Approximating The Cos Function Challenge

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  • Опубліковано 22 гру 2024

КОМЕНТАРІ • 277

  • @jawadbenbrahim5933
    @jawadbenbrahim5933 Місяць тому +1241

    Now prove that as R→∞, your function becomes the cos function.

    • @SilentALume
      @SilentALume  Місяць тому +195

      @@jawadbenbrahim5933 my function as a much smaller infinity then the Taylor series. But infinity is an Infinity, and yes it would make sense for me putting infinity to the equation and calling it complete but I'm also doing it for computer graphics.

    • @Tabu11211
      @Tabu11211 Місяць тому +23

      @@SilentALume ngl you still need that x/pi tho

    • @siddude8021
      @siddude8021 Місяць тому

      ​​@@SilentALumethe function you have created converges to the function (1/2 (-2 EllipticTheta(2, 0, 16/e^4) - EllipticTheta(4, 0, 2/e) + (2 EllipticTheta(3, -π x, e^(π^2/(-1 + log(2)))) - EllipticTheta(3, -(π x)/2, e^(π^2/(4 (-1 + log(2)))))) sqrt(π/(1 - log(2)))))/(EllipticTheta(4, 0, 2/e)), this function is different from cos(pi*x) as cos(pi*x) = 0 at x = n + 1/2 for n beeing a integer. While your aproximation is zero at x = n + 1/2 + epsilon where epsilon is the error, epsilon is on the order of 10^-(10), that beeing said it seems the two functions have the same tops and bottoms, basically as R goes to infinity the margin of error goes to 10^(-10)

  • @andrasfogarasi5014
    @andrasfogarasi5014 Місяць тому +1163

    Welcome back Ramanujan

  • @sNazzy_nazzy
    @sNazzy_nazzy Місяць тому +915

    I feel like this video went from "oh huh I see where he's going with this" to "what the fuck" in the span of 0.2 seconds.

    • @TriflingToad
      @TriflingToad 21 день тому

      2:40 went from "what the fuck" to "what the actual hell is wrong with your brain to think this is fun"

  • @epixel7897
    @epixel7897 Місяць тому +603

    Wait until he finds out about sin(x+π/2)

    • @blobthekat
      @blobthekat Місяць тому +16

      ☠️

    • @ivanb493
      @ivanb493 Місяць тому +7

      thats cheating ;p

    • @TotalTimoTime
      @TotalTimoTime Місяць тому +20

      @@ivanb493he‘s using imaginary exponents. Those are automatically trig functions. If you think this suggestion would be cheating then the video is cheated too.

    • @leeroyjenkins0
      @leeroyjenkins0 27 днів тому +7

      ​@@TotalTimoTimecheck again, i is the variable of the sum not an imaginary number. I guess it kind of looks like the Taylor expansion if you dissect it though. But that seems fair.

    • @LIKERorHATER
      @LIKERorHATER 18 днів тому

      @@leeroyjenkins0 ok now i see it

  • @itsdab2763
    @itsdab2763 Місяць тому +222

    This man went from watching 3blue1brown to graphing complex equations with custom colors in 3 seconds

  • @T3WI
    @T3WI Місяць тому +205

    2:45 where the trivial stuff begins

  • @SilentALume
    @SilentALume  Місяць тому +358

    I was not expecting to get even that close.

    • @ignaciosavi7739
      @ignaciosavi7739 Місяць тому

      I'll try to make something better

    • @ignaciosavi7739
      @ignaciosavi7739 Місяць тому +1

      import matplotlib.pyplot as plt
      import math
      def apsin(x):
      pi = math.pi
      multy = 1
      # in what interval is x in relation to the roots of cos(x)
      #inty = x / 2 # this gives an approximation of how many roots in front of x .
      sign = 1
      if(math.ceil((x)/pi)%2 == 0):
      sign = -1
      x = (x + pi/2 - (pi *math.ceil((x)/pi)))

      inty = 1
      return sign*(1/(9*pi*pi*pi*pi/16)) * (x+((inty+1)*pi)-pi/2)*(x+((inty)*pi)-pi/2)*(x-((inty)*pi)-pi/2)*(x-((inty)*pi)+pi/2)
      listy = []
      liste= []
      print(math.sin(360))
      print(apsin(360))
      for i in range(1000):
      print('ap')
      listy.append(apsin(i))
      liste.append(math.sin(i))

      print(apsin(i))
      print(math.sin(i))
      plt.plot(listy)
      plt.plot(liste)
      plt.show()

