Prove derivative of order $n$ at neighbourhood $x_0$












0












$begingroup$



If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19
















0












$begingroup$



If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19














0












0








0





$begingroup$



If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity







share|cite|improve this question









$endgroup$





If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity




derivatives proof-verification continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 17:11









weilam06weilam06

9511




9511












  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19


















  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19
















$begingroup$
Did induction failed?
$endgroup$
– Will M.
Dec 7 '18 at 17:14




$begingroup$
Did induction failed?
$endgroup$
– Will M.
Dec 7 '18 at 17:14












$begingroup$
I didn't try induction and I think there are some ways can prove it directly.
$endgroup$
– weilam06
Dec 7 '18 at 17:16




$begingroup$
I didn't try induction and I think there are some ways can prove it directly.
$endgroup$
– weilam06
Dec 7 '18 at 17:16




1




1




$begingroup$
Sure, you can find $n$th the derivative of a product... by induction.
$endgroup$
– Will M.
Dec 7 '18 at 17:17




$begingroup$
Sure, you can find $n$th the derivative of a product... by induction.
$endgroup$
– Will M.
Dec 7 '18 at 17:17












$begingroup$
By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
$endgroup$
– Will M.
Dec 7 '18 at 17:19




$begingroup$
By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
$endgroup$
– Will M.
Dec 7 '18 at 17:19










1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint: Note that
begin{align}
f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
end{align}
which means
begin{align}
frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
end{align}






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%2f3030140%2fprove-derivative-of-order-n-at-neighbourhood-x-0%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









    0












    $begingroup$

    Hint: Note that
    begin{align}
    f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
    end{align}
    which means
    begin{align}
    frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
    end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint: Note that
      begin{align}
      f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
      end{align}
      which means
      begin{align}
      frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint: Note that
        begin{align}
        f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
        end{align}
        which means
        begin{align}
        frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
        end{align}






        share|cite|improve this answer









        $endgroup$



        Hint: Note that
        begin{align}
        f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
        end{align}
        which means
        begin{align}
        frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 17:37









        Jacky ChongJacky Chong

        18.4k21128




        18.4k21128






























            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%2f3030140%2fprove-derivative-of-order-n-at-neighbourhood-x-0%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more