考研数学公式及其证明整理

CreateTime 2020-12-01
UpdateTime 2020-12-05

微积分

高阶导数

(a^x)^{(n)} = (\ln a)^n\cdot a^x

证明: 用数学归纳法

n = 1 时:

(a^x)' = (e^{x\ln a})' = \ln a \cdot e^{x\ln a} = \ln a \cdot a^x,成立

n = k 时,(a^x)^{(k)} = (\ln a)^k\cdot a^x 成立,则当 n = k + 1 时:

(a^x)^{(k+1)} = [(\ln a)^k\cdot a^x] = (\ln a)^{k+1}\cdot a^x

由第一数学归纳法,命题成立,证毕。


(e^x)^{(n)} = e^x

证明:

显然此命题为上一命题的简单推论

a = e,则 \ln e = 1,故

(e^x)^{(n)} = (\ln e)^n\cdot e^x = e^x


\displaystyle (\sin kx)^{(n)} = k^n\sin \left(\frac{n\pi}{2} + kx\right)

证明: 用数学归纳法

n = 1
$$
(\sin kx)’ =k \cos kx = k\sin \left(\frac{\pi}{2} + kx\right)
$$

n = 2 时(该步骤可以省略)

\begin{aligned} &(\sin kx)'' \\ = & \left[k\sin \left(\frac{\pi}{2} + kx\right)\right]' \\ =& k^2\cos \left(\frac{\pi}{2} + kx\right) \\ =& k^2\sin \left(\frac{2\pi}{2} + kx\right) \\ \end{aligned}

n = t 时,

\displaystyle (\sin kx)^{(t)} =k^t\sin \left(\frac{t\pi}{2} + kx\right) 成立

n = t + 1

\begin{aligned} &(\sin kx)^{(t + 1)} \\ = & \left[k^t\sin \left(\frac{t\pi}{2} + kx\right)\right]' \\ =& k^{t+1}\cos \left(\frac{t\pi}{2} + kx\right) \\ =& k^{t+1}\sin \left[\frac{(t + 1)\pi}{2} + kx\right] \\ \end{aligned}

故,由第一数学归纳法,命题成立,证毕。


\displaystyle (\cos kx)^{(n)} = k^n\cos \left(\frac{n\pi}{2} + kx\right)

证明:证明方法同上 (\sin kx)^{(n)},不再赘述。


\displaystyle (\ln x)^{(n)} = (-1)^{n-1}\frac{(n-1)!}{x^n}

证明: 用数学归纳法

n = 1

\displaystyle(\ln x)' = \frac{1}{x} = (-1)^{0}(1!)\frac{1}{x^1}

n = k

\displaystyle(\ln x)^{(k)} = (-1)^{k-1}(k-1)!\frac{1}{x^k} 成立

则当 n = k + 1

\begin{aligned} &(\ln x)^{(k+1)} \\ = &\left[(-1)^{k-1}(k-1)!\frac{1}{x^k}\right]' \\ = &(-1)^{k}k!\frac{1}{x^{(k+1)}} \end{aligned}

故,由第一数学归纳法,命题成立,证毕。


\displaystyle [\ln(1 + x)]^{(n)} = (-1)^{(n-1)}\frac{(n-1)!}{(1 + x)^n}

证明: 此命题为上一命题的简单推论

做变量替换 u = 1 + x 可得:

\displaystyle [\ln(1 + x)]^{(n)} = (-1)^{(n-1)}\frac{(n-1)!}{(1 + x)^n}


\displaystyle \left(\frac{1}{x}\right)^{(n)} = (-1)^{n}\frac{n!}{x^{n + 1}}

证明:

由于 \displaystyle(\ln x)' = \frac{1}{x},根据 \displaystyle (\ln x)^{(n)} = (-1)^{n-1}\frac{(n-1)!}{x^n} 可得:

\displaystyle \left(\frac{1}{x}\right)^{(n)} = (-1)^{n}\frac{n!}{x^{n + 1}}


\displaystyle \left(\frac{1}{x + a}\right)^{(n)} = (-1)^{n}\frac{n!}{(x + a)^{n + 1}}

