Math equations and formulas

Logarithms

The fundamental property of a logarithm (base b) is:

$$\displaystyle y = b^x \iff x = log_b y \iff b = \sqrt[x ] y \qquad \small{for} b \gt 1, y \gt 0

Properties:

log2 xy=logx+logylog2 x/y=logxlogylog2 xy=y log xlog2 1=0log2 2=1log_{2}\ xy = log x + log y \\ log_{2}\ {x/y} = log x - log y \\ log_{2}\ x^y = y\ log\ x \\ log_{2}\ 1 = 0 \\ log_{2}\ 2 = 1

Summations

i=1n i=n(n+1)2i=1n i2=2n3+3n2+n6i=0n 2i=2n+11i=0n ai=an+11a1i=1n i2i=(n1)2n+1+2i=1n log2 (i+1)=(n+1) log2 (n+1)ni=1n log2 i=(n+1) log2 (n+1)2n\displaystyle{ \sum_{i=1}^{n} \ i = \frac {n(n+1)} {2} \\ \sum_{i=1}^{n} \ i^2 = \frac {2n^3+3n^2+n} {6} \\ \sum_{i=0}^{n} \ 2^i = 2^{n+1} - 1 \\ \sum_{i=0}^{n} \ a^i = \frac {a^{n+1}-1} {a-1} \\ \sum_{i=1}^{n} \ i2^i = (n-1) 2^{n+1} + 2 \\ \sum_{i=1}^{n} \ \lceil log_2\ (i+1)\rceil = (n+1)\ log_2\ (n+1) - n \\ \sum_{i=1}^{n} \ \lfloor log_2\ i \rfloor = (n+1)\ log_2\ (n+1) - 2n \\ }

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