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Background

As background, I have found that taylor expansion provides poor estimates of a function at extreme parameter values. Indeed, the approximation at extreme values can get worse (more rapid exponential increase) as the order of the taylor series increases.

This seems intuitive, but I don't know that it is a rule... or if there is a proof.

Questions

  1. Do all polynomials with order $> 1$ go to $\pm$ infinity?
  2. Is there a good reference where I can find answers to a qustion such as this?
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    & lhf: It doesn't look like an *exact* duplicate, but it is indeed similar. I think it is only clear that the ideas are duplicated if you already understand the problem and the question. Since the OP might not, as he *is* asking this question, I don't think it should be closed.2011-02-24

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Yes all non-constant polynomials are unbounded. You could see Liouville's Theorem, or just notice that the leading coefficient of the polynomial will dominate. That is if I look at $p(x)=a_n x^n +a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ then as $x\rightarrow\infty$ we have $p(x)\sim a_nx^n$ which goes to infinity.

Edit: I will elaborate, but it may not be needed, and may not clarify things further. Factoring out $x^n$, we see $p(x)=x^n\cdot \left( a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right)$ taking $x\rightarrow \infty$ we find $ \left( a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right)\rightarrow a_n$ so that $p(x)\sim a_nx^n$. As $x^n\rightarrow \infty$ when $x\rightarrow \infty$ we see that for nonzero $n$ the polynomial will go to infinity.

Remark: The symbol $g(x)\sim f(x)$ is short hand for $\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=1$, and reads "f(x) is asymptotic to g(x)."

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    and thank you for your helpful elaboration.2011-02-24
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By scaling by $\rm\:x^k\:,\:$ a nonzero rational function (polynomial fraction) $\rm\:f(x)\:$ can be written uniquely in the form $\rm\ c\ x^n\: \bar f(x)\ $ where $\rm\ \lim_{x\to\infty}\:\bar f(x) = 1\:.\:$ The degree $\rm\:n\:$ is known as the order of $\rm\:f\:$ at $\rm\infty\:.\:$ These integer exponents serve as an (archimedean) scale to compare the asymptotic growth of rational functions in a neighborhood of $\rm\:\infty\:.\:$ More generally one can extend to wider classes of asymptotically well-behaved functions such as $\rm\ log(x),\ exp(x)\ $ etc to obtain more general (non-archimedean) scales that measure so-called "orders of infinity". For example see Hardy's classic textbook by that name, google "transseries", and see my post on boundaries of convergence.

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Even linear polynomials get larger (in absolute value) than any fixed number at the variable goes to $\pm \infty$ Please see the discussion cited by lhf in the comment.

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    thanks for pointing that out; I just formulated the question as such because first order are not relevant to my application (since I start with a minimum of three points).2011-02-24