Prove upper bound on derivative












0












$begingroup$


Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



a) if $I=[-a, a]$, then
$$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




  1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



$$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



Solutions for b) and c) were given here and here.



d) was solved here.



Now I am trying to solve a):



Using Taylor's expansion we have



$$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
for some $xi_1, xi_2$. From this



$$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



How to eliminate that $|x|cdot M_2$?






real-analysis derivatives taylor-expansion







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



    a) if $I=[-a, a]$, then
    $$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
    b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




    1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


    c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



    d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



    $$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



    Solutions for b) and c) were given here and here.



    d) was solved here.



    Now I am trying to solve a):



    Using Taylor's expansion we have



    $$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
    f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
    for some $xi_1, xi_2$. From this



    $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



    Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



    How to eliminate that $|x|cdot M_2$?






    real-analysis derivatives taylor-expansion







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      0



      $begingroup$


      Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



      a) if $I=[-a, a]$, then
      $$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
      b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




      1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


      c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



      d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



      $$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



      Solutions for b) and c) were given here and here.



      d) was solved here.



      Now I am trying to solve a):



      Using Taylor's expansion we have



      $$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
      f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
      for some $xi_1, xi_2$. From this



      $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



      Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



      How to eliminate that $|x|cdot M_2$?






      real-analysis derivatives taylor-expansion







      share|cite|improve this question









      $endgroup$




      Let $f$ be twice differentiable function on interval $I$. Let $M_0=sup_{xin I}|f(x)|, M_1=sup_{xin I}|f'(x)|, M_2=sup_{xin I}|f''(x)|$. Show that



      a) if $I=[-a, a]$, then
      $$|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2$$
      b) 1. if length of $I$ is not less than $2sqrt{M_0/M_2}$, then $M_1leqslant 2sqrt{M_0M_2}$




      1. if $I=mathbb{R}$, then $M_1leqslant sqrt{2M_0M_2}$


      c) Constants $2$ and $sqrt{2}$ from b) cannot be improved



      d) if $f$ is $p$ times differentiable on $mathbb{R}$ and $M_0$ and $M_p=sup_{xinmathbb{R}}|f^{(p)}(x)|$ are finite, then for $1leqslant kleqslant p$ $M_k=sup_{xinmathbb{R}}|f^{(k)}(x)|$ are finite and



      $$M_kleqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$



      Solutions for b) and c) were given here and here.



      d) was solved here.



      Now I am trying to solve a):



      Using Taylor's expansion we have



      $$f(a)=f(x)+f'(x)(a-x)+frac{1}{2}f''(xi_1)(a-x)^2\
      f(-a)=f(x)+f'(x)(-a-x)+frac{1}{2}f''(xi_2)(-a-x)^2,$$
      for some $xi_1, xi_2$. From this



      $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



      Using triangle inequality gives $|f'(x)|leqslant frac{M_0}{a}+frac{x^2+a^2}{2a}M_2+|x|cdot M_2$



      How to eliminate that $|x|cdot M_2$?






      real-analysis derivatives taylor-expansion




      real-analysis derivatives taylor-expansion






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 11 '18 at 6:37









      B. BekmaganbetovB. Bekmaganbetov

      214




      214






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Don't write $f'(x)$ as:



          $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



          but simply as:



          $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



          and so it follows that:



          $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



          that is the claim.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Taylor expansion about $x=x_0$ is given by
            $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
            So,
            $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
            and
            $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
            Subtracting $(2)$ from $(1)$, we get
            $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
            $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
            Using triangle inequality, we have
            $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
            $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
            Since $x_0$ was arbitrary, so
            $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
              $endgroup$
              – B. Bekmaganbetov
              Mar 11 '18 at 8:35











            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%2f2686081%2fprove-upper-bound-on-derivative%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












            $begingroup$

            Don't write $f'(x)$ as:



            $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



            but simply as:



            $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



            and so it follows that:



            $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



            that is the claim.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Don't write $f'(x)$ as:



              $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



              but simply as:



              $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



              and so it follows that:



              $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



              that is the claim.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Don't write $f'(x)$ as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



                but simply as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



                and so it follows that:



                $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



                that is the claim.






                share|cite|improve this answer









                $endgroup$



                Don't write $f'(x)$ as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}+frac{1}{4a}Bigl((x^2+a^2)(f''(xi_1)-f''(xi_2))-2ax(f''(xi_1)+f''(xi_2))Bigr)$$



                but simply as:



                $$f'(x)=frac{f(a)-f(-a)}{2a}-frac{1}{4a}f''(xi_1)(a-x_0)^2+frac{1}{4a}f''(xi_2)(a+x_0)^2$$



                and so it follows that:



                $$|f'(x)|leq frac{M_0}{a}+frac{M_2}{4a}(a-x_0)^2+frac{M_2}{4a}(a+x_0)^2=frac{M_0}{a}+frac{M_2}{2a}(a^2+x_0^2)$$



                that is the claim.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 24 at 20:34









                NamelessNameless

                788




                788























                    0












                    $begingroup$

                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35
















                    0












                    $begingroup$

                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35














                    0












                    0








                    0





                    $begingroup$

                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$






                    share|cite|improve this answer









                    $endgroup$



                    Taylor expansion about $x=x_0$ is given by
                    $$f(x)=f(x_0)+f'(x_0)(x-x_0)+frac{f''(x_0)}{2}(x-x_0)^2.$$
                    So,
                    $$f(a)=f(x_0)+f'(x_0)(a-x_0)+frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$
                    and
                    $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$
                    Subtracting $(2)$ from $(1)$, we get
                    $$f(a)-f(-a)=2a~f'(x_0)+frac{1}{2}(-4ax_0)f''(x_0)$$
                    $$implies f'(x_0)=frac{f(a)-f(-a)}{2a}+frac{1}{2a}(2ax_0)f''(x_0).$$
                    Using triangle inequality, we have
                    $$|f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(2ax_0)M_2$$
                    $$implies |f'(x_0)|leqfrac{M_0}{a}+frac{1}{2a}(x_0^2+a^2)M_2~~~(because (x_0-a)^2geq 0).$$
                    Since $x_0$ was arbitrary, so
                    $$|f'(x)|leqfrac{M_0}{a}+frac{1}{2a}(x^2+a^2)M_2.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 11 '18 at 7:48









                    GoldyGoldy

                    43414




                    43414












                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35


















                    • $begingroup$
                      (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                      $endgroup$
                      – B. Bekmaganbetov
                      Mar 11 '18 at 8:35
















                    $begingroup$
                    (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                    $endgroup$
                    – B. Bekmaganbetov
                    Mar 11 '18 at 8:35




                    $begingroup$
                    (1) and (2) are incorrect. See en.wikipedia.org/wiki/Taylor's_theorem. You cannot write $f''(x_0)$ because the point lies between $x_0$ and $a$ or $-a$, but it is not $x_0$. Consider $f(x)=x^3$ on $[-1, 1]$.
                    $endgroup$
                    – B. Bekmaganbetov
                    Mar 11 '18 at 8:35


















                    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%2f2686081%2fprove-upper-bound-on-derivative%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

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                    Read more

                    Antonio Litta Visconti Arese

                    Image
                    Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                    Read more