证明:x' = (x + a)' 易知,此命题为上一命题的简单推论。


\displaystyle (x^m)^{(n)} = m(m-1)\cdots(m - n + 1) x^{m - n}

证明: 用数学归纳法

n = 1

(x^m)' = m x^{m - 1}m - 1 + 1 = m 成立

n = k

(x^m)^{(k)} = m(m-1)\cdots(m - k + 1) x^{m - k} 成立

则当 n = k + 1

\begin{aligned} & (x^m)^{(k + 1)} \\ = &[m(m-1)\cdots(m - k + 1) x^{m - k}]' \\ = &m(m-1)\cdots(m - k + 1)(m - k) x^{m - k - 1} \\ = &m(m-1)\cdots[m - (k + 1) - 1] x^{m - (k + 1)} \\ \end{aligned}

故,由第一数学归纳法,命题成立,证毕。


\displaystyle (u \pm v)^{(n)} = u^{(n)} \pm v^{(n)}

证明: 用数学归纳法

n = 1 时:

\begin{aligned} & [u(x) \pm v(x)]' \\ = & \lim_{\Delta x\to 0} \frac{[u(x + \Delta x) \pm v(x + \Delta x)] - [u(x) \pm v(x)]}{\Delta x} \\ = & \lim_{\Delta x\to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \pm \lim_{\Delta x\to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} \\ = & u(x)' \pm v(x)' \end{aligned}

n = k 时:

\displaystyle (u \pm v)^{(k)} = u^{(k)} \pm v^{(k)}

n = k + 1 时:

\displaystyle (u \pm v)^{(k + 1)} = [u^{(k)} \pm v^{(k)}]' = u^{(k+1)} \pm v^{(k+1)}

故,由第一数学归纳法,命题成立,证毕。


莱布尼茨公式: \displaystyle (uv)^{(n)} = \sum_{k=0}^{n} \mathrm{C}_n^ku^{(n-k)}v^{(k)}

n = 1

\begin{aligned} & [u(x)v(x)]' \\ = & \lim_{\Delta x\to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \\ = & \lim_{\Delta x\to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x + \Delta x) + u(x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \\ = & \lim_{\Delta x\to 0} \frac{[u(x + \Delta x) - u(x)]v(x + \Delta x)}{\Delta x} + \lim_{\Delta x\to 0} \frac{u(x)[v(x + \Delta x) - v(x)]}{\Delta x} \\ = & \lim_{\Delta x\to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \pm \lim_{\Delta x\to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} \\ = & u(x)'v(x) + u(x)v(x)' \end{aligned}

n = u

\displaystyle (uv)^{(u)} = \sum_{k=0}^{u} \mathrm{C}_u^ku^{(u-k)}v^{(k)} 成立

n = u + 1 时:

\begin{aligned} & (uv)^{(u + 1)} \\ = & \left[\sum_{k=0}^{u} \mathrm{C}_u^ku^{(u-k)}v^{(k)}\right]' \\ = & \left[ \mathrm{C}_u^0 u^{(u)}v^{(0)} + \mathrm{C}_u^1 u^{(u-1)}v^{(1)} + \cdots + \mathrm{C}_u^u u^{(0)}v^{(u)}\right]' \\ = & \mathrm{C}_u^0 u^{(u + 1)}v^{(0)} + \mathrm{C}_u^0 u^{(u)}v^{(1)} + \\ & \mathrm{C}_u^1 u^{(u)}v^{(1)} + \mathrm{C}_u^1 u^{(u-1)}v^{(2)} + \cdots + \\ &\mathrm{C}_u^u u^{(1)}v^{(u)} + \mathrm{C}_u^u u^{(0)}v^{(u + 1)} \\ = & \mathrm{C}_{u+1}^0 u^{(u + 1)}v^{(0)} + \\ & [\mathrm{C}_u^0 u^{(u)}v^{(1)} + \mathrm{C}_u^1 u^{(u)}v^{(1)} ]+ \\ & [\mathrm{C}_u^1 u^{(u - 1)}v^{(2)} + \mathrm{C}_u^2 u^{(u-1)}v^{(2)}] + \cdots + \\ & [\mathrm{C}_u^1 u^{(1)}v^{(u)} + \mathrm{C}_u^u u^{(1)}v^{(u)}] + \\ = & \mathrm{C}_{u+1}^0 u^{(u + 1)}v^{(0)} + [\mathrm{C}_{u + 1}^1 u^{(u)}v^{(1)}]+ \\ & \mathrm{C}_{u + 1}^2 u^{(u - 1)}v^{(2)} + \cdots + \\ & \mathrm{C}_{u + 1}^u u^{(1)}v^{(u)} + \mathrm{C}_{u+1}^{u+1} u^{(0)}v^{(u + 1)} \\ = & \sum_{k=0}^{u + 1} \mathrm{C}_{u+1}^ku^{(u + 1 -k)}v^{(k)} \\ \end{aligned}

