| $y=f(x)$ | $dy/dx=f'(x)$ | $\int f(x)dx$ |
|---|---|---|
| $ax^n$ | $anx^{n-1}$ | $a(n+1)^{-1}x^{n+1}$ |
| $1/x$ | $-x^{-2}$ | $\ln|x|$ |
| $a$ | 0 | $ax$ |
| $a^x$ | $a^x\ln(a)$ | $a^x/\ln(a)$ |
| ${\rm e}^x$ | ${\rm e}^x$ | ${\rm e}^x$ |
| $^a\log(x)$ | $(x\ln(a))^{-1}$ | $(x\ln(x)-x)/\ln(a)$ |
| $\ln(x)$ | $1/x$ | $x\ln(x)-x$ |
| $\sin(x)$ | $\cos(x)$ | $-\cos(x)$ |
| $\cos(x)$ | $-\sin(x)$ | $\sin(x)$ |
| $\tan(x)$ | $\cos^{-2}(x)$ | $-\ln|\cos(x)|$ |
| $\sin^{-1}(x)$ | $-\sin^{-2}(x)\cos(x)$ | $\ln|\tan(\frac{1}{2} x)|$ |
| $\sinh(x)$ | $\cosh(x)$ | $\cosh(x)$ |
| $\cosh(x)$ | $\sinh(x)$ | $\sinh(x)$ |
| $\arcsin(x)$ | $1/\sqrt{1-x^2}$ | $x\arcsin(x)+\sqrt{1-x^2}$ |
| $\arccos(x)$ | $-1/\sqrt{1-x^2}$ | $x\arccos(x)-\sqrt{1-x^2}$ |
| $\arctan(x)$ | $(1+x^2)^{-1}$ | $x\arctan(x)-\frac{1}{2}\ln(1+x^2)$ |
| $(a+x^2)^{-1/2}$ | $-x(a+x^2)^{-3/2}$ | $\ln|x+\sqrt{a+x^2}|$ |
| $(a^2-x^2)^{-1}$ | $2x(a^2+x^2)^{-2}$ | $\displaystyle\frac{1}{2a}\ln|(a+x)/(a-x)|$ |
The theorem of De 'l H\^opital: if $f(a)=0$ and $g(a)=0$, then is $\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$
Products and quotients of complex numbers can be written as: \begin{eqnarray*} z_1\cdot z_2&=&|z_1|\cdot|z_2|(\cos(\varphi_1+\varphi_2)+i\sin(\varphi_1+\varphi_2))\\ \frac{z_1}{z_2}&=&\frac{|z_1|}{|z_2|}(\cos(\varphi_1-\varphi_2)+i\sin(\varphi_1-\varphi_2)) \end{eqnarray*} The following can be derived: \[ |z_1+z_2|\leq|z_1|+|z_2|~~,~~|z_1-z_2|\geq|~|z_1|-|z_2|~| \] And from $z=r\exp(i\theta)$ follows: $\ln(z)=\ln(r)+i\theta$, $\ln(z)=\ln(z)\pm2n\pi i$.
Further holds: \[ \frac{\tan(\frac{1}{2}(\alpha+\beta))}{\tan(\frac{1}{2}(\alpha-\beta))}=\frac{a+b}{a-b} \] The surface of a triangle is given by $\frac{1}{2} ab\sin(\gamma)=\frac{1}{2} ah_a=\sqrt{s(s-a)(s-b)(s-c)}$ with $h_a$ the perpendicular on $a$ and $s=\frac{1}{2}(a+b+c)$.
where $\varphi$ is the angle between $\vec{a}$ and $\vec{b}$. The external product is in $ I\hspace{-1mm}R^3$ defined by: \[ \vec{a}\times\vec{b}=\left(\begin{array}{c} a_yb_z-a_zb_y\\ a_zb_x-a_xb_z\\ a_xb_y-a_yb_x\end{array}\right)= \left|\begin{array}{ccc} \vec{e}_x&\vec{e}_y&\vec{e}_z\\ a_x&a_y&a_z\\ b_x&b_y&b_z\end{array}\right| \] Further holds: $|\vec{a}\times\vec{b}|=|\vec{a}|\cdot|\vec{b}|\sin(\varphi)$, and $\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$.
