草稿紙（練習 LaTeX 使用）

02-09

${ m e}^x=cosh x+sinh x$

${ m e}^{{ m i}x}=cos x+{ m i}sin x$

$sin x=frac{{ m e}^{{ m i}x}-{ m e}^{-{ m i}x}}{2{ m i}},sin { m i}x=frac{{ m e}^{-x}-{ m e}^{x}}{2{ m i}}=frac{-sinh x}{ m i}={ m i}sinh x$

$cos x=frac{{ m e}^{{ m i}x}+{ m e}^{-{ m i}x}}{2},cos{ m i}x=frac{{ m e}^{-x}+{ m e}^x}{2}=cosh x$

$sinh x=frac{{ m e}^x-{ m e}^{-x}}{2},sinh{ m i}x=frac{{ m e}^{{ m i}x}-{ m e}^{-{ m i}x}}{2}={ m i}cdotfrac{{ m e}^{{ m i}x}-{ m e}^{-{ m i}x}}{2{ m i}}={ m i}sin x$

$cosh x=frac{{ m e}^x+{ m e}^{-x}}{2},cosh{ m i}x=frac{{ m e}^{{ m i}x}+{ m e}^{-{ m i}x}}{2}=cos x$

$\$

$lg2=-lg0.5=-frac{ln0.5}{ln10}=frac{ln0.5}{ln0.1}$

$ln0.5=ln(1-0.5)=sum_{n=1}^{infty}{left(-frac{0.5^n}{n} ight)}$

$ln0.1=ln(1-0.9)=sum_{n=1}^{infty}{left(-frac{0.9^n}{n} ight)}$

$sqrt[10]{10}=10^{0.1}={ m e}^{0.1ln10}={ m e}^{-0.1ln0.1}$

$ln(1-x)=sum_{n=1}^{infty}{left(-frac{x^n}{n} ight)}$

$ln x=lim_{n ightarrow0}frac{x^n-1}n$

${ m e}^x=lim_{n ightarrowinfty}left(1+frac xn ight)^n=sum_{n=0}^{infty}{frac{x^n}{n!}}$

$pi=3.141 592 653 589 793...$

$z=Re z+{ m i}cdotIm z$

$Re(z^2)=(Re z)^2-(Im z)^2$

$Im(z^2)=2cdotRe zcdotIm z$

$|z|=sqrt{(Re z)^2+(Im z)^2}$

$設pm w 為複數 z 的兩個平方根,$

$當 Re z>0 且 Im z=0 時：Re w=sqrt{Re z} Im w=0$

$當 Re z<0 且 Im z=0 時：Re w=0 Im w=sqrt{-Re z}$

$當 Im z e0 時：Re w=sqrt{frac{|z|+Re z}{2}} Im w=frac{Im z}{|Im z|}cdotsqrt{frac{|z|-Re z}{2}}$

$(lg x)=left(frac{ln x}{ln10} ight)=frac{frac{ln10}{x}}{(ln10)^2}=frac{1}{xln10}=frac{lg{ m e}}{x}$

