Show that $f(a)=f(a-1).$












3












$begingroup$


Suppose that $f :[0,2]rightarrow mathbb{R}$ is a continuous function with $f(0)=f(2)$. Show that there is a real number $ain[1,2]$ with $f(a)=f(a-1)$.



For the answer I tried to apply the mean value theorem to the function $g(x)=int_0^2f(t)dt$ in the interval $[0,2]$ but it didn't lead any where.






calculus analysis







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 21:22






  • 3




    $begingroup$
    Define $g(x)=f(x)-f(x-1)$, Then consider $g$ at $x=1$ and $x=2$ $g(1)=f(1)$ $g(2)=-f(1)$ so they have opposite sign... (well... if they don't have opposite sign then it's even easier... )
    $endgroup$
    – Mason
    Dec 18 '18 at 21:24












  • $begingroup$
    .Got it. Thank you!
    $endgroup$
    – Charith Jeewantha
    Dec 18 '18 at 21:33
















3












$begingroup$


Suppose that $f :[0,2]rightarrow mathbb{R}$ is a continuous function with $f(0)=f(2)$. Show that there is a real number $ain[1,2]$ with $f(a)=f(a-1)$.



For the answer I tried to apply the mean value theorem to the function $g(x)=int_0^2f(t)dt$ in the interval $[0,2]$ but it didn't lead any where.






calculus analysis







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 21:22






  • 3




    $begingroup$
    Define $g(x)=f(x)-f(x-1)$, Then consider $g$ at $x=1$ and $x=2$ $g(1)=f(1)$ $g(2)=-f(1)$ so they have opposite sign... (well... if they don't have opposite sign then it's even easier... )
    $endgroup$
    – Mason
    Dec 18 '18 at 21:24












  • $begingroup$
    .Got it. Thank you!
    $endgroup$
    – Charith Jeewantha
    Dec 18 '18 at 21:33














3












3








3





$begingroup$


Suppose that $f :[0,2]rightarrow mathbb{R}$ is a continuous function with $f(0)=f(2)$. Show that there is a real number $ain[1,2]$ with $f(a)=f(a-1)$.



For the answer I tried to apply the mean value theorem to the function $g(x)=int_0^2f(t)dt$ in the interval $[0,2]$ but it didn't lead any where.






calculus analysis







share|cite|improve this question











$endgroup$




Suppose that $f :[0,2]rightarrow mathbb{R}$ is a continuous function with $f(0)=f(2)$. Show that there is a real number $ain[1,2]$ with $f(a)=f(a-1)$.



For the answer I tried to apply the mean value theorem to the function $g(x)=int_0^2f(t)dt$ in the interval $[0,2]$ but it didn't lead any where.






calculus analysis




calculus analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 21:23









user376343

3,8733829




3,8733829










asked Dec 18 '18 at 21:19









Charith JeewanthaCharith Jeewantha

898




898








  • 7




    $begingroup$
    You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 21:22






  • 3




    $begingroup$
    Define $g(x)=f(x)-f(x-1)$, Then consider $g$ at $x=1$ and $x=2$ $g(1)=f(1)$ $g(2)=-f(1)$ so they have opposite sign... (well... if they don't have opposite sign then it's even easier... )
    $endgroup$
    – Mason
    Dec 18 '18 at 21:24












  • $begingroup$
    .Got it. Thank you!
    $endgroup$
    – Charith Jeewantha
    Dec 18 '18 at 21:33














  • 7




    $begingroup$
    You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$.
    $endgroup$
    – Arthur
    Dec 18 '18 at 21:22






  • 3




    $begingroup$
    Define $g(x)=f(x)-f(x-1)$, Then consider $g$ at $x=1$ and $x=2$ $g(1)=f(1)$ $g(2)=-f(1)$ so they have opposite sign... (well... if they don't have opposite sign then it's even easier... )
    $endgroup$
    – Mason
    Dec 18 '18 at 21:24












