A RIDICULOUSLY AWESOME FRACTIONAL TRIGONOMETRIC INTEGRAL!!!

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  • Опубликовано: 6 сен 2024
  • WOW!!!
    WHAT AN INTEGRAL!!!
    Adding trigonometry to the fractional part mix was definitely an awesome idea.
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Комментарии • 61

  • @manstuckinabox3679
    @manstuckinabox3679 Год назад +15

    BRO THIS WAS FREAKING INSANE! each move was so unexpected lol! thank you for reminding me that I skippped the infinite product chapter from my complex analysis book.

  • @jonsmith8579
    @jonsmith8579 Год назад +19

    Thank you Maths 505 for solving the integrals

  • @mikecaetano
    @mikecaetano Год назад +5

    Logging the product to obtain a series was a really nice touch.

    • @maths_505
      @maths_505  Год назад +1

      Thanks. Playing around with infinite products often yields surprisingly beautiful results

  • @bhalchandraganeshbhatstude8290
    @bhalchandraganeshbhatstude8290 Год назад +15

    69.420? You sir are the Shelon musky of calculus. I wrote jee in 2002. Got a 4200 rank which wasn’t good enough to go to any iits. But I get a ton of joy watching your videos. Hats off to you!

    • @maths_505
      @maths_505  Год назад +4

      I figured I should use the fundamental constant of nice for a change😂

    • @arxalier2956
      @arxalier2956 Год назад

      Sir I am also JEE aspirant, was 4200 rank below cutoff in 2002?

    • @bhalchandraganeshbhatstude8290
      @bhalchandraganeshbhatstude8290 Год назад +1

      @@maths_505 I’d call it the bronze ration, if there isn’t one already

    • @bhalchandraganeshbhatstude8290
      @bhalchandraganeshbhatstude8290 Год назад +2

      @@arxalier2956 I’m so sorry, I have been out of touch with jee for 20 years, this channel brings back those good memories. Back in 2000, you had to have a sub 4000 rank to get anywhere, even in ism dhanbad as it was called then. I lived very close to iit Mumbai, and you needed a sub 1500 rank to get in the integrated ms in civil engineering.😅. So I went to a local engg college called Spce, 2 yrs with infosys, and pgp mba from Iim ahd. Then I emigrated to USA, now a citizen after 12 yrs. I wish you well on the journey. Remember, when you get there, know where you came from!

    • @Aditya_196
      @Aditya_196 3 месяца назад

      Damn awesome bro 😎

  • @user-nt7cg6ok6f
    @user-nt7cg6ok6f Год назад +4

    Your channel is like a breath of fresh air for me)
    Thanks!😄

  • @monikaherath7505
    @monikaherath7505 Год назад +7

    OMG I almost did this myself except I made the mistake of thinking you could telescope it instantly (mistaking the initial k as a constant) getting pi/2 -1/2 ln2, although I felt intuitivey that it was wrong lol thanks for the proper solution I feel I'm improving just by watching your vids

  • @hyperpsych6483
    @hyperpsych6483 Год назад +3

    It took me a solid few seconds to realize why you chose that particular decimal 😂

  • @martiribapons
    @martiribapons Год назад +3

    Simply brilliant!!

  • @cicik57
    @cicik57 8 месяцев назад

    right it is standard strategy when you have floor function integral_, worth to remember

  • @nickruffmath
    @nickruffmath Год назад +9

    Great solution from start to finish! Loved the sinh result for log..reminds me of using the sin version and taking the logarithmic derivative to get a cotangent generating series for even values of the Zeta function.
    Your approach to the frac/floor makes me wonder if it generalizes. It seems like a good strategy to take the inverse function of whatever's under the floor operator, then run thru the integer inputs to that inverse function.

    • @maths_505
      @maths_505  Год назад

      I thought the same while penning this down.

    • @immortalking.youtube
      @immortalking.youtube 8 месяцев назад

      @@maths_505 bro do there is any to evaluate this for every power of tanx

  • @vicentepoblete6627
    @vicentepoblete6627 Год назад +1

    brilliant example of the fractional part

  • @krishgarg2806
    @krishgarg2806 7 месяцев назад

    I was able to get to the ln series by myself, and thats where I got stuck. amazing integral!

  • @cristoduke8953
    @cristoduke8953 Год назад +1

    Love these interesting integrals

  • @fartoxedm5638
    @fartoxedm5638 Год назад

    did pretty the same with the original remainder function and bunch of telescoping sums and limits, but yours looks much cleaner!

  • @Noam_.Menashe
    @Noam_.Menashe Год назад +2

    Alternatively for the sum you could have used the sums you got for cot(x), put in an imaginary x and integrated.

