Interpreting Volterra Series CorrectlyWhat is $int frac{delta F}{delta u} frac{delta G}{delta v} , dx ;...

How many wives did king shaul have

Unlock My Phone! February 2018

How seriously should I take size and weight limits of hand luggage?

If a warlock makes a Dancing Sword their pact weapon, is there a way to prevent it from disappearing if it's farther away for more than a minute?

How dangerous is XSS

How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?

How does a dynamic QR code work?

Is this draw by repetition?

What is required to make GPS signals available indoors?

How to compactly explain secondary and tertiary characters without resorting to stereotypes?

My ex-girlfriend uses my Apple ID to log in to her iPad. Do I have to give her my Apple ID password to reset it?

Where would I need my direct neural interface to be implanted?

Can compressed videos be decoded back to their uncompresed original format?

OP Amp not amplifying audio signal

Was the old ablative pronoun "med" or "mēd"?

What Exploit Are These User Agents Trying to Use?

Why are UK visa biometrics appointments suspended at USCIS Application Support Centers?

Theorists sure want true answers to this!

How to install cross-compiler on Ubuntu 18.04?

In the UK, is it possible to get a referendum by a court decision?

What is the opposite of "eschatology"?

ssTTsSTtRrriinInnnnNNNIiinngg

How exploitable/balanced is this homebrew spell: Spell Permanency?

Is it possible to map the firing of neurons in the human brain so as to stimulate artificial memories in someone else?



Interpreting Volterra Series Correctly


What is $int frac{delta F}{delta u} frac{delta G}{delta v} , dx ; $?Doubt in the derivation of the field Euler-Lagrange equationsTwo Approaches Two Different Solutions: Optimal Controls vs. Different MethodFunctional derivative: How to obtain $delta F=int frac{delta F}{delta f}delta f dx$?Functional Taylor seriesDoes the chain rule need to account for derivatives as test functions?Functional derivative in QFTFunctional derivative of a functional that depends on antiderivativeDo variations obey the product rule?Euler-Lagrange equation in polar or cylindrical coordinates













0












$begingroup$


I was reading this and found on page 6 was a description of a generalization of Taylor series to linear functionals. Reproduced below as



$$ F[phi + lambda] = sum_{n=0}^{infty} left[frac{1}{n!} intint...int frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) } dx^0 dx^1 ... dx^{n-1}
right] $$



I don't understand exactly what the term $$ frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) }$$



Means.



And let me be extremely concrete. Consider an operator $F: phi rightarrow phi + x^2phi'$, quite literally what this means is that $F: (mathbb{R} rightarrow mathbb{R}) rightarrow (mathbb{R} rightarrow mathbb{R}) $



And $F:= phi(x) rightarrow phi(x) + x^2 frac{d}{dx}left[ phi(x)right] $ Can be viewed as an explicit "form".



So when we compute $$frac{delta F}{delta phi(x)}$$



Everything makes sense (since the argument of the $phi$ we are differentiating w.r.t matches the argument of the $phi$ in the original function and the corresponding $frac{d}{dx}$ is compatible with the $x$ arguments in both cases.) And we use our typical euler lagrange equations.



$$ frac{delta F}{delta phi(x)} = frac{partial F}{partial phi(x)} - frac{d}{dx} left[ frac{partial F}{partial left( frac{d}{dx}[phi(x)] right) } ...right] = 1+2x $$



If you make an expression like



$$frac{delta F}{delta phi(x^0)}$$



Now it's unclear what you mean, but we can guess. Let us substitute every instance of the string "$x$" with "$x^0$" in $F$ and then this statement can be still well defined. And we conclude that it is equal to



$$ 1 + 2x^0 $$



Now consider:



$$frac{delta^2 F}{delta phi(x^0) delta phi(x^1) }$$



This is now getting us into bad territory. Not only do we not have a compatible variable $x$ in either of arguments but we cannot implicitly "guess" what string substitution needs to be made here since any global level string subtitution will render the other derivative to 0. So I am completely lost as to what this could mean.










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I was reading this and found on page 6 was a description of a generalization of Taylor series to linear functionals. Reproduced below as



    $$ F[phi + lambda] = sum_{n=0}^{infty} left[frac{1}{n!} intint...int frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) } dx^0 dx^1 ... dx^{n-1}
    right] $$



    I don't understand exactly what the term $$ frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) }$$



    Means.



    And let me be extremely concrete. Consider an operator $F: phi rightarrow phi + x^2phi'$, quite literally what this means is that $F: (mathbb{R} rightarrow mathbb{R}) rightarrow (mathbb{R} rightarrow mathbb{R}) $



    And $F:= phi(x) rightarrow phi(x) + x^2 frac{d}{dx}left[ phi(x)right] $ Can be viewed as an explicit "form".



