User:TheEternalVortex

$\sum _{n=1}^{\infty }{\frac {n^{x}}{n!}}$

$X={{B\times A+B+A(A\cdot B)} \over {A\cdot A+1}}$

$a_{n-1}^{2}-2a_{n-2}=r_{1}^{2}+\cdots +r_{n}^{2}\geq n(r_{1}^{2}\cdots r_{n}^{2})^{1/n}=n(a_{0}^{2})^{1/n}=n$

$f_{n}(x)=\int _{0}^{1-x}f_{n-1}(t)dt,\,\,\,f_{0}(x)=1$

$\sum _{i=1}^{n}d(\mathbf {x} ,\mathbf {x_{i}} )w_{i}$

For the analysis we will count compares and arithmetic operations. There is no multiplication/division used so they should all be roughly the same time.

Sequential search does 2 compares per iteration and one increment per iteration. When searching for an item in the array we expect n/2 iterations, and n iterations if searching for an item not in the array. Thus the expected number of compares+increments is $E(C)=3(0.1(n)+0.9(n/2))=0.3n+1.35n=1.65n.$

Binary search always does lg n iterations (in this implementation). Each iteration it does 2 compares, 1 subtract, 1 add, 1 bitshift, and 1 assignment, for 6 total operations, regardless of whether the item is found or not. So we get 6 lg n. Binary search also does 2 assignments and 2 compares and 1 or no matter what n, so we actually have 5 + 6 lg n.

Empirically about n = 50, numerically ~18.

$\mathbf {P} (U+V\leq a)=\int _{-\infty }^{\infty }\mathbf {P} (U+V\leq a|V=x)dx=\int _{-\infty }^{\infty }\mathbf {P} (U\leq a-x)dx$