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增长趋势
n → + ∞ , ∀ p , q > 0 , a > 1 , ( ln n ) q ≪ n p ≪ a n ≪ n ! ≪ n n n\to+\infty,\forall p,q>0,a>1,{(\ln n)}^q\ll n^p\ll a^n\ll n!\ll n^n n → + ∞ , ∀ p , q > 0 , a > 1 , ( ln n ) q ≪ n p ≪ a n ≪ n ! ≪ n n
积分表
反读可得导数表,此处略。
∫ k d x = k x + C ∫ x a d x = x a + 1 a + 1 + C ∫ 1 x d x = ln ∣ x ∣ + C ∫ e x d x = e x + C ∫ a x d x = a x ln a + C ∫ cos x d x = sin x + C ∫ sin x d x = − cos x + C ∫ 1 c o s 2 x d x = ∫ sec 2 x d x = tan x + C ∫ 1 s i n 2 x d x = ∫ csc 2 x d x = − cot x + C ∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C ∫ 1 1 + x 2 d x = arctan x + C = − a r c c o t x + C ∫ sec x tan x d x = sec x + C ∫ csc x cot x d x = − csc x + C ∫ tan x d x = − ln ∣ cos x ∣ + C ∫ cot x d x = ln ∣ sin x ∣ + C ∫ sec x d x = ln ∣ sec x + tan x ∣ + C ∫ csc x d x = ln ∣ csc x − cot x ∣ + C ∫ s h x d x = c h x + C ∫ c h x d x = s h x + C ∫ 1 x 2 + a 2 d x = 1 a arctan x a + C ∫ 1 x 2 − a 2 d x = 1 2 a ln ∣ x − a x + a ∣ + C ∫ 1 a 2 − x 2 d x = arcsin x a + C ∫ 1 x 2 − a 2 d x = ln ∣ x + x 2 − a 2 ∣ + C ∫ 1 x 2 + a 2 d x = ln ∣ x + x 2 + a 2 ∣ + C \int k\,\mathrm{d}x=kx+C\\
\int x^a\,dx=\frac{x^{a+1}}{a+1}+C\\
\int\frac{1}{x}\,dx=\ln\mid x\mid +C\\
\int e^x\,dx=e^x + C\\
\int a^x\,dx=\frac{a^x}{\ln a}+C\\
\int\cos x\,dx=\sin x+C\\
\int\sin x\,dx=-\cos x+C\\
\int\frac{1}{cos^2x}\,dx=\int\sec^2 x\,dx=\tan x+C\\
\int\frac{1}{sin^2x}\,dx=\int\csc^2 x\,dx=-\cot x+C\\
\int\frac{1}{\sqrt{1-x^2}}\,dx=\arcsin x+C=-\arccos x+C\\
\int\frac{1}{1+x^2}\,dx=\arctan x+C=-arccot\,x+C\\
\int\sec x\tan x\,dx=\sec x+C\\
\int\csc x\cot x\,dx=-\csc x+C\\
\int\tan x\,dx=-\ln\mid \cos x\mid +C\\
\int\cot x\,dx=\ln\mid \sin x\mid +C\\
\int\sec x\,dx=\ln\mid \sec x+\tan x\mid +C\\
\int\csc x\,dx=\ln\mid \csc x-\cot x\mid +C\\
\int sh\,x\,dx=ch\,x+C\\
\int ch\,x\,dx=sh\,x+C\\
\int\frac{1}{x^2+a^2}\,dx=\frac{1}{a}\arctan\frac{x}{a}+C\\
\int\frac{1}{x^2-a^2}\,dx=\frac{1}{2a}\ln\mid \frac{x-a}{x+a}\mid +C\\
\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin\frac{x}{a}+C\\
\int\frac{1}{\sqrt{x^2-a^2}}\,dx=\ln\mid x+\sqrt{x^2-a^2}\mid +C\\
