Lecture 3: Two binomial tricks. Special numbers

Transcription

1 Lecture 3: Two biomial trics. Special umbers Trics with biomial coefficiets. Goig halves Begi with the duplicatio formula r (r ) r(r )(r )(r 3 ) (r +)(r + r (r + ) ) (r) /, iteger 0 ad divide by (!) r r / r (r + ) ( + ) / ( r ) / r gives / / Negate upper idex / ( /4) Usig Vadermode: ( r s ) ( )( r+s ) we get the corollary / / ( ), iteger 0 () By () we have ( / ) / ( ) ( /4) ( /4) ( /4) which sums to ( ) by the above, ad thus gives for the middle elemets of Pascal s triagle Chagig ( ( )( ) 4 ) ( to / ), for appropriate (usually 0, or ), ca give a simpler formula.. High-order differeces: Not part of the course. 3. Iversio g() ( ) f() iff f() ( ) g()

2 sice ( ) g() ( ) j ( ) j f(j) j ) j j j j f(j) ( ( ) +j j f(j) j ( ) +j j j j f(j) ( ) j f(j) [ j 0] f() j Applicatio: The football victory problem h(, ) exactly of hats fall o the right head. Example: 3 ABC (3 right), BAC ( right), CAB (0 right), ACB ( right), BCA (0 right), CBA ( right). Hece, h(3,3), h(3,) 0, h(3,) 3, h(3,0). I geeral: h(,) ( ) h(,0) ( ) ( ) We wat to express ( ) o closed form.! h(,) ( ) iteger 0 Iversio formula with g()! ad f() ( ) give Simplified ( ) 0 ( )!! ( )! ( )+! ( )! The last sum rapidly coverges to 0 ( ) /! /e, ad! e + + [ 0] is the umber of possibilities that o hat is beig right (for istace, 3 6 e + ) The probability is!!/e + O()! /e idepedet of if the hats are 0 or 0 6. We will retur to this problem.

3 Harmoic umbers H, iteger 0 Example: Stac cards to get the largest overhag. See figure o page 73. Oe card ca overhag half its legth. Ceter of gravity of two cards will be the middle of their commo part, a additioal half uit. The ceter of gravity of the top cards should the be at the edge of card +, that is where each card has legth. d + (d + ) + (d + ) + + (d + ), ad d 0 d + + d + d d, 0 () ( )d + d + d d, (3) () (3): d + ( )d + d, d + d + / H How may cards are eeded to get a overhag of oe cardlegth? < H 5, with 4 5 cards give a overhag of H 5 /.7 cardlegths Logarithmic bouds H is a discrete aalog of l. Figures o pages give dx x l < H < l +, > Harmoic umbers of higher orders H (r) r, iteger r Digressio: a famous result by Euler The Riema eta fuctio ζ(r) H (r) π 6 r 3

4 ca be used to approximate H (), sice l + ( ) + 3 ( ) 3 + ad but l( ) l l ( ) 33 l l ( ) l l Hece, rearragig the terms we have ( + + ) (H ) + (H() ) + 3 (H(3) ) + H l (H() ) 3 (H(3) ) (ζ() ) (ζ(3) ) 3 γ (Euler s costat) ad A fier approximatio lim (H l ) γ H l + γ + + ǫ 0 4, 0 < ǫ < The worm o the rubber bad Worm W crawls o a meter-log rubber bad at a speed of oe cm per miute. Each miute the bad is stretched oe meter, while W maitais its relative positio. Will W ever reach the ed of the bad? Yes! After miutes W has crawled 00 ( ) H 00 fractio of the bad H > 00 whe e 00 γ e 99.43, i.e. after years, whe the bad has bee stretched to a legth of 0 7 light years. The superworm crawls 50 cm per miute, ad thus ( ) H fractio of the bad after miutes It will reach the ed of the bad withi 4 miutes sice H 4 > (i 3 miutes ad 40 secods). 4

5 Beroulli umbers Patter? S m () 0 m S 0 () S () S () S 3 () S 4 () [ m+ ]S m () /(m + ), [ m ]S m () /, [ m ]S m () m/ S m () [ m ]S m () 0... [ m ]S m () costat m m + m + m + (B 0 m+ + B m + + B m ) m m m + B m+ (4) m + 0 The Beroulli umbers are defied implicitly by m m + B [m 0], m 0 0 For istace B B B There is o closed form for B. Usig geeratig fuctios we will show i lecture 7 that e 0 B, which gives (4) above! Digressio: Euler s costat γ + B 5

6 Fiboacci umbers 0 if 0 F if F + F if > Fiboacci umbers appear ofte i ature. Example: a suflower has F 0 55 spirals clocwise ad F 9 34 couterclocwise. Bee trees i the figure o page 9: The droe at the bottom has F gradmother ad F gradfather, ad i geeral, has F + great -gradmothers ad F + great -gradfathers. Light ray through two glass paes I how may ways a ca a ray pass through or be reflected after chagig directio times? if 0 a if a + a F + if > Cassii s idetity F + F F ( ), > 0 (5) F F + F implies F+ F + F F ( ) ad F + F F+ ( ) + (F + + F )F F+ (F+ F + F F) So Cassii() is true iff Cassii( + ) is true. Equatio (5) follows by iductio. Base case is for, whe F F 0 F 0 ( ). We preset a figure o OH where 64 squares become 65, justified by F 7 F 5 F ( ) 6 Fiboacci umber system Every > 0 has a uique represetatio F + F + + F r, +, 3 +,..., r Example: F + F 6 + F 4 (foud by a greedy method). Gives a umber system (b m b m b ) F m b F Example: (00) 0 ( ) F (0000). Requires more bits as there are o adjacet s. 6