高中 (secx)^2-1=(tanx)^2 csc2x-1=(cotx)^2

cos2x=cosx^2-sinx^2=1-(sinx)^2=(cosx)^2*2-1

2sinacosb=sin(a+b)-sin(a-b) 2sinasinb=cos(a-b)-cos(a+b)

cosacosb=(cos(a-b)+cos(a+b))/2?

sin(x+y)=sinxcosy+sinycosx cos(a+b)=cosacosb-sinasinb

平面夾角 cos=i1*i2+j1*j2+k1*k2/|i1*i1+j1*j1+k1*k1|^0.5/|i2*i2+j2*j2+k2*k2|^0.5

(e^(2x+y))'x=e^(2x+y)*(2cos(x-y)-sin(x-y))dx

(arcsinx)'=(1-x*x)^-0.5

(arctan)'=1/(1+x*x) $arctanx/(1+x*x)dx=(arctanx)^2/2

(tanx)'=(secx)^2 (sec)'=secxtanx

求導 帶根號 x*x為正用ln|x+(x*x+a*a)^0.5 為負用arcsin

不帶根號 兩項相加用arctan(x/a) 相減用1/2a*ln|(a+x)/(a-x)|等

e^x=1+x=1+x+x*x/2=1+x+x^2/(2!)+x^n/(n!)

ln(1+x)=x=x+(-1)^n*(x^n)/n

(1+x)^a=1+ax=1+ax+a(a-1)/(2!)+(a*(a-1)*(a-2)...(a-n+1))/(n!)

1/(1-x)=1+x+x^2+x^3+x^n

(a^x)'=a^xlna (log(a,x))'=1/x/(lna)

sinx=x=x-x^3/6=x-x^3/(3!)+(-1)^n*x^(2n+1)/((2n+1)!)

tanx=x=x+x^3/3 arctanx=x

cosx=1-x*x/2=1+x^(2n)/(2!)+(-1)^n*x^(2n)/((2n)!)

(1+x)^(1/n)=1+1/n

x^x=x^x*(1*lnx+x*(1/x))=x^x+x^x*lnx=e^(xlnx)

a^b=e^(alnb)

(e^x-1)^3=x^3

Taylor:(1-x*x)^0.5=1-x*x/2

(1+1/x)^0.5=1+1/2x-1/8/x/x

(1+2/x)^0.5=1+1/x-1/2/x/x

$xdln(x+(1+x*x)^0.5) t=(1+x*x)^0.5 =$x/(x+t)*(1+2x/2t)dx=$x/tdx=t

e^(tanx)=1+tanx+(tanx)^n/(n!)

研cosx=1-x*x/2+4^4/(4!)

arcsinx=x+x^3/6

(1-cos2x)'=2sin2x

|1+x*x|^-1=1-x

ln[(a/b)^(c/x)]=c/x*(lna-lnb)

$xcos(x+y)dy=xsin(x+y)

tcost=(cost+tsint)' tsint=(sint-tcost)'

包含乘數(shù)求導 (ln(x*x+y*y))'=2x/(x*x+y*y) (sinxy)'=ycosxy

偏導 &pz/&px=-F(x,y,z')/F(x',y,z) 全微分dz=f'(x)dx+f'(y)dy

曲線在某點x0,y0,z0法線 F(xyz)分別求導為f 求完代入?

法線方程F=f(x,y,z)分別求導代入得結果x0 y0 z0 (x-px)/x0=(y-py)/y0=(z-pz)/z0

法平面方程 f'(x)|x0*(x-px)+f'(y)|y0*(y-py)+f'(z)|z0*(z-pz)=0

函數(shù)沿方向導數(shù) &f分別偏導再代入求值 方向L分別求cos 然后每個偏導乘cos代入

極值就是x和y分別求導一次聯(lián)等求駐點，再導得三個二次導結果，delta=f(xy)^2-f(xx)*f(yy)

如果delta<0 f(x,y)<0是極大值，f(x,y)>0是極小值

gradf(x,y)=f(f'(x),f'(y)) grad點處梯度=i&pf/&px+j&pf/&py+k&pf/&pz微分后代入點值，結果是二維坐標

切平面 等式分別求導代入值 f(x)*(x-x0)+f(y)*(y-y0)+f(z)*(z-z0)=0

求體積且xy對稱 x*x+y*y=r*r $(0,2pi)dtheta$(0,1)dr$(r*r,1)f(x,y,z)dz

-------換元積分∫ x=t*t, dx=(t*t)'/(t*t)'dx=2tdt v=y*y 2ydy=dv

(a^2-x^2)^0.5,x=asint, dx=acostdt

(a^2+x^2)^0.5,x=atant

(x^2-a^2)^0.5,x=+-asect(0<t<pi/2)

