Relating the mean value theorem to the directional derivatives












0












$begingroup$


I am having some trouble understanding the process for proving the following statement:



If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.



In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:



$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$



I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?



Thanks for your help.






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:27












  • $begingroup$
    What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:31












  • $begingroup$
    Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:37












  • $begingroup$
    I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:38










  • $begingroup$
    Thanks famesyasd - that is clear now.
    $endgroup$
    – Christian
    Dec 6 '18 at 12:06
















0












$begingroup$


I am having some trouble understanding the process for proving the following statement:



If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.



In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:



$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$



I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?



Thanks for your help.






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:27












  • $begingroup$
    What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:31












  • $begingroup$
    Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:37












  • $begingroup$
    I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:38










  • $begingroup$
    Thanks famesyasd - that is clear now.
    $endgroup$
    – Christian
    Dec 6 '18 at 12:06














0












0








0





$begingroup$


I am having some trouble understanding the process for proving the following statement:



If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.



In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:



$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$



I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?



Thanks for your help.






calculus







share|cite|improve this question









$endgroup$




I am having some trouble understanding the process for proving the following statement:



If $f'(x,v) = 0$ is a closed ball $B(x_0,r)$ for all $x$, show that $f$ is constant.



In the proof, we take two elements $x_1, x_2$ in $B(x_0,r)$, then we define the vector $v$ between $x_1$ and $x_2$ as $v = x_2 - x_1$, so $x_2 = v + x_1$. Finally, we apply the mean value theorem so that:



$f(x_1 +v) -f(x_1) = f'(x_1 + theta v, v) = 0$



I can not see how the above relates to the mean value theorem. Don't we need to divide by the distance between $x_2$ and $x_1$ on the left hand side? Could someone help me to fill in the missing steps?



Thanks for your help.






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 0:49









ChristianChristian

388




388












  • $begingroup$
    You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:27












  • $begingroup$
    What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:31












  • $begingroup$
    Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:37












  • $begingroup$
    I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:38










  • $begingroup$
    Thanks famesyasd - that is clear now.
    $endgroup$
    – Christian
    Dec 6 '18 at 12:06


















  • $begingroup$
    You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:27












  • $begingroup$
    What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:31












  • $begingroup$
    Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:37












  • $begingroup$
    I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
    $endgroup$
    – famesyasd
    Dec 6 '18 at 2:38










  • $begingroup$
    Thanks famesyasd - that is clear now.
    $endgroup$
    – Christian
    Dec 6 '18 at 12:06
















$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27






$begingroup$
You see, the problem here is that it should have been $frac{f(x_1+v)-f(x_1)}{(x1+v)-x1} = f'(x_1+theta v,v) = 0$ thus $f(x_1+v) = f(x_1)$ right? However $(x1+v) - x1 = v$ is a vector now and not a real number so you can not divide by it and apply mean value theorem directly
$endgroup$
– famesyasd
Dec 6 '18 at 2:27














$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31






$begingroup$
What you should do in this case is to approach a little bit more carefully with the mean value theorem. For any given point $x_1$ in your ball define a new function: $f(t) = (1-t)cdot x_0 + tcdot x_1, t in [0,1]$ This function parametrises the segmen $[x_0,x_1]$ (the center of your ball and any other point inside of it.
$endgroup$
– famesyasd
Dec 6 '18 at 2:31














$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37






$begingroup$
Now, this function is diffrentiable on $[0,1]$ (can you see why?) Thus one can apply the mean vlaue theorem to this function at the points $0$ and $1$ for example as to get $f(x_1) - f(x_0) = f'(xi)(1 - 0) = 0$
$endgroup$
– famesyasd
Dec 6 '18 at 2:37














$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38




$begingroup$
I'm sorry in my first comment I mistyped the function it should have been $g(t) = f[(1-t)cdot x_0 + tcdot x_1], t in [0,1]$
$endgroup$
– famesyasd
Dec 6 '18 at 2:38












$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06




$begingroup$
Thanks famesyasd - that is clear now.
$endgroup$
– Christian
Dec 6 '18 at 12:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

$newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
$$
lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
$$
and note that
$$
f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
$$
provided that the limit exists, we have
$$
lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
$$
i.e.
$$
F'(k) = f'(bm x_1 +kbm v , bm v).
$$
So by Lagrange's MVT,
$$
F(1) - F(0) = 1 cdot F'(k),
$$
which is just
$$
f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
$$
as shown in the OP.



To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.






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%2f3027884%2frelating-the-mean-value-theorem-to-the-directional-derivatives%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












    $begingroup$

    $newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
    $$
    lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
    $$
    and note that
    $$
    f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
    $$
    provided that the limit exists, we have
    $$
    lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
    $$
    i.e.
    $$
    F'(k) = f'(bm x_1 +kbm v , bm v).
    $$
    So by Lagrange's MVT,
    $$
    F(1) - F(0) = 1 cdot F'(k),
    $$
    which is just
    $$
    f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
    $$
    as shown in the OP.



    To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
      $$
      lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
      $$
      and note that
      $$
      f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
      $$
      provided that the limit exists, we have
      $$
      lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
      $$
      i.e.
      $$
      F'(k) = f'(bm x_1 +kbm v , bm v).
      $$
      So by Lagrange's MVT,
      $$
      F(1) - F(0) = 1 cdot F'(k),
      $$
      which is just
      $$
      f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
      $$
      as shown in the OP.



      To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
        $$
        lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
        $$
        and note that
        $$
        f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
        $$
        provided that the limit exists, we have
        $$
        lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
        $$
        i.e.
        $$
        F'(k) = f'(bm x_1 +kbm v , bm v).
        $$
        So by Lagrange's MVT,
        $$
        F(1) - F(0) = 1 cdot F'(k),
        $$
        which is just
        $$
        f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
        $$
        as shown in the OP.



        To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.






        share|cite|improve this answer









        $endgroup$



        $newcommand{bm}[1]{boldsymbol {#1}}$ Let $F(t) = f(bm x_1 + t bm v)$ where $tin [0,1]$. Then $F(0)=f(bm x_1), F(1)= f(bm x_2)$ and $F$ is continuous on $[0, 1]$, since each $f'(bm y, bm v)$ exists and equals $0$. Now check the differentiability of $F$: for each $kin (0,1)$,
        $$
        lim_{hto 0} frac {F(k+h) - F(k)}h = lim_{hto 0} frac {f((bm x_1 +kbm v) + h bm v)-f(bm x_1 + kbm v)}h,
        $$
        and note that
        $$
        f'(bm y, bm v) := lim_{tto 0}frac {f(bm y + tbm v) - f(bm y)}t
        $$
        provided that the limit exists, we have
        $$
        lim_{h to 0}frac {F(k+h) - F(k)}h = f'(bm x + kbm v, bm v),
        $$
        i.e.
        $$
        F'(k) = f'(bm x_1 +kbm v , bm v).
        $$
        So by Lagrange's MVT,
        $$
        F(1) - F(0) = 1 cdot F'(k),
        $$
        which is just
        $$
        f(bm x_2) - f(bm x_1) = f'(bm x_1 + kbm v, bm v),
        $$
        as shown in the OP.



        To conclude, that "division by $bm x_1, bm x_2$" is included in the expression of the directional derivatives.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 5:24









        xbhxbh

        6,3001522




        6,3001522






























            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%2f3027884%2frelating-the-mean-value-theorem-to-the-directional-derivatives%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

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more