设f(x)在[a,b]上连续,则
定理1(有界定理)∣f(x)∣≤M (M>0)\lvert f(x) \rvert\leq M \ \ (M>0)∣f(x)∣≤M (M>0)
定理2(最值定理)m≤f(x)≤Mm \leq f(x) \leq Mm≤f(x)≤M,其中m,M分别为f(x)在 [a, b] 上的最小值与最大值
定理3(介值定理)当m≤μ≤M时,∃ξϵ[a,b],使得f(ξ)=μm \leq \mu \leq M\text{时,} \exists \xi \epsilon{[a,b]}\text{,使得}f(\xi)=\mum≤μ≤M时,∃ξϵ[a,b],使得f(ξ)=μ
定理4(零点定理)当f(a)⋅f(b)<0时,∃ξϵ(a,b),使得f(ξ)=0f(a)\cdot f(b)<0\text{时,}\exists \xi\epsilon(a,b)\text{,使得}f(\xi)=0f(a)⋅f(b)<0时,∃ξϵ(a,b),使得f(ξ)=0
定理5(费马定理)
设f(x)满足在x0点处{(1)可导,(2)取极值,则f′(x0)=0 \text{设f(x)满足在}x_0\text{点处} \begin{cases} \text{(1)可导,}\\ \text{(2)取极值,} \end{cases}\text{则}f'(x_0)=0 设f(x)满足在x0点处{(1)可导,(2)取极值,则f′(x0)=0
定理6(罗尔定理)
设f(x)满足{(1)[a, b]上连续,(2)(a, b)内可导,(3)f(a) = f(b),则,∃ξϵ(a,b),使得f′(ξ)=0 \text{设f(x)满足}\begin{cases} \text{(1)[a, b]上连续,} \\ \text{(2)(a, b)内可导,} \\ \text{(3)f(a) = f(b),} \end{cases}\text{则,}\exists\xi\epsilon(a,b),\text{使得}f'(\xi)=0 设f(x)满足⎩⎨⎧(1)[a, b]上连续,(2)(a, b)内可导,(3)f(a) = f(b),则,∃ξϵ(a,b),使得f′(ξ)=0
定理7(拉格朗日中值定理)
设f(x)满足{(1)[a, b]上连续,(2)(a, b)内可导,则,∃ξϵ(a,b),使得
\text{设f(x)满足}\begin{cases} \text{(1)[a, b]上连续,} \\ \text{(2)(a, b)内可导,} \end{cases}\text{则,}\exists\xi\epsilon(a,b)
\text{,使得}
设f(x)满足{(1)[a, b]上连续,(2)(a, b)内可导,则,∃ξϵ(a,b),使得
f(b)−f(a)=f′(ξ)(b−a)
f(b)-f(a)=f'(\xi)(b-a)
f(b)−f(a)=f′(ξ)(b−a)
或者写成
f′(ξ)=f(b)−f(a)b−a f'(\xi)=\frac{f(b)-f(a)}{b-a} f′(ξ)=b−af(b)−f(a)
定理8(柯西中值定理)
设f(x)满足{(1)[a, b]上连续,(2)(a, b)内可导,(3)g′(x)≠0,则,∃ξϵ(a,b),使得f(b)−f(a)g(b)−g(a)=f′(ξ)g′(ξ).\text{设}f(x)\text{满足}\begin{cases} \text{(1)[a, b]上连续,} \\ \text{(2)(a, b)内可导,} \\ \text{(3)}g'(x)\ne0, \end{cases}\text{则,}\exists\xi\epsilon(a,b),\text{使得} \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}. 设f(x)满足⎩⎨⎧(1)[a, b]上连续,(2)(a, b)内可导,(3)g′(x)=0,则,∃ξϵ(a,b),使得g(b)−g(a)f(b)−f(a)=g′(ξ)f′(ξ).
定理9(泰勒公式)
(1)带拉格朗日余项的n阶泰勒公式
设f(x)在点x0x_0x0的某个领域内有n+1阶导数存在,则对该领域内的任意点x均有
f(x)=f(x0)+f′(x0)(x−x0)+12!f′′(x0)(x−x0)2+⋯+1n!f(n)(x0)(x−x0)n+f(n+1)(ξ)(n+1)!(x−x0)n+1 ,
\begin{array}{l}
f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2!}f''(x_0)(x-x_0)^2+\cdots
+\frac{1}{n!}f^{(n)}(x_0)(x-x_0)^n
\\
+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}\text{ ,}
\end{array}
f(x)=f(x0)+f′(x0)(x−x0)+2!1f′′(x0)(x−x0)2+⋯+n!1f(n)(x0)(x−x0)n+(n+1)!f(n+1)(ξ)(x−x0)n+1 ,
其中ξ介于x,x0之间
\text{其中}\xi\text{介于}x,x_0\text{之间}
其中ξ介于x,x0之间
(2)带佩亚诺余项的n阶泰勒公式
设f(x)在点x0x_0x0处n阶可导,则存在x0x_0x0的一个领域,对于该领域中的任一点,成立
f(x)=f(x0)+f′(x0)(x−x0)+12!f′′(x0)(x−x0)2+⋯+1n!f(n)(x0)(x−x0)n+o((x−x0)n).
\begin{array}{l}
f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2!}f''(x_0)(x-x_0)^2+\cdots+\frac{1}{n!}f^{(n)}(x_0)(x-x_0)^n
\\
\\
+o((x-x_0)^n).