$ab=10^{lg a+lg b}$

$a=27, b=81$

$lg a=lg27=1.4314$

$lg b=lg 81=1.9085$

$1.4314+1.9085=3.3399$

$10^{3.3399}=2187$

$ab=frac{cos(arccos a+arccos b)+cos(arccos a-arccos b)}2$

$a=27, b=81$

$arccosleft(frac a{100} ight)=arccos0.27=74^{circ} 20 9$

$arccosleft(frac b{100} ight)=arccos0.81=35^{circ} 54 15$

$74^{circ} 20 9+35^{circ} 54 15=110^{circ} 14 24$

$74^{circ} 20 9-35^{circ} 54 15=38^{circ} 25 54$

$cos110^{circ} 14 24=-0.3460$

$cos38^{circ} 25 54=0.7834$

$frac{-0.3460+0.7834}2=0.2187$

$0.2187 imes10000=2187$

$lim_{n ightarrow0}frac{sin(x+n)-sin x}n$

$=lim_{n ightarrow0}frac{sin xcos n+cos xsin n-sin x}n$

$=lim_{n ightarrow0}frac{sin x(cos n-1)+cos xsin n}n$

$=lim_{n ightarrow0}frac{cos xsin n}n$

$=cos x$

$lim_{n ightarrow0}frac{cos(x+n)-cos x}n$

$=lim_{n ightarrow0}frac{cos xcos n-sin xsin n-cos x}n$

$=lim_{n ightarrow0}frac{cos x(cos n-1)-sin xsin n}n$

$=lim_{n ightarrow0}frac{-sin xsin n}n$

$=-sin x$

$lim_{n ightarrow0}frac{ an(x+n)- an x}n$

$=lim_{n ightarrow0}frac{frac{ an x+ an n}{1- an x an n}- an x}n$

$=lim_{n ightarrow0}frac{(1+ an^2 x) an n}{(1- an x an n)n}$

$=1+ an^2x=sec^2x$

$anx=left(frac{sin x}{cos x} ight)=frac{sinxcos x-sin xcosx}{cos^2 x}$

$=frac{cos^2x+sin^2x}{cos^2x}=frac{1}{cos^2x}=sec^2x$

$cotx=left(frac{cos x}{sin x} ight)=frac{cosxsin x-cos xsinx}{sin^2 x}$

$=frac{-sin^2x-cos^2x}{sin^2x}=-frac{1}{sin^2x}=-csc^2x$

$secx=left(frac1{cos x} ight)=frac{-cosx}{cos^2 x}$

$=frac{sin x}{cos^2x}=frac{ an x}{cos x}= an xsec x$

$cscx=left(frac1{sin x} ight)=frac{-sinx}{sin^2 x}$

$=-frac{cos x}{sin^2x}=-frac{cot x}{sin x}=-cot xcsc x$

$lim_{n ightarrow0}frac{arctan(x+n)-arctan x}{n}$

$=lim_{n ightarrow0}frac{arctanfrac{n}{1+(x+n)x}}{n}$

$=lim_{n ightarrow0}frac{frac{n}{1+x^2}}{n}$

$=frac1{1+x^2}$

$求半圓 y=sqrt{1-x^2} 的弧長。$

$解：k=y=left(sqrt{1-x^2} ight)cdotleft(1-x^2 ight)$

$=frac{1}{2sqrt{1-x^2}}cdot(-2x)=frac{-x}{sqrt{1-x^2}}$

$sqrt{1+k^2}=sqrt{1+left(frac{-x}{sqrt{1-x^2}} ight)^2}=sqrt{1+frac{x^2}{1-x^2}}$

$=sqrt{frac{1-x^2+x^2}{1-x^2}}=sqrt{frac{1}{1-x^2}}=frac{1}{sqrt{1-x^2}}$

$int_{-1}^{1}frac{{ m d}x}{sqrt{1-x^2}}=pi$

$設五角星或正五邊形的中心到頂點的距離為 r，$

$則這五個頂點的坐標為：$

$A (0,r)$

$B left(frac{sqrt{10+2sqrt5}}4r,frac{sqrt5-1}4r ight)$

$C left(frac{sqrt5+1}4r,-frac{sqrt{10-2sqrt5}}4r ight)$

$D left(-frac{sqrt5+1}4r,-frac{sqrt{10-2sqrt5}}4r ight)$

$E left(-frac{sqrt{10+2sqrt5}}4r,frac{sqrt5-1}4r ight)$

$現有矩陣 A，使得如下：$

$A=left(egin{matrix}1&15&14&4\12&6&7&9\8&10&11&5\13&3&2&16end{matrix} ight)$

$其每行、每列、兩條對角線之和皆為 34。有趣的是，其行列式的值為：$

$det(A)=0$

$left{egin{matrix}a+b+c+d=1\8a+4b+2c+d=5\27a+9b+3c+d=14\64a+16b+4c+d=30end{matrix} ight.$

$設 f(x)=sin(ax+b)，則 f(x)=sin(ax+b)cdot(ax+b)$

$=acos(ax+b)=asinleft(ax+b+frac{pi}2 ight)$

$ln(1-x)=sum_{n=1}^{infty}{left(-frac{x^n}{n} ight)}$

$frac11-frac12+frac13-frac14+frac15-frac16+frac17-frac18+ ...$

$=-left[frac{(-1)^1}1+frac{(-1)^2}2+frac{(-1)^3}3+frac{(-1)^4}4+frac{(-1)^5}5+frac{(-1)^6}6+frac{(-1)^7}7+frac{(-1)^8}8+ ... ight]$

$=sum_{n=1}^{infty}{left[-frac{(-1)^n}{n} ight]}=ln(1+1)=ln2$

$frac11+frac{ m i}2-frac13-frac{ m i}4+frac15+frac{ m i}6-frac17-frac{ m i}8+ ...$

$=-{ m i}left(frac{{ m i}^1}1+frac{{ m i}^2}2+frac{{ m i}^3}3+frac{{ m i}^4}4+frac{{ m i}^5}5+frac{{ m i}^6}6+frac{{ m i}^7}7+frac{{ m i}^8}8+ ... ight)$

$={ m i}cdotsum_{n=1}^inftyleft(-frac{{ m i}^n}{n} ight)={ m i}cdotln(1-{ m i})$

$frac11-frac{ m i}2-frac13+frac{ m i}4+frac15-frac{ m i}6-frac17+frac{ m i}8+ ...$

$={ m i}left[frac{(-{ m i})^1}1+frac{(-{ m i})^2}2+frac{(-{ m i})^3}3+frac{(-{ m i})^4}4+frac{(-{ m i})^5}5+frac{(-{ m i})^6}6+frac{(-{ m i})^7}7+frac{(-{ m i})^8}8+ ... ight]$