2022 MIT integration Bee Regular season problem # 15 (2nd method)

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  • เผยแพร่เมื่อ 11 ม.ค. 2025

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  • @toadjiang7626
    @toadjiang7626 11 ชั่วโมงที่ผ่านมา

    This integral is like tailor-made for the reciprocal beta function, it will directly get you the final result, only takes like 30 seconds.

    • @cipherunity
      @cipherunity  10 ชั่วโมงที่ผ่านมา +1

      I shall do that. Thanks for your advise.

    • @cipherunity
      @cipherunity  10 ชั่วโมงที่ผ่านมา +1

      Sir, it is done. Thanks again

  • @holyshit922
    @holyshit922 19 ชั่วโมงที่ผ่านมา

    Generally we can expand function in terms of Chebyshov polynomials
    f(x) = \sum_{n=0}^{\infty}{a_{n}T_{n}(x)}
    where
    a_{n} = \frac{2-\delta_{n0}}{\pi}\int_{0}^{\pi}{f(cos(t))cos(nt)dt}
    \delta_{nm} is Kronecker delta
    f(x) is polynomial here so we can calculate this integral without integration
    Suppose that you have function -2(1+x)^{5}
    and you want to calculate coefficient in front of T_{5}(x)
    where T_{n}(x) is Chebyshov T polynomial
    To expand function in terms of Chebyshov polynomials on the one side you can calculate it using integral
    \frac{(2-\delta_{n0})}{\pi}\int_{0}^{\pi}f(cos(t))cos(nt)dt , where n non-negative integer number
    but in the other side you can can calculate it without integral
    \delta_{nm} is Kronecker symbol
    So I suggest to expand function -2(1+x)^{5}
    in terms of Chebyshov polynomials without using integral
    -2(1+x)^{5} = a_{5}(16x^{5}-20x^{3}+5x)+a_{4}(8x^{4}-8x^{2}+1)+a_{3}(4x^{3}-3x)+a_{2}(2x^2-1)+a_{1}x+a_{0}
    but we look for coefficient in front of T_{5}(x) so
    -2 = 16a_{5}
    a_{5} = -1/8
    2/pi * I= -1/8
    I = pi/2*(-1/8)
    I = -pi/16
    To prepare this integral for use of Chebyshov polynomials
    use substitution t = pi - x , and fact that integrand is even on interval symmetric around zero

    • @holyshit922
      @holyshit922 16 ชั่วโมงที่ผ่านมา

      It may look like integral from calculating coefficient of Fourier series expansion with cosine
      but it expands function in terms of Chebyshov polynomials instead of monomials like in Taylor expansion

    • @cipherunity
      @cipherunity  13 ชั่วโมงที่ผ่านมา

      I need to read about Chebyshov polynomials

    • @holyshit922
      @holyshit922 13 ชั่วโมงที่ผ่านมา

      @@cipherunity Name of this guy in original Cyrylic is Пафнутий Львович Чебышёв
      And vowel ё is read as yo
      Львович means that first name of his father was Lev
      If you want to read about it in original
      you can type Многочлены Чебышёва
      Here expressing functions in terms of Chebyshov polynomials can be useful
      In numerical analysis there is a topic about approximating functions using Chebyshov polynomials

  • @yentungchung862
    @yentungchung862 8 ชั่วโมงที่ผ่านมา

    Integrate (1 - cos x)^5 cos(5*x) dx from 0 to 2π
    = Integrate (1 - 5cos (x)+10cos^2(x)-10cos^3(x)+5cos^4(x)-5*cos^5(x)) cos(5*x) dx from 0 to 2π
    = Integrate (63/8-105/8*cos(x)+15/2*cos(2*x)-45/16*cos(3*x)+5/8*cos(4*x)-1/16*cos(5*x))*cos(5*x) dx from 0 to 2π
    =Integrate (63/8-105/8*cos(x)+15/2*cos(2*x)-45/16*cos(3*x)+5/8*cos(4*x))*cos(5*x) dx from 0 to 2π - 1/16 Integrate cos^2(5*x) dx from 0 to 2π
    =0-1/16*π = -π/16
    Note: (i) Integrate cos (n*x) dx from 0 to 2π = 0 (ii) Integrate cos(m*x)*cos (n*x) dx from 0 to 2π =0 ,if m≠n
    (iii) Integrate cos(m*x)*cos (n*x) dx from 0 to 2π =π , if m=n

    • @cipherunity
      @cipherunity  6 ชั่วโมงที่ผ่านมา

      It is done. I just posted a new video using you solution

  • @chancia8990
    @chancia8990 8 ชั่วโมงที่ผ่านมา

    lol not sick but very long