A polynomial under radical












-1












$begingroup$


Is a polynomial expression under radical symbol say, $sqrt{(x+4)}$, considered to be a polynomial?
As it is under the radical symbol wouldn't affect its degree, so it might be.






polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What has this question to do with linear-algebra? Or with irreducible-polynomials?
    $endgroup$
    – José Carlos Santos
    Dec 27 '18 at 15:10










  • $begingroup$
    Why do you think $sqrt{x+4}$ is not a polynomial?
    $endgroup$
    – Prakhar Nagpal
    Dec 27 '18 at 15:14






  • 2




    $begingroup$
    No, it isn't. Look at any definition of polynomial.
    $endgroup$
    – saulspatz
    Dec 27 '18 at 15:29










  • $begingroup$
    But by definition, it should be
    $endgroup$
    – user629353
    Dec 27 '18 at 15:29






  • 1




    $begingroup$
    What is your definition of a polynomial?
    $endgroup$
    – bruderjakob17
    Dec 27 '18 at 15:31
















-1












$begingroup$


Is a polynomial expression under radical symbol say, $sqrt{(x+4)}$, considered to be a polynomial?
As it is under the radical symbol wouldn't affect its degree, so it might be.






polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What has this question to do with linear-algebra? Or with irreducible-polynomials?
    $endgroup$
    – José Carlos Santos
    Dec 27 '18 at 15:10










  • $begingroup$
    Why do you think $sqrt{x+4}$ is not a polynomial?
    $endgroup$
    – Prakhar Nagpal
    Dec 27 '18 at 15:14






  • 2




    $begingroup$
    No, it isn't. Look at any definition of polynomial.
    $endgroup$
    – saulspatz
    Dec 27 '18 at 15:29










  • $begingroup$
    But by definition, it should be
    $endgroup$
    – user629353
    Dec 27 '18 at 15:29






  • 1




    $begingroup$
    What is your definition of a polynomial?
    $endgroup$
    – bruderjakob17
    Dec 27 '18 at 15:31














-1












-1








-1


0



$begingroup$


Is a polynomial expression under radical symbol say, $sqrt{(x+4)}$, considered to be a polynomial?
As it is under the radical symbol wouldn't affect its degree, so it might be.






polynomials







share|cite|improve this question











$endgroup$




Is a polynomial expression under radical symbol say, $sqrt{(x+4)}$, considered to be a polynomial?
As it is under the radical symbol wouldn't affect its degree, so it might be.






polynomials




polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 15:15









amWhy

1




1










asked Dec 27 '18 at 15:04







user629353















  • 1




    $begingroup$
    What has this question to do with linear-algebra? Or with irreducible-polynomials?
    $endgroup$
    – José Carlos Santos
    Dec 27 '18 at 15:10










  • $begingroup$
    Why do you think $sqrt{x+4}$ is not a polynomial?
    $endgroup$
    – Prakhar Nagpal
    Dec 27 '18 at 15:14






  • 2




    $begingroup$
    No, it isn't. Look at any definition of polynomial.
    $endgroup$
    – saulspatz
    Dec 27 '18 at 15:29










  • $begingroup$
    But by definition, it should be
    $endgroup$
    – user629353
    Dec 27 '18 at 15:29






  • 1




    $begingroup$
    What is your definition of a polynomial?
    $endgroup$
    – bruderjakob17
    Dec 27 '18 at 15:31














  • 1




    $begingroup$
    What has this question to do with linear-algebra? Or with irreducible-polynomials?
    $endgroup$
    – José Carlos Santos
    Dec 27 '18 at 15:10










  • $begingroup$
    Why do you think $sqrt{x+4}$ is not a polynomial?
    $endgroup$
    – Prakhar Nagpal
    Dec 27 '18 at 15:14






  • 2




    $begingroup$
    No, it isn't. Look at any definition of polynomial.
    $endgroup$
    – saulspatz
    Dec 27 '18 at 15:29










  • $begingroup$
    But by definition, it should be
    $endgroup$
    – user629353
    Dec 27 '18 at 15:29






  • 1




    $begingroup$
    What is your definition of a polynomial?
    $endgroup$
    – bruderjakob17
    Dec 27 '18 at 15:31








