数学公式集

Posted 2020-11-28 lunar-ubuntu

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)} ]