  • $begingroup$
    .Got it. Thank you!
    $endgroup$
    – Charith Jeewantha
    Dec 18 '18 at 21:33








7




7




$begingroup$
You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$.
$endgroup$
– Arthur
Dec 18 '18 at 21:22




$begingroup$
You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$.
$endgroup$
– Arthur
Dec 18 '18 at 21:22




3




3




$begingroup$
Define $g(x)=f(x)-f(x-1)$, Then consider $g$ at $x=1$ and $x=2$ $g(1)=f(1)$ $g(2)=-f(1)$ so they have opposite sign... (well... if they don't have opposite sign then it's even easier... )
$endgroup$
– Mason
Dec 18 '18 at 21:24






$begingroup$
Define $g(x)=f(x)-f(x-1)$, Then consider $g$ at $x=1$ and $x=2$ $g(1)=f(1)$ $g(2)=-f(1)$ so they have opposite sign... (well... if they don't have opposite sign then it's even easier... )
$endgroup$
– Mason
Dec 18 '18 at 21:24














$begingroup$
.Got it. Thank you!
$endgroup$
– Charith Jeewantha
Dec 18 '18 at 21:33




$begingroup$
.Got it. Thank you!
$endgroup$
– Charith Jeewantha
Dec 18 '18 at 21:33










3 Answers
3






active

oldest

votes


















5












$begingroup$

Define a function $gcolon [1,2] to mathbb{R}$ by $g(x) = f(x) - f(x-1)$, so that you're looking for an $a$ such that $g(a) = 0$. Check that $g$ is continuous and apply the Intermediate Value Theorem.



You need to know something about $f(1)$ to do this properly, but you can consider separately the three cases where $f(1)$ is greater than, less than, or (the trivial case) equal to $f(0)$ and $f(2)$.



ETA: While I was writing this, two comments appeared suggesting the same approach.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Let $g(x) = f(x) - f(x-1)$



    $g(2) = f(2) - f(1) = f(0) - f(1) = -g(1)$



    if $g(1) = 0$ you are done.



    Otherwise $g(x)$ is a continuous function that is negative at one endpoint and positive at the other.



    By the IVT there must be an $ain [1,2]$ such that $g(a) = 0 $






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      "if g(x)=0 you are done." should be g(1)=0
      $endgroup$
      – Toby Bartels
      Dec 18 '18 at 21:29










    • $begingroup$
      @TobyBartels thanks.
      $endgroup$
      – Doug M
      Dec 18 '18 at 21:29



















    2












    $begingroup$

    Elementary Approach using Bolzano's Theorem and not IVT :



    Let $g(x) = f(x) - f(x-1)$. This function is continuous in $[0,2]$ as an operation of continuous functions.



    Now, $g(1) = f(1)- f(0)$ and $g(2) = f(2)-f(1)$.



    $$g(1)cdot g(2) =f(1)f(2)-f(1)^2 - f(2)f(0)+ f(0)f(1)=2f(0)f(1) - f(1)^2-f(0)^2 $$
    $$implies$$
    $$g(1)cdot g(2) = -big[f(1)-f(0)big]^2 leq 0 $$



    Thus, there exists $a in [0,2]$ such that $g(a) =0$ by Bolzano's Theorem.



    Functional Analysis (sorry for this) :



    Essentialy, we want to show that the equation



    $$f(x) - f(x-1) =0$$



    has a solution.



    Let $T$ be a linear operator $T:C[0,2] to C[0,2]$ such that $Tf(x) = -f(x-1)$. One can easily see that $T$ is a bounded linear operator $in Bbig(C[0,2]big)$. Also, one can take $0$ to be a function defined in $C[0,2]$. Then, the equation is re-written as :



    $$f + Tf = 0$$



    Note that $C[0,2]$ is a Banach Space and since $T in Bbig(C[0,2]big)$ then there exists a unique solution in $C[0,2]$, meaning that there exists such function $f(x) in C[0,2]$ which of course implies that the equation has a solution for $x in [0,2]$.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Define a function $gcolon [1,2] to mathbb{R}$ by $g(x) = f(x) - f(x-1)$, so that you're looking for an $a$ such that $g(a) = 0$. Check that $g$ is continuous and apply the Intermediate Value Theorem.