    • @ignaciosavi7739
      @ignaciosavi7739 Місяць тому

      i made a sin function by accident

    • @God-gi9iu
      @God-gi9iu Місяць тому

      Sigma

    • @MaIarky
      @MaIarky Місяць тому

      Cosine = e^(ix).real
      Sine = e^(ix).imag
      You can also convert this to:
      i^x.real = cos(2x/pi)
      i^x.imag = sin(2x/pi)

  • @kuzhy.
    @kuzhy. Місяць тому +404

    2:43 “this video is gonna take about 2π”
    turns out the video length is just about 6:28 haha

    • @troubledouble106
      @troubledouble106 Місяць тому +10

      Lol. Prolly intentional.

    • @uggupuggu
      @uggupuggu Місяць тому +8

      2pi minutes is around 6:17 as a youtube timestamp

    • @jesp9435
      @jesp9435 Місяць тому +4

      @@uggupugguwhat?

    • @xatnu
      @xatnu Місяць тому +6

      ​@@jesp9435let him cook

    • @thomasdemilio6164
      @thomasdemilio6164 Місяць тому

      ​@@uggupuggu mmmmh....

  • @esp-elagogisch-sozialepart9701
    @esp-elagogisch-sozialepart9701 Місяць тому +75

    sin(x+π/2) is a decent Approximation if you ask me

    • @plenus2017
      @plenus2017 Місяць тому

      nah no one asks

    • @syncradar
      @syncradar Місяць тому

      I don't want to use sine

    • @jamie31415
      @jamie31415 Місяць тому +1

      sin(x-π/2) = -cos(x), not cos(x)

    • @gilernt
      @gilernt 24 дні тому

      @@plenus2017 shut up,. thankyou

  • @BaukBauk9491
    @BaukBauk9491 Місяць тому +70

    I think part of the reason why the quadratic in the exponent helped in making the cos function approximation is because of the jacobi theta function. Basically the third order jacobi theta function is θ₃(z, q) = Σ[n=-∞,∞](q^(n²) * e^(2niz)). When z = 0, the imaginary part (e^(2niz)) disappears, so we get θ₃(0, q) = Σ[n=-∞,∞](q^(n²))., splitting this into two sums θ₃(0, q) = q^(0²) + Σ[n=1,∞](q^(n²)) +Σ[n=-1,-∞](q^(n²)) = 1 + Σ[n=1,∞](q^(n²)) +Σ[n=-1,-∞](q^(n²)), notice by symmetry of the sum ((-n)² = n²), Σ[n=1,∞](q^(n²)) = Σ[n=-1,-∞](q^(n²)), therefore we have θ₃(0, q) = 2*Σ[n=1,∞](q^(n²)) + 1, therefore Σ[n=1,∞](q^(n²)) = (θ₃(0, q) - 1)/2. Then for a lot of these sums that appear we can just express the sum in terms of the θ function. By substitution and rearranging we can express many if not all the sum terms in terms of the theta function. The jacobi θ function is essentially an elliptic analogue of the exponential and does exhibit quasi-double periodicity, basically it means that the periodicity goes out to two dimensions and only roughly follows the periodic nature so f(z+u) and f(z+v) may not equal f(z) exactly (for this u and v are linearly independent), but there still follows a trend. Though because the imaginary part is removed it is only singly quasi-periodic hence yielding the cos approximation. Sorry if I made any mistakes make sure to tell me. en.wikipedia.org/wiki/Doubly_periodic_function mathworld.wolfram.com/JacobiThetaFunctions.html en.wikipedia.org/wiki/Quasiperiodicity

    • @John-cl8iv2
      @John-cl8iv2 Місяць тому +9

      I literally just learned in my math class yesterday lol

    • @BaukBauk9491
      @BaukBauk9491 Місяць тому +6

      @@John-cl8iv2 Oh cool what class is that?

    • @John-cl8iv2
      @John-cl8iv2 Місяць тому +5

      @@BaukBauk9491 Wait never min I learned a Jacobian in calc 3

    • @thespiciestmeatball
      @thespiciestmeatball Місяць тому

      That was a nice digestible explanation. Well done

    • @linuxnoodle8682
      @linuxnoodle8682 Місяць тому +2

      @@BaukBauk9491 you learn about theta functions in complex analysis right?

