Problem 4: Evaluate by egatig actually u-egatig its upper idex We ow that Biomial coefficiet r { where r is a real umber, is a iteger The above defiitio ca be recast i terms of factorials i the commo case that the upper idex r is a positive iteger atural umber, that s greater tha or equal to the lower idex :! itegers 0! -!
Now to get the values of the coefficiets for differet values of ad, we have the Pascal s Triagle 0 0 2 3 4 5 6 7 8 9 0 2 2 3 3 3 4 4 6 4 5 5 0 0 5 6 6 5 20 5 6 7 7 2 35 35 2 7 8 8 28 56 70 56 28 8 9 9 36 84 26 26 84 36 9 0 0 45 20 20 252 20 20 45 0
Observatio: Every umber i Pascal s triagle is the sum of two umbers i the previous row, oe directly above it ad the other just to the left. So this gives the additio formula r r- r- + Text boo defiitio 5.8 Where r is a positive iteger ad is a iteger Now let s loo more closely at the values of whe is a egative iteger. Oe way to approach these values is to use the additio law ad obtai the below table. For example, we 0 must have, sice + ad 0 0; the we must have, 0 0 Ad + ; ad so o. 0 0 0
Pascal s Triagle exteded upwards -4-4 0-20 35-56 84 20 65-220 286-3 -3 6 0 5-2 28-36 45-55 66-2 -2 3-4 5-6 7-8 9 0 0 0 2 3 4 5 6 7 8 9 For egative we ca expad the biomial coefficiet as r r r -... r - + here we tae r to get a geeral case - - r r... - rwe have commo,as all factors are egative r whe 0, ad whe < 0 the value is 0
We ow that r > r r!! r iteger ; we get this by taig r as r i biomial coefficiet formula Text Boo Defiitio 5.4 Now to get value we tae r - iteger ; Sice So the solutio will be or depedig o the value of, where is a iteger greater tha 0 Whe <0 the value is 0
Problem 6: Fix up the text s derivatio i Problem 6, Sectio 5.2, by correctly applyig symmetry. Problem 6 is to fid a closed form for the summatio: + >0 2 2 +, iteger >0 Observatio: I this sum the idex of summatio appears six times. This ca be tricy to solve the summatio. But the 2 s i the summatio appear i a way that we ca mae use the idetity to simplify the product of two biomial coefficiets. r m m r r - m - where m, are itegers.
Applyig the idetity to our summatio we get: + >0 2 2 + + >0 Here r+, m2 ad + This step reduces the occurrece of by. So ow we have 5 more to go. The + factor i this above summatio ca be absorbed ito the followig idetity: This gives us: r + + r Rearragig terms we get: r - - + + + + + + Let us use this value of i our summatio equatio. by usig
Hece we get : + >0 + + + + + + + + + + + + Here >0 + + + 2 Now we have got rid of 2 occurreces of. To elimiate aother we ca either use symmetry o + or egate the upper idex + which elimiates ad also the factor. The rule of symmetry is as follows: -
The textboo uses symmetry idetity o equatio 2 as follows: + + + + + This elimiates aother occurrece of i the summatio. + + Now we ca apply the idetity for sums of products of biomial coefficiets to equatio 3. The idetity we ca mae use of here is: + 3 l s+ m+ l+m s-m -l l,m, are itegers ad l>0 Replacig l,m,,s i this idetity with +,,, we get: + + + + + Usig the fact 0 whe <0 0 4
Let us chec the solutio obtaied through symmetry taig 2. Equatio 4 ow yields: 2 2 3 2 0 0 - + 2 2 4 4 4 2-6 + 6 0 This example results i the same solutio as 2 3 i Equatio 4 Now let us explore the secod optio of egatig the upper idex of i Equatio 2 The idetity for upper egatio is give by: 3 + r -r Usig this idetity i Equatio 4 we get: + + + + - + + + 5
To the Equatio 5 we ca apply the followig idetity: l s m+ + l + s l-m+ Apply this to Equatio 5 with l,m,,s +,,0,-. This gives : - + + + 0 + This results i a 0 whe >0. But whe 0, we get 0 0 0 + This is a CONTRADICTION to the result obtaied by usig symmetry o The result was 0 for all cases. But egatig the upper idex +, the result is 0 whe >0 ad whe 0. The mistae happeed whe we applied symmetry whe the upper idex + could be egative. + + caot be replaced by whe rages over all itegers sice it coverts 0 ito a o-zero value whe <-.
+ Also whe <-, the factor + becomes 0. The oly exceptio is whe 0 ad. Hece the result becomes differet ad this did ot tur up whe we used the example with 2. So to rectify mistae doe while usig symmetry o Equatio 2, we eed to cosider the case whe 0 ad. Applyig this to our symmetry we get: + + + + + + >0 + + - + + + + + + 0 From Equatio 4 we ow that the first term i the above expressio is 0. So : + + - + + + + 0 Cotiued.
Cotiued. + + + + + + + 0 + 0 Closed Formula Hece we get the closed formula as : + >0 2 2 + + + 0 Verificatio : This sum is 0 for all values of other tha for 0. Ad for 0 we get the sum as. Hece the closed formula for the summatio is right.