Anti derivative notation?












3












$begingroup$


For derivatives, we use $f'(x), f''(x),$ etc. until that comes too unwieldy so we just use $f^{(n)} (x)$



What about for anti derivatives? I've seen using $F(x)$ to denote the first antiderivative of $f(x)$, but what would one do if you want to find the second anti derivative of $f(x)$? What would that be denoted by?






notation







share|cite|improve this question









$endgroup$












  • $begingroup$
    I've only very rarely seen $F$ denote an antiderivative of $f$ without it being explicitly mentioned it is so. So if you're gonna declare variables for a first antiderivative, you might as well do it for antiderivatives of all orders.
    $endgroup$
    – Git Gud
    Apr 25 '14 at 20:17






  • 1




    $begingroup$
    Let $D=frac{d}{dx}$ then $int f(x)dx=left(D^{-1}fright)(x)=f^{(-1)}(x)$
    $endgroup$
    – BPP
    Apr 25 '14 at 20:20








  • 1




    $begingroup$
    I' ve seen $ f^{(-1)} (x) $ and $ f^{(-2)} (x)$.
    $endgroup$
    – Américo Tavares
    Apr 25 '14 at 20:21












  • $begingroup$
    Note that antiderivatives are only defined up to a constant and, more generally, the $n^{th}$ antiderivative is only defined up to a polynomial of degree $n-1$. So such notation won't uniquely determine a function, unlike the case of derivatives.
    $endgroup$
    – Qiaochu Yuan
    Apr 26 '14 at 5:38
















3












$begingroup$


For derivatives, we use $f'(x), f''(x),$ etc. until that comes too unwieldy so we just use $f^{(n)} (x)$



What about for anti derivatives? I've seen using $F(x)$ to denote the first antiderivative of $f(x)$, but what would one do if you want to find the second anti derivative of $f(x)$? What would that be denoted by?






notation







share|cite|improve this question









$endgroup$












  • $begingroup$
    I've only very rarely seen $F$ denote an antiderivative of $f$ without it being explicitly mentioned it is so. So if you're gonna declare variables for a first antiderivative, you might as well do it for antiderivatives of all orders.
    $endgroup$
    – Git Gud
    Apr 25 '14 at 20:17






  • 1




    $begingroup$
    Let $D=frac{d}{dx}$ then $int f(x)dx=left(D^{-1}fright)(x)=f^{(-1)}(x)$
    $endgroup$
    – BPP
    Apr 25 '14 at 20:20








  • 1




    $begingroup$
    I' ve seen $ f^{(-1)} (x) $ and $ f^{(-2)} (x)$.
    $endgroup$
    – Américo Tavares
    Apr 25 '14 at 20:21












  • $begingroup$
    Note that antiderivatives are only defined up to a constant and, more generally, the $n^{th}$ antiderivative is only defined up to a polynomial of degree $n-1$. So such notation won't uniquely determine a function, unlike the case of derivatives.
    $endgroup$
    – Qiaochu Yuan
    Apr 26 '14 at 5:38














3












3








3


1



$begingroup$


For derivatives, we use $f'(x), f''(x),$ etc. until that comes too unwieldy so we just use $f^{(n)} (x)$



What about for anti derivatives? I've seen using $F(x)$ to denote the first antiderivative of $f(x)$, but what would one do if you want to find the second anti derivative of $f(x)$? What would that be denoted by?






notation







share|cite|improve this question









$endgroup$




For derivatives, we use $f'(x), f''(x),$ etc. until that comes too unwieldy so we just use $f^{(n)} (x)$



What about for anti derivatives? I've seen using $F(x)$ to denote the first antiderivative of $f(x)$, but what would one do if you want to find the second anti derivative of $f(x)$? What would that be denoted by?






notation




notation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 25 '14 at 20:11









MCTMCT

14.5k42668




14.5k42668












  • $begingroup$
    I've only very rarely seen $F$ denote an antiderivative of $f$ without it being explicitly mentioned it is so. So if you're gonna declare variables for a first antiderivative, you might as well do it for antiderivatives of all orders.
    $endgroup$
    – Git Gud
    Apr 25 '14 at 20:17






  • 1




    $begingroup$
    Let $D=frac{d}{dx}$ then $int f(x)dx=left(D^{-1}fright)(x)=f^{(-1)}(x)$
    $endgroup$
    – BPP
    Apr 25 '14 at 20:20








  • 1




    $begingroup$
    I' ve seen $ f^{(-1)} (x) $ and $ f^{(-2)} (x)$.
    $endgroup$
    – Américo Tavares
    Apr 25 '14 at 20:21












  • $begingroup$
    Note that antiderivatives are only defined up to a constant and, more generally, the $n^{th}$ antiderivative is only defined up to a polynomial of degree $n-1$. So such notation won't uniquely determine a function, unlike the case of derivatives.
    $endgroup$
    – Qiaochu Yuan
    Apr 26 '14 at 5:38


















  • $begingroup$
    I've only very rarely seen $F$ denote an antiderivative of $f$ without it being explicitly mentioned it is so. So if you're gonna declare variables for a first antiderivative, you might as well do it for antiderivatives of all orders.
    $endgroup$
    – Git Gud
    Apr 25 '14 at 20:17






  • 1




    $begingroup$
    Let $D=frac{d}{dx}$ then $int f(x)dx=left(D^{-1}fright)(x)=f^{(-1)}(x)$
    $endgroup$
    – BPP
    Apr 25 '14 at 20:20








