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





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      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






























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