\int\frac{1}{\sqrt{x^2+a^2}}\,dx=\ln\mid x+\sqrt{x^2+a^2}\mid +C\\ ∫ k d x = k x + C ∫ x a d x = a + 1 x a + 1 + C ∫ x 1 d x = ln ∣ x ∣ + C ∫ e x d x = e x + C ∫ a x d x = ln a a x + C ∫ cos x d x = sin x + C ∫ sin x d x = − cos x + C ∫ co s 2 x 1 d x = ∫ sec 2 x d x = tan x + C ∫ s i n 2 x 1 d x = ∫ csc 2 x d x = − cot x + C ∫ 1 − x 2 1 d x = arcsin x + C = − arccos x + C ∫ 1 + x 2 1 d x = arctan x + C = − a rcco t x + C ∫ sec x tan x d x = sec x + C ∫ csc x cot x d x = − csc x + C ∫ tan x d x = − ln ∣ cos x ∣ + C ∫ cot x d x = ln ∣ sin x ∣ + C ∫ sec x d x = ln ∣ sec x + tan x ∣ + C ∫ csc x d x = ln ∣ csc x − cot x ∣ + C ∫ s h x d x = c h x + C ∫ c h x d x = s h x + C ∫ x 2 + a 2 1 d x = a 1 arctan a x + C ∫ x 2 − a 2 1 d x = 2 a 1 ln ∣ x + a x − a ∣ + C ∫ a 2 − x 2 1 d x = arcsin a x + C ∫ x 2 − a 2 1 d x = ln ∣ x + x 2 − a 2 ∣ + C ∫ x 2 + a 2 1 d x = ln ∣ x + x 2 + a 2 ∣ + C
积分求几何量
弧长
若简单闭曲线
{ x = x ( t ) , y = y ( t ) , t ∈ [ α , β ] \begin{cases}
x=x(t),\\
y=y(t),
\end{cases}
t\in[\alpha,\beta] { x = x ( t ) , y = y ( t ) , t ∈ [ α , β ]
端点处重合(x ( α ) = x ( β ) , y ( α ) = y ( β ) x(\alpha)=x(\beta),y(\alpha)=y(\beta) x ( α ) = x ( β ) , y ( α ) = y ( β ) )且其他地方不自交,x ( t ) , y ( t ) x(t),y(t) x ( t ) , y ( t ) 连续且满足
[ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 ≠ 0 , ∀ t ∈ [ α , β ] [x'(t)]^2+[y'(t)]^2\ne0,\forall t\in[\alpha,\beta] [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 = 0 , ∀ t ∈ [ α , β ]
此时称曲线光滑,其长度
s = ∫ α β [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t s=\int_\alpha^\beta\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt s = ∫ α β [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t
此式可对称推广到高维空间曲线。
极坐标下,
r = r ( θ ) , θ ∈ [ α , β ] r=r(\theta),\theta\in[\alpha,\beta] r = r ( θ ) , θ ∈ [ α , β ]
的长度为
s = ∫ α β [ r ( θ ) ] 2 + [ r ′ ( θ ) ] 2 d θ s=\int_\alpha^\beta\sqrt{[r(\theta)]^2+[r'(\theta)]^2}\,d\theta s = ∫ α β [ r ( θ ) ] 2 + [ r ′ ( θ ) ] 2 d θ
面积
若简单闭曲线
{ x = x ( t ) , y = y ( t ) , t ∈ [ α , β ] \begin{cases}
x=x(t),\\
y=y(t),
\end{cases}
t\in[\alpha,\beta] { x = x ( t ) , y = y ( t ) , t ∈ [ α , β ]
端点处连续(x ( α ) = x ( β ) , y ( α ) = y ( β ) x(\alpha)=x(\beta),y(\alpha)=y(\beta) x ( α ) = x ( β ) , y ( α ) = y ( β ) )且其他地方不自交,x ( t ) , y ( t ) x(t),y(t) x ( t ) , y ( t ) 都逐段有连续微商,则此闭合曲线围起来的有界区域面积
S = − ∫ α β x ′ ( t ) y ( t ) d t = − ∫ α β y ( t ) d x ( t ) = − ∮ Γ y d x = ∮ Γ x d y S=-\int_\alpha^\beta x'(t)y(t)\,dt=-\int_\alpha^\beta y(t)\,dx(t)=-\oint_\Gamma y\,dx=\oint_\Gamma x\,dy S = − ∫ α β x ′ ( t ) y ( t ) d t = − ∫ α β y ( t ) d x ( t ) = − ∮ Γ y d x = ∮ Γ x d y
等式右边称为曲线Γ \Gamma Γ 上的积分,其计算方法是带入参数方程到定积分计算式中,积分上下限为始点与终点对应的参数值。下限并不总是小于上限,参数从下限到上限变化时对应曲线的正向(沿正向观察时,曲线所围的区域永远在左侧)。
极坐标下,连续非负曲线r = r ( θ ) r=r(\theta) r = r ( θ ) 与向径θ = α , θ = β \theta=\alpha,\theta=\beta θ = α , θ = β ,其中0 ≤ β − α ≤ 2 π 0\leq\beta-\alpha\leq2\pi 0 ≤ β − α ≤ 2 π 所围成的平面图形面积
