural numbers. Here is the math for the summand:
    n^3/n! = n^2/(n-1)! = (n^2-1+1)/(n-1)! = (n+1)(n-1)/(n-1)! + 1/(n-1)! = (n+1)/(n-2)! + 1/(n-1)! = (n-2+3)/(n-2)! + 1/(n-1)! = 1/(n-3)! + 3/(n-2)! + 1/(n-1)! = 1*e +3*e + 1*e = 5e.

    • @ars7595
      @ars7595 หลายเดือนก่อน

      Noiiiiiiiceee

  • @bot24032
    @bot24032 หลายเดือนก่อน +6

    The last one can also (somewhat tediously) be solved by changing the numerator such that it as much as possible with the denominator (n³=n(n-1)(n-2)+3n(n-1)+n lets you split the sum into multiple ones, all related to sum of 1/n! from 0 to infty; one should also take the first two terms out before doing that to not cause negative number shenanigans)

  • @Stranger-oy3iy
    @Stranger-oy3iy หลายเดือนก่อน +107

    there are too many e's

    • @sigmagamer111
      @sigmagamer111 หลายเดือนก่อน +1

      😂

    • @fifiwoof1969
      @fifiwoof1969 หลายเดือนก่อน +4

      5 of them

    • @kartikvyas3384
      @kartikvyas3384 หลายเดือนก่อน

      😂

    • @programmingpi314
      @programmingpi314 หลายเดือนก่อน

      My solution is several times faster and easier, and it only even mentions e in the final answer.

    • @Sphinxinator
      @Sphinxinator หลายเดือนก่อน

      @@programmingpi314 can I see it?

  • @Silver-cu5up
    @Silver-cu5up หลายเดือนก่อน +4

    YEEESSSSS, conjugating and using the area of a partial circle is the best and fastest way to solve it!!!!

    • @fxrce6929
      @fxrce6929 หลายเดือนก่อน +1

      hello Mr sexy speed integrator

    • @Silver-cu5up
      @Silver-cu5up หลายเดือนก่อน

      @@fxrce6929 oh shizz wasabi!! xD

  • @proximitygaming8253
    @proximitygaming8253 หลายเดือนก่อน +5

    Probably a useless observation, but #3 can be expressed as something like 1/k! times the convolution of the sum of (k-n)^3 and the sum of the gamma function, taking its limit as k --> ∞.

  • @megazebra228
    @megazebra228 หลายเดือนก่อน +1

    Thank you!

  • @yehonathanshapira7090
    @yehonathanshapira7090 หลายเดือนก่อน +1

    You can also solve the last sum using power series and solving the deferential equations you get from it

  • @scottleung9587
    @scottleung9587 หลายเดือนก่อน +2

    Nice job!

  • @taranmellacheruvu2504
    @taranmellacheruvu2504 หลายเดือนก่อน

    I did problem 1 a bit differently:
    For problem 1, if you take the cube out and write the limit as 1 over its reciprocal, you get the cube of the limit of 1/(1 + x/arctan(x)). We know tanx/x goes to 1 as x->0, i.e. tan(x) is approximately x. Then arctan(x) is also approximately x in the limit, so the original limit becomes (1/(1+1))^3 = 1/8.

  • @blauornuo7060
    @blauornuo7060 หลายเดือนก่อน

    thanks

  • @anarchosnowflakist786
    @anarchosnowflakist786 หลายเดือนก่อน +1

    great video thanks ! the solutions seem so obvious once you've shown them

  • @dayingale3231
    @dayingale3231 หลายเดือนก่อน

    In the first one i just divided up and down by arctan and use that x/arctan x tends to 1

  • @frankezendam5409
    @frankezendam5409 หลายเดือนก่อน

    Question nr 2 you can also use X=sin(t) which will lead to ∫cos(t)- (cos(t))^2 use the double formula for the second part and fill in π/2 and 0 which will give 1-π/4

  • @Metaverse-d9f
    @Metaverse-d9f หลายเดือนก่อน

    3. use the expansion of e^x to do(d/dx→*x)3 times, and plug-in x=1 will finish it.

  • @Dr.1.
    @Dr.1. หลายเดือนก่อน

    awesome video

  • @cdkw2
    @cdkw2 หลายเดือนก่อน +4

    man if only india had these, it would be so epic!

  • @wargreymon2024
    @wargreymon2024 หลายเดือนก่อน +1

    The last one is 🤯🤯🤯🤯

  • @alexkaralekas4060
    @alexkaralekas4060 หลายเดือนก่อน +2

    Number 3 can be solve using touchard polynomial

  • @A.Tripathi071
    @A.Tripathi071 หลายเดือนก่อน

    i saw wwangs solution few days ago, but this is brilliant

  • @ikarienator
    @ikarienator หลายเดือนก่อน

    note that sum(n(n-1)...(n-k)/n!) is always e, then write n^3 as n(n-1)(n-2) + 3n(n-1)+n, you immediately get the result 5e.