故,由第一数学归纳法,命题成立,证毕。


中值定理辅助函数

对于中值定理辅助函数的证明,只需要找到一个原函数即可,所以可以忽略不定积分的常数,题目中可能需要一个常数,这个需酌情添加。


f(x)f'(x) = 0 \to F(x) = f^2(x)

证明:

\begin{aligned} &f(x)f'(x) = 0 \\ &\int f(x)f'(x) dx \\ =& \int f(x)d f(x) \\ =& \frac{1}{2} f^2(x) \\ \end{aligned}

于是可以取 F(x) = f^2(x),证毕。


[f'(x)]^2 + f(x)f''(x)= 0 \to F(x) = f'(x)f(x)

证明:

\begin{aligned} &[f'(x)]^2 + f(x)f''(x)= 0 \\ & \frac{f'(x)}{f(x)} + \frac{f''(x)}{f'(x)}= 0 \\ & [\ln f(x) + \ln f'(x)]'= 0 \\ & \ln[f(x)f'(x)]'= 0 \\ \end{aligned}

于是可以取 F(x) = f(x)f'(x),证毕。

类似地,f(x)f''(x) - [f'(x)]^2= 0 \to \displaystyle F(x) = \frac{f'(x)}{f(x)}


f'(x) + f(x)\varphi'(x)= 0 \to F(x) = f(x)e^{\varphi(x)}

证明:

\begin{aligned} &f'(x) + f(x)\varphi'(x)= 0 \\ & \frac{f'(x)}{f(x)} + \varphi'(x)= 0 \\ & [\ln f(x) + \ln e^{\varphi(x)}]'= 0 \\ & [\ln f(x)e^{\varphi(x)}]'= 0 \\ \end{aligned}

于是可以取 F(x) = f(x)e^{\varphi(x)},证毕。

类似地,可知:

f'(x) - f(x)\varphi'(x)= 0 \to \displaystyle F(x) = \frac{f(x)}{e^{\varphi(x)}}


xf'(x) + f(x)= 0 \to F(x) = xf(x)

证明:

\begin{aligned} & xf'(x) + f(x)= 0 \\ & \frac{f'(x)}{f(x)} + \frac{1}{x}= 0 \\ & [\ln f(x) + \ln x]'= 0 \\ &[\ln xf(x)]'= 0 \end{aligned}

于是可以取 \displaystyle F(x) = xf(x),证毕。

类似地,可知:

xf'(x) - f(x)= 0 \to \displaystyle F(x) = \frac{f(x)}{x}


\displaystyle \int_a^b f(x) dx \to \displaystyle F(x) = \int_a^x f(t) dt


定积分常用公式

区间再线公式:\displaystyle \int_a^b f(x) dx = \int_a^b f(a + b - x) dx

证明:

\begin{aligned} & \int_a^b f(x) dx \\ \xlongequal{x = a+b-t}& \int_b^a f(a+b-t) d(-t) \\ =& \int_a^b f(a+b-t) dt \\ =& \int_a^b f(a+b-x) dx \end{aligned}

证毕。