How to check for polynomial [closed]












0












$begingroup$


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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closed as off-topic by Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho Dec 26 '18 at 10:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    What is your definition of polynomial? (Hint: no.)
    $endgroup$
    – Sean Roberson
    Dec 26 '18 at 2:55










  • $begingroup$
    My definition of polynomial is taken from Wikipedia
    $endgroup$
    – user629353
    Dec 26 '18 at 2:56






  • 4




    $begingroup$
    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...
    $endgroup$
    – LoveTooNap29
    Dec 26 '18 at 3:21
















0












$begingroup$


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.










share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho Dec 26 '18 at 10:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    What is your definition of polynomial? (Hint: no.)
    $endgroup$
    – Sean Roberson
    Dec 26 '18 at 2:55










  • $begingroup$
    My definition of polynomial is taken from Wikipedia
    $endgroup$
    – user629353
    Dec 26 '18 at 2:56






  • 4




    $begingroup$
    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...
    $endgroup$
    – LoveTooNap29
    Dec 26 '18 at 3:21














0












0








0


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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.







functions polynomials logarithms






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share|cite|improve this question













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edited Dec 26 '18 at 3:18









twnly

867213




867213










asked Dec 26 '18 at 2:53







user629353











closed as off-topic by Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho Dec 26 '18 at 10:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho Dec 26 '18 at 10:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord Shark the Unknown, Lord_Farin, user91500, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    What is your definition of polynomial? (Hint: no.)
    $endgroup$
    – Sean Roberson
    Dec 26 '18 at 2:55










  • $begingroup$
    My definition of polynomial is taken from Wikipedia
    $endgroup$
    – user629353
    Dec 26 '18 at 2:56






  • 4




    $begingroup$
    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...
    $endgroup$
    – LoveTooNap29
    Dec 26 '18 at 3:21














  • 2




    $begingroup$
    What is your definition of polynomial? (Hint: no.)
    $endgroup$
    – Sean Roberson
    Dec 26 '18 at 2:55










  • $begingroup$
    My definition of polynomial is taken from Wikipedia
    $endgroup$
    – user629353
    Dec 26 '18 at 2:56






  • 4




    $begingroup$
    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...
    $endgroup$
    – LoveTooNap29
    Dec 26 '18 at 3:21








2




2




$begingroup$
What is your definition of polynomial? (Hint: no.)
$endgroup$
– Sean Roberson
Dec 26 '18 at 2:55




$begingroup$
What is your definition of polynomial? (Hint: no.)
$endgroup$
– Sean Roberson
Dec 26 '18 at 2:55












$begingroup$
My definition of polynomial is taken from Wikipedia
$endgroup$
– user629353
Dec 26 '18 at 2:56




$begingroup$
My definition of polynomial is taken from Wikipedia
$endgroup$
– user629353
Dec 26 '18 at 2:56




4




4




$begingroup$
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...
$endgroup$
– LoveTooNap29
Dec 26 '18 at 3:21




$begingroup$
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...
$endgroup$
– LoveTooNap29
Dec 26 '18 at 3:21










7 Answers
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active

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4












$begingroup$

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 ${1,x,x^2,x^3,...}$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    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 $f:Rrightarrow R$ is a (univariate) polynomial if and only if there exists constants $a_0,ldots,a_nin R$ such that
    $$
    f(x)=sum_{k=0}^na_kx^k
    $$

    for all $xin R$.



    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 $lim_{xrightarrow 0^+}log(x)=-infty$, which is an impossibility for a polynomial.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        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 $x$.






        share|cite|improve this answer











        $endgroup$





















          0












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Good idea to keep this grounded in calculus and analysis!
            $endgroup$
            – ImNotTheGuy
            Dec 26 '18 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.
            $endgroup$
            – Eevee Trainer
            Dec 26 '18 at 3:14



















          0












          $begingroup$

          It's a polynomial in $log x$. It's not a polynomial in $x$, which is presumably what was meant by the question.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            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 $log x$.






            share|cite|improve this answer









            $endgroup$



















              7 Answers
              7






              active

              oldest

              votes








              7 Answers
              7






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4












              $begingroup$

              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 ${1,x,x^2,x^3,...}$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                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 ${1,x,x^2,x^3,...}$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  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 ${1,x,x^2,x^3,...}$.






                  share|cite|improve this answer









                  $endgroup$



                  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 ${1,x,x^2,x^3,...}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 2:58









                  D.B.D.B.

                  1,2438




                  1,2438























                      3












                      $begingroup$

                      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 $f:Rrightarrow R$ is a (univariate) polynomial if and only if there exists constants $a_0,ldots,a_nin R$ such that
                      $$
                      f(x)=sum_{k=0}^na_kx^k
                      $$

                      for all $xin R$.



                      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 $lim_{xrightarrow 0^+}log(x)=-infty$, which is an impossibility for a polynomial.






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        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 $f:Rrightarrow R$ is a (univariate) polynomial if and only if there exists constants $a_0,ldots,a_nin R$ such that
                        $$
                        f(x)=sum_{k=0}^na_kx^k
                        $$

                        for all $xin R$.



                        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 $lim_{xrightarrow 0^+}log(x)=-infty$, which is an impossibility for a polynomial.






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          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 $f:Rrightarrow R$ is a (univariate) polynomial if and only if there exists constants $a_0,ldots,a_nin R$ such that
                          $$
                          f(x)=sum_{k=0}^na_kx^k
                          $$

                          for all $xin R$.



                          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 $lim_{xrightarrow 0^+}log(x)=-infty$, which is an impossibility for a polynomial.