1




1




$begingroup$
What has this question to do with linear-algebra? Or with irreducible-polynomials?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 15:10




$begingroup$
What has this question to do with linear-algebra? Or with irreducible-polynomials?
$endgroup$
– José Carlos Santos
Dec 27 '18 at 15:10












$begingroup$
Why do you think $sqrt{x+4}$ is not a polynomial?
$endgroup$
– Prakhar Nagpal
Dec 27 '18 at 15:14




$begingroup$
Why do you think $sqrt{x+4}$ is not a polynomial?
$endgroup$
– Prakhar Nagpal
Dec 27 '18 at 15:14




2




2




$begingroup$
No, it isn't. Look at any definition of polynomial.
$endgroup$
– saulspatz
Dec 27 '18 at 15:29




$begingroup$
No, it isn't. Look at any definition of polynomial.
$endgroup$
– saulspatz
Dec 27 '18 at 15:29












$begingroup$
But by definition, it should be
$endgroup$
– user629353
Dec 27 '18 at 15:29




$begingroup$
But by definition, it should be
$endgroup$
– user629353
Dec 27 '18 at 15:29




1




1




$begingroup$
What is your definition of a polynomial?
$endgroup$
– bruderjakob17
Dec 27 '18 at 15:31




$begingroup$
What is your definition of a polynomial?
$endgroup$
– bruderjakob17
Dec 27 '18 at 15:31










3 Answers
3






active

oldest

votes


















1












$begingroup$

To answer whether $f(x) = sqrt{x+4}$ is a polynomial, we have to check if it can be written as a (finite) linear combination of powers $1,x,x^2,x^3,ldots$.



Assume that $f$ is a polynomial, i.e. $f(x) = sum_{i=0}^n a_kx^k$. Taking derivatives gives
$$sum_{i=1}^n k a_k x^{k-1} = f'(x) = frac12 (x+4)^{-1/2}$$
$$sum_{i=2}^n k(k-1) a_k x^{k-2} = f''(x) = frac14 (x+4)^{-3/2}$$
$$vdots$$
$$n! a_n = f^{(n)}(x)=(-1)^{n-1}frac{(2n-3)!!}{2^n} (x+4)^{1/2-n}$$
$$0 = f^{(n+1)}(x)=(-1)^{n}frac{(2n-1)!!}{2^{n+1}} (x+4)^{1/2-(n+1)}$$



which is a contradiction because the right hand side is clearly not zero.



Hence $sqrt{x+4}$ is not a polynomial.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the answer
    $endgroup$
    – user629353
    Dec 27 '18 at 15:41






  • 1




    $begingroup$
    @user629353 It you feel you've got the answer you were looking for, it's good to accept it by clicking the tick next to it—this tells people it's now answered, and earns reputation points for both you and the answerer.
    $endgroup$
    – timtfj
    Dec 27 '18 at 17:07



















2












$begingroup$

If $sqrt{x+4}$ were a polynomial $f(x)$, then $x+4=f(x)^2$, which is impossible since the LHS has odd degree and the RHS has even degree.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The definition of a polynomial I know is the following: Polynomials are the elements of $R[x]= {sum_{k=0}^n a_k x^k space | space a_0,dots a_n in R, nin mathbb{N} }$, where $R$ is a ring. Now since $sqrt{x+4} notin mathbb{R}[x]$, I wouldn't consider it a polynomial.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      To answer whether $f(x) = sqrt{x+4}$ is a polynomial, we have to check if it can be written as a (finite) linear combination of powers $1,x,x^2,x^3,ldots$.



      Assume that $f$ is a polynomial, i.e. $f(x) = sum_{i=0}^n a_kx^k$. Taking derivatives gives
      $$sum_{i=1}^n k a_k x^{k-1} = f'(x) = frac12 (x+4)^{-1/2}$$
      $$sum_{i=2}^n k(k-1) a_k x^{k-2} = f''(x) = frac14 (x+4)^{-3/2}$$
      $$vdots$$
      $$n! a_n = f^{(n)}(x)=(-1)^{n-1}frac{(2n-3)!!}{2^n} (x+4)^{1/2-n}$$
      $$0 = f^{(n+1)}(x)=(-1)^{n}frac{(2n-1)!!}{2^{n+1}} (x+4)^{1/2-(n+1)}$$



      which is a contradiction because the right hand side is clearly not zero.



