Summation of the polynomial and division

[Susanne217]

Homework Statement

Let p(z) = \sum_{j=0}^{n} a_{n-j}z^j be a polynomial of at least degree 1 thus
n \geq 1.
Show that if z\neq 0 then 1/z is a root of the polynomial p.

Homework Equations

Fundamental theorem of Algebra

The Attempt at a Solution

If a expand the polynomial above then

p(z) = \sum_{j=0}^{n} a_{n-j}z^j = a_n + a_{n-1}z + a_{n-2}z^2 + \cdots + a_0 z^n

to test if 1/z is a root of p then this must be equal too

p(1/z) = \sum_{j=0}^{n} a_{n-j}z^j = a_n + a_{n-1}(1/z) + a_{n-2}(1/z)^2 + \cdots + a_0 (1/z)^n = 0

Then as I see it its my task to show that the new sum

\sum_{n=0}^{\infty} a_n (\frac{1}{z})^n converges to zero. Then by the ratio-test

\frac{a_{n+1}(\frac{1}{z})^{n+1}}{a_n (\frac{1}{z})^{n}} \rightarrow 0 as z \rightarrow \infty therefore 1/z is a root of p.

Am I on the right track here?

 

[Count Iblis]

I think your problem should be formulated as follows. If

q(z) = \sum_{j=0}^{n}a_{j}z^{j}

and we define p(z) in the way you have written down, then if q(z)=0 for z not equal to zero, then we have that p(1/z) = 0.

 

[Susanne217]

I think your problem should be formulated as follows. If

q(z) = \sum_{j=0}^{n}a_{j}z^{j}

and we define p(z) in the way you have written down, then if q(z)=0 for z not equal to zero, then we have that p(1/z) = 0.

Hi Count Iblis and thank you for your answer,

So what you are saying (please correct if I missunderstand) that if I show that if q(z) doesn't converge to zero then p(1/z) = 0 is a root of p?

thats easy cause if n is of degree at least 1 then then in my understanding
\lim_{n \rightarrow 1} q(z) \rightarrow 1

 

[Count Iblis]

q(z) is assumed to be zero for some z. And then you can see that
p(1/z) must be zero as well. This, of course, for z itself not equal to zero, otherwise 1/z would not exist.

I don't see how convergence plays any role here, as we're dealing with polynomials...

 

[Susanne217]

q(z) is assumed to be zero for some z. And then you can see that
p(1/z) must be zero as well. This, of course, for z itself not equal to zero, otherwise 1/z would not exist.

I don't see how convergence plays any role here, as we're dealing with polynomials...

I see that now :) But is this really that easily? No need to show anything to show that p(1/z) is a root of p?

 

[Count Iblis]

You do need to show that p(1/z) is zero. You can take the expression for p(1/z) you wrote down:

a_n + a_{n-1} 1/z + ...+a_0 1/z^n

If you multiply this with z^n, you get...?

 

[Susanne217]

You do need to show that p(1/z) is zero. You can take the expression for p(1/z) you wrote down:

a_n + a_{n-1} 1/z + ...+a_0 1/z^n

If you multiply this with z^n, you get...?

I get 1, but if I can't use convergense how else am I suppose to show that its zero?

 

[Count Iblis]

What I mean is:

(a_n + a_{n-1} 1/z + ...+a_0 1/z^n) z^n =

a_n z^n+ a_{n-1} z^(n-1) + ...+a_0

 

[Susanne217]

I am such a dumb person Count...

If you multiply the above by z^n then you get the original p(z)...

where q(z) = z^n for a set not equal to zero then p(1/z) = 0 and thus 1/z is a root of p :) Right ?