  • @TheBoeingCompany-h9z
    @TheBoeingCompany-h9z Місяць тому +70

    Bro went from "sooo so close" to an entire mathematic documents that exist

  • @yariklukianenko4046
    @yariklukianenko4046 Місяць тому +30

    after you zoomed out at 0:12 ... i instantly went y=0 will do xD

  • @unflexian
    @unflexian Місяць тому +32

    4:36 bro the music is pi!!! that's how i memorize it so i recognized immediately, this is awesome!

  • @EnricoRodolico
    @EnricoRodolico Місяць тому +64

    You are implicitly using e^ix, which itself encodes the desired results from Euler's formula.

    • @CarlosRoxo
      @CarlosRoxo Місяць тому +28

      I thought so too, but 'i' is not being used as the imaginary unit. It comes from the summation.

  • @vaycoo
    @vaycoo Місяць тому +107

    cosinus truly was the euqations we made along the way

  • @faded_ace5144
    @faded_ace5144 Місяць тому +10

    Man was I a fool to think that when I clicked on this video it was gonna be about anything I understand.

  • @ataphelicopter5734
    @ataphelicopter5734 15 днів тому +5

    Close enough, welcome back Srinivasa Ramanujan

  • @ChaineekToaster
    @ChaineekToaster Місяць тому +26

    Absolute cinema

  • @eos_rf
    @eos_rf Місяць тому +19

    unbelievable work.
    im to dumb to understand the process but looks like you made a hard work on this one 🔥🔥🔥

  • @gamerboy7224
    @gamerboy7224 Місяць тому +213

    Wait until bro discovers taylor series 💀💀💀

    • @zakariachouhou1280
      @zakariachouhou1280 Місяць тому +2

      hahah literally what i thought

    • @raepiste8354
      @raepiste8354 Місяць тому

      Dumbass he literally said it in the beginning

    • @Tabu11211
      @Tabu11211 Місяць тому +16

      True but just think about how the numbers involved here don't blow up. This can produce an amazing aproximation with as little as 6 terms to infinity with wrapping.

    • @gamerboy7224
      @gamerboy7224 Місяць тому +13

      @@Tabu11211 6 terms only makes this approximation valid for around |x|

    • @Tabu11211
      @Tabu11211 Місяць тому

      @@gamerboy7224 please do because I might be missing something. What I did to extend it was this: (x - (2pi × floor(x/(2pi)))/pi. Replace x with that and it will use the single cycle for the whole domain.

  • @dproscripts1811
    @dproscripts1811 Місяць тому +2

    Great work here! To me, it seems that you've derived a quirky Fourier-Poisson approximation with a mainly hyperbolic cosine approximation. I think one of the more concrete places to start would be the complex exponential definition of trigonometry, and approximate that, instead of doing visual approximation. Overall, great job though!

  • @GavinSpitz
    @GavinSpitz Місяць тому +12

    YOU FINALLY GOT A VIDEO THAT WENT SEMI VIRAL YESSSS

  • @HarpanW
    @HarpanW Місяць тому +24

    Obviously, very trivial stuff really

  • @ZephRanAway
    @ZephRanAway Місяць тому +5

    close enough, welcome back Ramanujan

  • @MegoZ_
    @MegoZ_ 20 днів тому +1

    Nintendo (1996) hire this man

  • @toblobs
    @toblobs Місяць тому +15

    I was thinking this might have applications to like work out cos quickly without a calc until I saw the final equation XD
    it feels like a taylor expansion anyways but in the most roundabout way possible

  • @landsgevaer
    @landsgevaer Місяць тому +2

    Instead of adding parabolas in the beginning, you could multiply them
    y = (1-(x/0.5pi)²)*(1-(x/1.5pi)²)*..
    and that will be exact as you add infinite factors...

  • @James2210
    @James2210 Місяць тому +3

    In desmos, it's really easy to get an approximation of the cosine function: cos(x)

  • @3zk1i_93
    @3zk1i_93 22 дні тому

    I don't know much math and have no idea what you were doing but this is really entertaining

  • @lucastornado9496
    @lucastornado9496 Місяць тому +5

    your last term can just be simplified as "R" you don't need the sumation of 1 from 1 to R

  • @Tabu11211
    @Tabu11211 Місяць тому +2

    Rung the bell. Love this exploratory chaos.

