Product expression for roots of polynomial

53 Views Asked by At

I am working slowly through "Riemann's zeta function", HM Edwards, Dover Publications, 1974. At the top of page 18, I read (I have altered the notation from $p(s)$ to $P(x)$)

"any polynomial $P(x)$ can be expanded as a finite product $P(x) = P(0) \prod_\rho \bigl(1-\frac{x}{\rho } \bigr)$ where $\rho$ ranges over the roots of the equation $P(\rho) = 0$.

Really? Surely, $P(x) = \prod_\rho \bigl(x-\rho \bigr)$? I wondered if this was something to do with extracting the factor $P(0)$ in order to give the roots of a monomial, but of course this cannot be the case, since this would require division by the coefficient of highest order. And besides, a simple counter-example establishes that Edwards' assertion is incorrect. Define

$$ \begin{aligned} P(x) :=& (x+1) (x+2) (x+3) (x+4) (x+5) \\ =& x^5 + 15x^4 + 85x^3 + 225x^2 + 274x + 120 \end{aligned} $$

(I have chosen a polynomial with real roots for simplicity, but I see no reason why the same definitions should not apply to finite polynomials with complex coefficients.)

Applying $P(x) = P(0) \prod_\rho \bigl(1-\frac{x}{\rho } \bigr)$ gives

$$ \begin{aligned} P(x)=&120 (1-x) \left(1-\frac{x}{2}\right) \left(1-\frac{x}{3}\right) \left(1-\frac{x}{4}\right) \left(1-\frac{x}{5}\right) \\ =& -x^5 + 15x^4 - 85x^3 + 225x^2 - 274x + 120 \end{aligned}$$

So, in this case, the correct assertion would be $P(x) = P(0) \prod_\rho \bigl(1+\frac{x}{\rho } \bigr)$.

Could someone please explain what is going on here? Weierstrass and Hadamard's factorisation theorems don't seem to help, but perhaps I have understood them only poorly.

How is Edwards' original assertion arrived at, and why does the counter-example not work?

1

There are 1 best solutions below

5
On BEST ANSWER

Your simple example establishes that your assertion is incorrect.

The roots of

$$P(x) := (x+1) (x+2) (x+3) (x+4) (x+5)$$

being $-1,-2,-3,-4,-5$ we do have

$$P(x)=120\left(1+\dfrac x1\right) \left(1+\dfrac x2\right) \left(1+\dfrac x3\right) \left(1+\dfrac x4\right) \left(1+\dfrac x5\right).$$

The proof is very trivial by transforming all factors with

$$\rho-x=\rho\left(1-\dfrac x\rho\right).$$