Talk:Degree of a polynomial

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Formal definition[edit]

I'm new to this Wikipedia thing, so I'll just post my opinion here before editing. Hope that someone with more experience could help.

The point is that there is no formal (mathematically correct) definition given. What you have is just an explanation/illustration.

Generally a polynomial is defind as a sequence of elements (called coefficients) from a ring indexed by natural numbers with the following characteristic: there is a natural number n (called the degree) so that the n-th coefficient is non-zero and all coefficients with index higher than are zero.

i.e. it is something like P={a0,a1,...,an,0,0,...} where a0, are ellements of R and 0 is the zero in R.

More formally: a polynomial is a function P from N (natural numbers) to R (where R is a ring) where there is n in N (the degree) so that P(n) is not zero and for all m>n P(m)=0.

The thing is that the definition of the degree is embedded in the definition of the polynomial. What would be the best way to include here the formal/mathematically correct definition of the degree - should we repeat the polynomial definition here? - AdamSmithee 22:25, 17 November 2005 (UTC)

Two conflicting definitions seem to be given. The first paragraph says the degree is the sum of the powers of all terms, the second says it is the sum of powers in one term. That is confusing. Jewels Vern (talk) 05:23, 18 October 2013 (UTC)

Please, when you post a comment which does not belongs to an existing thread, put it in a new section at the end of the page. The "New section" button at the top of the talk page allow to do this easily. I so not understand you post: The first paragraph says correctly "The degree of a polynomial is the highest degree of its terms". "Highest" is not the same as "sum". D.Lazard (talk) 09:22, 18 October 2013 (UTC)

Degrees 0 and 1[edit]

I've had a bit of a debate with someone regarding degrees 0 and 1 of a polynomial. My argument is this: the degree of is 1, as the exponent on the variable is 1. But the degree of any constant term is 0. E.g: , as 0 is the exponent on x. To my surprise, they were arguing that , therefore the degree is 1. Surely the coefficients can't be used in determining the degree, otherwise the degree of would be 3?! I'm not a mathematician so I'm a bit reluctant to change the article, but would it be reasonable to clarify this in the article? Or maybe I was wrong? -- 15:55, 13 January 2006 (UTC)

The degree = the highest power of X with a non-zero coeff. Indeed, the degree of the polynomial is 0. The factorization of the coeff into primes (if available) doesn't change the degree of the polynomial AdamSmithee 09:07, 16 January 2006 (UTC)

fractional degree[edit]

degree of a function being "D", can |D|<1, while squreroots are 1/2 i'm not very sure, on this. but it isn't adressed in the artical

Thanks for asking. See Degree_of_a_polynomial#The_degree_computed_from_the_function_values. The degree of 3x1/2+5x is 1, while the degree of 3x1/2+17 is 1/2. I moved your question to the bottom of this discussion page where new contributions are expected to appear. You may sign your discussion entries automatically by four tildes '~'. Bo Jacoby 10:50, 9 August 2006 (UTC)

The degree of the zero polynomial is minus infinity[edit]

This is completely bogus. The degree of f(x) = 0 is 0. Ceroklis 20:17, 10 April 2007 (UTC)

The degree of f(x)=c is zero, but only if c is not 0. The degree of 0 is undefined (but many say that the degree is minus infinity so that it "sorts" below the constants). Here are a couple justifications for this statement:
  • In a Euclidean domain, the "norm" function (which is what "degree" is over the polynomial ring) is not defined at 0.
  • If f(x) has degree a and g(x) has degree b, we want the polynomial f(x)*g(x) to have degree a+b. Our polynomials should not "shrink" when we multiply them. If f(x)=0, then f(x)*g(x)=0, no matter what degree g(x) has. This will break many of our theorems. Therefore, we do not define the degree at 0. (This is a special case of the "norm" function for a euclidean domain which requires that deg(f*g)≥deg(f) and deg(f*g)≥deg(g).)
I'm not sure if this is enough reasoning for you... Does this make sense? - grubber 04:36, 11 April 2007 (UTC)

I does make sense, I have now figured it out. I guess the proper way to define this would go along those lines:
A polynomial over a ring is a function such that is finite.

  • Note that this definition, contrary to the one proposed by AdamSmithee above is correct in the sense that it doesn't exclude the zero polynomial.

