数学公式集
Mathematical Formula
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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.
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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.
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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.
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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
\[y = u\cdot v = \left[\int e^{(\int P(x)dx)}\cdot Q(x)\cdot dx + C \right]\cdot e^{(-\int P(x)dx)} \]
