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 - - 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 +
A + B + C = 180 A B C B A
Pythagoras a2 + b2 = c2
a b c a b a a b a b a
c2 b2
b b a c c c b
a2
+ + 90 = 180
x
A = 2∙B
y y A B x 180
- 2x
180
- 2y
B = x + y A = 360–(180–2x)–(180–2y) = 2x + 2y = 2B [Angle at the Center Theorem]
x
A = 2∙B
y y A B x 180–2x 360–(180–2y) B = x – y A = 360–(180–2x)–(180+2y) = 2x – 2y = 2B [Angle at the Center Theorem]
x x x A B C
A = B = C
2x [Angles Subtended by Same Arc Theorem]
90 A
A = 90
180
A + B = 180
2A 2B A B 2A + 2B = 360 [Cyclic Quadrilateral]
Sums
1 2 3 4 ∙∙∙ n 1 2 3 n
1 + 2 + ∙∙∙ + n = n2/2 + n/2 = n(n + 1)/2
20 + 21 + 22 + 23 + ∙∙∙ + 2k = 2k+1 - 1
20 22 21 23 24 2k 2k - 1 2 ∙ 2k - 1 = 2k+1 - 1 induction step
1 3 7 15 31
k i = 0
Σ
αi = for α 1
αk+1 – 1 α – 1
k i = 0
Σ
(α – 1)∙ αi = αi – αi = αk+1 – 1
k+1 i = 1
Σ
k i = 0
Σ
i 1 k k-1 ... k i = 0
Σ (k – i) ∙ 2i = 2k+1 – 2 – k
2i nodes k-i edges # nodes = 1 + 2 + 4 + ∙∙∙ + 2k = 2k+1 – 1 # edges = # nodes – 1 = 2k+1 – 2 (k – i)∙2i = # edges – k = 2k+1 – 2 – k
k i = 0
Σ
i 2i
k i = 0
Σ
= 0 + 1 + 2 + 3 + 4 + ∙∙∙ + k = 2 – 2+k 1 2 4 8 16 2k 2k ∙ 2k = 2k = i∙2k–i
k i = 0
Σ
i 2i
k i = 0
Σ
Proof by induction : n = 1 : i2 = 1 = n > 1 : i2 = n2 + i2 = n2 + = =
1 i = 1
Σ
= n(n+1)(2n+1) 6 1(1+1)(2·1+1) 6
n i = 1
Σ
i2
n i = 1
Σ
(n-1)((n-1)+1)(2(n-1)+1) 6 2n3+3n2+n 6 n(n+1)(2n+1) 6
n-1 i = 1
Σ
n-th Harmonic number Hn = 1/1 + 1/2 + 1/3 +∙∙∙+ 1/n = 1/i
n i = 1
Σ
1/n n 1 2 3 4 5 1/1 1/2 1/3 1/4 1/n
1/x dx = [ ln x ] = ln n – ln 1 = ln n
∫
n 1 n 1
Hn – 1 Hn – 1/n ln n + 1/n Hn ln n + 1
Approximations
ln (1 + ) 1 + e (1 + ) 1/ e (1 + 1/x) x e
for 0 and x large ”” is actually ””
1 1+ 1 x ln x = x x ln x
ln x dx
∫
n 1 n 1
ln n! – ln n =
n-1 i = 1
Σ
n i = 1
Σ
= n∙ln n – n + 1 ln n! ln i = = [ x∙ln x – x] ln i n∙ln n – n + 1 ln n! n∙ln n – n + 1 + ln n
n 1 2 3 4 5 ln 4
[Stirling’s Approximation]
ln x ln 2 ln n
Primes
) 2 log( ) 2 ( 2 log log 2 log 2 log 2 log
prime 2 prime 2
n n n p n n n n
n p n p p n
1 ) 1 ( 2 2 n n n n n
n p n p n
p n i i i
p2 log / /
2 log 1
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
(n) = |{ p | p prime and 2 p n }| = Θ(n/log n) Prime Number Theorem
(30) = 10
n n 2
n n p n n π n π
n n p n
2 2 ) ( ) 2 (
2 2
) / 2 ( 1 / 2 ) 2 ( )) 2 ( ) 2 ( ( ) 2 (
1 1 1 1 1 1
k O i π π π π
k k i i k i i i k
Tchebycheff 1850
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
) 1 log (ln 1 1 2 log / 2 2 1 1 1
log 1 log 1 1 1 log 1 2 prime 2 prime
1
n ce i ce c p p
n i n i i i i n i p n p
i i
∑
) 1 ( 2 1 2 2 log log 1 2 log 1 log 1 log 1
1 1 2 2 prime 2 prime 2 prime 2
1 1 2 1 2 2O i c c p p p p p
i i i i i p p i p
i i i i i
∑
n n O n n c n cn n p p
i i i i i i i n p n n p
i i
log 1 2 1 log 2 ) 2 log( / 2 ) 2 ( 1 1 1
1 1 2 2 prime 2 2 prime 2
1
∑
∑
prime
) log (log 1
n p
n O p
prime 2
log 1 1
n p
n n O p
prime
) 1 ( log 1
p
O p p
∑
n i
n O i
) (log 1
n i
n O i ) / 1 ( 1
2
) log (log log 1
2
n O i i
n i
Series for Primes
Sums not restricted to primes