  • @seaassasin1855
    @seaassasin1855 หลายเดือนก่อน +1

    The third one is the most intuitive tbh, you only need to know the fact that the sum from 0 to infinity of 1/n! =e

  • @lostrxses
    @lostrxses หลายเดือนก่อน +1

    8:15 so now we can take the derivative.. so now we can take the derivative

  • @saf-q1p
    @saf-q1p หลายเดือนก่อน

    Sir can you do integral tan^-1(sqrt(x+1)) using D I method

  • @Betterthanbaybay1234
    @Betterthanbaybay1234 หลายเดือนก่อน +2

    3:03 Please teach me how you solve that

  • @CamiKite
    @CamiKite หลายเดือนก่อน

    Another easy way to solve the last sum is to play with the developpement of exp(x) to show that sum(x^n*n^3/n!)=d/dx (x*d/dx (x*exp(x)))=exp(x)*(1+3x+x^2)

  • @antonello123able
    @antonello123able หลายเดือนก่อน

    For (1): recall arctan(x)~x, when X->0: you immediately get the result. 😊

  • @CalculusIsFun1
    @CalculusIsFun1 หลายเดือนก่อน

    I find it hard to believe these are Berkeley math tournament questions. I don’t even do math competitions (too stressful) and yet I still managed all three in a combined time of 4 minutes.
    The limit was easy, the integral was something an algebra student could do (if they understood what the question was asking, which they probably wouldn’t. My point being it’s mostly area of a circle.), and the final one looks so close to the Taylor series definition of e^x it’s bound to set off alarms for anyone who’s taken calculus II.

  • @ShenghuiYang
    @ShenghuiYang หลายเดือนก่อน

    Very interesting infinite sum. This is related to OEIS A000110.

  • @awoomywang
    @awoomywang หลายเดือนก่อน

    Banger questions

  • @paulveba6225
    @paulveba6225 หลายเดือนก่อน

    I know your channel is huge, but I thought it was so cool that you can be identified by Akinator(the game).
    Thanks for doing what you do like always!❤

  • @Metaverse-d9f
    @Metaverse-d9f หลายเดือนก่อน

    b/c when x goes to 0, x/arctanx=1, expression=[1/(1+1)]^3=1/8

  • @Leticia-un8sp
    @Leticia-un8sp หลายเดือนก่อน

    I didn't get why d(e^nx/n!)/dx = ne^nx/n! and the 1/n! factor wasn't part of the diferentiation

  • @wongmanwaihehe
    @wongmanwaihehe 25 วันที่ผ่านมา

    3:42 I just watched it literally 5 minutes ago 😂

  • @Denis-bu4ri
    @Denis-bu4ri หลายเดือนก่อน +2

    Would you like to spend 5 hours solving hard integrals?

  • @CubeRex_
    @CubeRex_ หลายเดือนก่อน +2

    Literally did in mind (JEE aspirant) first I simplified it as n^2/(n-1)!
    Then wrote it as n^2-1+1/(n-1)! Then it is n+1/(n-2)! +1/(n-1)! Which is 1/(n-3)! +3/(n-2)! +1/(n-1)! And its sum is 5e

  • @kushagraverma7158
    @kushagraverma7158 หลายเดือนก่อน

    Please clarify. dy/dx Can be written as Dy. That means D=d/dx. My questions: does D in general means d/dx or it means d/dx because y=f(x)

  • @Kotuseid
    @Kotuseid หลายเดือนก่อน

    which grade are these questions for

  • @ValidatingUsername
    @ValidatingUsername หลายเดือนก่อน

    Sum(n^3)*Sum(1/n!) = Sum 1/(factor)(n-3)!

  • @Mediterranean81
    @Mediterranean81 หลายเดือนก่อน

    The +C

  • @hectorminec
    @hectorminec หลายเดือนก่อน +1

    4:32 how is that true for x=0?

    • @blackpenredpen
      @blackpenredpen  หลายเดือนก่อน

      You would get 1=1, with the convention that 0^0=1 in power series and 0!=1

  • @netanelkomm5636
    @netanelkomm5636 หลายเดือนก่อน

    7:58 Siri trying to beatbox:

  • @evancisolomon2930
    @evancisolomon2930 หลายเดือนก่อน

    I've solved your equation like 1^x = 2 without seeing your video (by seeing just question) . But I got only principal value, i.e, without getting any "n" on my answer. Could you please tell me, is it necessary to get only general solutions?

  • @ShikharGupta-tn9wp
    @ShikharGupta-tn9wp หลายเดือนก่อน +3

    Bring some more fun and hard problems....
    HERE IS A QUESTION FOR YOU!
    Find the sum of all the subsets from the set {1,2,3,4........2020} for which sum of their elements is divisible by 5.
    IT'S SUPER INTERESTING PROBLEM!!!

