CHAPTER II DIMENSION In the present chapter we investigate ...

which is the require formula.
Now we have the following situation: given p(x), once we known that there is q(x)
such that q(x + 1) − q(x) = p(x), then we can find q(x). The question is, how do we
know such q(x) exists? Now Theorem 2.3.1 comes to help. Denote by Pd the space of all
polynomials of degree not exceeding d. Notice that if q(x) is in Pd+ 1, then the degree
of q(x + 1) − q(x) will be 1 less than that of q(x) (because the highest power terms in
q(x + 1) and q(x) are canceled out) and hence q(x + 1) − q(x) belongs to Pd. Thus we can
define a mapping from Pd+ 1 to Pd by putting T (q(x)) = q(x + 1) − q(x). Now we apply
dim ker T + dim T (V ) = dim V with V = Pd+ 1. As we know, dim V = dim Pd+ 1 = d + 2.
We have to find out dim ker T . Suppose q(x) belongs to ker T . Then q(x + 1) − q(x) = 0,
or q(x) = q(x + 1) (for all x). In particular, q(0) = q(1) = q(2) = q(3) = · · · . This
tells us that all positive integers are roots of q(x) − q(0). But a polynomial cannot
have infinitely many roots unless it is the zero polynomial. Hence q(x) − q(0) = 0, or
q(x) = q(0), that is, q(x) is a constant polynomial. We have shown that ker T is the
space of constant polynomials. Hence dim ker T = 1. Now dim ker T + dim T (V ) = dim V
becomes 1 + dim T (V ) = d + 2, or dim T (V ) = d + 1. Here T (V ) is a subspace of Pd with
dim T (V ) = dim Pd = d + 1 and hence T (V ) = Pd. In particular p(x) ∈ Pd is in the range
of T , showing that there exists q(x) in Pd+ 1 with q(x + 1) − q(x) = p(x).
2.6. Consider a variation of the general problem studied in the last subsection: given
a polynomial p(x) of degree d, and a constant a, how can we find a formula for
Sn = p(0) + p(1)a + p(2)a 2 + · · · + p(n)a n ≡ n
k= 1 p(k)ak
(2.6.1)
which is valid for all n? Certainly, when a = 1, we return to the problem which is dealt
with in the last subsection. So we assume a = 1.
We look for an expression f(x) such that f(k + 1) − f(k) = p(k)a k so that we can
use a “telescoping sum” as in the previous subsection to obtain Sn = f(n + 1) − f(0). Let
us try f(k) = q(k)a k , where q(x) is some polynomial. Now
f(k + 1) − f(k) = q(k + 1)a k+ 1 − q(k)a k = (aq(k + 1) − q(k))a k .
So f(k + 1) − f(k) = p(k)a k becomes (aq(k + 1) − q(k))a k = p(k)a k . Naturally we ask
if there is a polynomial q(x) such that aq(x + 1) − q(x) = p(x). Again, the question is
whether such q(x) exists. Once we know it does exist, we get the license to look for it!
Define a linear operator T on Pd by putting T (q(x)) = aq(x + 1) − q(x). By
Corollary 2.4.3, we know that, in order to show that T (q(x)) = p(x) has a solution (here
p(x) is given and q(x) is unknown), it suffices to show that ker T is the zero space. Let
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