Lagrange notation: $f^{(0)}(x)$?












1












$begingroup$


Using Lagrange notation, is $f^{(0)}(x)=f(x)$? Is this standard notation, or would one have to define $f^{(0)}(x)=f(x)$ first, before using it?
Aside: the context of the question is whether to include the first term within the summation when expressing the Taylor series and hence start at $n=0$, or to write it separately outside the summation and start the summation at $n=1$.
The former has been done here: https://en.wikipedia.org/wiki/Taylor_series#Definition






calculus functions derivatives notation taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    In Lagrange notation $f(x)$ is simply $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:35






  • 1




    $begingroup$
    But in the Taylor expansion notation using the bigsum, $f^{(0)}(x)$ is clearly $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:36






  • 2




    $begingroup$
    I have even seen $f^{(-n)}$ for the iterated integral (or anti-derivative). If the context is clear with usage of parenthesis there is no problem identifying $f^{(0)}$ and $f$ rather than exponentiation which uses no parenthesis $f^n$. Only issue would be to differentiate from iterated composition which has no consensus scripting so either $f^n, f^{(n)}, f^{[n]}, f^{circ^n}$ could be encountered.
    $endgroup$
    – zwim
    Dec 2 '18 at 11:59






  • 1




    $begingroup$
    I think there's a case to be made for $f^{circ n}$ for the iterated function; I can't recall where I first saw this, or if I just made it up myself at some point.
    $endgroup$
    – John Hughes
    Dec 9 '18 at 17:20
















1












$begingroup$


Using Lagrange notation, is $f^{(0)}(x)=f(x)$? Is this standard notation, or would one have to define $f^{(0)}(x)=f(x)$ first, before using it?
Aside: the context of the question is whether to include the first term within the summation when expressing the Taylor series and hence start at $n=0$, or to write it separately outside the summation and start the summation at $n=1$.
The former has been done here: https://en.wikipedia.org/wiki/Taylor_series#Definition






calculus functions derivatives notation taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    In Lagrange notation $f(x)$ is simply $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:35






  • 1




    $begingroup$
    But in the Taylor expansion notation using the bigsum, $f^{(0)}(x)$ is clearly $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:36






  • 2




    $begingroup$
    I have even seen $f^{(-n)}$ for the iterated integral (or anti-derivative). If the context is clear with usage of parenthesis there is no problem identifying $f^{(0)}$ and $f$ rather than exponentiation which uses no parenthesis $f^n$. Only issue would be to differentiate from iterated composition which has no consensus scripting so either $f^n, f^{(n)}, f^{[n]}, f^{circ^n}$ could be encountered.
    $endgroup$
    – zwim
    Dec 2 '18 at 11:59






  • 1




    $begingroup$
    I think there's a case to be made for $f^{circ n}$ for the iterated function; I can't recall where I first saw this, or if I just made it up myself at some point.
    $endgroup$
    – John Hughes
    Dec 9 '18 at 17:20














1












1








1





$begingroup$


Using Lagrange notation, is $f^{(0)}(x)=f(x)$? Is this standard notation, or would one have to define $f^{(0)}(x)=f(x)$ first, before using it?
Aside: the context of the question is whether to include the first term within the summation when expressing the Taylor series and hence start at $n=0$, or to write it separately outside the summation and start the summation at $n=1$.
The former has been done here: https://en.wikipedia.org/wiki/Taylor_series#Definition






calculus functions derivatives notation taylor-expansion







share|cite|improve this question











$endgroup$




Using Lagrange notation, is $f^{(0)}(x)=f(x)$? Is this standard notation, or would one have to define $f^{(0)}(x)=f(x)$ first, before using it?
Aside: the context of the question is whether to include the first term within the summation when expressing the Taylor series and hence start at $n=0$, or to write it separately outside the summation and start the summation at $n=1$.
The former has been done here: https://en.wikipedia.org/wiki/Taylor_series#Definition






calculus functions derivatives notation taylor-expansion




calculus functions derivatives notation taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 11:37