    So when we compute $$frac{delta F}{delta phi(x)}$$



    Everything makes sense (since the argument of the $phi$ we are differentiating w.r.t matches the argument of the $phi$ in the original function and the corresponding $frac{d}{dx}$ is compatible with the $x$ arguments in both cases.) And we use our typical euler lagrange equations.



    $$ frac{delta F}{delta phi(x)} = frac{partial F}{partial phi(x)} - frac{d}{dx} left[ frac{partial F}{partial left( frac{d}{dx}[phi(x)] right) } ...right] = 1+2x $$



    If you make an expression like



    $$frac{delta F}{delta phi(x^0)}$$



    Now it's unclear what you mean, but we can guess. Let us substitute every instance of the string "$x$" with "$x^0$" in $F$ and then this statement can be still well defined. And we conclude that it is equal to



    $$ 1 + 2x^0 $$



    Now consider:



    $$frac{delta^2 F}{delta phi(x^0) delta phi(x^1) }$$



    This is now getting us into bad territory. Not only do we not have a compatible variable $x$ in either of arguments but we cannot implicitly "guess" what string substitution needs to be made here since any global level string subtitution will render the other derivative to 0. So I am completely lost as to what this could mean.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I was reading this and found on page 6 was a description of a generalization of Taylor series to linear functionals. Reproduced below as



      $$ F[phi + lambda] = sum_{n=0}^{infty} left[frac{1}{n!} intint...int frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) } dx^0 dx^1 ... dx^{n-1}
      right] $$



      I don't understand exactly what the term $$ frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) }$$



      Means.



      And let me be extremely concrete. Consider an operator $F: phi rightarrow phi + x^2phi'$, quite literally what this means is that $F: (mathbb{R} rightarrow mathbb{R}) rightarrow (mathbb{R} rightarrow mathbb{R}) $



      And $F:= phi(x) rightarrow phi(x) + x^2 frac{d}{dx}left[ phi(x)right] $ Can be viewed as an explicit "form".



      So when we compute $$frac{delta F}{delta phi(x)}$$



      Everything makes sense (since the argument of the $phi$ we are differentiating w.r.t matches the argument of the $phi$ in the original function and the corresponding $frac{d}{dx}$ is compatible with the $x$ arguments in both cases.) And we use our typical euler lagrange equations.



      $$ frac{delta F}{delta phi(x)} = frac{partial F}{partial phi(x)} - frac{d}{dx} left[ frac{partial F}{partial left( frac{d}{dx}[phi(x)] right) } ...right] = 1+2x $$



      If you make an expression like



      $$frac{delta F}{delta phi(x^0)}$$



      Now it's unclear what you mean, but we can guess. Let us substitute every instance of the string "$x$" with "$x^0$" in $F$ and then this statement can be still well defined. And we conclude that it is equal to



      $$ 1 + 2x^0 $$



      Now consider:



      $$frac{delta^2 F}{delta phi(x^0) delta phi(x^1) }$$



      This is now getting us into bad territory. Not only do we not have a compatible variable $x$ in either of arguments but we cannot implicitly "guess" what string substitution needs to be made here since any global level string subtitution will render the other derivative to 0. So I am completely lost as to what this could mean.










      share|cite|improve this question











      $endgroup$




      I was reading this and found on page 6 was a description of a generalization of Taylor series to linear functionals. Reproduced below as



      $$ F[phi + lambda] = sum_{n=0}^{infty} left[frac{1}{n!} intint...int frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) } dx^0 dx^1 ... dx^{n-1}
      right] $$



      I don't understand exactly what the term $$ frac{delta ^n F[phi] }{delta phi(x^0) delta phi(x^1) ... delta phi(x^{n-1}) }$$



      Means.



      And let me be extremely concrete. Consider an operator $F: phi rightarrow phi + x^2phi'$, quite literally what this means is that $F: (mathbb{R} rightarrow mathbb{R}) rightarrow (mathbb{R} rightarrow mathbb{R}) $



      And $F:= phi(x) rightarrow phi(x) + x^2 frac{d}{dx}left[ phi(x)right] $ Can be viewed as an explicit "form".