By subtracting the series $\sum\limits_{k=0}^n r^k$ and $r\sum\limits_{k=0}^n r^k$ one finds: \[ \sum_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r} \] and for $|r|<1$ this gives the geometric series: $\displaystyle\sum_{k=0}^\infty r^k=\frac{1}{1-r}$.
The arithmetic series is given by: $\displaystyle\sum_{n=0}^N(a+nV)=a(N+1)+\frac{1}{2} N(N+1)V$.
The expansion of a function around the point $a$ is given by the Taylor series: \[ f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2}f''(a)+\cdots+\frac{(x-a)^n}{n!}f^{(n)}(a)+R \] where the remainder is given by: \[ R_n(h)=(1-\theta)^n\frac{h^n}{n!}f^{(n+1)}(\theta h) \] and is subject to: \[ \frac{mh^{n+1}}{(n+1)!}\leq R_n(h)\leq\frac{Mh^{n+1}}{(n+1)!} \] From this one can deduce that \[ (1-x)^\alpha=\sum_{n=0}^\infty{\alpha\choose n}x^n \] One can derive that: \[ \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}~,~~ \sum_{n=1}^\infty\frac{1}{n^4}=\frac{\pi^4}{90}~,~~ \sum_{n=1}^\infty\frac{1}{n^6}=\frac{\pi^6}{945} \] \[ \sum_{k=1}^nk^2=\mbox{$\frac{1}{6}$}n(n+1)(2n+1)~,~~ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}~,~~ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\ln(2) \] \[ \sum_{n=1}^\infty\frac{1}{4n^2-1}=\mbox{$\frac{1}{2}$}~,~~ \sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}~,~~ \sum_{n=1}^\infty\frac{1}{(2n-1)^4}=\frac{\pi^4}{96}~,~~ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n-1)^3}=\frac{\pi^3}{32} \]
If $\lim\limits_{n\rightarrow\infty}u_n\neq0$ then $\sum\limits_n u_n$ is divergent.
An alternating series of which the absolute values of the terms drop monotonously to 0 is convergent (Leibniz).
If $\int_p^{\infty}f(x)dx<\infty$, then $\sum\limits_n f_n$ is convergent.
If $u_n>0~\forall n$ then is $\sum\limits_n u_n$ convergent if $\sum\limits_n\ln(u_n+1)$ is convergent.
If $u_n=c_nx^n$ the radius of convergence $\rho$ of $\sum\limits_n u_n$ is given by: $\displaystyle\frac{1}{\rho}=\lim_{n\rightarrow\infty}\sqrt[n]{|c_n|}= \lim_{n\rightarrow\infty}\left|\frac{c_{n+1}}{c_n}\right|$.
The series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent if $p>1$ and divergent if $p\leq1$.
If: $\displaystyle\lim_{n\rightarrow\infty}\frac{u_n}{v_n}=p$, than the following is true: if $p>0$ than $\sum\limits_{n}u_n$ and $\sum\limits_{n}v_n$ are both divergent or both convergent, if $p=0$ holds: if $\sum\limits_{n}v_n$ is convergent, than $\sum\limits_{n}u_n$ is also convergent.
If $L$ is defined by: $\displaystyle L=\lim_{n\rightarrow\infty}\sqrt[n]{|n_n|}$, or by: $\displaystyle L=\lim_{n\rightarrow\infty}\left|\frac{u_{n+1}}{u_n}\right|$, then is $\sum\limits_{n}u_n$ divergent if $L>1$ and convergent if $L<1$.
If $f(x)$ is continuous in $a$ and: $\lim\limits_{x\uparrow a}f'(x)=\lim\limits_{x\downarrow a}f'(x)$, then $f(x)$ is differentiable in $x=a$.
We define: $\|f\|_W:={\rm sup}(|f(x)|~|x\in W)$, and $\lim\limits_{x\rightarrow\infty}f_n(x)=f(x)$. Than holds: $\{f_n\}$ is uniform convergent if $\lim\limits_{n\rightarrow\infty}\|f_n-f\|=0$, or: $\forall(\varepsilon>0)\exists(N)\forall(n\geq N)\|f_n-f\|<\varepsilon$.
Weierstrass' test: if $\sum\|u_n\|_W$ is convergent, than $\sum u_n$ is uniform convergent.
