Is my solution correct? Something feels wrong with it.












0












$begingroup$


I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.



Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.

I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$



I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!






calculus limits derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Treating $x$ as a constant here is fine, because it is, in fact, constant.
    $endgroup$
    – user3482749
    Jan 9 at 20:16










  • $begingroup$
    Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
    $endgroup$
    – Paramanand Singh
    Jan 10 at 3:08


















0












$begingroup$


I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.



Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.

I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$



I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!






calculus limits derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Treating $x$ as a constant here is fine, because it is, in fact, constant.
    $endgroup$
    – user3482749
    Jan 9 at 20:16










  • $begingroup$
    Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
    $endgroup$
    – Paramanand Singh
    Jan 10 at 3:08
















0












0








0





$begingroup$


I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.



Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.

I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$



I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!






calculus limits derivatives







share|cite|improve this question











$endgroup$




I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.



Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.

I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$



I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!






calculus limits derivatives




calculus limits derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 20:28







Omer

















asked Jan 9 at 20:07









OmerOmer

3138




3138








  • 2




    $begingroup$
    Treating $x$ as a constant here is fine, because it is, in fact, constant.
    $endgroup$
    – user3482749
    Jan 9 at 20:16










  • $begingroup$
    Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
    $endgroup$
    – Paramanand Singh
    Jan 10 at 3:08
















  • 2




    $begingroup$
    Treating $x$ as a constant here is fine, because it is, in fact, constant.
    $endgroup$
    – user3482749
    Jan 9 at 20:16










  • $begingroup$
    Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
    $endgroup$
    – Paramanand Singh
    Jan 10 at 3:08










2




2




$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16




$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16












$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08






$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08












2 Answers
2






active

oldest

votes


















1












$begingroup$

To elaborate on a comment:



The premise of the theorem is that you are given a value of $x$ such that
$f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
for that value of $x$.



It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.



So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.



Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
then while you are evaluating $f''(x)$ at any value of $x$
it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
That's generally true when you attempt take the derivative of any function
(in this case $f'$) at any point in its domain using this definition of the derivative.
Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.



Hence it is correct to treat $x$ as a constant while taking your limit.



It's a good thing that you're thinking about things like this, however,
because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    hint



    $f''(x)$ exists, thus by Taylor-Young formula,



    $$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
    with $$lim_{yto x}epsilon(y)=0$$



    replace $y$ by $x+h$ and by $x-h$ and sum.






    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%2f3067901%2fis-my-solution-correct-something-feels-wrong-with-it%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












      $begingroup$

      To elaborate on a comment:



      The premise of the theorem is that you are given a value of $x$ such that
      $f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
      for that value of $x$.



      It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.



      So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
      from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.



      Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
      then while you are evaluating $f''(x)$ at any value of $x$
      it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
      That's generally true when you attempt take the derivative of any function
      (in this case $f'$) at any point in its domain using this definition of the derivative.
      Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.



      Hence it is correct to treat $x$ as a constant while taking your limit.



      It's a good thing that you're thinking about things like this, however,
      because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        To elaborate on a comment:



        The premise of the theorem is that you are given a value of $x$ such that
        $f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
        for that value of $x$.



        It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.



        So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
        from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.



        Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
        then while you are evaluating $f''(x)$ at any value of $x$
        it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
        That's generally true when you attempt take the derivative of any function
        (in this case $f'$) at any point in its domain using this definition of the derivative.
        Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.



        Hence it is correct to treat $x$ as a constant while taking your limit.



        It's a good thing that you're thinking about things like this, however,
        because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          To elaborate on a comment:



          The premise of the theorem is that you are given a value of $x$ such that
          $f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
          for that value of $x$.



          It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.



          So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
          from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.



          Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
          then while you are evaluating $f''(x)$ at any value of $x$
          it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
          That's generally true when you attempt take the derivative of any function
          (in this case $f'$) at any point in its domain using this definition of the derivative.
          Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.



          Hence it is correct to treat $x$ as a constant while taking your limit.



          It's a good thing that you're thinking about things like this, however,
          because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.






          share|cite|improve this answer









          $endgroup$



          To elaborate on a comment:



          The premise of the theorem is that you are given a value of $x$ such that
          $f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
          for that value of $x$.



          It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.



          So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
          from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.



          Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
          then while you are evaluating $f''(x)$ at any value of $x$
          it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
          That's generally true when you attempt take the derivative of any function
          (in this case $f'$) at any point in its domain using this definition of the derivative.
          Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.



          Hence it is correct to treat $x$ as a constant while taking your limit.



          It's a good thing that you're thinking about things like this, however,
          because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 20:59









          David KDavid K

          53.2k341115




          53.2k341115























              1












              $begingroup$

              hint



              $f''(x)$ exists, thus by Taylor-Young formula,



              $$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
              with $$lim_{yto x}epsilon(y)=0$$



              replace $y$ by $x+h$ and by $x-h$ and sum.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                hint



                $f''(x)$ exists, thus by Taylor-Young formula,



                $$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
                with $$lim_{yto x}epsilon(y)=0$$



                replace $y$ by $x+h$ and by $x-h$ and sum.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  hint



                  $f''(x)$ exists, thus by Taylor-Young formula,



                  $$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
                  with $$lim_{yto x}epsilon(y)=0$$



                  replace $y$ by $x+h$ and by $x-h$ and sum.






                  share|cite|improve this answer









                  $endgroup$



                  hint



                  $f''(x)$ exists, thus by Taylor-Young formula,



                  $$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
                  with $$lim_{yto x}epsilon(y)=0$$



                  replace $y$ by $x+h$ and by $x-h$ and sum.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 20:17









                  hamam_Abdallahhamam_Abdallah

                  38k21634




                  38k21634






























                      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%2f3067901%2fis-my-solution-correct-something-feels-wrong-with-it%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

                      The Binding of Isaac: Rebirth/Afterbirth

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                      Read more

                      Dobbiaco

                      Image
                      Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
                      Read more