BINOMIAL COEFFICIENT AND THE GAUSSIAN

Transcription

1 BINOMIAL COEFFICIENT AND THE GAUSSIAN The biomial coefficiet is defied as-! k!(! ad ca be writte out i the form of a Pascal Triagle startig at the zeroth row with elemet 0,0) ad followed by the two umbers, 0) ad, ) i the first row. The secod row produces the umbers [,, ] ad the third row the four umbers [, 3, 3, ]. The row umber is here give by the value of ad the umber of elemets i a row will be +. From these first few results oe ca state that- The Pascal Triagle thus reads- C ( + k + ) +, k + ) It is kow that the umbers i the idividual rows of this triagle also give the coefficiets i the biomial expasio of (a+b)^. Thus, for example, for 4, we have ( a + b) a + 4a b + 6a b + 4ab + b Oe also otices that the biomial coefficiets for fixed are symmetric about their cetral maximum value. Plottig their ormalized values at 7 with -5<k<5 we get the graph-

2 We have superimposed o these values the Gaussia exp(-k^/36). The results are almost idetical. Ideed as is allowed to become large oe has the result- [( / )!] exp( k / ) k! [( / )!] + / ) ( / +!( /! with >k. At 8 ad k8 oe fids- 8! exp() [64! C (8,7)] which is close to the correct value of A eve better approximatio occurs whe k is ear zero sice the two sides of the equatio are ormalized at k0. So keepig k ad k<< we fid- exp( ) N N + N

3 This yields the approximatio for N0,000. Note that oe ca also use the Stirlig Approximatio for large. It gives, for example, the estimate+ / [ π k k + / ( k + / ] for >> which yields the approximatio 000,500) ^300 compared to the exact value of 000,5000) ^300. Goig back to the Pascal Triagle give above oe otices the followig-, , , From this it follows that-. Oe also otices that if ) is prime the /) is a iteger provided 0<k<. Furthermore a violatio of this coditio idicates that the umber is composite. Thus oe has that- ( N )! k N is prime if T ( ( N ) ) k!( N! k! is a iteger for all values of k ragig from k through kn/. Ay violatio will idicate the umber N is composite. Cosider the umber N77. O ruig the oe lie MAPLE program- for k from to 60 do {k,(/k!)*product((77-),..k-)}od; oe fids T(3),T(6),T(9),T(),T(5),T(59) up to k60 to be o-iteger ad hece the umber is composite. A extra beefit is that oe actually fids the factors to be the lowest values of k ot multiples of a lower value. Thus k3 ad k59 i this case, so that Ufortuately this approach for determiig the primeess of a large umber

4 will be quite cumbersome ad is actually more time cosumig tha just lookig at the ratio N/(k+) for k+<sqrt(n). Ruig the oe-lier- for k from to 4 do {*k+,77/(*k+)}od; already produces a iteger for k yieldig 77/357 so that 773(57). Goig back to the Pascal Triagle we observe that the sequece of umbers lyig o the vertical lie dividig the triagle i half is I terms of the coefficiet this meas the sequece goes as- 0,0)-,)-4,)-6,3)-8,4)- ) Addig the terms together through the m th row, we fid the Sum(m) to be- Sum( m) m m ()! ) 0 0! For m6 this sum equals Sum(6) Oe ca also ask for the value of the reciprocal series which coverges to the fiite value -! + 0 ()! We ca also work out the series- S( ) [ ] It yields S(), S()6, S(3)0, ad S(4)70. From this oe sees that S() sums to ). Fially we poit out that oe ca geeralize the Pascal Triagle startig with m oes i the first row ad the addig the m earest values i the (-) row to fid the elemet i the th row. For m3 this will lead to the triagle-

5 etc. with the geeratig formula- D( D(, k ) + D(, k ) + D(, Thus D(5,5) ad D(6,7) Note that for this m3 case that the umber of elemets i the th row is + ad they sum to 3^. The elemets i the th row agai have the character of a Gaussia distributio with very small discrepacies oticed ear the tails as show o the followig graph-. Comparig this result ad that for the classic Pascal triagle, we ca coclude that-

6 m oes i the first row of a modified Pascal Triagle will lead to a triagle whose elemets i the th row umber [(m-)]+ ad add up to m^. The elemet F( of such a modified Pascal Triagle will be give by- F( F(-,k-m+)+F(-,k-m+)+..+F(-, Also oe otes that there is a geeratig fuctio for the coefficiets F( for a give value of m. For example at m3 the geeratig fuctio is G+x+x^ which has coefficiets [,,] ad correspods to the first row of D(,. G^+x+3x^+x^3+x^4 has the coefficiets [,,3,,] ad thus yields the coefficiets D(,. For m3 ad 6 we fid- ( + x + x ) 6 D(6, x 7 + 6x k + 90x 8 + 6x + x + 50x 9 + x x + 6x x + x 4 + 6x 5 + 4x 6 December 009