Geometry

A + B + C = 180

A B

CBA

Pythagoras a2 + b2 = c2

a

b c

a

b a

a

b

a

b a

c2

b2

b

b

a

cc

c

b

a2

+ + 90 = 180

x

A = 2∙B

y

y

A

B

x

180-2x

180 -2y

B = x + yA = 360–(180–2x)–(180–2y) = 2x + 2y = 2B [Angle at the Center Theorem]

x

A = 2∙B

y

y

A

B

x

180–2x360–(180–2y)

B = x – yA = 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

B2A + 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 + + 2∙∙∙ k = 2k+1 - 1

20 2221 23 24

2k2k - 1

2 2∙ k - 1 = 2k+1 - 1

inductionstep

1 3 7 15 31

k

i = 0Σ αi = for α 1αk+1 – 1α – 1

k

i = 0Σ(α – 1) ∙ αi = αi – αi = αk+1 – 1k+1

i = 1Σk

i = 0Σ

i

01

kk-1

...

k

i = 0Σ (k – i) 2∙ i = 2k+1 – 2 – k

2i nodesk-i edges

# nodes = 1 + 2 + 4 + + 2∙∙∙ k = 2k+1 – 1

# edges = # nodes – 1 = 2k+1 – 2

(k – i)∙2i = # edges – k = 2k+1 – 2 – k k

i = 0Σ

i2i

k

i = 0Σ =

0+

1+

2+

3+

4+ ∙∙∙

+

k= 2 –

2+k1 2 4 8 16 2k 2k

∙ 2k = 2k = i 2∙ k–ik

i = 0Σi2i

k

i = 0Σ

Proof by induction :

n = 1 : i2 = 1 =

n > 1 : i2 = n2 + i2

= n2 +

= =

1

i = 1Σ

=

n(n+1)(2n+1)6

1(1+1)(2·1+1)6

n

i = 1Σ i2

n

i = 1Σ(n-1)((n-1)+1)(2(n-1)+1)

6

2n3+3n2+n6

n(n+1)(2n+1)6

n-1

i = 1Σ

n-th Harmonic number Hn = 1/1 + 1/2 + 1/3 + + 1/∙∙∙ n = 1/in

i = 1Σ

1/n

n1 2 3 4 5

1/1

1/2

1/31/4

1/n

1/x dx = [ ln x ] = ln n – ln 1 = ln n∫ n

1

n

1Hn – 1 Hn – 1/n

ln n + 1/n Hn ln n + 1

Approximations

ln (1 + ) 1 + e

(1 + ) 1/ e

(1 + 1/x) x efor 0 and x large”” is actually ””

1 1+

1x

ln x = x

x

ln x

ln x dx∫ n

1

n

1ln n! – ln n =n-1

i = 1Σ

n

i = 1Σ= 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

n1 2 3 4 5

ln 4

[Stirling’s Approximation]

ln x

ln 2

ln n

Primes

)2log()2(2loglog2log2

log2logprime 2prime 2

nnnpnn

nn

npnpp

n

1)1(22

n

nnnn

npnpn p

n

i

iip

2log// 2log

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

nn2 n

npn

nπnπ

nn

pn 2

2

)()2( 22

)/2(1/2)2())2()2(()2(1

1

11

1

1 kOiππππ kk

i

ik

i

iik

Tchebycheff 1850

Upper BoundAll primes p, n < p 2n, divide . Fromwe have (2n)-(n) 2n/log n, implying

Lower BoundConsider 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

)1log(ln1

12log/2

2111 log

1

log

1

11log

1 2prime 2prime 1

ncei

cecpp

n

i

n

i

iii

n

i pnp ii∑

)1(2

12

2loglog12log

1log1

log1

11

2

0 2prime 2

prime 0 2prime 2

1

12122

Oi

cc

ppppp ii

ii

i pp i p

i

ii

ii

∑

nnO

nnc

ncnnpp i

ii

iii

i npnnp ii log1

21

log2

)2log(/2)2(

111

00

112

0 2prime 22

prime 2

1∑

∑prime

)log(log1

np

nOp

prime 2 log

11

np nnO

p

prime

)1(log1

p

Opp

∑ni

nOi

)(log1

ni

nOi

)/1(12

)log(loglog1

2

nOii

n

i

Series for PrimesSums not restricted to primes