      You need to know something about $f(1)$ to do this properly, but you can consider separately the three cases where $f(1)$ is greater than, less than, or (the trivial case) equal to $f(0)$ and $f(2)$.



      ETA: While I was writing this, two comments appeared suggesting the same approach.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Define a function $gcolon [1,2] to mathbb{R}$ by $g(x) = f(x) - f(x-1)$, so that you're looking for an $a$ such that $g(a) = 0$. Check that $g$ is continuous and apply the Intermediate Value Theorem.



        You need to know something about $f(1)$ to do this properly, but you can consider separately the three cases where $f(1)$ is greater than, less than, or (the trivial case) equal to $f(0)$ and $f(2)$.



        ETA: While I was writing this, two comments appeared suggesting the same approach.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Define a function $gcolon [1,2] to mathbb{R}$ by $g(x) = f(x) - f(x-1)$, so that you're looking for an $a$ such that $g(a) = 0$. Check that $g$ is continuous and apply the Intermediate Value Theorem.



          You need to know something about $f(1)$ to do this properly, but you can consider separately the three cases where $f(1)$ is greater than, less than, or (the trivial case) equal to $f(0)$ and $f(2)$.



          ETA: While I was writing this, two comments appeared suggesting the same approach.






          share|cite|improve this answer









          $endgroup$



          Define a function $gcolon [1,2] to mathbb{R}$ by $g(x) = f(x) - f(x-1)$, so that you're looking for an $a$ such that $g(a) = 0$. Check that $g$ is continuous and apply the Intermediate Value Theorem.



          You need to know something about $f(1)$ to do this properly, but you can consider separately the three cases where $f(1)$ is greater than, less than, or (the trivial case) equal to $f(0)$ and $f(2)$.



          ETA: While I was writing this, two comments appeared suggesting the same approach.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 21:24









          Toby BartelsToby Bartels

          669515




          669515























              4












              $begingroup$

              Let $g(x) = f(x) - f(x-1)$



              $g(2) = f(2) - f(1) = f(0) - f(1) = -g(1)$



              if $g(1) = 0$ you are done.



              Otherwise $g(x)$ is a continuous function that is negative at one endpoint and positive at the other.



              By the IVT there must be an $ain [1,2]$ such that $g(a) = 0 $






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                "if g(x)=0 you are done." should be g(1)=0
                $endgroup$
                – Toby Bartels
                Dec 18 '18 at 21:29










              • $begingroup$
                @TobyBartels thanks.
                $endgroup$
                – Doug M
                Dec 18 '18 at 21:29
















              4












              $begingroup$

              Let $g(x) = f(x) - f(x-1)$



              $g(2) = f(2) - f(1) = f(0) - f(1) = -g(1)$



              if $g(1) = 0$ you are done.



              Otherwise $g(x)$ is a continuous function that is negative at one endpoint and positive at the other.



              By the IVT there must be an $ain [1,2]$ such that $g(a) = 0 $






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                "if g(x)=0 you are done." should be g(1)=0
                $endgroup$
                – Toby Bartels
                Dec 18 '18 at 21:29










              • $begingroup$
                @TobyBartels thanks.
                $endgroup$
                – Doug M
                Dec 18 '18 at 21:29














              4












              4








              4





              $begingroup$

              Let $g(x) = f(x) - f(x-1)$



              $g(2) = f(2) - f(1) = f(0) - f(1) = -g(1)$



              if $g(1) = 0$ you are done.



              Otherwise $g(x)$ is a continuous function that is negative at one endpoint and positive at the other.