Mauro ALLEGRANZA

64.7k448112




64.7k448112










asked Dec 2 '18 at 11:32









o co c

171




171












  • $begingroup$
    In Lagrange notation $f(x)$ is simply $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:35






  • 1




    $begingroup$
    But in the Taylor expansion notation using the bigsum, $f^{(0)}(x)$ is clearly $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:36






  • 2




    $begingroup$
    I have even seen $f^{(-n)}$ for the iterated integral (or anti-derivative). If the context is clear with usage of parenthesis there is no problem identifying $f^{(0)}$ and $f$ rather than exponentiation which uses no parenthesis $f^n$. Only issue would be to differentiate from iterated composition which has no consensus scripting so either $f^n, f^{(n)}, f^{[n]}, f^{circ^n}$ could be encountered.
    $endgroup$
    – zwim
    Dec 2 '18 at 11:59






  • 1




    $begingroup$
    I think there's a case to be made for $f^{circ n}$ for the iterated function; I can't recall where I first saw this, or if I just made it up myself at some point.
    $endgroup$
    – John Hughes
    Dec 9 '18 at 17:20


















  • $begingroup$
    In Lagrange notation $f(x)$ is simply $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:35






  • 1




    $begingroup$
    But in the Taylor expansion notation using the bigsum, $f^{(0)}(x)$ is clearly $f(x)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 2 '18 at 11:36






  • 2




    $begingroup$
    I have even seen $f^{(-n)}$ for the iterated integral (or anti-derivative). If the context is clear with usage of parenthesis there is no problem identifying $f^{(0)}$ and $f$ rather than exponentiation which uses no parenthesis $f^n$. Only issue would be to differentiate from iterated composition which has no consensus scripting so either $f^n, f^{(n)}, f^{[n]}, f^{circ^n}$ could be encountered.
    $endgroup$
    – zwim
    Dec 2 '18 at 11:59






  • 1




    $begingroup$
    I think there's a case to be made for $f^{circ n}$ for the iterated function; I can't recall where I first saw this, or if I just made it up myself at some point.
    $endgroup$
    – John Hughes
    Dec 9 '18 at 17:20
















$begingroup$
In Lagrange notation $f(x)$ is simply $f(x)$.
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 11:35




$begingroup$
In Lagrange notation $f(x)$ is simply $f(x)$.
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 11:35




1




1




$begingroup$
But in the Taylor expansion notation using the bigsum, $f^{(0)}(x)$ is clearly $f(x)$.
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 11:36




$begingroup$
But in the Taylor expansion notation using the bigsum, $f^{(0)}(x)$ is clearly $f(x)$.
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 11:36




2




2




$begingroup$
I have even seen $f^{(-n)}$ for the iterated integral (or anti-derivative). If the context is clear with usage of parenthesis there is no problem identifying $f^{(0)}$ and $f$ rather than exponentiation which uses no parenthesis $f^n$. Only issue would be to differentiate from iterated composition which has no consensus scripting so either $f^n, f^{(n)}, f^{[n]}, f^{circ^n}$ could be encountered.
$endgroup$
– zwim
Dec 2 '18 at 11:59




$begingroup$
I have even seen $f^{(-n)}$ for the iterated integral (or anti-derivative). If the context is clear with usage of parenthesis there is no problem identifying $f^{(0)}$ and $f$ rather than exponentiation which uses no parenthesis $f^n$. Only issue would be to differentiate from iterated composition which has no consensus scripting so either $f^n, f^{(n)}, f^{[n]}, f^{circ^n}$ could be encountered.
$endgroup$
– zwim
Dec 2 '18 at 11:59




1




1




$begingroup$
I think there's a case to be made for $f^{circ n}$ for the iterated function; I can't recall where I first saw this, or if I just made it up myself at some point.
$endgroup$
– John Hughes
Dec 9 '18 at 17:20




$begingroup$
I think there's a case to be made for $f^{circ n}$ for the iterated function; I can't recall where I first saw this, or if I just made it up myself at some point.
$endgroup$
– John Hughes
Dec 9 '18 at 17:20