\end{array}
f(x)=f(x0)+f′(x0)(x−x0)+2!1f′′(x0)(x−x0)2+⋯+n!1f(n)(x0)(x−x0)n+o((x−x0)n).
定理10(导数零点定理)
设f(x)在[a,b]上可导,当f+′(a)⋅f−′(b)<0 时,∃ξϵ(a,b),使得f′(ξ)=0f'_+(a)\cdot f'_-(b)< 0 \text{ 时,}\exists \xi\epsilon(a,b)\text{,使得}f'(\xi)=0f+′(a)⋅f−′(b)<0 时,∃ξϵ(a,b),使得f′(ξ)=0
定理11(导数介值定理)
设f(x)在[a,b]上可导,若f+′(a)≠f−′(b), 则∀μ介于f+′(a)与f−′(b)之间,∃ξϵ(a,b),使得f′(ξ)=μf'_+(a)\ne f'_-(b)\text{, 则}\forall \mu\text{介于}f'_+(a)\text{与}f'_-(b)\text{之间,} \exists \xi\epsilon(a,b)\text{,使得}f'(\xi)=\muf+′(a)=f−′(b), 则∀μ介于f+′(a)与f−′(b)之间,∃ξϵ(a,b),使得f′(ξ)=μ
麦克劳林公式(x0=0x_0=0x0=0时)
f(x)=f(0)+f′(0)x+f′′(0)2!x2+⋯+f(n)(0)n!xn+f(n+1)(ξ)(n+1)!xn+1f(x)=f(0)+f′(0)x+f′′(0)2!x2+⋯+f(n)(0)n!xn+o(xn) f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n+\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}\\ f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n+o(x^n) f(x)=f(0)+f′(0)x+2!f′′(0)x2+⋯+n!f(n)(0)xn+(n+1)!f(n+1)(ξ)xn+1f(x)=f(0)+f′(0)x+2!f′′(0)x2+⋯+n!f(n)(0)xn+o(xn)
几个重要函数的麦克劳林展开式
eu=1+u+12u2+⋯+1n!un+o(un)
e^u=1+u+\frac{1}{2}u^2+\cdots+\frac{1}{n!}u^n+o(u^n)
eu=1+u+21u2+⋯+n!1un+o(un)
sinu=u−u33!+⋯+(−1)nu2n+1(2n+1)!+o(u2n+1)
sinu=u-\frac{u^3}{3!}+\cdots+(-1)^n\frac{u^{2n+1}}{(2n+1)!}+o(u^{2n+1})
sinu=u−3!u3+⋯+(−1)n(2n+1)!u2n+1+o(u2n+1)
cosu=u−u22!+u44!+⋯+(−1)nu2n(2n)!+o(u2n)
cosu=u-\frac{u^2}{2!}+\frac{u^4}{4!}+\cdots+(-1)^n\frac{u^{2n}}{(2n)!}+o(u^{2n})
cosu=u−2!u2+4!u4+⋯+(−1)n(2n)!u2n+o(u2n)
11−u=1+u+u2+⋯+un+o(un)
\frac{1}{1-u}=1+u+u^2+\cdots+u^n+o(u^n)
1−u1=1+u+u2+⋯+un+o(un)
11+u=1−u+u2−⋯+(−1)nun+o(un)
\frac{1}{1+u}=1-u+u^2-\cdots+(-1)^nu^n+o(u^n)
1+u1=1−u+u2−⋯+(−1)nun+o(un)
ln(1+u)=u−u22+u33−⋯+(−1)nun+1(n+1)+o(un+1)
ln(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots+(-1)^n\frac{u^{n+1}}{(n+1)}+o(u^{n+1})
ln(1+u)=u−2u2+3u3−⋯+(−1)n(n+1)un+1+o(un+1)
(1+u)α=1+αu+α(α−1)2!u2+⋯+α(α−1)⋯(α−n+1)n!un+o(un)
(1+u)^\alpha=1+\alpha u+\frac{\alpha(\alpha-1)}{2!}u^2+\cdots+\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}u^n+o(u^n)
(1+u)α=1+αu+2!α(α−1)u2+⋯+n!α(α−1)⋯(α−n+1)un+o(un)