  • @luigav5663
    @luigav5663 Місяць тому +2

    I have no clue how you did anything, but this is the type of smart I aspire to be

  • @0679-Janitza
    @0679-Janitza Місяць тому +4

    Very nice!!

  • @stresswaves01
    @stresswaves01 16 днів тому +1

    this is the math analogy of "doing a little mining of camera"

  • @badamson
    @badamson Місяць тому +2

    the (D^ix + D^-ix) /2 type stuff you have going on in there is literally just the definition of cos for D=e. its not exactly what you have but something like that is going on there

  • @willmckelvey5337
    @willmckelvey5337 11 днів тому

    you can actually perfectly recreate cos(x) 1 to 1 via taking the real of i^(pi*x/2), which would look like real(i^(pi*x/2)), but you have to make sure you toggle on complex mode in the settings first

  • @Superhirn
    @Superhirn Місяць тому +3

    well done!

  • @coopervr1975
    @coopervr1975 24 дні тому

    There is a channel that made a really fast sim function for the n64. The channel is called Kaze emanuar. His isn’t as accurate but I think he used what he calls a folded polynomial to get different parts of it

  • @yuyuyu0201
    @yuyuyu0201 Місяць тому +1

    Truly remarkable 👏🏻

  • @ClementinesmWTF
    @ClementinesmWTF Місяць тому +88

    You might have just accidentally created an actual expansion of cos, especially with using e in your constant (cos(x) = (exp(ix)+exp(-ix))/2 after all).
    Also, the second sum of 1 from 1 to R is just equal to R, the sum is unnecessary.

    • @SilentALume
      @SilentALume  Місяць тому +18

      Its because I wanted to be a whole number

    • @xxd4rk_f1ngerxx89
      @xxd4rk_f1ngerxx89 Місяць тому +15

      Then the floor function might be your go-to ig

    • @SilentGamer._
      @SilentGamer._ Місяць тому +3

      @@SilentALume you can make the slider go up by 1s

    • @ClementinesmWTF
      @ClementinesmWTF Місяць тому +1

      @@SilentALume use the “step” option when you get into the slider range editor and make it 1 is what the above meant.

  • @TheOddPolymath
    @TheOddPolymath Місяць тому +1

    Subscribed just because of this xD

  • @transcendenceistaken
    @transcendenceistaken Місяць тому +4

    The design of the cosmos (simplified)

  • @lowenheim
    @lowenheim Місяць тому +3

    another video with some more explanation of the process/your thinking would be awesome!

  • @Houshalter
    @Houshalter Місяць тому +1

    Divide the input by 2pi and take the remainder. Then you only need to approximate that little range from 0 to 2pi, and every other input will work too. It's what computers actually do when they calculate these functions. It's called modular division and range reduction, and it's used everywhere.
    You can actually do more than 2pi because between 0 and pi is symmetrical to the part between pi and 2pi. And between 0 and pi/2 is also symmetrical to the part between pi/2 and pi. That leaves you with a tiny little piece of curve. And if you can approximate it, you get the rest of the values everywhere else.
    I tried a quadratic and got -0.35*x^2 -0.1*x -1. Looks pretty close by eye. I'm sure it's possible to do infinitely better of course. Computers can break it up into many segments, with lookup tables for the best fitting polynomial for each segment. You can do even better than that, but lookup tables and polynomials are very fast to compute.

  • @origimed5162
    @origimed5162 Місяць тому +10

    Bro just bruteforce the taylor series

    • @lucastornado9496
      @lucastornado9496 Місяць тому +3

      no. the taylor series is much less efficient than this

  • @thomasbeaumont3668
    @thomasbeaumont3668 Місяць тому +2

    plugging this instead of trig functions, to avoid trig in pre calc

  • @lalityt07
    @lalityt07 Місяць тому

    Thats impressive and also genius

  • @Cool_Bungle
    @Cool_Bungle Місяць тому +10

    The co-sine function

    • @acuriousmind6217
      @acuriousmind6217 Місяць тому +1

      "tHe cOS fUnCtiOn" it sent vibrations down my spine

  • @lillegitimate
    @lillegitimate Місяць тому +1

    2:44 this is when shit gets serious
    u should do some exploration on y=sqrt(24x+1) there are beautiful patterns in the primes of the function.

