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I am trying to understand the logic behind solving this problem from youtube.

Given the equation:

$p_n-e_1p_{n-1}+\dots+n(-1)^ne_n = 0$

And the Newton-Girard identity:

$ke_k = e_{k-1}p_1-e_{k-2}p_2+\dots+(-1)^{k-1}e_0p_k$

How are these 2 equations related ?

Does it mean $ke_k = 0$ ? That makes no sense, yet looking at the right hand side of both the equations they both look the same.

What is going on? I cant seem to understand.

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ng.newbie
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    I noticed the first equation on the left hand side ends with $n(-1)^ne_n$. That seems to be the term on the left hand side of the second equation. It doesn't follow the pattern suggested by the first two terms. – Matt Samuel Dec 05 '19 at 19:35
  • @MattSamuel could explain how $n(-1)^ne_n = ke_k$ ? I cant follow how you are arriving to that conclusion. – ng.newbie Dec 07 '19 at 18:12

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Consider specific values. Can you go between

$$ p_5-e_1p_4+e_2p_3-e_3p_2+e_4p_1-5e_5=0 $$

and

$$ 5e_5=e_4p_1-e_3p_2+e_2p_3-e_1p_4+p_4? $$

Sure you can; just solve for $5e_5$ by adding it to both sides (and reordering the terms on the one side). If instead $k$ is even then you could subtract $ke_k$ and multiply by $-1$.

(Note $e_0=1$ is the sum of the empty product and if $k=n$ then $p_n=n$ so the identity would have the even more symmetric form $\sum_{i=0}^n (-1)^ie_ip_{n-i}=0$.)

anon
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  • Oh I see. It's actually $p_0=n$. – Matt Samuel Dec 05 '19 at 21:41
  • @runway44 how is $n(-1)^ne_n = ke_k$ – ng.newbie Dec 07 '19 at 18:12
  • @ng They're not necessarily equal because of the sign, but it's really the same equation with the sign flipped and one term moved. – Matt Samuel Dec 07 '19 at 23:44
  • @ng.newbie In order to go between the two identities you have to replace $n$ with $k$. And then you have to solve for $ke_k$ by getting it to one side. So you either add $ke_k$ over to the lone side of the equation, or you subtract it over and then multiply both sides by $-1$. Did you read my answer and try to understand the $k=5$ example? – anon Dec 08 '19 at 01:00