Why $v_amapsto D_{v}|_a$ (the directional derivative) is a linear map $mathbb R^n_ato T_amathbb R^n$?












0












$begingroup$


Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.



Why $v_amapsto D_v|_a$ is a linear map ?




Attempts



Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$



Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$






derivatives manifolds







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
    $endgroup$
    – Ian
    Jan 12 at 16:32












  • $begingroup$
    @Ian: Thank you. It's not possible like I tried to ?
    $endgroup$
    – user623855
    Jan 12 at 16:34










  • $begingroup$
    Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
    $endgroup$
    – Ian
    Jan 12 at 16:34
















0












$begingroup$


Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.



Why $v_amapsto D_v|_a$ is a linear map ?




Attempts



Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$



Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$






derivatives manifolds







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
    $endgroup$
    – Ian
    Jan 12 at 16:32












  • $begingroup$
    @Ian: Thank you. It's not possible like I tried to ?
    $endgroup$
    – user623855
    Jan 12 at 16:34










  • $begingroup$
    Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
    $endgroup$
    – Ian
    Jan 12 at 16:34














0












0








0





$begingroup$


Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.



Why $v_amapsto D_v|_a$ is a linear map ?




Attempts



Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$



Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$






derivatives manifolds







share|cite|improve this question









$endgroup$




Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.



Why $v_amapsto D_v|_a$ is a linear map ?




Attempts



Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$



Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$






derivatives manifolds




derivatives manifolds






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 16:28









user623855user623855

1507




1507












  • $begingroup$
    It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
    $endgroup$
    – Ian
    Jan 12 at 16:32












  • $begingroup$
    @Ian: Thank you. It's not possible like I tried to ?
    $endgroup$
    – user623855
    Jan 12 at 16:34










  • $begingroup$
    Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
    $endgroup$
    – Ian
    Jan 12 at 16:34


















  • $begingroup$
    It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
    $endgroup$
    – Ian
    Jan 12 at 16:32












  • $begingroup$
    @Ian: Thank you. It's not possible like I tried to ?
    $endgroup$
    – user623855
    Jan 12 at 16:34










  • $begingroup$
    Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
    $endgroup$
    – Ian
    Jan 12 at 16:34
















$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32






$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32














$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34




$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34












$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34




$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint



If $vin mathbb R^n$,
$$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$






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%2f3071081%2fwhy-v-a-mapsto-d-v-a-the-directional-derivative-is-a-linear-map-mathbb%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint



    If $vin mathbb R^n$,
    $$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint



      If $vin mathbb R^n$,
      $$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint



        If $vin mathbb R^n$,
        $$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$






        share|cite|improve this answer









        $endgroup$



        Hint



        If $vin mathbb R^n$,
        $$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 16:39









        SurbSurb

        38k94375




        38k94375






























            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%2f3071081%2fwhy-v-a-mapsto-d-v-a-the-directional-derivative-is-a-linear-map-mathbb%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

            Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

            Image
            28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
            Read more

            Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

            Image
            up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
            Read more

            A Topological Invariant for $pi_3(U(n))$

            Image
            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
            Read more