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  • Polynomials of degree 2, if the first term is positive, starts descreasing.
  • Polynomials of degree 3, if the first term is positive, starts increasing.

is this constant? like..

  • Polynomials of degree 4, if the first term is positive, starts descreasing.
  • Polynomials of degree 5, if the first term is positive, starts increasing.
  • ...
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1  
What increase/decrease are you talking about? –  user17762 Feb 22 '11 at 23:42
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@Sivaram Ambikasaran in a 2d graph, drawing from left to right, the line starts increasing or decreasing. –  Tom Brito Feb 23 '11 at 0:13
    
@jasper maybe I need some sleep, but doesn't the term "term" include the variable and the coefficient (the product)? –  Tom Brito Feb 23 '11 at 0:42
    
@Jasper Loy if the coefficient is negative, doesn't that makes the whole term negative (and vice versa)? –  Tom Brito Feb 23 '11 at 22:25

3 Answers 3

up vote 3 down vote accepted

Assuming you mean if we start drawing it from $-\infty$ and proceed towards $+\infty$, then yes that is correct.

This is because $a_m x^m$ is the dominating term in the polynomial $P(x) = \sum_{k=0}^{m} a_k x^k$.

Now for even degree $Q(x)$ (with positive leading coefficient), it's derivative $P(x) = Q'(x)$ is of odd degree and so is negative as we start out from $-\infty$, and thus $Q(x)$ is decreasing.

For odd degree $Q(x)$, the derivative has even degree and so is positive near $-\infty$ (and also $+\infty$) and is thus increasing.

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Your second sentence is misleading, see my comment on Matt E's answer. –  Did Feb 23 '11 at 12:42
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@Didier: Dear Didier, I think that Moron is using the second sentence to make deductions about the derivative of the polynomial (so the "because" is alluding to a slightly more subtle deduction than in my answer). Regards, –  Matt E Feb 23 '11 at 13:44
    
@Didier: Matt is correct. The statement is used for the derivatives, not the original polynomial itself (if it was, there would be no mention of derivatives in the answer!). –  Aryabhata Feb 23 '11 at 15:27
    
Yes, to work on the derivative is indeed the heart of the matter. –  Did Feb 23 '11 at 15:42

As already noted, the answer is "yes". Here is one way to see it: factor out the leading term, so the polynomial has the form $P(x) = a x^n( 1 + \text{ expression in powers of } \frac{1}{x}).$

If $x$ is very positive or very negative, i.e. if $|x| \gg 0$, then the expression in powers of $\frac{1}{x}$ will be negligible, and so $P(x) \sim a x^n$ for $| x | \gg 0.$

So now just look at the behaviour of $a x^n$. If $n$ is odd this is very negative when $x$ is very negative, becoming more so as $x\to -\infty$, while it is very positive when $x$ is very positive, becoming more so as $x \to \infty$. If $n$ is even then it is very positive when $x$ is either very negative or very positive, and increases both as $x \to -\infty$ and as $x \to \infty$.

Edit: As Didier Piau notices, this argument with crude asymptotics is not precise enough to conclude true monotonicity for $|x| \gg 0$; Moron's argument with derivatives is better for that. However, it does give an explanation for the rough behaviour of $P(x)$ for $|x| \gg 0$.

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Not sure the behaviour of $x\mapsto ax^n$ proves anything about the monotonicity of $P$... Note that there exists functions $f$ such that $f(x)\sim ax^n$ when $|x|\to\infty$ although on no interval $(-\infty,x_0)$ nor $(x_0,+\infty)$ the restriction of $f$ is monotonous. –  Did Feb 23 '11 at 12:40
    
@Didier: Dear Didier, You are correct; I will add a disclaimer to this effect. Best wishes, –  Matt E Feb 23 '11 at 13:38
    
@Didier: but the question asked only about polynomials. For them, the argument holds when $|x|$ is large enough. –  Ross Millikan Feb 23 '11 at 13:54
    
@Ross: Dear Ross, But probably a little more argument is required than what I wrote; one has to say something about why the leading term swamps the other terms for large $|x|$. Rather than add the necessary estimates, I'm happy to defer to Moron's answer if the OP wants a proof of this more precise statement. Best wishes, –  Matt E Feb 23 '11 at 14:11

If by "starts", you mean the behaviour in a $(-\infty, -M)$ where $M$ some large positive real number, then yes.

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