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Help is needed in explaining the following (partial) proof:-

Let $Q(x) = ax^4 + bx^3 + cx^2 + dx + e$.

Suppose “that Q(x) = 0 has no real roots. Thus, Q(x) is always positive or negative for all real x. WLOG, (we can) assume that Q(x) > 0 for all real x, in which case a > 0.”

My question is:- Is the “a > 0” part a further assumption? Or is it a direct consequence of the only assumption. If it is yes to the latter, is there any supporting reason for that?

Mick
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  • The behaviour of $f$ for really large values of $x$ is more or less only decided by $a$. So if the function is positive there, it means that $a$ is positive. – Arthur Jul 26 '16 at 09:11
  • @Arthur In layman’s term, can I argue in the following way? For some positive X, (b is very large) but (a is extremely small and negative), we can still get Q(X) > 0. If this is so, then we have no right in claiming a > 0 always. – Mick Jul 26 '16 at 10:54
  • When $x$ passes $|b/a|$ then $ax^4$ will start dominating $bx^3$. Go a bit further beyond that to account for $c,d$ and $e$, and $ax^4$ dominates the entire expression, and will keep doing so for any bigger $x$. – Arthur Jul 26 '16 at 11:14
  • @Arthur Agree that the 4th degree term will start to dominate when X > |b/a|. What about if X is to the left of it? – Mick Jul 26 '16 at 11:38
  • That's not important. Since $ax^4$ dominates somewhere and we assume that $f$ is positive everywhere, that means that $ax^4$ must be positive. – Arthur Jul 26 '16 at 12:14

3 Answers3

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It follows from the fact that $Q(x)$ is assumed to always be positive. Note that as $x \to \infty$, the term $ax^4$ will become dominant for the value of $Q(x)$.

To be more specific, choose some $x > \max\left\{ \left|4\frac{b}{a}\right| ,\left|4\frac{c}{a}\right| ^{1/2},\left|4\frac{d}{a}\right| ^{1/3},\left|4\frac{e}{a}\right| ^{1/4}\right\}$. Then $$|ax^4| = |a||x|^4 = \frac{|a|}{4}\left(|x||x|^3+|x|^2|x|^2+|x|^3|x|+|x|^4\right) > \frac{|a|}{4}\left(\left|4\frac{b}{a}\right||x|^3+\left|4\frac{c}{a}\right||x|^2+\left|4\frac{d}{a}\right||x|+\left|4\frac{e}{a}\right|\right) = |bx^3|+|cx^2|+|dx|+|e| > |bx^3+cx^2+dx+e|$$ Hence the sign of $a$ determines the sign of $Q(x)$. Since we assumed $Q(x)>0$, we must have $a>0$.

sTertooy
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  • In layman’s term, can I argue in the following way? For some positive X, (b is very large) but (a is extremely small and negative), we can still get Q(X) > 0. If this is so, then we have no right in claiming a > 0 always. – Mick Jul 26 '16 at 10:54
  • For a general $4$-th order polynomial, yes, but the polynomial must have a root then. In the case you describe ($Q(x)$ has no roots) this will not work. – sTertooy Jul 26 '16 at 11:22
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The acronym Without Loss of Generality means that you can hypothetise some property without weakening the generality of the solution, because you can always transform the given problem into one that verifies the property.

In this particular case, if $Q(x)<0$ for all $x$, you can replace $Q(x)$ by $Q'(x)=-Q(x)$ so that $Q'(x)>0$ for all $x$.

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There are 3 cases to consider: $a>0$, $a<0$ and possibly $a=0$. If the polynomial is of degree 4, the last case is excluded. It then suffices, as already mentioned, to consider $a>0$ (say, of positive type). For non-zero $a$ this is equivalent to ask that $Q>0$ as the leading term dominates when $|x|$ goes to infinity. The map $Q \mapsto -Q$ is a bijection from solutions of positive type to the solutions of negative type.

However, if the polynomial need not be of degree 4, then other solutions are the non-zero constants and quadratic polynomials without roots.

H. H. Rugh
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