S = 1 2 ∫ α β r 2 ( θ ) d θ S=\frac{1}{2}\int_\alpha^\beta r^2(\theta)\,d\theta S = 2 1 ∫ α β r 2 ( θ ) d θ
体积
记立体过 x 点且垂直于 x 轴的截面面积为S ( x ) S(x) S ( x ) ,则其体积
V = ∫ a b S ( x ) d x V=\int_a^bS(x)\,dx V = ∫ a b S ( x ) d x
连续曲线y = f ( x ) ≥ 0 , x ∈ [ a , b ] y=f(x)\ge 0,x\in[a,b] y = f ( x ) ≥ 0 , x ∈ [ a , b ] 绕 x 轴旋转一周产生的旋转体体积
V = π ∫ a b y 2 d x V=\pi\int_a^by^2\,dx V = π ∫ a b y 2 d x
旋转体侧面积
若曲线由参数方程
{ x = x ( t ) , y = y ( t ) , t ∈ [ α , β ] \begin{cases}
x=x(t),\\
y=y(t),
\end{cases}
t\in[\alpha,\beta] { x = x ( t ) , y = y ( t ) , t ∈ [ α , β ]
给出,则其绕 x 轴旋转体的侧面积
s = 2 π ∫ α β y ( t ) [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t s=2\pi\int_\alpha^\beta y(t)\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt s = 2 π ∫ α β y ( t ) [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t
方向导数
设三元函数u = f ( x , y , z ) u=f(x,y,z) u = f ( x , y , z ) 在点P 0 ( x 0 , y 0 , z 0 ) P_0(x_0,y_0,z_0) P 0 ( x 0 , y 0 , z 0 ) 的某邻域内有定义,任意给定始于点P 0 P_0 P 0 的射线l l l ,P ( x , y , z ) P(x,y,z) P ( x , y , z ) 为 l 上且含于定义域内的点。若极限
lim r ( p , p 0 ) → 0 + f ( P ) − f ( P 0 ) r ( P , P 0 ) = lim r ( p , p 0 ) → 0 + Δ l f ( P 0 ) r ( P , P 0 ) \lim_{r(p,p_0)\to0^+}\frac{f(P)-f(P_0)}{r(P,P_0)}=\lim_{r(p,p_0)\to0^+}\frac{\Delta_lf(P_0)}{r(P,P_0)} r ( p , p 0 ) → 0 + lim r ( P , P 0 ) f ( P ) − f ( P 0 ) = r ( p , p 0 ) → 0 + lim r ( P , P 0 ) Δ l f ( P 0 )
存在,则称该极限值为函数f f f 在点P 0 P_0 P 0 沿方向l l l 的方向导数,记为
∂ f ∂ l ∣ P 0 \frac{\partial f}{\partial l}\mid _{P_0} ∂ l ∂ f ∣ P 0 或∂ f ( P 0 ) ∂ l \frac{\partial f(P_0)}{\partial l} ∂ l ∂ f ( P 0 ) ,Δ l f ( P 0 ) r ( P , P 0 ) \frac{\Delta_lf(P_0)}{r(P,P_0)} r ( P , P 0 ) Δ l f ( P 0 ) 称为函数在P 0 P_0 P 0 点沿l l l 方向的增量。
特别地,∂ f ( P 0 ) ∂ x \frac{\partial f(P_0)}{\partial x} ∂ x ∂ f ( P 0 ) 就是函数在P 0 P_0 P 0 点沿x x x 轴正向的方向导数,y , z y,z y , z 轴上的方向导数同理。若函数在P 0 P_0 P 0 点可微,则其在P 0 P_0 P 0 沿任何方向l l l 的方向导数都存在,则有以下公式
∂ f ( P 0 ) ∂ l = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) ∣ P 0 ⋅ l 0 ⃗ \frac{\partial f(P_0)}{\partial l}=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})\mid _{P_0}\cdot\vec{l_0} ∂ l ∂ f ( P 0 ) = ( ∂ x ∂ f , ∂ y ∂ f , ∂ z ∂ f ) ∣ P 0 ⋅ l 0
其中l 0 ⃗ = ( cos α , cos β , c o s γ ) = 1 ρ ( Δ x , Δ y , Δ z ) \vec{l_0}=(\cos\alpha,\cos\beta,cos\gamma)=\frac{1}{\rho}(\Delta x,\Delta y,\Delta z) l 0 = ( cos α , cos β , cos γ ) = ρ 1 ( Δ x , Δ y , Δ z ) 为l l l 的方向余弦。
曲率
若曲线由参数方程