We define $\displaystyle S(x)=\sum_{n=N}^\infty u_n(x)$ and $\displaystyle F(y)=\int\limits_a^bf(x,y)dx:=F$. Than it can be proved that:
| Theorem | For | Demands on $W$ | Then holds on $W$ |
|---|---|---|---|
| rows | $f_n$ continuous, $\{f_n\}$ uniform convergent | $f$ is continuous | |
| C | series | $S(x)$ uniform convergent, $u_n$ continuous | $S$ is continuous |
| integral | $f$ is continuous | $F$ is continuous | |
| rows | $f_n$ can be integrated, $\{f_n\}$ uniform convergent | $f_n$ can be integrated, $\int f(x)dx=\lim\limits_{n\rightarrow\infty}\int f_ndx$ | |
| I | series | $S(x)$ is uniform convergent, $u_n$ can be integrated | $S$ can be integrated, $\int Sdx=\sum\int u_ndx$ |
| integral | $f$ is continuous | $\int Fdy=\iint f(x,y)dxdy$ | |
| rows | $\{f_n\}\in$C$^{-1}$; $\{f_n'\}$ unif.conv $\rightarrow\phi$ | $f'=\phi(x)$ | |
| D | series | $u_n\in$C$^{-1}$; $\sum u_n$ conv; $\sum u_n'$ u.c. | $S'(x)=\sum u_n'(x)$ |
| integral | $\partial f/\partial y$ continuous | $F_y=\int f_y(x,y)dx$ | |
Further holds: \[ \frac{a^{2n}-b^{2n}}{a\pm b}=a^{2n-1}\pm a^{2n-2}b+a^{2n-3}b^2\pm\cdots\pm b^{2n-1}~~~,~~~ \frac{a^{2n+1}-b^{2n+1}}{a+b}=\sum_{k=0}^n a^{2n-k}b^{2k} \] \[ (a\pm b)(a^2\pm ab+b^2)=a^3\pm b^3~,~~(a+b)(a-b)=a^2+b^2~,~~ \frac{a^3\pm b^3}{a+b}=a^2\mp ba+b^2 \]
Rules: $\log(x^n)=n\log(x)$, $\log(a)+\log(b)=\log(ab)$, $\log(a)-\log(b)=\log(a/b)$.
For $a,b,c\in I\hspace{-1mm}R$ and $a\neq0$ holds: the 2nd order equation $ax^2+bx+c=0$ has the general solution: \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] For $a,b,c,d\in I\hspace{-1mm}R$ and $a\neq0$ holds: the 3rd order equation $ax^3+bx^2+cx+d=0$ has the general analytical solution: \begin{eqnarray*} x_1&=&~K-\frac{3ac-b^2}{9a^2K}-\frac{b}{3a}\\ x_2=x_3^*&=&-\frac{K}{2}+\frac{3ac-b^2}{18a^2K}-\frac{b}{3a}+i\frac{\sqrt{3}}{2}\left(K+\frac{3ac-b^2}{9a^2K}\right)\\ \end{eqnarray*} with $\displaystyle K=\left(\frac{9abc-27da^2-2b^3}{54a^3}+ \frac{\sqrt{3}\,\sqrt{4ac^3-c^2b^2-18abcd+27a^2d^2+4db^3}}{18a^2}\right)^{1/3}$
If $\pi(x)$ is the number of primes $\leq x$, than holds: \[ \lim_{x\rightarrow\infty}\frac{\pi(x)}{x/\ln(x)}=1~~~\mbox{and}~~~ \lim_{x\rightarrow\infty}\frac{\pi(x)}{\int\limits_2^x\frac{dt}{\ln(t)}}=1 \] For each $N\geq2$ there is a prime between $N$ and $2N$.
The numbers $F_k:=2^k+1$ with $k\in I\hspace{-1mm}N$ are called Fermat numbers. Many Fermat numbers are prime.
The numbers $M_k:=2^k-1$ are called Mersenne numbers. They occur when one searches for perfect numbers, which are numbers $n\in I\hspace{-1mm}N$ which are the sum of their different dividers, for example $6=1+2+3$. There are 23 Mersenne numbers for $k<12000$ which are prime: for $k\in\{2,3,5,7,13,17,19,31,61,89,107,127,521,$ $607,1279,2203,2281,3217,4253,4423,9689,9941,11213\}$.
To check if a given number $n$ is prime one can use a sieve method. The first known sieve method was developed by Eratosthenes. A faster method for large numbers are the 4 Fermat tests, who don't prove that a number is prime but give a large probability.