Newton's Identity

Question

While reading about quadratic equations, I came across Newton's Identity formula which said we can express $\alpha^n+\beta^n$ in simpler forms but not given any explanation. They wrote $S_n=\alpha^n+\beta^n$ and plugged in the quadratic equation $f(x)=ax^2+bx+c$ to write: $$f(x)=aS_{n+1}+bS_n+cS_{n-1}$$ Also what if we want to find for any polynomial the value: $$\alpha^n-\beta^n$$ I went through articles on internet but they have some very complicated proofs. I want to understand how do we derive this in a simple manner and what have they exactly done with the quadratic equation? Some examples would make it very clear to understand. Thanks

Answer 1

$\newcommand{\a}{\alpha}\newcommand{\b}{\beta}$Consider the quadratic equation $$x^2 - (\a + \b)x + \a\b = 0.$$ It is easy to see that $\a$ and $\b$ both satisfy the equation. In particular, this means that $$x^2 = (\a + \b)x - \a\b$$ for $x = \a, \b$. Multiplying by $x^n$ for $n \geqslant 0$, we get $$x^{n + 2} = (\a + \b) x^{n + 1} - \a\b x^n.$$

The above is also satisfied by $\a$ and $\b$ both. Thus, adding the corresponding equations for $\a$ and $\b$, we get $$S_{n + 2} = (\a + \b) S_{n + 1} - \a\b S_n$$ for all $n \geqslant 0$. (With the convention that $S_0 := 2$.)

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This technique similarly gives you the recursion for $\a^n - \b^n$ as well. I leave that to you.