Geometry

A + B + C = 180 B A C A B

Pythagoras a2 + b2 = c2 a2 c2 b2 b a b a a a c  b c  b c b c a  a b  +  + 90 = 180

A = 2∙B x B y 180 -2x 180 -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 Opgave 2 Tegn en cirkel med radius 7 cm. Tegn en korde. Tegn to punkter A og B på cirklen på samme side af korden. Tegn trekanten med hjørne A og korden som en side. Mål vinklen A. Tegn trekanten med hjørne B og korden som en side. Mål vinklen B. C x [Angles Subtended by Same Arc Theorem]

A = 90 A 90 180 Opgave 1 Tegn en cirkel med radius 7 cm. Tegn diameteren. Tegn et punkt A på cirklen. Tegn trekanten med hjørne A og diameteren som en side. Mål vinklen A.

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

Sums

1 + 2 + ∙∙∙ + n = n2/2 + n/2 = n(n + 1)/2 3 4 ∙∙∙ n 1 + 2 + ∙∙∙ + n = n2/2 + n/2 = n(n + 1)/2

Σ Σ Σ Σ 20 + 21 + 22 + 23 + ∙∙∙ + 2k = 2k+1 - 1 αi = for α 1 7 15 31 20 21 22 23 24 induction step 2k 2k - 1 2 ∙ 2k - 1 = 2k+1 - 1 Σ k i = 0 αk+1 – 1 α – 1 αi = for α 1 Σ Σ Σ k i = 0 k+1 i = 1 k i = 0 (α – 1)∙ αi = αi – αi = αk+1 – 1

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

Σ Σ Σ Σ = n(n+1)(2n+1) 6 i2 Proof by induction : n = 1 : i2 = 1 = n > 1 : i2 = n2 + i2 = n2 + = = Σ 1 i = 1 1(1+1)(2·1+1) 6 Σ n i = 1 Σ n-1 i = 1 (n-1)((n-1)+1)(2(n-1)+1) 6 2n3+3n2+n 6 n(n+1)(2n+1) 6

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

Approximations

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

n∙ln n – n + 1  ln n!  n∙ln n – n + 1 + ln n ∫ n 1 Σ n-1 i = 1 n 1 ln n! – ln n = ln i  ln x dx = [ x∙ln x – x] Σ n i = 1 = n∙ln n – n + 1  ln i = ln n! 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

(n) = |{ p | p prime and 2  p  n }| = Θ(n/log n) Prime Number Theorem Tchebycheff 1850 (n) = |{ p | p prime and 2  p  n }| = Θ(n/log n) Upper Bound All primes p, n < p  2n, divide . From we have (2n)-(n)  2n/log n, implying Lower Bound Consider prime power pm dividing . Since pi divides between n/pi and n/pi factors in both denominator and numerator, we have m bounded by , implying (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

Sums not restricted to primes Series for Primes Sums not restricted to primes

Master theorem Pick’s theorem Euler’s formula: E=O(V) for connected graphs