{ x = x ( t ) , y = y ( t ) , t ∈ [ α , β ] \begin{cases}
x=x(t),\\
y=y(t),
\end{cases}
t\in[\alpha,\beta] { x = x ( t ) , y = y ( t ) , t ∈ [ α , β ]
给出且有二阶微商,则其在一点的曲率
K = ∣ y ′ ′ x ′ − y ′ x ′ ′ ∣ [ x ′ 2 + y ′ 2 ] 3 2 K=\frac{\mid y''x'-y'x''\mid }{[x'^2+y'^2]^{\frac{3}{2}}} K = [ x ′2 + y ′2 ] 2 3 ∣ y ′′ x ′ − y ′ x ′′ ∣
若y = f ( x ) y=f(x) y = f ( x ) ,则
K = ∣ y ′ ′ ∣ ( 1 + y ′ 2 ) 3 2 K=\frac{\mid y''\mid }{(1+y'^2)^\frac{3}{2}} K = ( 1 + y ′2 ) 2 3 ∣ y ′′ ∣
同时记1 K \frac{1}{K} K 1 为曲率半径。
空间曲线的切线与法平面
若已知曲线上一点P ( x 0 , y 0 , z 0 ) P(x_0,y_0,z_0) P ( x 0 , y 0 , z 0 ) 处的切向量为τ ( x 0 , y 0 , z 0 ) = ( A , B , C ) \tau(x_0,y_0,z_0)=(A,B,C) τ ( x 0 , y 0 , z 0 ) = ( A , B , C ) 则曲线在该点的切线方程为
x − x 0 A = y − y 0 B = z − z 0 C \frac{x-x_0}A=\frac{y-y_0}B=\frac{z-z_0}C A x − x 0 = B y − y 0 = C z − z 0
法平面方程为
A ( x − x 0 ) + B ( y − y 0 ) + C ( z − z 0 ) = 0 A(x-x_0)+B(y-y_0)+C(z-z_0)=0 A ( x − x 0 ) + B ( y − y 0 ) + C ( z − z 0 ) = 0
当曲线由参数方程
{ x = x ( t ) , y = y ( t ) , z = z ( t ) , t ∈ [ α , β ] \begin{cases}
x=x(t),\\
y=y(t),\\
z=z(t),
\end{cases}
t\in[\alpha,\beta] ⎩ ⎨ ⎧ x = x ( t ) , y = y ( t ) , z = z ( t ) , t ∈ [ α , β ]
给出时,曲线在 P 点的切向量为
τ = ± ( x ′ ( t 0 ) , y ′ ( t 0 ) , z ′ ( t 0 ) ) \tau=\pm(x'(t_0),y'(t_0),z'(t_0)) τ = ± ( x ′ ( t 0 ) , y ′ ( t 0 ) , z ′ ( t 0 ))
更一般地,若曲线用两曲面的交线给出
{ F ( x , y , z ) = 0 , G ( x , y , z ) = 0 , \begin{cases}
F(x,y,z)=0,\\
G(x,y,z)=0,
\end{cases} { F ( x , y , z ) = 0 , G ( x , y , z ) = 0 ,
且在 P 点的某邻域能确定函数组y = y ( x ) , z = z ( x ) y=y(x),z=z(x) y = y ( x ) , z = z ( x ) 满足y 0 = y ( x 0 ) , z 0 = z ( x 0 ) y_0=y(x_0),z_0=z(x_0) y 0 = y ( x 0 ) , z 0 = z ( x 0 ) ,且y ’ ( x ) , z ’ ( x ) y’(x),z’(x) y ’ ( x ) , z ’ ( x ) 存在,则曲线在 P 点的切向量
τ = ± ( ∂ ( F , G ) ∂ ( y , z ) , ∂ ( F , G ) ∂ ( z , x ) , ∂ ( F , G ) ∂ ( x , y ) ) \tau=\pm(\frac{\partial(F,G)}{\partial(y,z)},\frac{\partial(F,G)}{\partial(z,x)},\frac{\partial(F,G)}{\partial(x,y)}) τ = ± ( ∂ ( y , z ) ∂ ( F , G ) , ∂ ( z , x ) ∂ ( F , G ) , ∂ ( x , y ) ∂ ( F , G ) )
空间曲面的切平面与法线
若已知曲面上一点P ( x 0 , y 0 , z 0 ) P(x_0,y_0,z_0) P ( x 0 , y 0 , z 0 ) 处的切平面的法向量为n ⃗ = ( A ’ , B ’ , C ’ ) \vec n=(A’,B’,C’) n = ( A ’ , B ’ , C ’ ) 则曲线在该点的法线方程为
x − x 0 A ′ = y − y 0 B ′ = z − z 0 C ′ \frac{x-x_0}{A'}=\frac{y-y_0}{B'}=\frac{z-z_0}{C'} A ′ x − x 0 = B ′ y − y 0 = C ′ z − z 0
切平面方程为
A ′ ( x − x 0 ) + B ′ ( y − y 0 ) + C ′ ( z − z 0 ) = 0 A'(x-x_0)+B'(y-y_0)+C'(z-z_0)=0 A ′ ( x − x 0 ) + B ′ ( y − y 0 ) + C ′ ( z − z 0 ) = 0