    • @zimamalmuntazir6258
      @zimamalmuntazir6258 หลายเดือนก่อน +1

      I know that you got this question from 3b1b the only difference is yours is upto 2022 and he's was 2000 and i believe he's answer was
      1/5(2^2000+4×2^400) so yours will be 1/5(2^2020+4×2^404)

    • @ShikharGupta-tn9wp
      @ShikharGupta-tn9wp หลายเดือนก่อน +1

      @@zimamalmuntazir6258 yeahh...
      RIGHTT!!
      IT WAS A WONDERFUL SOLUTION!

  • @giuseppemalaguti435
    @giuseppemalaguti435 หลายเดือนก่อน

    3)...derivo 2 volte e^x=Σx^n/n...S=2e+6...no,ho rifatto i calcoli S=5e

  • @vata7_
    @vata7_ หลายเดือนก่อน +2

    i wonder will you be able to do it, but, let's say we have f(x)=2^x, and g(g(x))=f(x), what's g(x)?

    • @Anmol_Sinha
      @Anmol_Sinha หลายเดือนก่อน

      Let f^1/2(x) = g(x)
      To find g(x)
      So f(x) = g^2(x)
      Put in equation you provided:
      g(g(x)) = g^2(x)
      Put t = g(x)
      g(t) = t²
      Reuse variable x (cuzwe need g(x))
      Put t = x
      g(x) = x²
      I am assuming that g(x) is bijective here

    • @vata7_
      @vata7_ หลายเดือนก่อน +1

      @@Anmol_Sinha f(x) is equal to 2^x, not x^2, and if you'll put x² into f^1/2(f^1/2(x)), you'll get f^1/2(x²) and then x⁴, not 2^x

    • @chinmay1958
      @chinmay1958 หลายเดือนก่อน

      first of all what is f^1/2(x)? is it the square root of f(x)? or just a different function altogether?

    • @vata7_
      @vata7_ หลายเดือนก่อน +2

      @@chinmay1958 yeah, you can consider it as different function, as i said, f^1/2(f^1/2(x))=f(x), how √f(√f(x)) can be always equal to f(x)? lol

    • @deinauge7894
      @deinauge7894 หลายเดือนก่อน

      i've atempted a similar question for a while: finding a continous version of f^n(x) where f(x)=e^x. (continous in n).
      Then i found that some guy named Kneser did this decades ago, but the solution is behind paywalls...

  • @tongabonga
    @tongabonga หลายเดือนก่อน +1

    😁😁 interesting problems

  • @AlyoshaK
    @AlyoshaK หลายเดือนก่อน

    I went to Cal. The math professors there are murder. They do this kind of thing for fun and most problems were much worse than these, and took quite some time for the teaching assistants to explain. A lot more time than this video. We poor science and engineering students never had it so rough.

  • @zhabiboss
    @zhabiboss หลายเดือนก่อน +1

    The last one is just EEEEEEEE

  • @henry55
    @henry55 หลายเดือนก่อน +2

    hello !!!!

  • @biscuit_6081
    @biscuit_6081 หลายเดือนก่อน

    For the first question I just said "sin=x, tanx=x and tan inverse(x) =x. Hence (x)/(x+x) so 1/2 and 1/2^3 = 1/8 😂

  • @ferashamdan4252
    @ferashamdan4252 หลายเดือนก่อน

    رائع

  • @holyshit922
    @holyshit922 หลายเดือนก่อน

    3.
    sum(x^n/n!,n=0..infinity) = exp(x)
    sum(nx^n/n!,n=0..infinity) = x*exp(x) (by differentiation once and shifting)
    sum(n^2x^n/n!,n=0..infinity) = x*d/dx(x*exp(x)) = x*(x+1)exp(x)
    sum(n^3x^n/n!,n=0..infinity) = x*d/dx((x^2+x)*exp(x))
    sum(n^3x^n/n!,n=0..infinity) = x*((2x+1)*exp(x)+(x^2+x)exp(x))
    sum(n^3x^n/n!,n=0..infinity) = x*(x^2+3x+1)exp(x)
    and then if we plug in x=1 we will get 5e

  • @robertszuba3382
    @robertszuba3382 หลายเดือนก่อน

    👍

  • @programmingpi314
    @programmingpi314 หลายเดือนก่อน

    Why did he solve such an easy problem in a ridiculous overcomplicated way?

  • @chitralekhakumari2842
    @chitralekhakumari2842 หลายเดือนก่อน +1

    INDIANS ATTENDANCE HERE 😂😂
    👇

  • @Kishblockpro
    @Kishblockpro 29 วันที่ผ่านมา

    1.3 million subs but 23k views... sad

  • @TlSoul-x1t
    @TlSoul-x1t หลายเดือนก่อน +1

    1st view 😊