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Is $\log x$, $x$ is a variable and greater than zero, a polynomial or not? Ambiguity is there because $\log x$ can be treated as a variable and single variable is a polynomial.

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    $\begingroup$ What is your definition of polynomial? (Hint: no.) $\endgroup$ Dec 26, 2018 at 2:55
  • $\begingroup$ My definition of polynomial is taken from Wikipedia $\endgroup$
    – user629353
    Dec 26, 2018 at 2:56
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    $\begingroup$ These types of questions, which have more to do with definitions than with routines computations, are savagely degraded by lack of context—specifically you should at least provide the definition of $\log x$ and polynomial that you are working with... $\endgroup$ Dec 26, 2018 at 3:21

7 Answers 7

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To check whether a given function is a polynomial, you need to check that it can be written exactly as a finite linear combination of functions in the basis $\{1,x,x^2,x^3,...\}$.

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Polynomials have a quite rigid definition. In the most general setting, we need some ambient ring $R$. Please note that this need not be the reals, complexes, or anything else you are familiar with.

Definition: A function $f:R\rightarrow R$ is a (univariate) polynomial if and only if there exists constants $a_0,\ldots,a_n\in R$ such that $$ f(x)=\sum_{k=0}^na_kx^k $$ for all $x\in R$.

For your case, $\log(x)$ is defined from $\mathbb{R}^+$ to $\mathbb{R}$, and you can use various methods to show that it is not a polynomial; my favorite is that $\lim_{x\rightarrow 0^+}\log(x)=-\infty$, which is an impossibility for a polynomial.

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The only way $\log x$ could be treated as a polynomial is if we make the change of variable $u = \log x$. Then this is linear with some restrictions.

We typically, however, don't think like this.

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We can always set $u = \ln x$ and consider polynomials in $u$ to be polynomials in $\ln x$:

$q(u) = \displaystyle \sum_0^m q_j u^j = \sum_0^m q_j (\ln x)^j, \tag 0$

but we can't take $\ln x$ to be a polynomial in $x$ itself, to wit:

If we accept the definition

$\ln x = \displaystyle \int_1^x \dfrac{ds}{s}, \tag 1$

and $\ln x$ is a polynomial

$p(x) = \displaystyle \sum_0^n p_k x^k \in \Bbb R[x], \tag 2$

we have

$\ln x = p(x) = \displaystyle \sum_0^n p_k x^k, \tag 3$

whence, differentiating,

$\dfrac{1}{x} = p'(x) = \displaystyle \sum_1^n p_k k x^{k - 1}, \tag 4$

and therefore

$xp'(x) = \displaystyle \sum_1^n p_k k x^k = 1, \tag 5$

which is impossible since $xp'(x)$ has no constant term, and $1$ has no variable terms.

We conclude that $\ln x$ is not a polynomial in $x$.

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We can loosely define a polynomial as follows: a polynomial is of the form, for some nonnegative integer $n$,

$$f(x) = a_0 + a_1x^1 + a_2x^2 + ... + a_nx^n$$

There are a few subtle nuances this definition implies to keep in mind:

  • $a_i$ is some sort of constant for all $i$. What this constant is, i.e. a real or complex number or whatever, depends on the context we're looking at. Point being, it's a fixed value for the polynomial.
  • $x$ is a singular input, presumably from the same set of numbers that the $a_i$'s can be from (but again, context dependent). Point being, it's variable.
  • This represents a polynomial of a finite degree, which in this case is $n$ if $a_n \neq 0$. Notice how there isn't some $(n+1)^{th}$ term, or whatever, for example. There are finitely many terms, in other words.

(There is a bit more of a technical definition about the space of values in which the constants/variable can come from as touched on in ImNotTheGuy's answer. I'm not going to elaborate on that because (1) I suspect that the study of algebraic structures is probably above your head and beyond the scope of what you want to discuss and (2) my foundations in polynomials and rings is extremely weak so I don't want to say anything that's probably wrong or misleading in that respect.)

In this light, at least at first glance $\log(x)$ is not a polynomial. It's a function in its own right, but it's not expressible as a linear combination of finitely many of $1, x, x^2, ..., x^n$ (which is a more fundamental definition of what a polynomial really is, aside from those already presented).

You can also make the argument that the $\log(x)$ function has a power series representation,

$$\log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} \cdot x^k$$

and thus might fit the form of a polynomial with $a_k = (-1)^{k-1}/k$, but this has infinitely many, not finitely many, terms so this is still not a polynomial.

So, either just by itself or as a power series, the $\log(x)$ function is not a polynomial.

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  • $\begingroup$ Good idea to keep this grounded in calculus and analysis! $\endgroup$ Dec 26, 2018 at 3:11
  • $\begingroup$ Eh, your answer is better in its own way, I just feel like OP is probably looking at this from the perspective of a high school student. In a way I feel like one could almost look at the power series as an "infinite degree" polynomial if looking at it naively, which would probably have to turn the discussion into something more fundamental a la your approach. $\endgroup$ Dec 26, 2018 at 3:14
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It's a polynomial in $\log x$. It's not a polynomial in $x$, which is presumably what was meant by the question.

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This is Wikipedia's definition of a variable.

In elementary mathematics, a variable is a symbol, commonly a single letter,
that represents a number, called the value of the variable, which is either
arbitrary, not fully specified, or unknown. Making algebraic computations with
variables as if they were explicit numbers allows one to solve a range of problems
in a single computation. A typical example is the quadratic formula, which allows
one to solve every quadratic equation by simply substituting the numeric values of
the coefficients of the given equation to the variables that represent them.

....In more advanced mathematics, a variable is a symbol that denotes a mathematical
object, which could be a number, a vector, a matrix, or even a function. In this case, 
the original property of "variability" of a variable is not kept (except, sometimes,
for informal explanations).

Since $\log x$ represents the value of a function, a real number, if you want to think of it also as a symbol, then you are allowed to think of it as a variable. In that case, $\log x$ is a polynomial (more specifically, a monomial) in terms of $\log x$.

Sometimes it helps to think of such things as variables. For example, the problem

$$\text{Solve $(\log x)^2 - 5 \log x + 4 = 0$ for $x$.}$$

is most easily solved if $(\log x)^2 - 5 \log x + 4$ is treated as a quadratic polynomial in terms of $\log x$.

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