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Polynomials in one variable

A polynomial P in one variable x is formally defined as a follows

P(x) = p0 + p1x + ... + pnxn

where the pi are constants. If n = 0 we identify the polynomial with the constant p0. If pn $ \neq$ 0 then we say the polynomial has degree n. If pn = 0 then we drop the corresponding term unless n = 0; the degree of the constant polynomial 0 is considered undefined.

Exercise 9   Define the arithmetic operations on polynomials algorithmically so that polynomial manipulations can be implemented on a computer.

Polynomials can be ``evaluated'' to give functions; for any choice of a of constant b, we can substitute x by b to obtain the ``value'' of the polynomial; this gives us the function associated with a polynomial. However, note that when the field of constants is finite (e. g. the field with two elements F2) a non-constant polynomial might induce a constant function.

Exercise 10   Give an example of a polynomial that is not constant but gives a constant function on the field with three elements.

The points where the function associated with the polynomial vanishes are called solutions or roots of the polynomial.

Exercise 11   Let P(x) be a polynomial of degree n in one variable. The constant b is a root of P (i. e. P(b) = 0) if and only if P(x) can be written as a product (x - b)Q(x) where Q(x) has degree n - 1. Hence or otherwise show that P(x) has at most n roots.

The polynomial x2 + 1 has no roots over the field of real numbers. The field of complex numbers is collection of numbers of the form a + b$ \sqrt{-1}$; where a and b are real numbers.

Exercise 12   Define the arithmetic operations on complex numbers algorithmically in terms of the arithmetic operations on real numbers. Show that any quadratic polynomial x2 + ax + b is a product (x - d )(x - c) over the field of complex numbers (d and c need not be distinct).

The Fundamental Theorem of Algebra states that any polynomial over the field of real numbers is a product of linear and quadratic terms upto a non-zero constant multiple. Thus any polynomial with real coefficients has all its roots over complex numbers.

Exercise 13   Assuming the fundamental theorem of algebra show that any polynomial with complex coefficients has all its roots over complex numbers.

In the above discussion, we used the phrase ``has all its roots'' as a synonym for writing the polynomial as a product of linear terms. Now if some of these factors repeat then we say we have repeated or multiple roots. In particular, we can say that (x - b)k vanishes k times at b.

Exercise 14   Use the Binomial theorem to write a polynomial of P degree n as follows

P(x) = $\displaystyle \tilde{p}_{0}^{}$ + $\displaystyle \tilde{p}_{1}^{}$(x - b) + ... + $\displaystyle \tilde{p}_{n}^{}$(x - b)n

for some constants $ \tilde{p}_{i}^{}$.

Thus we can say that a polynomial vanishes to order k at b if the terms in the above expression for it satisfy $ \tilde{p}_{i}^{}$ = 0 whenever i < n.

Exercise 15   If P and Q vanish to order k at b, then so does R . P + Q for any polynomial R.

Note that $ \tilde{p}_{0}^{}$ is the value P(b) of P at b. Moreover, the term $ \tilde{p}_{1}^{}$ depends on P and on b that we will now to determine. Suppose that we have
P(x) = $\displaystyle \tilde{p}_{0}^{}$ + $\displaystyle \tilde{p}_{1}^{}$(x - b) + terms that vanish to order 2 at b  
Q(x) = $\displaystyle \tilde{q}_{0}^{}$ + $\displaystyle \tilde{q}_{1}^{}$(x - b) + terms that vanish to order 2 at b  
R(x) = $\displaystyle \tilde{r}_{0}^{}$ + $\displaystyle \tilde{r}_{1}^{}$(x - b) + terms that vanish to order 2 at b  

The above algebraic property of vanishing to order n shows us that

R\cdot P + Q = (\tilde{p}_0\tilde{q}_0+\tilde{r}_0)
+ (\tild...
...q}_1)(x-b) \\
+ \text{ terms that vanish to order 2 at $b$ }

Thus, if we denote the rule that associates the constant $ \tilde{p}_{1}^{}$ with P as (d /dx)| b, then this satisfies

(d /dx)| b(R . P + Q) = (d /dx)| b(R)P(b) + R(b)(d /dx)| b(P) + (d /dx)| b(Q)

Such a rule (which takes polynomial to constants) is called a (constant) derivation. If we formally replace b by the variable x we obtain the requirement for a polynomial derivation (which takes polynomial to polynomial). A polynomial derivation is a rule D which associates to every polynomial P another polynomial D(P) so that

D(R . P + Q) = R . D(P) + D(R) . P + D(Q).

and D(c) = 0 for a constant polynomial c.

Exercise 16   For any derivation D and any polynomial P show that D(Pn) = nPn - 1D(P). (Hint: Use induction). Hence or otherwise show that a derivation is determined on all polynomials once one knows what it does to the variable x.

In particular, for there is a derivation which takes x to 1. This is called the derivative with respect to x and is denoted by dP/dx.

Exercise 17   Show that the value of dP/dx at b is (d /dx)| b(P).

The relation between derivations and the order of vanishing is given by

Exercise 18   If P vanishes to order n at b then D(P) vanishes to order n - 1 (for any derivation D).

One of the aims of calculus is to find a larger class of functions which can be studied in a manner analogous to that given above for polynomials. To do this we need to generalise the notion of ``vanishing to order n'' and derivations.

A simple way to enlarge the class is to consider ``rational functions'', constructed from polynomials the same way as fractions are from natural numbers. A rational function is of the form P/Q where P and Q are polynomials with Q $ \neq$ 0.

Exercise 19   Extend the definitions of the arithmetic operations and d /dx to rational functions.

Let P and Q be any polynomials. The usual division algorithm allows us to write an expression P = RQ + S , where R and S are polynomials and the degree of S is less than that of Q or S is zero. Using this and the fundamental theorem of algebra it is not difficult to show

Exercise 20   Every rational function over real numbers is the sum of terms of the form

$\displaystyle {\frac{a}{(x-b)^n}}$ and/or $\displaystyle {\frac{a x + b}{((x-c)^2 + d^2)^n }}$

This the called the partial fraction expansion.

Given a polynomial P consider the problem of trying to find a polynomial Q so that dQ/dx = P. This is quite easily solved using the fact (proved above) that d (xn)/dx = nxn - 1. When the problem is posed for rational functions it becomes a bit harder.

Exercise 21   Show that
$\displaystyle {\frac{d}{dx}}$$\displaystyle {\frac{1}{(x-b)^n}}$ = $\displaystyle {\frac{-n}{(x-b)^{n+1}}}$  
$\displaystyle {\frac{d}{dx}}$$\displaystyle {\frac{1}{((x-c)^2 + d^2)^n}}$ = $\displaystyle {\frac{-2n(x-c)}{((x-c)^2 + d^2)^{n+1}}}$  

Hence given any rational function P/Q over reals the only hurdle to solving the problem of finding a function f so that df /dx = P/Q is to solve this when P/Q is either 1/(x - b) or (ax + b)/((x - c) + d2). In the section on integration we will see how these problems can be solved.

next up previous
Next: Polynomials in more than Up: Polynomials and polynomial functions Previous: Polynomials and polynomial functions
Kapil H. Paranjape 2001-01-20