  • @HejHejda-hh3wl
    @HejHejda-hh3wl Місяць тому +4

    Taylor series: Really bro?

  • @ЯківБайдук
    @ЯківБайдук 28 днів тому

    Bro spent so much time approximatting cosine function, while I easily got an ideal aprroximation with sin(x + π/2)

  • @epicflyingfalco7753
    @epicflyingfalco7753 27 днів тому

    W bro that's wild as hell

  • @erez2417
    @erez2417 Місяць тому +2

    bro forgot that adding parabolas gives you back a parabola

  • @mhm6421
    @mhm6421 Місяць тому +4

    4:55 You never needed pi to go over the circle though...

  • @Hg-201
    @Hg-201 27 днів тому

    Me when I can't figure out how to simplify my answer in an exam

  • @livwithpeacee
    @livwithpeacee 14 днів тому

    could've put down a sine function and shifted the phase and that would've been a perfect "approximation" lol

  • @JamesMcCullough-lu9gf
    @JamesMcCullough-lu9gf 21 день тому +2

    sin(x+pi/2)

  • @sirsamiboi
    @sirsamiboi 22 дні тому

    So that went from 0 to 100 real fast

  • @lpzmarkus564
    @lpzmarkus564 22 дні тому

    idk if im high or something but the formula in the thumbnail looks like a 50 bmg sniper rifle

  • @mcmint233
    @mcmint233 18 днів тому +1

    idk man i think using cos would've been easier

  • @Blockitjames
    @Blockitjames Місяць тому +3

    bros video actually blew up

  • @luccazafado
    @luccazafado 20 днів тому

    not trying to be rude, but you can click the home button under the zoom out button to go back to default center zoom, y'know? not sure if you knew but seeing you struggle to zoom back in (specifically between 0:18 and 0:30) was kinda painful lol

  • @infrieser
    @infrieser Місяць тому +2

    Ok this is cool, but by using the exponential function to approximate a trigonometric function (cosine), aren't you effectively approximating a trigonometric function with a trigonometric function? Since cos(x)=(1/2)(e^(ix)+e^-(ix)) and your equation looks oddly similar, in the sense that it is in the form of (the sum of) (e^f(x) - e^g(x))/c + some error, where the error decreases as I increases. I don't feel like attempting to prove it, but this looks like some sort of taylor expansion of Euler's formula.
    I might be wrong of course.

  • @richielickie
    @richielickie Місяць тому

    as people probably already said. the cos function is very closely related to the exponential function. And you probably made the vanishing part of the gaussian go to zero with the limits

  • @coolcarl2232
    @coolcarl2232 Місяць тому +1

    this is insane

  • @plane9182
    @plane9182 Місяць тому +8

    One thing i made to approximate cosine with fast computationally is. This works because its fitted to the first quartile range of a cosine function it's accurate to 0.1%
    a = mod(x,1)-0.5
    c = mod(x,2)-a-1
    b = a^2(4.897-3.588*a^2)-1
    Then graph 2b*c

  • @mihaleben6051
    @mihaleben6051 Місяць тому

    good luck. i couldnt even find out the sin function

  • @koalakid3609
    @koalakid3609 Місяць тому +2

    "if R goes to infinity what is D?"- SilentALume

  • @Waterlord2.0
    @Waterlord2.0 Місяць тому

    Fun desmos tip is you can make functions like
    T(x. y, y) = x * y * u
    Or without the static-like function
    x1 = 2
    y1 = 5
    u = 4
    T = x1 * y1 * u

  • @DJT_2024_
    @DJT_2024_ Місяць тому +1

    Good job

  • @hugurs2396
    @hugurs2396 Місяць тому +1

    from math import pi
    def cos_approximations(x):
    R=10
    total=0

    e=2.71828182846
    D=e/2
    for n in range(1,R+1):
    d1=-D**(-(x/pi+1-2*n)**2)
    d2=D**(-(-x/pi+1-2*n)**2)
    d3=2*D**(-(2*n-1)**2)
    denom=0
    for i in range(1,R+1):
    NUM=D**(8*i-4)+1-2*D**(4*i-1)
    DENOM=2*D**(4*i*i)
    denom+=NUM/DENOM
    total+=(d1-d2+d3)/denom+1/R
    return total
    print(cos_approximations(1)) #--> 0.5403023059153995
    translating bros formula in python code 💀