      Hence $sqrt{x+4}$ is not a polynomial.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer
        $endgroup$
        – user629353
        Dec 27 '18 at 15:41






      • 1




        $begingroup$
        @user629353 It you feel you've got the answer you were looking for, it's good to accept it by clicking the tick next to it—this tells people it's now answered, and earns reputation points for both you and the answerer.
        $endgroup$
        – timtfj
        Dec 27 '18 at 17:07
















      1












      $begingroup$

      To answer whether $f(x) = sqrt{x+4}$ is a polynomial, we have to check if it can be written as a (finite) linear combination of powers $1,x,x^2,x^3,ldots$.



      Assume that $f$ is a polynomial, i.e. $f(x) = sum_{i=0}^n a_kx^k$. Taking derivatives gives
      $$sum_{i=1}^n k a_k x^{k-1} = f'(x) = frac12 (x+4)^{-1/2}$$
      $$sum_{i=2}^n k(k-1) a_k x^{k-2} = f''(x) = frac14 (x+4)^{-3/2}$$
      $$vdots$$
      $$n! a_n = f^{(n)}(x)=(-1)^{n-1}frac{(2n-3)!!}{2^n} (x+4)^{1/2-n}$$
      $$0 = f^{(n+1)}(x)=(-1)^{n}frac{(2n-1)!!}{2^{n+1}} (x+4)^{1/2-(n+1)}$$



      which is a contradiction because the right hand side is clearly not zero.



      Hence $sqrt{x+4}$ is not a polynomial.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer
        $endgroup$
        – user629353
        Dec 27 '18 at 15:41






      • 1




        $begingroup$
        @user629353 It you feel you've got the answer you were looking for, it's good to accept it by clicking the tick next to it—this tells people it's now answered, and earns reputation points for both you and the answerer.
        $endgroup$
        – timtfj
        Dec 27 '18 at 17:07














      1












      1








      1





      $begingroup$

      To answer whether $f(x) = sqrt{x+4}$ is a polynomial, we have to check if it can be written as a (finite) linear combination of powers $1,x,x^2,x^3,ldots$.



      Assume that $f$ is a polynomial, i.e. $f(x) = sum_{i=0}^n a_kx^k$. Taking derivatives gives
      $$sum_{i=1}^n k a_k x^{k-1} = f'(x) = frac12 (x+4)^{-1/2}$$
      $$sum_{i=2}^n k(k-1) a_k x^{k-2} = f''(x) = frac14 (x+4)^{-3/2}$$
      $$vdots$$
      $$n! a_n = f^{(n)}(x)=(-1)^{n-1}frac{(2n-3)!!}{2^n} (x+4)^{1/2-n}$$
      $$0 = f^{(n+1)}(x)=(-1)^{n}frac{(2n-1)!!}{2^{n+1}} (x+4)^{1/2-(n+1)}$$



      which is a contradiction because the right hand side is clearly not zero.



      Hence $sqrt{x+4}$ is not a polynomial.






      share|cite|improve this answer









      $endgroup$



      To answer whether $f(x) = sqrt{x+4}$ is a polynomial, we have to check if it can be written as a (finite) linear combination of powers $1,x,x^2,x^3,ldots$.



      Assume that $f$ is a polynomial, i.e. $f(x) = sum_{i=0}^n a_kx^k$. Taking derivatives gives
      $$sum_{i=1}^n k a_k x^{k-1} = f'(x) = frac12 (x+4)^{-1/2}$$
      $$sum_{i=2}^n k(k-1) a_k x^{k-2} = f''(x) = frac14 (x+4)^{-3/2}$$
      $$vdots$$
      $$n! a_n = f^{(n)}(x)=(-1)^{n-1}frac{(2n-3)!!}{2^n} (x+4)^{1/2-n}$$
      $$0 = f^{(n+1)}(x)=(-1)^{n}frac{(2n-1)!!}{2^{n+1}} (x+4)^{1/2-(n+1)}$$



      which is a contradiction because the right hand side is clearly not zero.