                          share|cite|improve this answer









                          $endgroup$



                          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 $f:Rrightarrow R$ is a (univariate) polynomial if and only if there exists constants $a_0,ldots,a_nin R$ such that
                          $$
                          f(x)=sum_{k=0}^na_kx^k
                          $$

                          for all $xin R$.



                          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 $lim_{xrightarrow 0^+}log(x)=-infty$, which is an impossibility for a polynomial.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 26 '18 at 3:02









                          ImNotTheGuyImNotTheGuy

                          38516




                          38516























                              1












                              $begingroup$

                              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.






                              share|cite|improve this answer









                              $endgroup$


















                                1












                                $begingroup$

                                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.






                                share|cite|improve this answer









                                $endgroup$
















                                  1












                                  1








                                  1





                                  $begingroup$

                                  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.






                                  share|cite|improve this answer









                                  $endgroup$



                                  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.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Dec 26 '18 at 2:58









                                  Sean RobersonSean Roberson

                                  6,38931327




                                  6,38931327























                                      1












                                      $begingroup$

                                      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 $x$.






                                      share|cite|improve this answer











                                      $endgroup$


















                                        1












                                        $begingroup$

                                        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 $x$.






                                        share|cite|improve this answer











                                        $endgroup$
















                                          1












                                          1








                                          1





                                          $begingroup$

                                          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 $x$.






                                          share|cite|improve this answer











                                          $endgroup$



                                          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 $x$.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited Dec 26 '18 at 3:20

























                                          answered Dec 26 '18 at 3:13









                                          Robert LewisRobert Lewis

                                          46.3k23066




                                          46.3k23066























                                              0












                                              $begingroup$

                                              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.






                                              share|cite|improve this answer











                                              $endgroup$













                                              • $begingroup$
                                                Good idea to keep this grounded in calculus and analysis!
                                                $endgroup$
                                                – ImNotTheGuy
                                                Dec 26 '18 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.
                                                $endgroup$
                                                – Eevee Trainer
                                                Dec 26 '18 at 3:14
















                                              0












                                              $begingroup$

                                              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.






                                              share|cite|improve this answer











                                              $endgroup$













                                              • $begingroup$
                                                Good idea to keep this grounded in calculus and analysis!
                                                $endgroup$
                                                – ImNotTheGuy
                                                Dec 26 '18 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.
                                                $endgroup$
                                                – Eevee Trainer
                                                Dec 26 '18 at 3:14














                                              0












                                              0








                                              0





                                              $begingroup$

                                              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.






                                              share|cite|improve this answer











                                              $endgroup$



                                              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.







                                              share|cite|improve this answer














                                              share|cite|improve this answer



                                              share|cite|improve this answer








                                              edited Dec 26 '18 at 3:12

























                                              answered Dec 26 '18 at 3:11









                                              Eevee TrainerEevee Trainer

                                              5,9761936




                                              5,9761936












                                              • $begingroup$
                                                Good idea to keep this grounded in calculus and analysis!
                                                $endgroup$
                                                – ImNotTheGuy
                                                Dec 26 '18 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.
                                                $endgroup$
                                                – Eevee Trainer
                                                Dec 26 '18 at 3:14


















                                              • $begingroup$
                                                Good idea to keep this grounded in calculus and analysis!
                                                $endgroup$
                                                – ImNotTheGuy
                                                Dec 26 '18 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.
                                                $endgroup$
                                                – Eevee Trainer
                                                Dec 26 '18 at 3:14
















                                              $begingroup$
                                              Good idea to keep this grounded in calculus and analysis!
                                              $endgroup$
                                              – ImNotTheGuy
                                              Dec 26 '18 at 3:11




                                              $begingroup$
                                              Good idea to keep this grounded in calculus and analysis!
                                              $endgroup$
                                              – ImNotTheGuy
                                              Dec 26 '18 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.
                                              $endgroup$
                                              – Eevee Trainer
                                              Dec 26 '18 at 3:14




                                              $begingroup$
                                              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.
                                              $endgroup$
                                              – Eevee Trainer
                                              Dec 26 '18 at 3:14











                                              0












                                              $begingroup$

                                              It's a polynomial in $log x$. It's not a polynomial in $x$, which is presumably what was meant by the question.






                                              share|cite|improve this answer









                                              $endgroup$


















                                                0












                                                $begingroup$

                                                It's a polynomial in $log x$. It's not a polynomial in $x$, which is presumably what was meant by the question.






                                                share|cite|improve this answer









                                                $endgroup$
















                                                  0












                                                  0








                                                  0





                                                  $begingroup$

                                                  It's a polynomial in $log x$. It's not a polynomial in $x$, which is presumably what was meant by the question.






                                                  share|cite|improve this answer









                                                  $endgroup$



                                                  It's a polynomial in $log x$. It's not a polynomial in $x$, which is presumably what was meant by the question.







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered Dec 26 '18 at 3:21









                                                  Daniel McLauryDaniel McLaury

                                                  15.7k32980




                                                  15.7k32980























                                                      0












                                                      $begingroup$

                                                      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 $log x$.






                                                      share|cite|improve this answer









                                                      $endgroup$


















                                                        0












                                                        $begingroup$

                                                        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 $log x$.






                                                        share|cite|improve this answer









                                                        $endgroup$
















                                                          0












                                                          0








                                                          0





                                                          $begingroup$

                                                          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 $log x$.






                                                          share|cite|improve this answer









                                                          $endgroup$



                                                          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 $log x$.







                                                          share|cite|improve this answer












                                                          share|cite|improve this answer



                                                          share|cite|improve this answer










                                                          answered Dec 26 '18 at 3:50









                                                          steven gregorysteven gregory

                                                          18.1k32258




                                                          18.1k32258















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