Geometry A + B + C = 180 B A C A B Pythagoras a 2 + b 2 = c - PowerPoint PPT Presentation

Geometry

 A +  B +  C = 180  B A C A B

Pythagoras a 2 + b 2 = c 2 b a b a a a a 2 c  b c c 2  b c b b 2 c a  a a b b  +  + 90  = 180 

 A = 2∙  B x B y 180  180  -2x -2y x A y B = x + y A = 360  – (180  – 2x) – (180  – 2y) = 2x + 2y = 2B [ Angle at the Center Theorem]

 A = 2∙  B x y B 360  – (180  – 2y) 180  – 2x x A y B = x – y A = 360  – (180  – 2x) – (180  +2y) = 2x – 2y = 2B [ Angle at the Center Theorem]

 A =  B =  C B x  A x  2x  C x  [Angles Subtended by Same Arc Theorem]

 A = 90  A 90  180 

 A +  B = 180  A 2  B 2  A B 2  A + 2  B = 360  [ Cyclic Quadrilateral ]

Sums

n 1 + 2 + ∙∙∙ + n = n 2 /2 + n /2 = n ( n + 1)/2 3 2 1 1 2 3 4 ∙∙∙ n

2 0 + 2 1 + 2 2 + 2 3 + ∙∙∙ + 2 k = 2 k +1 - 1 3 7 15 1 31 2 0 2 1 2 2 2 3 2 4 induction 2 k - 1 2 k step 2 ∙ 2 k - 1 = 2 k +1 - 1 k α k +1 – 1 Σ α i = for α  1 α – 1 i = 0 k k +1 k Σ Σ Σ ( α – 1)∙ α i = α i – α i = α k +1 – 1 i = 0 i = 1 i = 0

# nodes = 1 + 2 + 4 + ∙∙∙ + 2 k = 2 k +1 – 1 k Σ ( k – i ) ∙ 2 i = 2 k +1 – 2 – k # edges = # nodes – 1 = 2 k +1 – 2 k Σ i = 0 ( k – i )∙ 2 i = # edges – k = 2 k +1 – 2 – k i = 0 i k k Σ Σ ∙ 2 k = 2 k = i ∙2 k – i 2 i i = 0 i = 0 k 0 + 1 + 2 + 3 + 4 + ∙∙∙ + k = 2 – 2+ k i Σ = 2 i 2 k 2 k 1 2 4 8 16 i = 0 0 1 2 i nodes i k - i edges ... k -1 k

n n ( n +1)(2 n +1) Σ i 2 = 6 i = 1 Proof by induction : 1 1(1+1)(2·1+1) Σ n = 1 : i 2 = 1 = 6 i = 1 n n -1 Σ Σ n > 1 : i 2 = n 2 + i 2 i = 1 i = 1 ( n -1)(( n- 1)+1)(2( n -1)+1) = n 2 + 6 2 n 3 +3 n 2 + n n ( n +1)(2 n +1) = = 6 6

n Σ n -th Harmonic number H n = 1/1 + 1/2 + 1/3 +∙∙∙+ 1/ n = 1/ i i = 1 ∫ n n H n – 1   H n – 1/n 1/ x dx = [ ln x ] = ln n – ln 1 = ln n 1 1 ln n + 1/ n  H n  ln n + 1 1/ n 1/1 1/2 1/3 1/4 1/ n 1 2 3 4 5 n

Approximations

ln (1 +  )   1 +   e  (1 +  ) 1/   e (1 + 1/ x ) x  e x for   0 and x large ”  ” is actually ”  ” ln x  1 1+   1 ln x =  x x

∫ n n -1 Σ n ln i  ln x dx = [ x∙ln x – x] ln n ! – ln n = 1 i = 1 1 n Σ  = n∙ ln n – n + 1 ln i = ln n ! i = 1 n∙ ln n – n + 1  ln n !  n∙ ln n – n + 1 + ln n ln n ln x ln 4 ln 2 1 2 3 4 5 n [ Stirling’s Approximation ]

Primes

Prime Number Theorem Tchebycheff 1850  ( n ) = |{ p | p prime and 2  p  n }| = Θ (n/log n) Upper Bound     2 n   2 n All primes p , n < p  2 n , divide . From      π ( 2 n ) π ( n ) 2 n   n p 2      n   n    we have  (2 n )-  ( n )  2 n /log n , implying n p 2 n   k 1 k 1           k i 1 i i 1 k π ( 2 ) ( π ( 2 ) π ( 2 )) π ( 2 ) 2 / i 1 O ( 2 / k )   i 1 i 1 Lower Bound    2 n 2 n  ( n 1 )    Consider prime power p m dividing . Since p i divides    n   n 1 between  n / p i  and  n / p i  factors in both denominator and     log 2 n p        numerator, we have m bounded by , implying   i i n / p n / p log 2 n p    i 1 2 n               n log 2 log log 2 log log 2 ( 2 ) log( 2 ) n   n p n n n p   n   p 2 n prime p 2 n prime  (30) = 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

Series for Primes Sums not restricted to primes 1 1 ∑ ∑   O (log log n ) O (log n ) p i   p n prime i n n 1 1     O (log log n ) O ( 1 ) i log i p log p  i 2 p prime   1 1  1      O   O ( 1 / n ) 2   p n log n 2 i  p n prime  i n       log n log n log n 1 1 1 1     ∑           i 1 i 1 c 2 / log 2 ce ce (ln log n 1 )  i p p 2 i 1      i   i 1 i 1 i 1 i 1 p n prime 2 p prime 2  i 1      2 1 1 1 1 c log log 2 i 1       ∑      c O ( 1 ) i i i p log p p log p 2 p 2 2 log 2       i i 1 i 1 i 0 i 0 i 1 i 1 p prime 2   2  2 2 p prime 2 p prime 2      1 1 1 2 c 1 1     ∑          i 1 i 1 cn 2 / log( n 2 ) O   2 2 i 2 i   p p ( n 2 ) n log n 2 n log n      i   i 1 i 0 i 0 i 0 p n prime n 2 p prime n 2