Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

PRELIMINARY ALGEBRA


The first is a further application of the method of induction. Consider the

proposal that, for anyn≥1andk≥0,


∑n−^1

s=0

k+sC
k=

n+kC
k+1. (1.53)

Notice that heren, the number of terms in the sum, is the parameter that varies,


kis a fixed parameter, whilstsis a summation index and does not appear on the


RHS of the equation.


Now we suppose that the statement (1.53) about the value of the sum of the

binomial coefficientskCk,k+1Ck,...,k+n−^1 Ckis true forn=N.Wenextwritedown


a series with an extra term and determine the implications of the supposition for


the new series:


N∑+1− 1

s=0

k+sC
k=

N∑− 1

s=0

k+sC
k+

k+NC
k

=N+kCk+1+N+kCk

=N+k+1Ck+1.

But this is just proposal (1.53) withnnow set equal toN+ 1. To obtain the last


line, we have used (1.52), withnset equal toN+k.


It only remains to consider the casen= 1, when the summation only contains

one term and (1.53) reduces to


kC
k=

1+kC
k+1.

This is trivially valid for anyksince both sides are equal to unity, thus completing


the proof of (1.53) for all positive integersn.


The second result, which gives a formula for combining terms from two sets

of binomial coefficients in a particular way (a kind of ‘convolution’, for readers


who are already familiar with this term), is derived by applying the binomial


expansion directly to the identity


(x+y)p(x+y)q≡(x+y)p+q.

Written in terms of binomial expansions, this reads


∑p

s=0

pC
sx

p−sys

∑q

t=0

qC
tx

q−tyt=

∑p+q

r=0

p+qC
rx

p+q−ryr.

We now equate coefficients ofxp+q−ryron the two sides of the equation, noting


that on the LHS all combinations ofsandtsuch thats+t=rcontribute. This


gives as an identity that


∑r

t=0

pC
r−t

qC
t=

p+qC
r=

∑r

t=0

pC
t

qC
r−t. (1.54)
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