12
$\begingroup$

I'm playing around with some equations in Geogebra and the degree of the polynomials I am using has become a variable. Is there an exact method for determining the degree of a polynomial?

I am currently approximating the degree of $f(x)$ by using:

$$\text{Degree}(f(x)) = \lfloor\log_{1000}(f(1000))\rfloor$$

This has been a decent enough approximation, but it got me wondering whether or not there exists a precise method? One that could be applied to non-integer degree polynomials.

$\endgroup$
  • 4
    $\begingroup$ Ah I see, in most contexts this is called the degree. $\endgroup$ – helloworld112358 May 16 '17 at 3:47
  • 1
    $\begingroup$ Now that you've mentioned it that does ring a bell. Rapidly changes all notation $\endgroup$ – Ben Crossley May 16 '17 at 3:49
  • 2
    $\begingroup$ Are you asking whether there is a function in Geogebra or in general? $\endgroup$ – martin.koeberl May 16 '17 at 3:56
  • 3
    $\begingroup$ There's something of an imprecision here. Sure, there's a function like this: $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 \mapsto n$, when we have access to the usual representation of our polynomial as a linear combination of powers of $x$. But really, you're interested in knowing if there's a way to find the degree of a polynomial when you only have access to the polynomial function's values, which is a seemingly more subtle problem. $\endgroup$ – pjs36 May 16 '17 at 4:01
  • 2
    $\begingroup$ Use @username to ping certain people. If you start typing a username, it should suggest you the options where you can select using arrow keys and use tab completion (You're automatically pinged because it's your question). $\endgroup$ – martin.koeberl May 16 '17 at 4:19
23
$\begingroup$

According to the GeoGebra wiki, the function you're looking for is Degree[<Polynomial]. If you have a polynomial in several variables, you can also use Degree[<Polynomial>,<Variable>] to get the degree of the polynomial in the specified variable.

$\endgroup$
  • 1
    $\begingroup$ I feel like I should have stumbled upon that. I guess that's what happens when you use the term order instead of degree! $\endgroup$ – Ben Crossley May 16 '17 at 4:17
  • $\begingroup$ Well, for next time you know :) $\endgroup$ – martin.koeberl May 16 '17 at 4:19
38
$\begingroup$

In general, if $f(x)$ is a polynomial of degree $n$, then $$\lim_{x\to\infty} \frac{\log(|f(x)|)}{\log(x)} = n$$

This works because, for large values of $x$, we would have $|f(x)| \approx |a_n| x^n$, where $a_n$ is the leading coefficient, and therefore $\log(|f(x)|) \approx \log(|a_n|) + n\log(x)$. Divide this by $\log(x)$, and we get $\frac{\log(|f(x)|)}{\log(x)} \approx \frac{\log(|a_n|)}{\log(x)} + n$. In the limit as $x \to \infty$, the first term goes to $0$.

Moreover, even if $f(x)$ is not a polynomial, but (say for example) something like $f(x) = x^{1/2} - 3x^{1/3}$, this method works, in that it returns the largest non-negative exponent among the terms; in this case, we get $1/2$.

$\endgroup$
  • $\begingroup$ This is how I arrived at my approximation. Since I can't use limits in geogebra I think I'm just going to have to work with my approximation. If you know of any alternatives please let me know! $\endgroup$ – Ben Crossley May 16 '17 at 4:15
  • 1
    $\begingroup$ @BenCrossley-hobbyist, why can you not use limits in GeoGebra? LimitBelow[log(abs(f(x)))/log(x), ∞] seems to work just fine. $\endgroup$ – N3buchadnezzar May 16 '17 at 9:53
  • 3
    $\begingroup$ Because I was unaware that putting $\infty$ was valid! I've learned some great stuff in this question! $\endgroup$ – Ben Crossley May 16 '17 at 11:00
  • $\begingroup$ This is essentially taking the logarithm base $x$. $\endgroup$ – Simon Kuang May 17 '17 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.