×

# 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
1 year, 3 months ago

Sort by:

@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 · 1 year, 3 months ago

thank you sir! · 1 year, 3 months ago

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, $$p_1 = p_2 = \cdots = p_k = 0$$. · 1 year, 3 months ago

what about $x^{100}-x^2+x+1=0$ i meant for that · 1 year, 3 months ago

I don't follow. What about that polynomial? · 1 year, 3 months ago

look at this · 1 year, 3 months ago

That one only applies for $$s_1 = s_2 = \ldots = s_{98}= 0$$. And $$s_{99} = -99 \ne 0$$. · 1 year, 3 months ago

i meant that see, both variables are "n". · 1 year, 3 months ago

I don't understand what you're saying. · 1 year, 3 months ago

i gave wrong variables in note, just realized now. edited · 1 year, 3 months ago

nice way.... · 1 year, 3 months ago

×