数学公式集

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 $\xi\in(a,b)$ 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

   $$\frac{dy}{dx} + P(x)y = Q(x)\tag{1}$$

   Let $y = u\cdot v$, $u$ and $v$ are all functions about $x$, so we get

   $$\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$$

   Then the differential equation become

   $$\frac{du}{dx}\cdot v + u\cdot\left(\frac{dv}{dx} + P(x)\cdot v \right) = Q(x)\tag{2}$$

   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

   $$v = C_1e^{-\int P(x)dx}\tag{3}$$

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

   $$\frac{du}{dx}\cdot C_1e^{-\int P(x)dx} = Q(x)\tag{4}$$

   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

   $$y = u\cdot v = \left[\int e^{(\int P(x)dx)}\cdot Q(x)\cdot dx + C \right]\cdot e^{(-\int P(x)dx)}$$