    • @hugurs2396
      @hugurs2396 Місяць тому

      also, this is what the original cos sin functions look like in python code:
      from math import pi, factorial
      def cos(x):
      x=x%(2*pi)
      total=0
      for n in range(10):
      total+=((-1)**n * (x**(2*n)))/(factorial(2*n))
      return total
      def sin(x):
      x=x%(2*pi)
      total=0
      for n in range(10):
      total+=((-1)**n * (x**(2*n+1)))/factorial(2*n+1)
      return total

  • @RedLamentations
    @RedLamentations 19 днів тому

    Beautiful

  • @krishnabasavaraju2926
    @krishnabasavaraju2926 23 дні тому +1

    now integrate it

  • @Elec-citrus
    @Elec-citrus Місяць тому

    2:48 Yoo, is that the fibonacci music?

  • @mihaleben6051
    @mihaleben6051 Місяць тому

    bro was DESPERATE

  • @YOGURT1
    @YOGURT1 Місяць тому +1

    hey! great vid, just want to ask what is the program you are using to annotate/draw and make text boxes during the timelapse?
    Thank you.

  • @ImperialFold
    @ImperialFold Місяць тому

    wait until this guy finds out about taylor series

  •  Місяць тому

    instead of writing the sum of 1 for "I" that goes from 1 to R (in the denominator of the last term), you could've just written R

  • @youdontneedtoknowmyname2753
    @youdontneedtoknowmyname2753 24 дні тому

    Wait until he discovers Taylor series

  • @watip1234
    @watip1234 Місяць тому

    also another interesting way to do this kind of thing is with the bell curve. i may be wrong am an engineer student

  • @Curryocity
    @Curryocity Місяць тому +6

    that wasn’t obvious for me

  • @DamDamPow
    @DamDamPow 12 днів тому

    taylor expansion enters the chat

  • @周品宏-o7w
    @周品宏-o7w Місяць тому

    1:52
    Bell-shaped function
    en.wikipedia.org/wiki/Bell-shaped_function

  • @claudiodai3791
    @claudiodai3791 22 дні тому

    bro has the weirdets time clock

  • @TerjeMathisen
    @TerjeMathisen Місяць тому +1

    It seemed quite obvious that what you was actually doing was to generate an alternative (and much more complicated) way to express the Taylor series?
    I clicked on this video expecting to see a fast approximation, useful for things like games/computer graphics, and in the beginning, this seemed like what you were doing, but then wham! 🙂

  • @ilovejesusandilovegod8803
    @ilovejesusandilovegod8803 Місяць тому +1

    Here's a fun question: How can you prove your function is similar to cos(x) without desmos? 🙂

  • @APotatoWT
    @APotatoWT 28 днів тому

    anything but the taylor series

  • @atom1kcreeper605
    @atom1kcreeper605 Місяць тому

    This is what i do for fun

  • @cookingandjava7574
    @cookingandjava7574 Місяць тому +1

    Isn't having e^i in your equation kinda cheating xD

  • @blackholegamer9
    @blackholegamer9 Місяць тому

    'use the long method'

  • @DarknessIsaGoober
    @DarknessIsaGoober 2 дні тому

    I love how the video was 2pi long (2:44)

  • @DaMonster
    @DaMonster Місяць тому +1

    Are you familiar with CORDIC? It is much faster on a computer than this series, even though it is a very cool series!

  • @jerpidude8416
    @jerpidude8416 Місяць тому

    approximate it with a forrier series

  • @moocatmeow
    @moocatmeow Місяць тому

    personally i like real(i^(x/π/2)) as an approximation of cos

  • @yaruzai6268
    @yaruzai6268 Місяць тому

    Whats the song called? Cant find it with Shazam

  • @SilverGoldYT
    @SilverGoldYT Місяць тому +1

    You can't just use pi as a unit for determining your video lengths 😭😭😭😭

  • @programmingpi314
    @programmingpi314 Місяць тому +1

    Why do you have the sum from I=1 to R of 1? That is just R.

  • @sovenok-hacker
    @sovenok-hacker Місяць тому

    Maclaurin and Taylor: 💀

  • @3ddruck410
    @3ddruck410 Місяць тому

    U must me genius af

  • @dIancaster
    @dIancaster Місяць тому +1

    all I need to approximate the sin function is to pull up to the crib w some baddies. The whole function be sinning.