  • 1




    $begingroup$
    I' ve seen $ f^{(-1)} (x) $ and $ f^{(-2)} (x)$.
    $endgroup$
    – Américo Tavares
    Apr 25 '14 at 20:21












  • $begingroup$
    Note that antiderivatives are only defined up to a constant and, more generally, the $n^{th}$ antiderivative is only defined up to a polynomial of degree $n-1$. So such notation won't uniquely determine a function, unlike the case of derivatives.
    $endgroup$
    – Qiaochu Yuan
    Apr 26 '14 at 5:38
















$begingroup$
I've only very rarely seen $F$ denote an antiderivative of $f$ without it being explicitly mentioned it is so. So if you're gonna declare variables for a first antiderivative, you might as well do it for antiderivatives of all orders.
$endgroup$
– Git Gud
Apr 25 '14 at 20:17




$begingroup$
I've only very rarely seen $F$ denote an antiderivative of $f$ without it being explicitly mentioned it is so. So if you're gonna declare variables for a first antiderivative, you might as well do it for antiderivatives of all orders.
$endgroup$
– Git Gud
Apr 25 '14 at 20:17




1




1




$begingroup$
Let $D=frac{d}{dx}$ then $int f(x)dx=left(D^{-1}fright)(x)=f^{(-1)}(x)$
$endgroup$
– BPP
Apr 25 '14 at 20:20






$begingroup$
Let $D=frac{d}{dx}$ then $int f(x)dx=left(D^{-1}fright)(x)=f^{(-1)}(x)$
$endgroup$
– BPP
Apr 25 '14 at 20:20






1




1




$begingroup$
I' ve seen $ f^{(-1)} (x) $ and $ f^{(-2)} (x)$.
$endgroup$
– Américo Tavares
Apr 25 '14 at 20:21






$begingroup$
I' ve seen $ f^{(-1)} (x) $ and $ f^{(-2)} (x)$.
$endgroup$
– Américo Tavares
Apr 25 '14 at 20:21














$begingroup$
Note that antiderivatives are only defined up to a constant and, more generally, the $n^{th}$ antiderivative is only defined up to a polynomial of degree $n-1$. So such notation won't uniquely determine a function, unlike the case of derivatives.
$endgroup$
– Qiaochu Yuan
Apr 26 '14 at 5:38




$begingroup$
Note that antiderivatives are only defined up to a constant and, more generally, the $n^{th}$ antiderivative is only defined up to a polynomial of degree $n-1$. So such notation won't uniquely determine a function, unlike the case of derivatives.
$endgroup$
– Qiaochu Yuan
Apr 26 '14 at 5:38










2 Answers
2






active

oldest

votes


















1












$begingroup$

I'm not aware of a standard notation for this. If I were to make something up, $f^{(-1)}(x)$, $f^{(-2)}(x)$, and so on would be consistent with the existing notation.



One thing to remember is that unlike $f^{(n)}(x)$, the antiderivative $f^{(-n)}(x)$ will represent a family of functions, not a single function.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I've never come across this before, but a conventional notation could be:



    For the first antiderivative $F_1(x)$ and for the second $F_2(x)$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
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      1












      $begingroup$

      I'm not aware of a standard notation for this. If I were to make something up, $f^{(-1)}(x)$, $f^{(-2)}(x)$, and so on would be consistent with the existing notation.



      One thing to remember is that unlike $f^{(n)}(x)$, the antiderivative $f^{(-n)}(x)$ will represent a family of functions, not a single function.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        I'm not aware of a standard notation for this. If I were to make something up, $f^{(-1)}(x)$, $f^{(-2)}(x)$, and so on would be consistent with the existing notation.



        One thing to remember is that unlike $f^{(n)}(x)$, the antiderivative $f^{(-n)}(x)$ will represent a family of functions, not a single function.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          I'm not aware of a standard notation for this. If I were to make something up, $f^{(-1)}(x)$, $f^{(-2)}(x)$, and so on would be consistent with the existing notation.



          One thing to remember is that unlike $f^{(n)}(x)$, the antiderivative $f^{(-n)}(x)$ will represent a family of functions, not a single function.






          share|cite|improve this answer









          $endgroup$



          I'm not aware of a standard notation for this. If I were to make something up, $f^{(-1)}(x)$, $f^{(-2)}(x)$, and so on would be consistent with the existing notation.



          One thing to remember is that unlike $f^{(n)}(x)$, the antiderivative $f^{(-n)}(x)$ will represent a family of functions, not a single function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 25 '14 at 20:19









          Umberto P.Umberto P.

          40.1k13368




          40.1k13368























              0












              $begingroup$

              I've never come across this before, but a conventional notation could be:



              For the first antiderivative $F_1(x)$ and for the second $F_2(x)$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I've never come across this before, but a conventional notation could be:



                For the first antiderivative $F_1(x)$ and for the second $F_2(x)$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I've never come across this before, but a conventional notation could be:



                  For the first antiderivative $F_1(x)$ and for the second $F_2(x)$






                  share|cite|improve this answer









                  $endgroup$



                  I've never come across this before, but a conventional notation could be:



                  For the first antiderivative $F_1(x)$ and for the second $F_2(x)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 25 '14 at 20:16









                  EllyaEllya

                  9,61711326




                  9,61711326






























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