2 Answers
2






active

oldest

votes


















2












$begingroup$

Writing $f^{(0)}(x)=f(x)$ just to be sure that we all agree on the same definitions is always smart to do. When in doubt specify unless it might even make things more confusing. Interpreting a function as its own zeroth derivative is not a weird thing to think about, but just mention it in case the reader is very fussy and wants to make a point about everything (knowing myself - I would). I think your reasoning is valid.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    A definition of something is always good before using it, as happend in the wikipedia article:




    The derivative of order zero of $f$ is defined to be $f$ itself.




    So you can start summation at $n=0$ in Taylor series after refering to this definition. But you can put your mind at rest. Most mathematicians would expect that $f^{(0)}(x)=f(x)$ without definition.






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022524%2flagrange-notation-f0x%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Writing $f^{(0)}(x)=f(x)$ just to be sure that we all agree on the same definitions is always smart to do. When in doubt specify unless it might even make things more confusing. Interpreting a function as its own zeroth derivative is not a weird thing to think about, but just mention it in case the reader is very fussy and wants to make a point about everything (knowing myself - I would). I think your reasoning is valid.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Writing $f^{(0)}(x)=f(x)$ just to be sure that we all agree on the same definitions is always smart to do. When in doubt specify unless it might even make things more confusing. Interpreting a function as its own zeroth derivative is not a weird thing to think about, but just mention it in case the reader is very fussy and wants to make a point about everything (knowing myself - I would). I think your reasoning is valid.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Writing $f^{(0)}(x)=f(x)$ just to be sure that we all agree on the same definitions is always smart to do. When in doubt specify unless it might even make things more confusing. Interpreting a function as its own zeroth derivative is not a weird thing to think about, but just mention it in case the reader is very fussy and wants to make a point about everything (knowing myself - I would). I think your reasoning is valid.






          share|cite|improve this answer









          $endgroup$



          Writing $f^{(0)}(x)=f(x)$ just to be sure that we all agree on the same definitions is always smart to do. When in doubt specify unless it might even make things more confusing. Interpreting a function as its own zeroth derivative is not a weird thing to think about, but just mention it in case the reader is very fussy and wants to make a point about everything (knowing myself - I would). I think your reasoning is valid.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 16:38









          Wesley StrikWesley Strik

          1,639423




          1,639423























              2












              $begingroup$

              A definition of something is always good before using it, as happend in the wikipedia article:




              The derivative of order zero of $f$ is defined to be $f$ itself.




              So you can start summation at $n=0$ in Taylor series after refering to this definition. But you can put your mind at rest. Most mathematicians would expect that $f^{(0)}(x)=f(x)$ without definition.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                A definition of something is always good before using it, as happend in the wikipedia article:




                The derivative of order zero of $f$ is defined to be $f$ itself.




                So you can start summation at $n=0$ in Taylor series after refering to this definition. But you can put your mind at rest. Most mathematicians would expect that $f^{(0)}(x)=f(x)$ without definition.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  A definition of something is always good before using it, as happend in the wikipedia article:




                  The derivative of order zero of $f$ is defined to be $f$ itself.




                  So you can start summation at $n=0$ in Taylor series after refering to this definition. But you can put your mind at rest. Most mathematicians would expect that $f^{(0)}(x)=f(x)$ without definition.






                  share|cite|improve this answer









                  $endgroup$



                  A definition of something is always good before using it, as happend in the wikipedia article:




                  The derivative of order zero of $f$ is defined to be $f$ itself.




                  So you can start summation at $n=0$ in Taylor series after refering to this definition. But you can put your mind at rest. Most mathematicians would expect that $f^{(0)}(x)=f(x)$ without definition.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 17:15









                  FakemistakeFakemistake

                  1,692815




                  1,692815






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022524%2flagrange-notation-f0x%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Image
                      0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                      Read more

                      Aardman Animations

                      Image
                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                      Read more

                      Are they similar matrix

                      Image
                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                      Read more