  • @antonyrodriguez5621
    @antonyrodriguez5621 Год назад +1

    Cannot lie...need an integral a day now!

    • @maths_505
      @maths_505  Год назад +1

      Inshallah. I try to upload everyday or atleast 5 videos a week.

  • @Maths_3.1415
    @Maths_3.1415 Год назад +1

    Nice integrals bro 🎉

  • @nimrodkadish
    @nimrodkadish Год назад

    So cool! Now do the same integral, just with the integrand being {cotx}, please

  • @MrWael1970
    @MrWael1970 Год назад

    Really Smart solution. Thank you.

  • @palestinemorocco1920
    @palestinemorocco1920 7 месяцев назад

    Amazing ❤

  • @Mephisto707
    @Mephisto707 Год назад

    Beautiful result indeed!

  • @slavinojunepri7648
    @slavinojunepri7648 Год назад

    This is so cool, my goodness!

  • @suvosengupta4657
    @suvosengupta4657 Год назад

    Thats some legit example for floor function

  • @jonathandawson3091
    @jonathandawson3091 Год назад

    Nice video. Had to speed up to 1.5x to follow.

  • @user-uf8qy1ur4c
    @user-uf8qy1ur4c Год назад +1

    Quite enlightening !

  • @niom9446
    @niom9446 Год назад

    crazy

  • @neomooooo
    @neomooooo Год назад

    Beautiful problem, elegant solution!
    I notice the numerical value is strikingly close to (1/2) ln(2 pi). Is this a coincidence?

    • @rafiihsanalfathin9479
      @rafiihsanalfathin9479 Год назад +1

      If we subtract 1/2*ln(2pi) from the result and uses the sinhx expontential form, we will get the difference is pi-1/2*ln(e^(2pi)-1), the reason why they are so close is because if we remove -1 in the ln (which is very small difference because it is inside a logarithm), we get the difference is exactly 0 because pi-1/2*2pi=0

    • @maths_505
      @maths_505  Год назад

      Noice

  • @anestismoutafidis4575
    @anestismoutafidis4575 Год назад

    Intg => π/4• dx

  • @i1C4RU5
    @i1C4RU5 Год назад

    What level of calculus is this? The most I've taken is AP Calc BC in high school, I'm wondering what this compares to that, because this ALMOST seems like something I could do, but there are a few things I don't recognize.

  • @sagbansal
    @sagbansal Год назад

    can you solve e^(1/x) ln(x) Already wasted 3 hrs, got nothing.

    • @maths_505
      @maths_505  Год назад

      I think you'll need a special function for that

  • @roshanmadhav8876
    @roshanmadhav8876 Год назад +1

    Can you solve integral of x^(1/x) from 0 to 1 ?

    • @maths_505
      @maths_505  Год назад

      That's a Bernoulli integtal. I think Dr. Trefor already made a video on that.

    • @roshanmadhav8876
      @roshanmadhav8876 Год назад

      Are you able to link the video having trouble finding it ?

    • @Maths_3.1415
      @Maths_3.1415 Год назад

      @@roshanmadhav8876
      ruclips.net/video/PxyK_Tsnz10/видео.html

    • @roshanmadhav8876
      @roshanmadhav8876 Год назад

      So is there anyway to solve the integral x^(1/x) 0 to 1?

    • @maths_505
      @maths_505  Год назад

      ruclips.net/video/PxyK_Tsnz10/видео.html

  • @achrafsaadali7459
    @achrafsaadali7459 Год назад

    Can you try to make an axproximation for the catalan constant ?

  • @trampoukisma4531
    @trampoukisma4531 Год назад +1

    My dumbass thought it was smart to cancel the tanx💀💀💀

  • @endersteph
    @endersteph Год назад

    I do not understand your explanation for sin arctan x = x/sqrt(1+x²)

    • @gamersocks2741
      @gamersocks2741 Год назад +1

      If tan(φ) = y, then this means that we can represent this as a right-angled triangle with height y and base 1 (remember that tan(x) = OPP/ADJ).
      From this we can conclude that the hypothenuse needs to have lenght L = sqrt(1²+y²) (from phytagorean theorem).
      Thus sin(φ) = sin(arctan(y))= OPP/HYP = y/sqrt(1+y²)

  • @alvarolouzi
    @alvarolouzi Год назад +1

    69.420? Really? 😂

    • @maths_505
      @maths_505  Год назад +1

      Whaaaat....
      Its nice 😂

  • @sagbansal
    @sagbansal Год назад

    I am amazed