If I have a polynomial, for example, $$ x^8 + x^2 + \dfrac{1}{x} $$ would this be considered to be of degree 8? I am working on a question involving the function $ \frac{1}{x} $ and I am wondering how this term affects the degree of the entire polynomial, if at all.

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    $\begingroup$ it is not polynomial , try to use Taylor expansion to get the polynomial $\endgroup$ – user133607 Mar 6 '14 at 21:09
  • $\begingroup$ @AhmedEssamTawfik: The Taylor expansion doesn't give a polynomial (also, considering the singularity at $0$, I'm not really sure what this would do for the problem). Can you clarify what you mean? $\endgroup$ – user61527 Mar 6 '14 at 21:26
  • $\begingroup$ See (math.stackexchange.com/questions/697838/degree-of-a-function) for a general degree of a function. Using such definition, the degree of your function is $8$. $\endgroup$ – pisoir Mar 6 '14 at 22:11
  • $\begingroup$ @pisoir: This is just plain wrong. This is a rational function and it has degree $9$. See my answer below. $\endgroup$ – MPW Mar 7 '14 at 12:02
  • $\begingroup$ @MPW Why do you think it is plain wrong? I think it depends on the definition of degree. If you compute degree as $\lim_{x\to\infty}\frac{xf^\prime(x)}{f(x)}=\lim_{x\to\infty}\frac{8x^8+2x^2-1/x}{x^8+x^2+1/x}=8$. Your answer is based on the definition of degree for rational functions. I don't think you can say "one is better than the other". $\endgroup$ – pisoir Mar 7 '14 at 12:44

That is not a polynomial because of the $1/x$. A polynomial in $x$ is of the form $a_nx^n+a_{n-1}x^{n-1}...+a_1x+a_0$ where $a_0,a_1,...,a_{n-1},a_n$ are constants.

  • $\begingroup$ 1 / x = x ^ -1. You never said n >= 0 $\endgroup$ – Alvaro Mar 6 '14 at 22:31
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    $\begingroup$ @Alvaro True, the definition isn't as clear as it could be, but if n < 0 then there is no such formula, since repeatedly subtracting 1 from a negative number will never get you to 1 or 0, which are the last two terms here. $\endgroup$ – Kyle Strand Mar 6 '14 at 22:37
  • $\begingroup$ Yes, you are right $\endgroup$ – Alvaro Mar 6 '14 at 22:40

An infinitely-differentiable function $f(x)$ defined for all real $x$ is a polynomial of degree $n$ if and only if $f^{(n+1)}(x)$ is identically zero, while $f^{(n)}(x)$ is not.

(Here $f^{(k)}(x)$ denotes the $k$-th derivative of $f(x)$.)

  • $\begingroup$ No. Functions and polynomials are entirely different things. This is a source of a lot confusion for students. It is true that the natural map $\mathbb{R}[X]$ to $C^{\infty}(\mathbb{R}) \subset \mathrm{Map}(\mathbb{R}, \mathbb{R})$ is injective, but when moving to polynomials with coefficients in other rings it is no longer true that $A[X] \to \mathrm{Map}(A,A)$ is injective. Otherwise one could easily prove that $\mathbb{F}_{2}$ is algebraically closed. [cntd] $\endgroup$ – jmc Mar 7 '14 at 5:50
  • $\begingroup$ (After all, a field is algebraically closed if every non-constant polynomial has a zero. Well, over $\mathbb{F}_{2}$, any polynomial that does not have zero's is constant $1$, right?) $\endgroup$ – jmc Mar 7 '14 at 5:50
  • $\begingroup$ So, what you can say, is that such a function $f(x)$ is the image of a polynomial under the map that I described. But I would still think it is a horrible definition, even though I agree that it is an immensely useful theorem. $\endgroup$ – jmc Mar 7 '14 at 5:52
  • $\begingroup$ This was not meant to be a definition of a polynomial, only a necessary and sufficient condition for a real-valued function to be a polynomial one. $\endgroup$ – LeoTheKub Mar 7 '14 at 17:04

The notion of degree is well-defined for rational functions. In this case, it is a rational function of degree $9$. The degree of a rational function $P(x)/Q(x)$ is $\max(\deg P,\deg Q)$ (where $P$ and $Q$ are polynomials with no common (nontrivial) factor and Q not identically zero).

You can see this because your function can be written as $$f(x)=\frac{x^9 + x^3 + 1}{x}$$

  • $\begingroup$ Your definition tells me that his $f$ has degree $9$. Don't you want to take the difference of the degrees? $\endgroup$ – jmc Mar 7 '14 at 5:53
  • $\begingroup$ No, it's definitely the larger, not the difference. The degree describes how many preimages the typical point has. Besides, if you subtracted, the you would get $0$ for something like $(x+1)/(x-1)$. You can think of it like this: if you tried to solve something like $f(x)=2$, what would you do? You would clear fractions and end up with a degree $9$ polynomial equation. $\endgroup$ – MPW Mar 7 '14 at 11:54
  • $\begingroup$ Hmm, well, I think it also makes sense to take the difference. But I see what your reasons are. (In algebraic geometry, one definitely takes the difference.) $\endgroup$ – jmc Mar 7 '14 at 19:03

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