The degree of a polynomial is the highest n such that . Obviously this would then not be defined for the zero polynomial.
We should have these formal definitions somewhere, and then everything follows, that R[X] is a ring, that deg is a valuation, etc... Right now there are bits and pieces spread among the polynomial, degree of a polynomial and polynomial ring articles. Any idea on how to unify these articles ?
grubber: I have noted your new section on abstract algebra. This is a good idea but as said above it should be based on formal definitions and united with the other articles. In particular, the last sentence is misleading. deg(0) is not undefined because the norm is undefined at zero, it is undefined due to the way deg was defined. Deriving stuff in the proper order is important. Ceroklis 23:19, 11 April 2007 (UTC)

A ring is euclidean if there is some norm that satisfies the appropriate axioms. Finding such a norm is often very difficult, and there are often (perhaps always??) numerous norms whenever there is one (for example, if deg(f) is defined as we would expect and then we create a new function D, defined as D(f) = 42*deg(f), this is a valid norm). So, I say that degree is not defined at 0 because degree is a norm and norms are not defined at 0. But, it is something of a circular argument, so there probably is a better way to state it.
As far as your way of defining a polynomial, I'm not following. As I read your definition, if we let R be the real numbers, then how is exp(x) not a polynomial by your definition? - grubber 22:17, 12 April 2007 (UTC)
Your are confused by the distinction between a polynomial (an abstract object) and a polynomial function.
In my definition gives the coefficient of the nth power of the formal variable .
To a polynomial over you associate a polynomial function defined by . We often identify and because on or , the mapping is injective, but it is important to understand that they are different objects. is not a polynomial function, because there is no polynomial over with .
Another thing that may confuse you is that I define the polynomial as a function, instead of as a sequence, but that is equivalent. I find this way of doing things more elegant because you don't have bounds on the size of the sequence appearing anywhere. Ceroklis 00:51, 13 April 2007 (UTC)
Aha.. I thought the function P was evaluating the polynomial in the ring, and I didn't understand what N was (now I see it). OK, now that I understand the setup, notice that even in your definition, for the zero polynomial, "the highest n such that " is not defined, because is an empty set. What would the maximum (or even supremum) of an empty set look like? You could make a theorem that if , then and if P1 represents the zero polynomial, we can conclude that the degree of the zero polynomial is any arbitrary natural number (with the proper choice of P2). The joys of vacuously true statements on the empty set! I think the only way to resolve this conflict is to set the degree of the polynomial zero to be either "undefined" or explicitly defined as some negative number (since the empty set is contained in every one of these degree "sets", it must be strictly smaller than all of them). - grubber 16:39, 13 April 2007 (UTC)
Perhaps you should read more carefully; I wrote above: "Obviously this would then not be defined for the zero polynomial".Ceroklis 18:06, 13 April 2007 (UTC)
So you did, my mistake. So, I guess we really are in agreement on all this? - grubber 19:00, 13 April 2007 (UTC)
Yes. The question is now how to best integrate these formal definitions (or equivalent) in the various articles I mentioned. Ceroklis 21:49, 13 April 2007 (UTC)

Degree of i[edit]

i is the squart root of -1. So can someone explain what degree i and its factors are? --pizza1512 Talk Autograph 20:02, 13 April 2007 (UTC)

Over the complex, is just a non-zero constant, like 2 for example. i.e. . So . Furthermore, it is already factored. Ceroklis 21:49, 13 April 2007 (UTC)

what is the sum of a polynomials[edit] (talk) 17:04, 15 March 2010 (UTC)

Naming a 100th degree polynomial[edit]

Why is hectic preferred over centic?? Georgia guy (talk) 00:01, 4 March 2011 (UTC)

For that matter, why there is a name for them at all? Even in the rare circumstance of one actually appearing, it would probably just get named "100th degree polynomial". They don't deserve a name any more than (say) 73rd degree ones do (and much less than 11th)... -- (talk) 20:31, 7 June 2011 (UTC)
I think 'hectic' is supposed to be math humor. We'll use 'bajillionic' if we really want, until the 100th degree polynomial actually appears in regular contexts, and then we'll name it appropriately. Sure, not everything gets a name (see 'once', 'twice', 'thrice', 4-ce? quince? ...), but those that do are named humorously at first (see 'velocity', 'acceleration', 'jerk', 'snap', 'crackle', and 'pop'.) Sobeita (talk) 20:50, 28 October 2011 (UTC)

Hectic's link[edit]

Hectic links to itself - the link should be removed or a stub should be made! Sobeita (talk) 20:50, 28 October 2011 (UTC)

What's the justification for this statement?[edit]

"To determine the degree of a polynomial that is not in standard form (for example (y − 3)(2y + 6)( − 4y − 21)) it is easier to expand or express the polynomial into a sum or difference of terms;"

