| 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\}\inC^{-1}; \{f_n'\} unif.conv \rightarrow\phi | f'=\phi(x) | |
| D | series | u_n\inC^{-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.