当曲面方程为π : F ( x , y , z ) = 0 \pi:F(x,y,z)=0 π : F ( x , y , z ) = 0 在曲面上任取一条过 P 的曲线,设其方程为
{ x = x ( t ) , y = y ( t ) , z = z ( t ) , t ∈ [ α , β ] \begin{cases}
x=x(t),\\
y=y(t),\\
z=z(t),
\end{cases}
t\in[\alpha,\beta] ⎩ ⎨ ⎧ x = x ( t ) , y = y ( t ) , z = z ( t ) , t ∈ [ α , β ]
此时有F ( x ( t ) , y ( t ) , z ( t ) ) = 0 F(x(t),y(t),z(t))=0 F ( x ( t ) , y ( t ) , z ( t )) = 0 令t = t 0 t=t_0 t = t 0 两边对 t 求导,并写成向量的内积式,得
( F x , F y , F z ) P ⋅ ( x ′ ( t 0 ) , y ′ ( t 0 ) , z ′ ( t 0 ) ) = 0 (F_x,F_y,F_z)_P\cdot(x'(t_0),y'(t_0),z'(t_0))=0 ( F x , F y , F z ) P ⋅ ( x ′ ( t 0 ) , y ′ ( t 0 ) , z ′ ( t 0 )) = 0
则曲线在 P 点的法向量为
n ⃗ = ± ( F x , F y , F z ) P \vec{n}=\pm(F_x,F_y,F_z)_P n = ± ( F x , F y , F z ) P
若曲线由参数方程给出
{ x = x ( u , v ) , y = y ( u , v ) , z = z ( u , v ) , \begin{cases}
x=x(u,v),\\
y=y(u,v),\\
z=z(u,v),
\end{cases} ⎩ ⎨ ⎧ x = x ( u , v ) , y = y ( u , v ) , z = z ( u , v ) ,
则曲线在 P 点的法向量
n ⃗ = ± ( ∂ ( y , z ) ∂ ( u , v ) , ∂ ( z , x ) ∂ ( u , v ) , ∂ ( x , y ) ∂ ( u , v ) ) \vec{n}=\pm(\frac{\partial(y,z)}{\partial(u,v)},\frac{\partial(z,x)}{\partial(u,v)},\frac{\partial(x,y)}{\partial(u,v)}) n = ± ( ∂ ( u , v ) ∂ ( y , z ) , ∂ ( u , v ) ∂ ( z , x ) , ∂ ( u , v ) ∂ ( x , y ) )
高阶导数与泰勒公式
用f ( n ) ( x ) f^{(n)}(x) f ( n ) ( x ) 表示 f(x)的 n 阶导数,只要让余项<EPS
即可计算指定函数到任意精确度,特别取 a=0 时称为麦克劳林公式。
f ( x ) = f ( a ) + f ( 1 ) ( a ) ( x − a ) + f ( 2 ) ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( n ) ( a ) n ! ( x − a ) n + R n ( x ) f(x)=f(a)+f^{(1)}(a)(x-a)+\frac{f^{(2)}(a)}{2!}(x-a)^2+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n+R_n(x) f ( x ) = f ( a ) + f ( 1 ) ( a ) ( x − a ) + 2 ! f ( 2 ) ( a ) ( x − a ) 2 + ⋯ + n ! f ( n ) ( a ) ( x − a ) n + R n ( x )
佩亚诺余项
R n ( x ) = o ( ( x − a ) n ) R_n(x)=o((x-a)^n) R n ( x ) = o (( x − a ) n )
积分余项
R n ( x ) = 1 n ! ∫ a x ( x − t ) n f ( n + 1 ) ( t ) d t R_n(x)=\frac{1}{n!}\int_a^x(x-t)^nf^{(n+1)}(t)\,dt R n ( x ) = n ! 1 ∫ a x ( x − t ) n f ( n + 1 ) ( t ) d t
拉格朗日余项
R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − a ) n + 1 , a < ξ < x R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1},a<\xi<x R n ( x ) = ( n + 1 )! f ( n + 1 ) ( ξ ) ( x − a ) n + 1 , a < ξ < x
柯西余项
R n ( x ) = ( x − a ) n + 1 n ! ( 1 − θ ) n f ( n + 1 ) ( a + θ ( x − a ) ) , 0 < θ < 1 R_n(x)=\frac{(x-a)^{n+1}}{n!}(1-\theta)^nf^{(n+1)}(a+\theta(x-a)),0<\theta<1 R n ( x ) = n ! ( x − a ) n + 1 ( 1 − θ ) n f ( n + 1 ) ( a + θ ( x − a )) , 0 < θ < 1
对数函数
