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200 6. Symmetric Functions of the Zeros

Another approach to obtaining the recursion relation for the pk is to

make use of infinite series expansions. Observe that the derivative satisfies

” p(t)

Iw) = z t - ti

-+lfp>-‘.

Make a change of variable t = l/s. Then

S "-'#(l/S) = S"p(l/S) fJl+ tiS+tfS2 +tfS3 + **')

i=l

or

nc” + (n - l)c,-1s + (n - 2)~“~~s~ +... + 2c2sne2 + c~s”-~

= (C” + cn-1s + C”4S2 + "'+COS")~(l+tiS+...).

i=l

On collecting terms in the various powers of s, the right side becomes

(C” + C”-1s + c,4s2 +.. .+c0s")(n+pls +p2s2 +p3s3+ -e-)

= nC"+(nC"-~+p~C")s+(nC,-2+plc"-l+p2~~)~~+~~-.

Bring all nonzero terms to one side of the equation. On the basis that

a power series (like a polynomial) vanishes iff all its coefficients vanish,

obtain the recursion relations.

E.53. A Recursion Relation. Fix the values of cl, cz, cs,... and consider

the sequence {PI, ~2, ~3,... , P”,.. .} whose first n terms are given and whose

remaining terms satisfy the equation

Pn+r + cn-1Pn+r-1 + Cn-2Pn+r-2 +... + qpr+1 + cop, = 0

forr= 1,2,3,4,5 ,.... Assign the values to the first n entries according to

the equations

Pk = -Cn-@k-l - C,-2Pk-2 - ***- kc”-+.

Use the theory of Exploration E.50 to find a formula for pk as a sum of

kth powers. Look at the particular cases n = 1,2,3.

E.54. Sum of the First n kth Powers. What is the formula for the sum

lk + 2k +. -. + nk, the sum of the first n kth powers? Consider the manic

polynomial

S”(t) = (t-- l)(t - 2)(t - 3)+..(t -n) = t” + b(n, l)t”-’

+ b(n, 2)tnv2 +... + b(n, n)

whose roots are the integers 1,2,... , n. Verify that the coefficients are given
in the following table: