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 \(\xi\in(a,b)\) 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

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

数学公式集

标签：line   solution   first   str   exit   tween   span   rac   method   

原文地址：https://www.cnblogs.com/lunar-ubuntu/p/12896651.html