              By the IVT there must be an $ain [1,2]$ such that $g(a) = 0 $






              share|cite|improve this answer











              $endgroup$



              Let $g(x) = f(x) - f(x-1)$



              $g(2) = f(2) - f(1) = f(0) - f(1) = -g(1)$



              if $g(1) = 0$ you are done.



              Otherwise $g(x)$ is a continuous function that is negative at one endpoint and positive at the other.



              By the IVT there must be an $ain [1,2]$ such that $g(a) = 0 $







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 18 '18 at 21:29

























              answered Dec 18 '18 at 21:26









              Doug MDoug M

              45.3k31954




              45.3k31954












              • $begingroup$
                "if g(x)=0 you are done." should be g(1)=0
                $endgroup$
                – Toby Bartels
                Dec 18 '18 at 21:29










              • $begingroup$
                @TobyBartels thanks.
                $endgroup$
                – Doug M
                Dec 18 '18 at 21:29


















              • $begingroup$
                "if g(x)=0 you are done." should be g(1)=0
                $endgroup$
                – Toby Bartels
                Dec 18 '18 at 21:29










              • $begingroup$
                @TobyBartels thanks.
                $endgroup$
                – Doug M
                Dec 18 '18 at 21:29
















              $begingroup$
              "if g(x)=0 you are done." should be g(1)=0
              $endgroup$
              – Toby Bartels
              Dec 18 '18 at 21:29




              $begingroup$
              "if g(x)=0 you are done." should be g(1)=0
              $endgroup$
              – Toby Bartels
              Dec 18 '18 at 21:29












              $begingroup$
              @TobyBartels thanks.
              $endgroup$
              – Doug M
              Dec 18 '18 at 21:29




              $begingroup$
              @TobyBartels thanks.
              $endgroup$
              – Doug M
              Dec 18 '18 at 21:29











              2












              $begingroup$

              Elementary Approach using Bolzano's Theorem and not IVT :



              Let $g(x) = f(x) - f(x-1)$. This function is continuous in $[0,2]$ as an operation of continuous functions.



              Now, $g(1) = f(1)- f(0)$ and $g(2) = f(2)-f(1)$.



              $$g(1)cdot g(2) =f(1)f(2)-f(1)^2 - f(2)f(0)+ f(0)f(1)=2f(0)f(1) - f(1)^2-f(0)^2 $$
              $$implies$$
              $$g(1)cdot g(2) = -big[f(1)-f(0)big]^2 leq 0 $$



              Thus, there exists $a in [0,2]$ such that $g(a) =0$ by Bolzano's Theorem.



              Functional Analysis (sorry for this) :



              Essentialy, we want to show that the equation



              $$f(x) - f(x-1) =0$$



              has a solution.



              Let $T$ be a linear operator $T:C[0,2] to C[0,2]$ such that $Tf(x) = -f(x-1)$. One can easily see that $T$ is a bounded linear operator $in Bbig(C[0,2]big)$. Also, one can take $0$ to be a function defined in $C[0,2]$. Then, the equation is re-written as :



              $$f + Tf = 0$$



              Note that $C[0,2]$ is a Banach Space and since $T in Bbig(C[0,2]big)$ then there exists a unique solution in $C[0,2]$, meaning that there exists such function $f(x) in C[0,2]$ which of course implies that the equation has a solution for $x in [0,2]$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Elementary Approach using Bolzano's Theorem and not IVT :



                Let $g(x) = f(x) - f(x-1)$. This function is continuous in $[0,2]$ as an operation of continuous functions.



                Now, $g(1) = f(1)- f(0)$ and $g(2) = f(2)-f(1)$.



                $$g(1)cdot g(2) =f(1)f(2)-f(1)^2 - f(2)f(0)+ f(0)f(1)=2f(0)f(1) - f(1)^2-f(0)^2 $$
                $$implies$$
                $$g(1)cdot g(2) = -big[f(1)-f(0)big]^2 leq 0 $$



                Thus, there exists $a in [0,2]$ such that $g(a) =0$ by Bolzano's Theorem.