[ ln ( 1 + x ) ] ( n ) = ( − 1 ) n − 1 ( n − 1 ) ! ( 1 + x ) − n ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + ⋯ + ( − 1 ) n − 1 x n n + R n ( x ) [\ln(1+x)]^{(n)}=(-1)^{n-1}(n-1)!(1+x)^{-n}
\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots+(-1)^{n-1}\frac{x^n}{n}+R_n(x) [ ln ( 1 + x ) ] ( n ) = ( − 1 ) n − 1 ( n − 1 )! ( 1 + x ) − n ln ( 1 + x ) = x − 2 x 2 + 3 x 3 − 4 x 4 + ⋯ + ( − 1 ) n − 1 n x n + R n ( x )
幂函数
[ ( 1 + x ) a ] ( n ) = a ( a − 1 ) … ( a − n + 1 ) ( 1 + x ) a − n ( 1 + x ) a = 1 + a x + a ( a − 1 ) 2 ! x 2 + ⋯ + a ( a − 1 ) … ( a − n + 1 ) n ! x n + R n ( x ) [(1+x)^a]^{(n)}=a(a-1)\dots(a-n+1)(1+x)^{a-n}\\
(1+x)^a=1+ax+\frac{a(a-1)}{2!}x^2+\dots+\frac{a(a-1)\dots(a-n+1)}{n!}x^n+R_n(x) [( 1 + x ) a ] ( n ) = a ( a − 1 ) … ( a − n + 1 ) ( 1 + x ) a − n ( 1 + x ) a = 1 + a x + 2 ! a ( a − 1 ) x 2 + ⋯ + n ! a ( a − 1 ) … ( a − n + 1 ) x n + R n ( x )
三角函数
( sin x ) ( n ) = sin ( x + n π 2 ) sin x = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + ⋯ + ( − 1 ) k − 1 x 2 k − 1 ( 2 k − 1 ) ! + R ∗ 2 k ( x ) R ∗ 2 k ( x ) = ( − 1 ) k cos θ x ( 2 k + 1 ) ! x 2 k + 1 ( cos x ) ( n ) = cos ( x + n π 2 ) cos x = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! + ⋯ + ( − 1 ) k − 1 x 2 k − 2 ( 2 k − 2 ) ! + R ∗ 2 k − 1 ( x ) R ∗ 2 k − 1 ( x ) = ( − 1 ) k cos θ x ( 2 k ) ! x 2 k (\sin x)^{(n)}=\sin(x+\frac{n\pi}{2})\\
\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots+(-1)^{k-1}\frac{x^{2k-1}}{(2k-1)!}+R*{2k}(x)
\\R*{2k}(x)=(-1)^k\frac{\cos\theta x}{(2k+1)!}x^{2k+1}\\
(\cos x)^{(n)}=\cos(x+\frac{n\pi}{2})\\
\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots+(-1)^{k-1}\frac{x^{2k-2}}{(2k-2)!}+R*{2k-1}(x)\\
R*{2k-1}(x)=(-1)^k\frac{\cos\theta x}{(2k)!}x^{2k} ( sin x ) ( n ) = sin ( x + 2 nπ ) sin x = x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + ⋯ + ( − 1 ) k − 1 ( 2 k − 1 )! x 2 k − 1 + R ∗ 2 k ( x ) R ∗ 2 k ( x ) = ( − 1 ) k ( 2 k + 1 )! cos θ x x 2 k + 1 ( cos x ) ( n ) = cos ( x + 2 nπ ) cos x = 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + ⋯ + ( − 1 ) k − 1 ( 2 k − 2 )! x 2 k − 2 + R ∗ 2 k − 1 ( x ) R ∗ 2 k − 1 ( x ) = ( − 1 ) k ( 2 k )! cos θ x x 2 k
指数函数
( e x ) ( n ) = e x e x = 1 + x + x 2 2 ! + x 3 3 ! + ⋯ + x n n ! + R n ( x ) R n ( x ) = e θ x ( n + 1 ) ! x n + 1 , ξ = θ x , 0 < θ < 1 (e^x)^{(n)}=e^x\\
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!}+R_n(x)\\
R_n(x)=\frac{e^{\theta x}}{(n+1)!}x^{n+1},\xi=\theta x,0<\theta<1 ( e x ) ( n ) = e x e x = 1 + x + 2 ! x 2 + 3 ! x 3 + ⋯ + n ! x n + R n ( x ) R n ( x ) = ( n + 1 )! e θ x x n + 1 , ξ = θ x , 0 < θ < 1
