Writing a polynomial as a product of it's roots

Let be a polynomial. We're trying to show that we can write:

.

If has a root (and it does by the FTA) then we can write:

for some entire function .

Next, in my notes it says "Clearly,

I really don't see how that is "clear", could anyone help?

Thanks.

Re: Writing a polynomial as a product of it's roots

Quote:

Originally Posted by

**Ant** Let

be a polynomial. We're trying to show that we can write:

.

If

has a root

(and it does by the FTA) then we can write:

for some entire function

.

Next, in my notes it says "Clearly,

I really don't see how that is "clear", could anyone help?

Thanks.

Did you mean to have subscript of k instead of m_k in the last factor? As in:

Are there further details for the z_i's? For instance, ?

Re: Writing a polynomial as a product of it's roots

yeah I did, I'll edit that now. Thanks.

There are no further restrictions on the z_i's, they are the roots distinct roots of P(Z) in

Re: Writing a polynomial as a product of it's roots

Okay, I think I might have it.

I think is the degree of our polynomial and so is a polynomial of degree .

Then, I think the inequality follows? Could anyone confirm this?

Re: Writing a polynomial as a product of it's roots

Quote:

Originally Posted by

**Ant** Okay, I think I might have it.

I think

is the degree of our polynomial

and so

is a polynomial of degree

.

Then, I think the inequality follows? Could anyone confirm this?

Since , wouldn't it make sense that deg(P(z))=k? Not sure how "n" fits into this. And I still think that would be too "handwavy" to say it follows thereafter. Regardless,

The reason why I asked about the z_i earlier was because it would follow nicely that if for each z_i, then

Otherwise, if this restriction on z_i isn't as above, then more will have to be done to bound the polynomial. Thoughts?

Re: Writing a polynomial as a product of it's roots

Another attempt here, bypassing the need for :

let then

where