Proving n+1 th differential as zero given lower differentials are 0If $f(x)=f'(x)+f''(x)$ then show that $f(x)=0$Derivative changes sign for continuous and differentiable functionA problem on Mean Value TheoremCalculus Three time differentiableTrue or false: If $f(x)$ is continuous on $[0, 2]$ and $f(0)=f(2)$, then there exists a number $cin [0, 1]$ such that $f(c) = f(c + 1)$.Rolle's theorem $beta cdot f(x)+f'(x)=0$Let $f(x)=4x^3-3x^2-2x+1,$ use Rolle's theorem to prove that there exist $c,0<c<1$ such that $f(c)=0$If $f(a)=f(b)=0$, then $f'(c)+f(c)g'(c)=0,$ for some $cin(a,b)$Let f be a continuous and differentiable function such that f(a)=f(b)=0Justify differentiability for a parametric function

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Proving n+1 th differential as zero given lower differentials are 0


If $f(x)=f'(x)+f''(x)$ then show that $f(x)=0$Derivative changes sign for continuous and differentiable functionA problem on Mean Value TheoremCalculus Three time differentiableTrue or false: If $f(x)$ is continuous on $[0, 2]$ and $f(0)=f(2)$, then there exists a number $cin [0, 1]$ such that $f(c) = f(c + 1)$.Rolle's theorem $beta cdot f(x)+f'(x)=0$Let $f(x)=4x^3-3x^2-2x+1,$ use Rolle's theorem to prove that there exist $c,0<c<1$ such that $f(c)=0$If $f(a)=f(b)=0$, then $f'(c)+f(c)g'(c)=0,$ for some $cin(a,b)$Let f be a continuous and differentiable function such that f(a)=f(b)=0Justify differentiability for a parametric function













4












$begingroup$


Following is a question I am stuck in.




Let $f : Bbb R to Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
$$f(1) = f(0) = f^(1)(0) = f^(2)(0) = · · · = f^(n)(0) = 0$$
Prove that there exists $x in (0, 1)$ such that $f^(n+1)(x) = 0$.




It is a past question of an entrance exam.



I thought to use Rolle's Theorem, but this requires information about $f^(n) (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$










share|cite|improve this question









New contributor




A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
















    4












    $begingroup$


    Following is a question I am stuck in.




    Let $f : Bbb R to Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
    $$f(1) = f(0) = f^(1)(0) = f^(2)(0) = · · · = f^(n)(0) = 0$$
    Prove that there exists $x in (0, 1)$ such that $f^(n+1)(x) = 0$.




    It is a past question of an entrance exam.



    I thought to use Rolle's Theorem, but this requires information about $f^(n) (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$










    share|cite|improve this question









    New contributor




    A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      4












      4








      4


      3



      $begingroup$


      Following is a question I am stuck in.




      Let $f : Bbb R to Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
      $$f(1) = f(0) = f^(1)(0) = f^(2)(0) = · · · = f^(n)(0) = 0$$
      Prove that there exists $x in (0, 1)$ such that $f^(n+1)(x) = 0$.




      It is a past question of an entrance exam.



      I thought to use Rolle's Theorem, but this requires information about $f^(n) (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$










      share|cite|improve this question









      New contributor




      A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Following is a question I am stuck in.




      Let $f : Bbb R to Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
      $$f(1) = f(0) = f^(1)(0) = f^(2)(0) = · · · = f^(n)(0) = 0$$
      Prove that there exists $x in (0, 1)$ such that $f^(n+1)(x) = 0$.




      It is a past question of an entrance exam.



      I thought to use Rolle's Theorem, but this requires information about $f^(n) (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$







      real-analysis calculus






      share|cite|improve this question









      New contributor




      A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question




      share|cite|improve this question








      edited 34 mins ago









      Ivo Terek

      47.2k954147




      47.2k954147






      New contributor




      A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 46 mins ago









      A.S. GhoshA.S. Ghosh

      232




      232




      New contributor




      A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read




          If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.




          How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
            $endgroup$
            – A.S. Ghosh
            30 mins ago


















          2












          $begingroup$

          Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            Love the answers. Another method.



            Taylor expansion:
            $$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$



            $c$ is in the open interval between $0$ and $x$.






            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









              3












              $begingroup$

              Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read




              If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.




              How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?






              share|cite|improve this answer









              $endgroup$








              • 1




                $begingroup$
                Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
                $endgroup$
                – A.S. Ghosh
                30 mins ago















              3












              $begingroup$

              Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read




              If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.




              How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?






              share|cite|improve this answer









              $endgroup$








              • 1




                $begingroup$
                Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
                $endgroup$
                – A.S. Ghosh
                30 mins ago













              3












              3








              3





              $begingroup$

              Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read




              If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.




              How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?






              share|cite|improve this answer









              $endgroup$



              Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read




              If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.




              How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 36 mins ago









              Ivo TerekIvo Terek

              47.2k954147




              47.2k954147







              • 1




                $begingroup$
                Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
                $endgroup$
                – A.S. Ghosh
                30 mins ago












              • 1




                $begingroup$
                Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
                $endgroup$
                – A.S. Ghosh
                30 mins ago







              1




              1




              $begingroup$
              Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
              $endgroup$
              – A.S. Ghosh
              30 mins ago




              $begingroup$
              Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
              $endgroup$
              – A.S. Ghosh
              30 mins ago











              2












              $begingroup$

              Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.






                  share|cite|improve this answer









                  $endgroup$



                  Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 35 mins ago









                  saulspatzsaulspatz

                  18.1k41636




                  18.1k41636





















                      2












                      $begingroup$

                      Love the answers. Another method.



                      Taylor expansion:
                      $$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$



                      $c$ is in the open interval between $0$ and $x$.






                      share|cite|improve this answer









                      $endgroup$

















                        2












                        $begingroup$

                        Love the answers. Another method.



                        Taylor expansion:
                        $$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$



                        $c$ is in the open interval between $0$ and $x$.






                        share|cite|improve this answer









                        $endgroup$















                          2












                          2








                          2





                          $begingroup$

                          Love the answers. Another method.



                          Taylor expansion:
                          $$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$



                          $c$ is in the open interval between $0$ and $x$.






                          share|cite|improve this answer









                          $endgroup$



                          Love the answers. Another method.



                          Taylor expansion:
                          $$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$



                          $c$ is in the open interval between $0$ and $x$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 27 mins ago









                          FnacoolFnacool

                          5,496612




                          5,496612




















                              A.S. Ghosh is a new contributor. Be nice, and check out our Code of Conduct.









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