二元函数
设f ( x , y ) f(x,y) f ( x , y ) 在P 0 ( x 0 , y 0 ) P_0(x_0,y_0) P 0 ( x 0 , y 0 ) 的某邻域O ( P 0 ) O(P_0) O ( P 0 ) 内有直到n + 1 n+1 n + 1 阶连续偏导数,则对O ( P 0 ) O(P_0) O ( P 0 ) 内∀ ( x 0 + Δ x , y 0 + Δ y ) , ∃ θ ∈ ( 0 , 1 ) \forall(x_0+\Delta x,y_0+\Delta y),\exists\theta\in(0,1) ∀ ( x 0 + Δ x , y 0 + Δ y ) , ∃ θ ∈ ( 0 , 1 ) ,使得
f ( x ∗ 0 + Δ x , y 0 + Δ y ) = ∑ ∗ k = 0 n 1 k ! ( ∂ ∂ x Δ x + ∂ ∂ y Δ y ) k f ( x 0 , y 0 ) + R n f(x*0+\Delta x,y_0+\Delta y)=\sum*{k=0}^n\frac{1}{k!}(\frac{\partial}{\partial x}\Delta x+\frac{\partial}{\partial y}\Delta y)^kf(x_0,y_0)+R_n f ( x ∗ 0 + Δ x , y 0 + Δ y ) = ∑ ∗ k = 0 n k ! 1 ( ∂ x ∂ Δ x + ∂ y ∂ Δ y ) k f ( x 0 , y 0 ) + R n
其中
R n = 1 ( n + 1 ) ! ( ∂ ∂ x Δ x + ∂ ∂ y Δ y ) n + 1 f ( x 0 + θ Δ x , y 0 + θ Δ y ) R_n=\frac{1}{(n+1)!}(\frac{\partial}{\partial x}\Delta x+\frac{\partial}{\partial y}\Delta y)^{n+1}f(x_0+\theta\Delta x,y_0+\theta\Delta y) R n = ( n + 1 )! 1 ( ∂ x ∂ Δ x + ∂ y ∂ Δ y ) n + 1 f ( x 0 + θ Δ x , y 0 + θ Δ y )
级数部分和
幂级数
快速计算幂级数的部分和∑ i = 1 n i k m o d M \sum_{i=1}^ni^k\mod M ∑ i = 1 n i k mod M 可借助伯努利数,详见组合数学模板。
∑ i = 1 n i 1 = 1 2 n ( n + 1 ) ∑ i = 1 n i 2 = 1 6 n ( n + 1 ) ( 2 n + 1 ) ∑ i = 1 n i 3 = 1 4 [ n ( n + 1 ) ] 2 ∑ i = 1 n i 4 = 1 30 n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) ∑ i = 1 n i 5 = 1 12 [ n ( n + 1 ) ] 2 ( 2 n 2 + 2 n − 1 ) ∑ i = 1 n i 6 = 1 42 n ( n + 1 ) ( 2 n + 1 ) ( 3 n 4 + 6 n 3 − 3 n + 1 ) \sum_{i=1}^ni^1=\frac 1 2n(n+1)\\
\sum_{i=1}^ni^2=\frac 1 6n(n+1)(2n+1)\\
\sum_{i=1}^ni^3=\frac 1 4[n(n+1)]^2\\
\sum_{i=1}^ni^4=\frac 1{30}n(n+1)(2n+1)(3n^2+3n-1)\\
\sum_{i=1}^ni^5=\frac 1{12}[n(n+1)]^2(2n^2+2n-1)\\
\sum_{i=1}^ni^6=\frac 1{42}n(n+1)(2n+1)(3n^4+6n^3-3n+1) i = 1 ∑ n i 1 = 2 1 n ( n + 1 ) i = 1 ∑ n i 2 = 6 1 n ( n + 1 ) ( 2 n + 1 ) i = 1 ∑ n i 3 = 4 1 [ n ( n + 1 ) ] 2 i = 1 ∑ n i 4 = 30 1 n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) i = 1 ∑ n i 5 = 12 1 [ n ( n + 1 ) ] 2 ( 2 n 2 + 2 n − 1 ) i = 1 ∑ n i 6 = 42 1 n ( n + 1 ) ( 2 n + 1 ) ( 3 n 4 + 6 n 3 − 3 n + 1 )
调和级数
n → ∞ , ∑ _ i = 1 n 1 i → ln n + r , r ≈ 0.5772156649015328 … n\to\infty,\sum\_{i=1}^n\frac 1 i\to\ln n+r,r\approx0.5772156649015328\dots n → ∞ , ∑ _ i = 1 n i 1 → ln n + r , r ≈ 0.5772156649015328 …
二分求零点、三分求极值点
需要f ( x ) f(x) f ( x ) 在区间[ l , r ] [l,r] [ l , r ] 上单调/凹凸性唯一。
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lf bs ( lf l , lf r , lf f ( lf x ))
{
