数学公式集

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Mathematical Formula

  1. Taylor expansion

    [g(x) = g(x_0) + sum_{k = 1}^{n}frac{f^k(x-x_0)^k}{k!}(x-x_0)^k + R_n(x) ]

    (R_n(x)) refers to the Lagrange remainder, which is

    [R_n(x) = frac{f^{n+1}(xi)}{(n+1)!}(x-x_0)^{n+1} ]

    Lagrange remainder is derived from Cauchy mean value theorem.

  2. Cauchy mean value theorem

    Between a and b always exits a (xiin(a,b)) to make

    [frac{f(b) - f(a)}{g(b) - g(b)} = frac{f^{‘}(xi)}{g^{‘}(xi)} ]

    true.

  3. Why all the functions represent to the sum of the odd and even functions?

    For a regular function (f(x))

    [f(x) = frac{f(x) + f(-x)}{2} + frac{f(x)-f(-x)}{2} ]

    even function on the left and odd function on the right.

  4. A method for solving first order nonlinear differential equations.

    For a regular first order nonlinear differential equation

    [frac{dy}{dx} + P(x)y = Q(x) ag{1} ]

    Let (y = ucdot v), (u) and (v) are all functions about (x), so we get

    [frac{dy}{dx} = vfrac{du}{dx} + ufrac{dv}{dx} ]

    Then the differential equation become

    [frac{du}{dx}cdot v + ucdotleft(frac{dv}{dx} + P(x)cdot v ight) = Q(x) ag{2} ]

    We want solve the (v) to make the formula inside the parenthesis to be zero, that is

    [frac{dv}{dx} + P(x)cdot v = 0 ]

    This is a first order linear differential equation, it‘s general solution is

    [v = C_1e^{-int P(x)dx} ag{3} ]

    Let‘s substitute v into the formula (2), we get

    [frac{du}{dx}cdot C_1e^{-int P(x)dx} = Q(x) ag{4} ]

    This is a linear differential equation, we can easily solve it, the general solution of (u) is

    [u = frac1{C_1}int Q(x)cdot e^{int P(x)dx}dx + C_2 ]

    So our target (y) is

    [y = ucdot v = left[int e^{(int P(x)dx)}cdot Q(x)cdot dx + C ight]cdot e^{(-int P(x)dx)} ]

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