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

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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 +


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SLIDE 1

Geometry

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SLIDE 2

A + B + C = 180 A B C B A

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SLIDE 3

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

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SLIDE 4

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]

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SLIDE 5

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]

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SLIDE 6

x x x A B C

A = B = C

2x [Angles Subtended by Same Arc Theorem]

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SLIDE 7

90 A

A = 90

180

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SLIDE 8

A + B = 180

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

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SLIDE 9

Sums

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SLIDE 10

1 2 3 4 ∙∙∙ n 1 2 3 n

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

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SLIDE 11

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

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SLIDE 12

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

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SLIDE 13

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

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SLIDE 14

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

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SLIDE 15

Approximations

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SLIDE 16

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

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SLIDE 17

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

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SLIDE 18

Primes

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SLIDE 19

 

) 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

p

2 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

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SLIDE 20

     

 

) 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 2

O 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