For a given polynomial of degree n with roots r1,r2,…rn, let sk be the kth symmetric sum and pk=∑i=1nrik is the kth power sum
Claim: If s1=s2=s3=...=sk=0,then p1=p2=p3=.,..=pk=0. where
Proof: By induction on k.
Base case: Since s1=p1, hence if s1=0 then p1=0.
Induction step: Suppose that for some j, we know that p1,p2,…pj=0. Then, by newtons sum, we have
pk=s1pk−1−s2pk−2+....+(−1)k−1sk−2p2+(−1)ksk−1p1+(−1)k+1ksk=0+0+…+0.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestThere's no need to apply Newton's Sum.
If s1=s2=⋯=sk=0, then the roots satisfy the equation, xk=0. So all its roots are 0, then by definition, p1=p2=⋯=pk=0.
Log in to reply
what about x100−x2+x+1=0 i meant for that
Log in to reply
I don't follow. What about that polynomial?
Log in to reply
this
look atLog in to reply
s1=s2=…=s98=0. And s99=−99=0.
That one only applies forLog in to reply
Log in to reply
Log in to reply
Log in to reply
@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 pj=0 is up to sj=0.
By improving the format of your presentation, you make it immediately clear what you want to show, and why it's interesting.
Log in to reply
thank you sir!
Log in to reply
nice way....
Log in to reply