      So when we compute $$frac{delta F}{delta phi(x)}$$



      Everything makes sense (since the argument of the $phi$ we are differentiating w.r.t matches the argument of the $phi$ in the original function and the corresponding $frac{d}{dx}$ is compatible with the $x$ arguments in both cases.) And we use our typical euler lagrange equations.



      $$ frac{delta F}{delta phi(x)} = frac{partial F}{partial phi(x)} - frac{d}{dx} left[ frac{partial F}{partial left( frac{d}{dx}[phi(x)] right) } ...right] = 1+2x $$



      If you make an expression like



      $$frac{delta F}{delta phi(x^0)}$$



      Now it's unclear what you mean, but we can guess. Let us substitute every instance of the string "$x$" with "$x^0$" in $F$ and then this statement can be still well defined. And we conclude that it is equal to



      $$ 1 + 2x^0 $$



      Now consider:



      $$frac{delta^2 F}{delta phi(x^0) delta phi(x^1) }$$



      This is now getting us into bad territory. Not only do we not have a compatible variable $x$ in either of arguments but we cannot implicitly "guess" what string substitution needs to be made here since any global level string subtitution will render the other derivative to 0. So I am completely lost as to what this could mean.







      physics euler-lagrange-equation functional-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 23:22







      frogeyedpeas

















      asked Feb 2 at 23:05









      frogeyedpeasfrogeyedpeas

      7,66872054




      7,66872054






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          I'll run through your example and hopefully that makes it clear.



          $$phimapsto F[phi]= phi+x^2phi'$$
          Vary this
          $$phi+deltaphimapsto F[phi]+deltaphi+x^2deltaphi'$$
          So we have formally
          $$frac{F[phi+deltaphi](x)-F[phi](x)}{deltaphi(y)}=frac{deltaphi(x)}{deltaphi(y)}+x^2frac{deltaphi'(x)}{deltaphi(y)}$$



          Then we take as our definition that (and more generally would have ignored $mathcal O(deltaphi)$ terms if we have things like $F[phi]=phi^2$)
          $$
          frac{deltaphi(x)}{deltaphi(y)}:=delta(x-y),
          $$

          the Dirac delta function. Similarly,
          $$
          frac{deltaphi'(x)}{deltaphi(y)}:=delta'(x-y).
          $$

          One can make this rigorous with compactly supported test functions and the like.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            FWIW, in OP's example OP is considering the functional
            $$ F[phi]~:=~ phi(x_0) + x_0^2phi^{prime}(x_0), $$
            which depends on the parameter $x^0inmathbb{R}$.



            Then the functional/variational derivative is
            $$ frac{delta F[phi]}{delta phi(x_1)}~=~delta(x_0!-!x_1) + x_0^2delta^{prime}(x_0!-!x_1), $$
            which doesn't depend on $phi$. Therefore the higher functional derivatives vanish.






            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%2f3097917%2finterpreting-volterra-series-correctly%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









              1












              $begingroup$

              I'll run through your example and hopefully that makes it clear.



              $$phimapsto F[phi]= phi+x^2phi'$$
              Vary this
              $$phi+deltaphimapsto F[phi]+deltaphi+x^2deltaphi'$$
              So we have formally
              $$frac{F[phi+deltaphi](x)-F[phi](x)}{deltaphi(y)}=frac{deltaphi(x)}{deltaphi(y)}+x^2frac{deltaphi'(x)}{deltaphi(y)}$$



              Then we take as our definition that (and more generally would have ignored $mathcal O(deltaphi)$ terms if we have things like $F[phi]=phi^2$)
              $$
              frac{deltaphi(x)}{deltaphi(y)}:=delta(x-y),
              $$

              the Dirac delta function. Similarly,
              $$
              frac{deltaphi'(x)}{deltaphi(y)}:=delta'(x-y).
              $$

              One can make this rigorous with compactly supported test functions and the like.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                I'll run through your example and hopefully that makes it clear.



                $$phimapsto F[phi]= phi+x^2phi'$$
                Vary this
                $$phi+deltaphimapsto F[phi]+deltaphi+x^2deltaphi'$$
                So we have formally
                $$frac{F[phi+deltaphi](x)-F[phi](x)}{deltaphi(y)}=frac{deltaphi(x)}{deltaphi(y)}+x^2frac{deltaphi'(x)}{deltaphi(y)}$$



                Then we take as our definition that (and more generally would have ignored $mathcal O(deltaphi)$ terms if we have things like $F[phi]=phi^2$)
                $$
                frac{deltaphi(x)}{deltaphi(y)}:=delta(x-y),
                $$

                the Dirac delta function. Similarly,
                $$
                frac{deltaphi'(x)}{deltaphi(y)}:=delta'(x-y).
                $$

                One can make this rigorous with compactly supported test functions and the like.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I'll run through your example and hopefully that makes it clear.



                  $$phimapsto F[phi]= phi+x^2phi'$$
                  Vary this
                  $$phi+deltaphimapsto F[phi]+deltaphi+x^2deltaphi'$$
                  So we have formally
                  $$frac{F[phi+deltaphi](x)-F[phi](x)}{deltaphi(y)}=frac{deltaphi(x)}{deltaphi(y)}+x^2frac{deltaphi'(x)}{deltaphi(y)}$$



