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A textbook I'm reading says that $f(x)=0$ is NOT a polynomial function, yet $g(x)=8$ IS a polynomial function since $g(x)=8=8x^0$ which satisfies the non-negative integer degree requirement. Yet, it's still a monomial! Or, can it be considered $g(x)=8=8x^0=8x^0+0$ which is now technically a POLYnomial ?

Also, $f(x)=0$ is not a poly by the same logic since it CAN be written as $f(x)=0=0x^{-3}$ which doesn't fit the poly definition above. (Counterexample)

Edit: The book says the zero function is not assigned a degree, while the nonzero constant function has a degree of 0.

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    $\begingroup$ Out of curiosity, what is the book, and where is this statement in the book? Claiming that the zero function is not a polynomial function seems pretty odd to me... $\endgroup$ – user64687 Oct 18 '13 at 19:36
  • $\begingroup$ The book says the zero function is not assigned a degree, while the nonzero constant function has a degree of 0. $\endgroup$ – JackOfAll Oct 18 '13 at 20:13
  • $\begingroup$ Sullican Precalculus 7th Edition, p145 (ch3.2) $\endgroup$ – JackOfAll Oct 18 '13 at 20:14
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    $\begingroup$ The question begins by claiming that the book says $f(x)=0$ is not a polynomial function, but the OP's comment and answer claim only that the book says this function is not assigned a degree. That's quite different. In particular, I consider not assigning it a degree to be reasonable, but I consider not calling it a polynomial to be unreasonable. $\endgroup$ – Andreas Blass Oct 18 '13 at 21:39
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The problem with your book's choice of definition is that, by that definition, the sum of two polynomial functions may not be a polynomial function. This is, generally speaking, a property one would really like to hold. I'm not going to go so far as to say the book is wrong (any author can define things however they like), but it seems a very unfortunate choice.

Edit: Most agree that the zero function is a polynomial function. The matter of degree is another story. It may be considered to have no degree or to have a degree of $-\infty$. The latter choice makes some theorems easier to state.

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OK. But $f(x) = 0 = 0x^{-2} + 0x^{-1} + 0x^0 + 0x + 0x^2$ just as well as $g(x) = 8 = 0x^{-2} + 0x^{-1} + 8x^0 + 0x + 0x^2$. So this definition of polynomial seems difficult to justify.

The part about having no non-zero coefficients to negative powers of x is important because it creates discontinuity at x = 0. But $f(x)$, as defined, has no such discontinuity.

As you point out, monomials are generally considered a subset of polynomials. Rather than poly meaning many, as in two or more, polynomial is generally considered to be "some number of terms" and one is an OK number. This brings me to the one justification I can think of for excluding $f(x) = 0$. Because all coefficients are 0, this function really has 0 terms (It's the unnomial!) and you could say that zero is not quite enough terms to join the polynomial club.

Re-edit: It has come to light that the original question is a misread of the text book. The zero function is considered by the author (as by the rest of the world) to be a polynomial, but one with no degree, or an undefined degree. This makes much better sense. The unnomial is a member of the polynomial club.

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    $\begingroup$ I feel this is a very misleading answer. All working mathematicians take the zero polynomial to be a polynomial (else the ring of polynomials has no zero element!). One definition that clears away much rubbish is: define the ring of polynomials to be sequences $a = (a_0,\ldots,a_m,\ldots)$ of elements drawn from a given coefficient ring, for which we have $a_i=0$ for all sufficiently large $i$, added componentwise and multiplied by the rule $ab=(\sum_{i+j=k}a_ib_j)_{k\in\mathbb{N}}$. Here $a$ represents the polynomial $\sum_ia_ix^i$, and the zero polynomial is represented by $(0,0,\ldots)$. $\endgroup$ – user43208 Oct 18 '13 at 21:55
  • $\begingroup$ I agree that including x = 0 as a polynomial is a better definition. But I think the author of this textbook has given an acceptable argument for why it is excluded in his definition of polynomial. $\endgroup$ – Mark Bailey Oct 22 '13 at 19:42
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    $\begingroup$ Are you sure that's what the author does (i.e., have you checked yourself)? Cf. Andreas Blass's comment below the OP. $\endgroup$ – user43208 Oct 22 '13 at 20:18
  • $\begingroup$ I do not have a copy of the textbook. No. The OP stated very clearly that the textbook said it is NOT a polynomial, so I'm going on an assumption that he is not mistaken about that. I'm certainly interested in hearing a second interpretation if anyone else has easy access to the text. $\endgroup$ – Mark Bailey Oct 22 '13 at 20:26
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    $\begingroup$ I don't have a copy either, but here apparently is a pdf extract of an edition of Sullivan's Precalculus: psportal.hbuhsd.org/Public%20Content/TA105478/… If you look at page 216 of the text (page 28 of 60 of the pdf, if I'm not mistaken), you will see the zero function described as a polynomial function which is not, however, assigned a degree. This is consistent with JackOfAll's Edit in the OP, the comments below the OP, and JackOfAll's answer. I therefore have cause to believe that the opening paragraph of the OP doesn't report accurately what the text actually said. $\endgroup$ – user43208 Oct 22 '13 at 21:16
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The book says the zero function is not assigned a degree, while the nonzero constant function has a degree of 0.

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