Prove that $f_n$ with $f_n :[0,a]rightarrowmathbb{R},ain mathbb{R}, f_n(x)=x^n$ is continous $forall n in...












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I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?



Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $






real-analysis continuity induction







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    1












    $begingroup$


    I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?



    Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $






    real-analysis continuity induction







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?



      Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $






      real-analysis continuity induction







      share|cite|improve this question











      $endgroup$




      I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?



      Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $






      real-analysis continuity induction




      real-analysis continuity induction






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      edited Jan 8 at 21:31









      José Carlos Santos

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      158k22126228










      asked Jan 8 at 20:59









      RM777RM777

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          $begingroup$

          Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.






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            $begingroup$

            Using the binomial theorem yields:



            $$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$



            because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.






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              $begingroup$

              For any $x,yin [0,a]$ we have
              begin{eqnarray}
              |x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
              &le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
              &le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
              &=&na^{n-1}|x-y|
              end{eqnarray}



              Hence $f_n$ is continuous.






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                3 Answers
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                $begingroup$

                Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.






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                  4












                  $begingroup$

                  Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.






                  share|cite|improve this answer









                  $endgroup$
















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                    4





                    $begingroup$

                    Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.






                    share|cite|improve this answer









                    $endgroup$



                    Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 8 at 21:04









                    José Carlos SantosJosé Carlos Santos

                    158k22126228




                    158k22126228























                        0












                        $begingroup$

                        Using the binomial theorem yields:



                        $$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$



                        because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Using the binomial theorem yields:



                          $$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$



                          because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Using the binomial theorem yields:



                            $$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$



                            because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.






                            share|cite|improve this answer









                            $endgroup$



                            Using the binomial theorem yields:



                            $$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$



                            because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 8 at 21:14









                            mechanodroidmechanodroid

                            27.3k62446




                            27.3k62446























                                0












                                $begingroup$

                                For any $x,yin [0,a]$ we have
                                begin{eqnarray}
                                |x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                &le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                &le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
                                &=&na^{n-1}|x-y|
                                end{eqnarray}



                                Hence $f_n$ is continuous.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  For any $x,yin [0,a]$ we have
                                  begin{eqnarray}
                                  |x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                  &le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                  &le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
                                  &=&na^{n-1}|x-y|
                                  end{eqnarray}



                                  Hence $f_n$ is continuous.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    For any $x,yin [0,a]$ we have
                                    begin{eqnarray}
                                    |x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                    &le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                    &le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
                                    &=&na^{n-1}|x-y|
                                    end{eqnarray}



                                    Hence $f_n$ is continuous.






                                    share|cite|improve this answer









                                    $endgroup$



                                    For any $x,yin [0,a]$ we have
                                    begin{eqnarray}
                                    |x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                    &le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
                                    &le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
                                    &=&na^{n-1}|x-y|
                                    end{eqnarray}



                                    Hence $f_n$ is continuous.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 8 at 21:31









                                    Mercy KingMercy King

                                    14k11428




                                    14k11428






























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