$f(x)adx=$f(x)*(x')au'/(u')dx=$f(x)a*(x)'/(u')du

$tanxdx=-ln|cosx| $cotxdx=ln|sinx|

$secxdx=ln|secx+tanx|dx $cscxdx=ln|cscx-cotx|dx

$siny/ydx=(1-y)siny $-du/sinu=ln|cot(u/2)|

$f(x^0.5)*x^-0.5dx=2$fx^0.5dx^0.5

$f(1/x)/x/xdx=-$f(1/x)d(1/x)

$lnxdx=xlnx-x $f(lnx)/xdx=$f(lnx)dlnx $xln

(1+x*x)^0.5 or arctanx -> x=tant

$(x*x+a*a)^-1dx=arctan(x/a)/a $(x*x-a*a)^-1dx=ln|(x-a)/(x+a)|/2a $(a*a-x*x)^-0.5dx=arcsin(x/a)

$(bx*x+a*a)^-0.5dx=ln|x+(x*x+a*a)^0.5|*b' $(bx*x-a*a)^-0.5dx=ln|x+(x*x-a*a)^0.5|*b'

$(x*x+-a*a)^0.5dx=+-(a*a/2)ln|x+(x*x+a*a)^0.5|+x/2*(x*x+-a*A)^0.5

$tanx(tanx^2+1)=$tanxdtanx

(tanx)^2+1=((sinx)^2+1)/cosx/cosx

$cscxdx=ln|cscx-cotx| $secxdx=$sec(x+pi/2)d(x+pi/2)

$x/(x+1)/(x+1)dx=ln|x+1|+1/(x+1)

(1/sinx)'=cosx/sinx/sinx? $cosx/sinx/sinxdx=$1/sinxdx

-fxd(x*x)=fxd(4-x*x)

$((x-1)^2+4)^0.5d(x-1)=ln|x-1+((x-1)^2+4)^0.5|

x=asect dx=asecttantdt $arctanxdx=xarctanx-0.5ln(1+x*x)

uv'=(uv)'-u'v $udv=uv-$vdu 定 $u'vdx=u*v|-$v'udx

$xcosxdx=xsinx-$sinxdx=xsinx+cosx

$e^xd(x*x)=2$x(e)^xdx=2$xd(x^e)

$xdsin2x=xsin2x+(cos2x)/2

3$x*xd(sinx/x)=3x*x*sinx/x-6$sinxdx

x=sint,dx=costdt

定積分 $f'(x)dx|(a,b)=f(a)-f(b)

$x*x*ln(1+x)dx=1/3*$ln(x+1)dx^3

$e^(2x)cos3xdx=1/2*$cos3xd(e^x)

$f(x)/x/xdx=f(x)d(1/x)

sinx=2tan(x/2)/(1+tan(x/2)^2) cosx=2tan(x/2)/(1-tan(x/2)^2)

$1/(u*u-1)du=ln|(u-1)/(u+1)|=ln|u*u-1|-2ln(u+1)

$(0,0.5)xdarcsinx=$(0,0.5)x/((1-x*x)^0.5)dx=0.5*$(0,0.5)1/((1-x*x)^0.5)d(x*x)

$1/(x-a)^2+b*b, set x-a=btanu |u|<pi/2

$0,x,arcsintdt=$0,x,arctantdt=$0,x,sintdt=x*x/2

0.5$1/(x*x-x+1)=0.5$dx/((x-0.5)^2+3/4)=1/(3^0.5)*arctan((2x-1)/(3^0.5))

重積分 $(0,1)e^(-y*y)dy$(0,y)x*xdx=$(0,1)e^(-y*y)/3*x^3|(0,y)dy=1/3$(0,1)y^3e^(-y*y)dy

極坐標 $$f(x,y)dxdy=$$f(rcosθ,rsinθ)rdrdt=$(a,b)dθ$(φ1(θ),φ2(θ)f(rcosθ,rsinθ)rdr

二重 $(0,2pi)θdθ$(0,max(f(x,y)))f(x,y)rdr 三重 $(0,2pi)θdθ$(0,1)rdr$(min,max)f(x,y)dxdy

三重積分體積 $(xmin,xmax)dx*$(ymin,ymax)dy*(zmax-zmin)

