Differential of scalar product












1












$begingroup$


Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    $endgroup$
    – manooooh
    Dec 25 '18 at 17:53












  • $begingroup$
    Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    $endgroup$
    – Mark
    Dec 25 '18 at 17:55








  • 1




    $begingroup$
    Particular case of math.stackexchange.com/questions/1120430/….
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 25 '18 at 18:24
















1












$begingroup$


Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    $endgroup$
    – manooooh
    Dec 25 '18 at 17:53












  • $begingroup$
    Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    $endgroup$
    – Mark
    Dec 25 '18 at 17:55








  • 1




    $begingroup$
    Particular case of math.stackexchange.com/questions/1120430/….
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 25 '18 at 18:24














1












1








1





$begingroup$


Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus







share|cite|improve this question











$endgroup$




Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 18:08









APC89

2,341620




2,341620










asked Dec 25 '18 at 17:49









LenaLena

335




335












  • $begingroup$
    I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    $endgroup$
    – manooooh
    Dec 25 '18 at 17:53












  • $begingroup$
    Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    $endgroup$
    – Mark
    Dec 25 '18 at 17:55








  • 1




    $begingroup$
    Particular case of math.stackexchange.com/questions/1120430/….
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 25 '18 at 18:24


















  • $begingroup$
    I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    $endgroup$
    – manooooh
    Dec 25 '18 at 17:53












  • $begingroup$
    Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    $endgroup$
    – Mark
    Dec 25 '18 at 17:55








  • 1




    $begingroup$
    Particular case of math.stackexchange.com/questions/1120430/….
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 25 '18 at 18:24
















$begingroup$
I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
$endgroup$
– manooooh
Dec 25 '18 at 17:53






$begingroup$
I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
$endgroup$
– manooooh
Dec 25 '18 at 17:53














$begingroup$
Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
$endgroup$
– Mark
Dec 25 '18 at 17:55






$begingroup$
Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
$endgroup$
– Mark
Dec 25 '18 at 17:55






1




1




$begingroup$
Particular case of math.stackexchange.com/questions/1120430/….
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 25 '18 at 18:24




$begingroup$
Particular case of math.stackexchange.com/questions/1120430/….
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 25 '18 at 18:24










2 Answers
2






active

oldest

votes


















2












$begingroup$

The value of the differential at a point $(x,y)$ is the linear part of the difference
$;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
$$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
      $endgroup$
      – Lena
      Dec 25 '18 at 18:21










    • $begingroup$
      It is a bit fishy, yes, I edit the answer.
      $endgroup$
      – gangrene
      Dec 25 '18 at 18:35











    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%2f3052310%2fdifferential-of-scalar-product%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












    $begingroup$

    The value of the differential at a point $(x,y)$ is the linear part of the difference
    $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
    $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
    so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      The value of the differential at a point $(x,y)$ is the linear part of the difference
      $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
      $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
      so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        The value of the differential at a point $(x,y)$ is the linear part of the difference
        $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
        $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
        so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






        share|cite|improve this answer











        $endgroup$



        The value of the differential at a point $(x,y)$ is the linear part of the difference
        $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
        $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
        so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 25 '18 at 19:26

























        answered Dec 25 '18 at 18:35









        BernardBernard

        123k741116




        123k741116























            0












            $begingroup$

            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              $endgroup$
              – Lena
              Dec 25 '18 at 18:21










            • $begingroup$
              It is a bit fishy, yes, I edit the answer.
              $endgroup$
              – gangrene
              Dec 25 '18 at 18:35
















            0












            $begingroup$

            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              $endgroup$
              – Lena
              Dec 25 '18 at 18:21










            • $begingroup$
              It is a bit fishy, yes, I edit the answer.
              $endgroup$
              – gangrene
              Dec 25 '18 at 18:35














            0












            0








            0





            $begingroup$

            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






            share|cite|improve this answer











            $endgroup$



            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 25 '18 at 18:36

























            answered Dec 25 '18 at 17:54









            gangrenegangrene

            915514




            915514












            • $begingroup$
              I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              $endgroup$
              – Lena
              Dec 25 '18 at 18:21










            • $begingroup$
              It is a bit fishy, yes, I edit the answer.
              $endgroup$
              – gangrene
              Dec 25 '18 at 18:35


















            • $begingroup$
              I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              $endgroup$
              – Lena
              Dec 25 '18 at 18:21










            • $begingroup$
              It is a bit fishy, yes, I edit the answer.
              $endgroup$
              – gangrene
              Dec 25 '18 at 18:35
















            $begingroup$
            I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
            $endgroup$
            – Lena
            Dec 25 '18 at 18:21




            $begingroup$
            I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
            $endgroup$
            – Lena
            Dec 25 '18 at 18:21












            $begingroup$
            It is a bit fishy, yes, I edit the answer.
            $endgroup$
            – gangrene
            Dec 25 '18 at 18:35




            $begingroup$
            It is a bit fishy, yes, I edit the answer.
            $endgroup$
            – gangrene
            Dec 25 '18 at 18:35


















            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%2f3052310%2fdifferential-of-scalar-product%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