An application of newtons sum

For a given polynomial of degree $n$ with roots $r_1, r_2, \ldots r_n$, let $s_k$ be the kth symmetric sum and $p_k = \sum_{i=1}^n r_i ^k$ is the kth power sum

Claim: If $s_1=s_2=s_3=...=s_k=0$,then $p_1=p_2=p_3=.,..=p_k=0$. where

Proof: By induction on $k$. Base case: Since $s_1 = p_1$, hence if $s_1 = 0$ then $p_1 = 0$.

Induction step: Suppose that for some $j$, we know that $p_1, p_2, \ldots p_j = 0$. Then, by newtons sum, we have

$p_k=s_1p_{k-1}-s_2p_{k-2}+....+(-1)^{k-1}s_{k-2}p_2+(-1)^{k}s_{k-1}p_1+(-1)^{k+1}ks_k = 0 + 0 + \ldots + 0.$

Note by Aareyan Manzoor
5 years, 1 month ago

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There's no need to apply Newton's Sum.

If $s_1 = s_2 = \cdots = s_k = 0$, then the roots satisfy the equation, $x^k = 0$. So all its roots are 0, then by definition, $p_1 = p_2 = \cdots = p_k = 0$.

- 5 years, 1 month ago

what about $x^{100}-x^2+x+1=0$ i meant for that

- 5 years, 1 month ago

I don't follow. What about that polynomial?

- 5 years, 1 month ago

look at this

- 5 years, 1 month ago

That one only applies for $s_1 = s_2 = \ldots = s_{98}= 0$. And $s_{99} = -99 \ne 0$.

- 5 years, 1 month ago

i meant that see, both variables are "n".

- 5 years, 1 month ago

I don't understand what you're saying.

- 5 years, 1 month ago

i gave wrong variables in note, just realized now. edited

- 5 years, 1 month ago

@Aareyan Manzoor Great note! I've cleaned it up so that it's much easier to understand what you are saying.

Presenting it via a proof by induction isn't necessary, but helps build the idea that all we need to show $p_j = 0$ is up to $s_j = 0$.

By improving the format of your presentation, you make it immediately clear what you want to show, and why it's interesting.

Staff - 5 years, 1 month ago

thank you sir!

- 5 years, 1 month ago

nice way....

- 5 years, 1 month ago