曲線積分 每條線段相加 不直的就 ds=((x')^2+(y')^2)^0.5dx=(1+(dy/dx)^2)^0.5dx

if y=x*x $曲線xydx=$x^3dx 兩邊換元到一起比如dx然后積分

$(0,max(t))f(x)ds=$(0,max(t))f(x)*((x*x)'+(y*y)'+(z*z)')^0.5dt

$Lf(x,y)ds=$(a,b)f(fx(t),fy(t))*((fx'(x))^0.5+(fy'(y))^0.5)^0.5dt

如y=x*x ds=(1+(y')^2)^0.5dx I=$(0,1)x^2^0.5*ds=1/8$(0,1)(1+4xx)^0.5d(1+4xx)

xy=1 1/2,2->1,1 $lxds=$(1,2)1/y*(1+(x')^2)^0.5dy

ds=(1+(y')^2)dx $f(x)(y=x(y))*dx

$Lyds=$y(t)x'(t)dt

Green公式 $oPdx+Qdy=$$(Q'(x)/dx-P'(y)/dy)dxdy

微分方程 dy/dx=3

ln|y|=x^3+c y=e^(x^3+c)=ce^(x^3) c和e^c都是常數(shù)

齊次微分 ux=y dy/dx=u+xdu/dx dy/dx=f(y/x)=ux'+xu'=u+x*du/dx=f(u)

一階線性微分 dy/dx+P(x)y=Q(x) dy/y=-P(x)dx lny=-$P(x)dx+lnC y=Ce^(-$P(x)dx

y=($Q(x)e^($P(x)dx)dx+C)e^(-$P(x)dx)

dy/y=-P(x)dx 兩邊反求導 ln|y|=-$(P(x)dx+C1 y=Ce^(-$P(x)dx) C=+/-e^C1

p/p'=x->dp/p=dx/x->lnp=lnx+c

y'=p y''=dp/dx=(dp/dy)*(dy/dx)=p*dp/dy 分離dp與x 積dp/p=lnp 積分=(u/v)'=(u'v-uv')/v/v

y''=yx p'-p=x dp/p=dx lnp=x+lnc1 p=c1(e^x) p=u(x)*e^x p'=u'(x)*e^x+u(x)*e^x u'(x)=xe^-x

y''+by'+cy=0 r*r+br+c=0 y=c1*e^(r1*x)+c2*e^(r2*x)?

if r1=r2 y=(c1+c2x)*e^-rx if b*b<4ac y=e^(real*x)*(c1*cos(|image|*x)+c2*sin(|image|*x))

1/n發(fā)散 f(x)的發(fā)散性和f(x)^n的發(fā)散性相同

|u[n]|收斂是絕對收斂 u[n]收斂而|u[n]|發(fā)散是條件收斂 (-1)^n/n 條件收斂

a*q^n: |q|<1收斂 和為a/(1-q) 否則發(fā)散 1/(n^p) p>1收斂 p<=1發(fā)散 p=1調和級數(shù)

冪級數(shù)? t=x-1 (x-1)^n=t^n |t|<k 1-t<k k+1<t<k-1

sigma(n=0,max)a(x^n) (|x|<1)的和函數(shù)=a/(1-x)

sigma(n=0,max)x^n/(n!)=e^x

1/(1-x)=sigma(0)x^n 1/(1+x)=sigma(0)(-1)^n*x^n |x|<1

ln(1+x)=sigma(1)(-1)^(n-1)*x^n/n 1<x<=1 ln(1-x)=-sigma(1)x^n/n -1<=x<1

e^x=sigma(0)x^n/(n!) arctanx=x+(-1)^n/(2n+1)*x^(2n+1)

sinx=sigma(1)(-1)^(n-1)*x^(2n-1)/(2n-1)! cosx=sigma(0)-1^n*x^2n/(2n)!

1/(1+(x-a)/b)=sigma(0)(-1)^n/(b^n)*(x-a)^n (-b<|x-a|<b)

Taylor展開：=sigma(f(x)導n次/n!)

Euler: e^(ix)=cosx+isinx

Fourier間斷點是左右極限平均數(shù) a[0]=1/pi*$(-pi,pi)f(x)dx

a[n]=1/pi*$(-pi,pi)f(x)cosnxdx=2/pi*$(0,pi)f(x)cosnxdx=2/npi*f(x)sinnx u'v=uv-v'u

b[n]=1/pi*$(-pi,pi)f(x)sinnxdx

a<=x<=a+2l a[n]=1/l*$(a,a+2l)f(x)cos(n*pi*x/l)dx b[n]=1/l*$(a,a+2l)f(x)dins(n*pi*x/l)dx