Complex roots problem (a proof by induction)

[TomMe]

Problem:

If c_{1}, ..., c_{n} are the complex roots of a_{n}z^n + a_{n-1}z^{n-1} + ... + a_{1}z + a_{0} = 0 with a_{n} not 0

and S_{k} is the sum of the products of these roots taken k by k,

then S_{k} = (-1)^k . \frac{a_{n-k}}{a_{n}}. Prove this by using induction.

So for n = 3 this will give:

S_{1} = c_{1} + c_{2} + c_{3}
S_{2} = c_{1}c_{2} + c_{1}c_{3} + c_{2}c_{3}
S_{3} = c_{1}c_{2}c_{3}

Now I am stuck just at the point where I need to prove that

S_{p} = (-1)^p . \frac{a_{n-p}}{a_{n}} => S_{p+1} = (-1)^{p+1} . \frac{a_{n-(p+1)}}{a_{n}}

I thought that by looking at the n = 3 case I would see this connection between S1 and S2, but I can't see it.

Hopefully someone can help me gain some insight. Thanks!

 

[NateTG]

Can you rewrite the a_n in terms of the c_i?

 

[TomMe]

The only way I can see is a_{n} = (-1)^p . \frac{a_{n-p}}{S_{p}} with S_{p} in terms of the c_{i}.
But I can't do that since S_{p} could be zero.

No, still don't see it.

 

[NateTG]

You know that:
(z-c_1)(z-c_2)(z-c_3)...(z-c_n)=a_nz^n+a_{n-1}z^{n-1}+...a_0z^0

right?

 

[snoble]

close

a_n\cdot (z-c_1)(z-c_2)(z-c_3)...(z-c_n)=a_nz^n+a_{n-1}z^{n-1}+...a_0z^0

 

[NateTG]

close

a_n\cdot (z-c_1)(z-c_2)(z-c_3)...(z-c_n)=a_nz^n+a_{n-1}z^{n-1}+...a_0z^0

Oh, yeah, oops. I suppose that explains where the \frac{}{a_n} comes from...

 

[TomMe]

a_n\cdot (z-c_1)(z-c_2)(z-c_3)...(z-c_n)=a_nz^n+a_{n-1}z^{n-1}+...a_0z^0

Yes I know this. That's what I used with the n = 3 case.
Are you saying I just need to compare both sides and that's my proof? The point of the excercise is to prove it using induction, but I don't see how this helps me.

If k = 1 then yes, I can see that S_{1} = c_{1} + c_{2} + ... + c_{n} = -1 . \frac{a_{n-1}}{a_{n}}
But why does this mean that S_{2} = \frac{a_{n-2}}{a_{n}}?
That's where I'm stuck.

 

[TomMe]

Anyone??

 

[AKG]

Expand a_n(z - c_1)\dots(z - c_n). Let a_n = 1 for now, for simplicity.

n=1:

z - c_1 = z + a_0

c_1 = -a_0 = (-1)^1a_{1-1}

n = 2:

(z - c_1)(z - c_2) = z^2 + (-c_1 - c_2)z + (c_1c_2) = z^2 + a_1z + a_0

-c_1 - c_2 = a_1

c_1 + c_2 = -a_1 = (-1)^1a_{2-1}

c_1c_2 = a_0 = (-1)^2a_{2-2}

Does this help?

 

[shmoe]

It looks like you're trying to do induction on k, which I think is doomed to fail. Try induction on n, the degree of the polynomial. It wouldn't hurt to add another index to your S, S_{k,n}.

The idea here is that your degree n polynomial is the product of (z-c_n) and a polynomial of degree n-1. You then know the coefficients of this degree n-1 polynomial in terms of it's roots.

You could also save some writing by assuming that a_n=1. If you know the result with 1, it shouldn't be a problem to jump to the general case.

 

[TomMe]

Well, it has crossed my mind that I was doing induction on the wrong parameter here but was convinced it had to be done this way.

Thanks for your advice. I'll work this out first thing tomorrow morning and let you know how it turns out.

 

[TomMe]

Okay, I think I have it now by doing induction on n. (with a_{n} = 1)

One last question though: is it really necessary to prove this using induction? In my induction step I assumed that S_{k} = (-1)^k . a_{p-k} and so I can rewrite my p-degree polynomial with coefficients in terms of the roots. But I KNOW this is true because if I work out (z-c_{1})...(z-c_{p}) I get the same answer.
Does this mean I can prove this theorem just by looking at coefficients of the n-degree polynomial?

 

[NateTG]

Okay, I think I have it now by doing induction on n. (with a_{n} = 1)

One last question though: is it really necessary to prove this using induction? In my induction step I assumed that S_{k} = (-1)^k . a_{p-k} and so I can rewrite my p-degree polynomial with coefficients in terms of the roots. But I KNOW this is true because if I work out (z-c_{1})...(z-c_{p}) I get the same answer.
Does this mean I can prove this theorem just by looking at coefficients of the n-degree polynomial?

Well, that really depends on what you start with. I've seen proofs that addition is commutative, but, most of the time, people don't bother with stuff like that.