                  Then we take as our definition that (and more generally would have ignored $mathcal O(deltaphi)$ terms if we have things like $F[phi]=phi^2$)
                  $$
                  frac{deltaphi(x)}{deltaphi(y)}:=delta(x-y),
                  $$

                  the Dirac delta function. Similarly,
                  $$
                  frac{deltaphi'(x)}{deltaphi(y)}:=delta'(x-y).
                  $$

                  One can make this rigorous with compactly supported test functions and the like.






                  share|cite|improve this answer











                  $endgroup$



                  I'll run through your example and hopefully that makes it clear.



                  $$phimapsto F[phi]= phi+x^2phi'$$
                  Vary this
                  $$phi+deltaphimapsto F[phi]+deltaphi+x^2deltaphi'$$
                  So we have formally
                  $$frac{F[phi+deltaphi](x)-F[phi](x)}{deltaphi(y)}=frac{deltaphi(x)}{deltaphi(y)}+x^2frac{deltaphi'(x)}{deltaphi(y)}$$



                  Then we take as our definition that (and more generally would have ignored $mathcal O(deltaphi)$ terms if we have things like $F[phi]=phi^2$)
                  $$
                  frac{deltaphi(x)}{deltaphi(y)}:=delta(x-y),
                  $$

                  the Dirac delta function. Similarly,
                  $$
                  frac{deltaphi'(x)}{deltaphi(y)}:=delta'(x-y).
                  $$

                  One can make this rigorous with compactly supported test functions and the like.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 18 at 12:02

























                  answered Feb 3 at 0:06









                  Alec B-GAlec B-G

                  52019




                  52019























                      1












                      $begingroup$

                      FWIW, in OP's example OP is considering the functional
                      $$ F[phi]~:=~ phi(x_0) + x_0^2phi^{prime}(x_0), $$
                      which depends on the parameter $x^0inmathbb{R}$.



                      Then the functional/variational derivative is
                      $$ frac{delta F[phi]}{delta phi(x_1)}~=~delta(x_0!-!x_1) + x_0^2delta^{prime}(x_0!-!x_1), $$
                      which doesn't depend on $phi$. Therefore the higher functional derivatives vanish.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        FWIW, in OP's example OP is considering the functional
                        $$ F[phi]~:=~ phi(x_0) + x_0^2phi^{prime}(x_0), $$
                        which depends on the parameter $x^0inmathbb{R}$.



                        Then the functional/variational derivative is
                        $$ frac{delta F[phi]}{delta phi(x_1)}~=~delta(x_0!-!x_1) + x_0^2delta^{prime}(x_0!-!x_1), $$
                        which doesn't depend on $phi$. Therefore the higher functional derivatives vanish.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          FWIW, in OP's example OP is considering the functional
                          $$ F[phi]~:=~ phi(x_0) + x_0^2phi^{prime}(x_0), $$
                          which depends on the parameter $x^0inmathbb{R}$.



                          Then the functional/variational derivative is
                          $$ frac{delta F[phi]}{delta phi(x_1)}~=~delta(x_0!-!x_1) + x_0^2delta^{prime}(x_0!-!x_1), $$
                          which doesn't depend on $phi$. Therefore the higher functional derivatives vanish.






                          share|cite|improve this answer









                          $endgroup$



                          FWIW, in OP's example OP is considering the functional
                          $$ F[phi]~:=~ phi(x_0) + x_0^2phi^{prime}(x_0), $$
                          which depends on the parameter $x^0inmathbb{R}$.



                          Then the functional/variational derivative is
                          $$ frac{delta F[phi]}{delta phi(x_1)}~=~delta(x_0!-!x_1) + x_0^2delta^{prime}(x_0!-!x_1), $$
                          which doesn't depend on $phi$. Therefore the higher functional derivatives vanish.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 3 at 12:05









                          QmechanicQmechanic

                          5,17711858




                          5,17711858






























                              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%2f3097917%2finterpreting-volterra-series-correctly%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

                              Nidaros erkebispedøme

                              Birsay

                              Where did Arya get these scars? Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara Favourite questions and answers from the 1st quarter of 2019Why did Arya refuse to end it?Has the pronunciation of Arya Stark's name changed?Has Arya forgiven people?Why did Arya Stark lose her vision?Why can Arya still use the faces?Has the Narrow Sea become narrower?Does Arya Stark know how to make poisons outside of the House of Black and White?Why did Nymeria leave Arya?Why did Arya not kill the Lannister soldiers she encountered in the Riverlands?What is the current canonical age of Sansa, Bran and Arya Stark?