      Hence $sqrt{x+4}$ is not a polynomial.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 27 '18 at 15:31









      mechanodroidmechanodroid

      27.6k62447




      27.6k62447












      • $begingroup$
        Thanks for the answer
        $endgroup$
        – user629353
        Dec 27 '18 at 15:41






      • 1




        $begingroup$
        @user629353 It you feel you've got the answer you were looking for, it's good to accept it by clicking the tick next to it—this tells people it's now answered, and earns reputation points for both you and the answerer.
        $endgroup$
        – timtfj
        Dec 27 '18 at 17:07


















      • $begingroup$
        Thanks for the answer
        $endgroup$
        – user629353
        Dec 27 '18 at 15:41






      • 1




        $begingroup$
        @user629353 It you feel you've got the answer you were looking for, it's good to accept it by clicking the tick next to it—this tells people it's now answered, and earns reputation points for both you and the answerer.
        $endgroup$
        – timtfj
        Dec 27 '18 at 17:07
















      $begingroup$
      Thanks for the answer
      $endgroup$
      – user629353
      Dec 27 '18 at 15:41




      $begingroup$
      Thanks for the answer
      $endgroup$
      – user629353
      Dec 27 '18 at 15:41




      1




      1




      $begingroup$
      @user629353 It you feel you've got the answer you were looking for, it's good to accept it by clicking the tick next to it—this tells people it's now answered, and earns reputation points for both you and the answerer.
      $endgroup$
      – timtfj
      Dec 27 '18 at 17:07




      $begingroup$
      @user629353 It you feel you've got the answer you were looking for, it's good to accept it by clicking the tick next to it—this tells people it's now answered, and earns reputation points for both you and the answerer.
      $endgroup$
      – timtfj
      Dec 27 '18 at 17:07











      2












      $begingroup$

      If $sqrt{x+4}$ were a polynomial $f(x)$, then $x+4=f(x)^2$, which is impossible since the LHS has odd degree and the RHS has even degree.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        If $sqrt{x+4}$ were a polynomial $f(x)$, then $x+4=f(x)^2$, which is impossible since the LHS has odd degree and the RHS has even degree.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          If $sqrt{x+4}$ were a polynomial $f(x)$, then $x+4=f(x)^2$, which is impossible since the LHS has odd degree and the RHS has even degree.






          share|cite|improve this answer









          $endgroup$



          If $sqrt{x+4}$ were a polynomial $f(x)$, then $x+4=f(x)^2$, which is impossible since the LHS has odd degree and the RHS has even degree.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 16:36









          lhflhf

          165k10171396




          165k10171396























              1












              $begingroup$

              The definition of a polynomial I know is the following: Polynomials are the elements of $R[x]= {sum_{k=0}^n a_k x^k space | space a_0,dots a_n in R, nin mathbb{N} }$, where $R$ is a ring. Now since $sqrt{x+4} notin mathbb{R}[x]$, I wouldn't consider it a polynomial.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The definition of a polynomial I know is the following: Polynomials are the elements of $R[x]= {sum_{k=0}^n a_k x^k space | space a_0,dots a_n in R, nin mathbb{N} }$, where $R$ is a ring. Now since $sqrt{x+4} notin mathbb{R}[x]$, I wouldn't consider it a polynomial.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The definition of a polynomial I know is the following: Polynomials are the elements of $R[x]= {sum_{k=0}^n a_k x^k space | space a_0,dots a_n in R, nin mathbb{N} }$, where $R$ is a ring. Now since $sqrt{x+4} notin mathbb{R}[x]$, I wouldn't consider it a polynomial.






                  share|cite|improve this answer











                  $endgroup$



                  The definition of a polynomial I know is the following: Polynomials are the elements of $R[x]= {sum_{k=0}^n a_k x^k space | space a_0,dots a_n in R, nin mathbb{N} }$, where $R$ is a ring. Now since $sqrt{x+4} notin mathbb{R}[x]$, I wouldn't consider it a polynomial.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 27 '18 at 15:39

























                  answered Dec 27 '18 at 15:30









                  bruderjakob17bruderjakob17

                  416110




                  416110






























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