$$x^4 + x^3 + x + 1$$
Notice how I skipped $x^2$. Do "polynomials" need to have a sequence of exponents that start from $1$ and go up by $1$ and only $1$ each time? Thanks
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ en.wikipedia.org/wiki/Polynomial#Definition $\endgroup$– mathseJun 16, 2014 at 6:23
-
9$\begingroup$ Yes. The co-efficent is just zero for $x^3$ $\endgroup$– JoaoJun 16, 2014 at 7:54
-
7$\begingroup$ Actually, the coefficient is zero for $x^2$. $\endgroup$– GEdgarJun 16, 2014 at 20:52
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
9
A polynomial is any sum of the type
$$a_0 + a_1x + a_2x^2 + \cdots + a_n x^n.$$
By taking $n=4$ and setting the coefficients at $a_0=1, a_1=1, a_2=0, a_3=1, a_4=1$, what do you get?
-
4$\begingroup$ Oh that's right! I totally forgot about constants being 0. My polynomial would then be (x^4 + x^3 + 0x^2 + x + 1), so in a sense, it still follows the general polynomial form with exponents of "x" increasing by 1 at each new term. $\endgroup$ Jun 16, 2014 at 6:28
-
2$\begingroup$ Strictly speaking a polynomial is not a function, but rather a finite formal sum. The distinction between a polynomial and a polynomial function comes to the fore in the context of abstract algebra over rings of a finite characteristic. In that context $x$ is an indeterminate, and you do not necessarily think of it as a variable ranging over a set. Thus the distinction can be ignored by high school/calculus students. $\endgroup$ Jun 16, 2014 at 6:30
-
8$\begingroup$ @JyrkiLahtonen judging by the fact that the question is really really simple (obviously not one to be asked by university students), I think an answer which does not involve words like "context of abstract algebra over rings of a finite characteristic" is a better answer than one that does. $\endgroup$– 5xumJun 16, 2014 at 6:35
-
2$\begingroup$ Undoubtedly you judged the needs of the asker well. It's just that the use of the word function raised an inner flag of mine. The autopilot cannot be stopped :-) +1 $\endgroup$ Jun 16, 2014 at 6:37
-
1$\begingroup$ @5xum, I'd be careful to so quickly judge the OP. He's asked some rather sophisticated questions in the past according to his profile page, which suggests he is a university student. We all have "those moments" from time to time. $\endgroup$ Jun 16, 2014 at 6:43
$\begingroup$
$\endgroup$
2
That is indeed a polynomial! You do not need the exponents increasing by $1$. For example, $x^{100} + x^2 + 2$ is a polynomial. By definition, any function $p(x)$ is a polynomial if it can be written in the form:
$$p(x) = \sum_{k=0}^n c_kx^k = c_0 + c_1x + c_2x^2 + ... + c_nx^n$$
Where the $c_k$'s are arbitrary elements of a particular ring. If you haven't seen rings before, there's no need to get caught up in too much theory. Outside of abstract algebra, the ring is usually the integers, the rational numbers, the real numbers, or the complex numbers. The specific reason why the exponents do not need to increase by $1$ each time is because $0$ is an element of all rings ($0$ is an integer, rational, real, and complex number). Thus, $c_k$ can be $0$ for any $k$.
Even if $c_k$ is $0$ for all $k$, then $p(x)$ is still a polynomial! Algebraists often call $p(x) = 0$ the "zero polynomial".
There is one last restriction: For $p(x)$ to be a polynomial, the $n$ in $\displaystyle \sum_{k = 0}^n c_kx^k$ must be finite. Otherwise, $\displaystyle \sum_{k=0}^\infty c_kx^k$ would be called a power series.
answered Jun 16, 2014 at 6:19
-
$\begingroup$ Still you should formulate more carefully. $x^{100}+x^2+2$ is neither of the form $p(x)$ (obviously), nor $\sum_{i=0}^nc_kx^k$ (even more so), nor even $c_0+c_1x+\cdots+c_nx^n$. An equivalent expression that is of the third form would be $2+0x+1x^2+0x^3+\cdots+0x^{99}+1x^{100}$. When talking about form, it is not just the contents that matters. You can get away a bit better with the formulation "can be written in the form". $\endgroup$ Jun 17, 2014 at 4:13
-
$\begingroup$ Point taken. I should perhaps be more careful with that phrase. $\endgroup$ Jun 17, 2014 at 4:32
$\begingroup$
$\endgroup$
7
Yes that is a polynomial. Any integer is a polynomial. Any combination of multiplcation/addition of numbers and variables will be a polynomial.
-
1$\begingroup$ Well, multiplication of the same variable—$y + 1 = xy$ is not a polynomial. $\endgroup$– wcharginJun 16, 2014 at 21:10
-
$\begingroup$ @WChargin "Not a polynomial" might be a little too strong. It's standard to call it a polynomial in the polynomial ring $R[x,y]$, even if it is a 'multivariable polynomial.' $\endgroup$– rschwiebJun 16, 2014 at 21:28
-
$\begingroup$ @rschwieb even though it represents a hyperbola? $\endgroup$– wcharginJun 16, 2014 at 21:30
-
$\begingroup$ @WChargin Sure: the shape is not really the source of the word. You can get a lot of information from the wiki article on polynomials. $\endgroup$– rschwiebJun 16, 2014 at 22:44
-
1$\begingroup$ @WChargin: An equation is not a polynomial, so "$y+1=xy$" is definitely not a polynomial. However both members in that equation are polynomials, though not polynomials in (just) $x$. When doing algebra, there is really nothing special about the names $x,y$, in particular there is no convention that $x$ represents input and $y$ the output; $xy$ is just the product of two indeterminates. Don't mix the notions of graph/function/polynomial! $\endgroup$ Jun 17, 2014 at 4:23