The example given in parentheses contradicts the statement made. If you're in an integral domain in the example, you can take the highest order term in each factor, and add them up. In terms of total number of operations, this is much "easier" than multiplying it all out first. For more general rings, you have to look for zero divisors and so on, but it's not obvious whether or not there is a more efficient method for doing this than doing the entire expansion. In any case, unless there is a definite computational idea in mind, "it is easier" sounds more like an opinion than a fact. — Preceding unsigned comment added by (talk) 03:51, 7 January 2012 (UTC)

I have corrected the statement and changed the example. D.Lazard (talk) 11:12, 20 October 2012 (UTC)

Degree of zero[edit]

Spencerleet has removed two times from the article that -1 is a common convention for the degree of 0. His argument is that he finds this convention not convenient. I agree with him that this convention breaks the formula for the degree of a product. However, the problem is not there. The article asserts (asserted) that some authors use this convention. Undoubtedly, the choice of -1 for the degree of zero is very common, when programming Euclidean division of polynomials, Euclid's algorithm for polynomials, etc., because using -∞ in a program is difficult. For this reason, I'll revert again Spencerleet's edit until he provides, if any, a reliable source asserting that this convention is not common. D.Lazard (talk) 21:30, 7 July 2014 (UTC)

Here is a mathematics book that defines the degree of the zero polynomial to be -1: "By convention, the zero polynomial has degree -1." (p. 233)
-- (talk) 04:39, 16 September 2014 (UTC)
Childs uses -1 in the 2nd edition, but in the third edition Childs uses −∞ and explains the advantages of using −∞ (p. 288). Since this is strong evidence that mathematicians regard -1 as a poor choice, I have cited both editions in the article. Do you have a reliable source that explains why -1 is preferable in some cases? -- (talk) 20:59, 16 September 2014 (UTC)
Ease of writing programs is no reason for a mathematical convention: it is only a reason for a choice of representation by a programmer, where −1 can represent −∞. I doubt whether there will be any solid reason to prefer −1 in a general context. What is important is whether it is used by convention; here mention of both conventions appears to be appropriate, but the explanation shows why −∞ "works" better mathematically. —Quondum 21:29, 16 September 2014 (UTC)
There is reason to believe that -1 is the degree of 1/x (see the later section). TomS TDotO (talk) 17:59, 17 September 2014 (UTC)
I agree, but 1/x is not a polynomial, and this article is about the "Degree of a polynomial".D.Lazard (talk) 19:07, 17 September 2014 (UTC)
I think the article covers this all perfectly adequately. —Quondum 19:09, 17 September 2014 (UTC)

Link to term[edit]

This edit changed the link for "term" from Term (mathematics) to Addition#Notation. Term (mathematics) was merged into Term (logic) and changed to a redirect. The section Term_(logic)#Elementary_mathematics discusses "term" in the context of polynomials. -- (talk) 16:29, 18 September 2014 (UTC)

What does "associated rule" refer to?[edit]

This edit added the phrase "associated rule" three times. The section gives two rules for computing with , but those rules do not use inequalities. -- (talk) 16:52, 18 September 2014 (UTC)

The wording to which you refer (which I inserted) is not clear and needs to be changed. What is being referred to is the behaviour of the degree of polynomials from the previous section: that the sum or difference of two polynomials has a degree that is less than or equal to the largest of the degree of each of the polynomials, and the product of two polynomials has a degree that is the sum of the degrees of each of the polynomials. —Quondum 17:30, 18 September 2014 (UTC)
Thanks for your clarification. The previous section lists propositions, but there are no citations saying so. The article should refer to propositions and definitions, not behavior and rules. -- (talk) 18:26, 18 September 2014 (UTC)
Actually, deg is a function from non-zero polynomials to non-negative integers, so the section on "Behaviour" might be better called "Properties of the deg function". A problem with the "Behaviour" section is that it doesn't say that the zero polynomial is not in the domain of the deg function. With that in mind, the section on the degree of the zero polynomial can say that the domain of deg can be extended to include the zero polynomial by defining deg 0 = −∞. -- (talk) 19:11, 18 September 2014 (UTC)
That is, by implication, what is already being stated. But you put it more clearly; cleaning up the presentation in the article would help. —Quondum 20:27, 18 September 2014 (UTC)

Degree is a non-negative integer[edit]

The lead does not explicitly say that the degree of a (non-zero) polynomial is a non-negative integer. -- (talk) 18:10, 21 September 2014 (UTC)