                Functional Analysis (sorry for this) :



                Essentialy, we want to show that the equation



                $$f(x) - f(x-1) =0$$



                has a solution.



                Let $T$ be a linear operator $T:C[0,2] to C[0,2]$ such that $Tf(x) = -f(x-1)$. One can easily see that $T$ is a bounded linear operator $in Bbig(C[0,2]big)$. Also, one can take $0$ to be a function defined in $C[0,2]$. Then, the equation is re-written as :



                $$f + Tf = 0$$



                Note that $C[0,2]$ is a Banach Space and since $T in Bbig(C[0,2]big)$ then there exists a unique solution in $C[0,2]$, meaning that there exists such function $f(x) in C[0,2]$ which of course implies that the equation has a solution for $x in [0,2]$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Elementary Approach using Bolzano's Theorem and not IVT :



                  Let $g(x) = f(x) - f(x-1)$. This function is continuous in $[0,2]$ as an operation of continuous functions.



                  Now, $g(1) = f(1)- f(0)$ and $g(2) = f(2)-f(1)$.



                  $$g(1)cdot g(2) =f(1)f(2)-f(1)^2 - f(2)f(0)+ f(0)f(1)=2f(0)f(1) - f(1)^2-f(0)^2 $$
                  $$implies$$
                  $$g(1)cdot g(2) = -big[f(1)-f(0)big]^2 leq 0 $$



                  Thus, there exists $a in [0,2]$ such that $g(a) =0$ by Bolzano's Theorem.



                  Functional Analysis (sorry for this) :



                  Essentialy, we want to show that the equation



                  $$f(x) - f(x-1) =0$$



                  has a solution.



                  Let $T$ be a linear operator $T:C[0,2] to C[0,2]$ such that $Tf(x) = -f(x-1)$. One can easily see that $T$ is a bounded linear operator $in Bbig(C[0,2]big)$. Also, one can take $0$ to be a function defined in $C[0,2]$. Then, the equation is re-written as :



                  $$f + Tf = 0$$



                  Note that $C[0,2]$ is a Banach Space and since $T in Bbig(C[0,2]big)$ then there exists a unique solution in $C[0,2]$, meaning that there exists such function $f(x) in C[0,2]$ which of course implies that the equation has a solution for $x in [0,2]$.






                  share|cite|improve this answer











                  $endgroup$



                  Elementary Approach using Bolzano's Theorem and not IVT :



                  Let $g(x) = f(x) - f(x-1)$. This function is continuous in $[0,2]$ as an operation of continuous functions.



                  Now, $g(1) = f(1)- f(0)$ and $g(2) = f(2)-f(1)$.



                  $$g(1)cdot g(2) =f(1)f(2)-f(1)^2 - f(2)f(0)+ f(0)f(1)=2f(0)f(1) - f(1)^2-f(0)^2 $$
                  $$implies$$
                  $$g(1)cdot g(2) = -big[f(1)-f(0)big]^2 leq 0 $$



                  Thus, there exists $a in [0,2]$ such that $g(a) =0$ by Bolzano's Theorem.



                  Functional Analysis (sorry for this) :



                  Essentialy, we want to show that the equation



                  $$f(x) - f(x-1) =0$$



                  has a solution.



                  Let $T$ be a linear operator $T:C[0,2] to C[0,2]$ such that $Tf(x) = -f(x-1)$. One can easily see that $T$ is a bounded linear operator $in Bbig(C[0,2]big)$. Also, one can take $0$ to be a function defined in $C[0,2]$. Then, the equation is re-written as :



                  $$f + Tf = 0$$



                  Note that $C[0,2]$ is a Banach Space and since $T in Bbig(C[0,2]big)$ then there exists a unique solution in $C[0,2]$, meaning that there exists such function $f(x) in C[0,2]$ which of course implies that the equation has a solution for $x in [0,2]$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 18 '18 at 21:40

























                  answered Dec 18 '18 at 21:34









                  RebellosRebellos

                  15k31249




                  15k31249






























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