if ( r - l < EPS )
return l ;
lf m = ( l + r ) / 2 ;
return sgn ( f ( l ) * f ( m )) < 0 ? bs ( l , m , f ) : ts ( m , r , f );
}
lf ts ( lf l , lf r , lf f ( lf x ))
{
if ( r - l < EPS )
return l ;
lf d = ( r - l ) / 3 , lm = l + d , rm = r - d ;
return f ( lm ) < f ( rm ) ? ts ( l , rm , f ) : ts ( lm , r , f ); //极小值
}
自适应辛普森求积分
使用示例
这篇论文 论证了加一个十五分之一的偏移收敛会比较快…
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struct Simpson
{
lf simpson ( lf a , lf b , lf f ( lf x ))
{
return ( f ( a ) + 4 * f (( a + b ) / 2 ) + f ( b )) * ( b - a ) / 6 ;
}
lf ask ( lf a , lf b , lf f ( lf x ), lf e = EPS )
{
lf c = ( a + b ) / 2 , L = simpson ( a , c , f ), R = simpson ( c , b , f ), delta = ( L + R - simpson ( a , b , f )) / 15 ;
return fabs ( delta ) < e ? L + R + delta : ask ( a , c , f , e / 2 ) + ask ( c , b , f , e / 2 );
}
};
插值法
拉格朗日插值法:插值多项式和插值基函数的形式对称,容易编程。但是,增加节点时,需要重新计算每一个插值基函数。要在( m o d p ) \pmod p ( mod p ) 意义下进行的话,那么 p 只能是质数。
牛顿插值法:当插值节点增加时,之前已计算的结果仍然能用,每增加一个节点,只要再增加一项即可,从而避免了重复性计算。如果要 mod 非质数的话,那么就要用牛顿插值法。
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typedef complex < lf > Coord ;
#define X real()
#define Y imag()
struct Lagrange
{
lf ask ( const vector < Coord > & p , lf x ) //返回p确定的多项式函数在x处的值
{
lf ret = 0 ;
for ( int i = 0 ; i < p . size (); ++ i )
{
lf tmp = p [ i ]. Y ;
for ( int j = 0 ; j < p . size (); ++ j )
if ( i != j )
tmp *= ( x - p [ j ]. X ) / ( p [ i ]. X - p [ j ]. X );
ret += tmp ;
}
return ret ;
}
vector < lf > ask ( vector < Coord > p ) //返回p确定的多项式系数向量
{
vector < lf > ret ( p . size ()), sum ( p . size ());
ret [ 0 ] = p [ 0 ]. Y , sum [ 0 ] = 1 ;
for ( int i = 1 ; i < p . size (); ++ i )
{
for ( int j = p . size () - 1 ; j >= i ; -- j )
p [ j ]. Y = ( p [ j ]. Y - p [ j - 1 ]. Y ) / ( p [ j ]. X - p [ j - i ]. X );
for ( int j = i ; ~ j ; -- j )
sum [ j ] = ( j ? sum [ j - 1 ] : 0 ) - sum [ j ] * p [ i - 1 ]. X ,
ret [ j ] += sum [ j ] * p [ i ]. Y ;
}
return ret ;
}
};
struct Newton
{
lf differenceQuotient ( const vector < Coord > & p , int k ) //计算差商
{
lf ret = 0 ;
for ( int i = 0 ; i <= k ; ++ i )
{
lf tmp = p [ i ]. Y ;
for ( int j = 0 ; j <= k ; ++ j )
if ( i != j )
tmp /= p [ i ]. X - p [ j ]. X ;
ret += tmp ;
}
return ret ;
}
lf ask ( const vector < Coord > & p , lf x )
{
lf ret = p [ 0 ]. Y ;
for ( int i = 1 ; i < p . size (); ++ i )
{
lf tmp = differenceQuotient ( p , i ); //多次求,可O(n^3)预处理优化
for ( int j = 0 ; j < i ; ++ j )
tmp *= x - p [ j ]. X ;
ret += tmp ;
}
return ret ;
}
};