Question on Infinitely Differentiable Functions [duplicate]












3















This question already has an answer here:




  • Let $f:(-1,1) rightarrow mathbb{R}$ be smooth on $(-1,1)$, $f(0) = 1$. Find $f(x)$.

    1 answer




Let $f:(-1,1)rightarrowmathbb R$ be infinitely differentiable on $(-1,1)$ with $f(0)=1$ and has the following properties:



(1) $vert f^{(n)}(x)vert le n!$ for every $xin (-1, 1)$ and for every $nin mathbb N$



(2) $f'(frac{1}{m+1})=0$ for every $min mathbb N$



I need to:



Find the value of $f^{(n)}(0)$ for each $nin mathbb N$



Determine the value of $f(x)$ for every $xin(-1,1)$



I claim that $f^{(n)}(0)=0$ for all $nin mathbb N$ and I wanted to prove this by induction. When $n=1$, we note that by (2), $f'(0)=0$ by the sequential criterion for limits with $mrightarrow infty$. However, I am not sure on how to prove the inductive step.



I note that from (1), we have $f(x)lefrac{1}{1-x}$ by Taylor's Theorem.



Any help is greatly appreciated!










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marked as duplicate by Martin R, Servaes, Community Nov 23 '18 at 11:38


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.




















    3















    This question already has an answer here:




    • Let $f:(-1,1) rightarrow mathbb{R}$ be smooth on $(-1,1)$, $f(0) = 1$. Find $f(x)$.

      1 answer




    Let $f:(-1,1)rightarrowmathbb R$ be infinitely differentiable on $(-1,1)$ with $f(0)=1$ and has the following properties:



    (1) $vert f^{(n)}(x)vert le n!$ for every $xin (-1, 1)$ and for every $nin mathbb N$



    (2) $f'(frac{1}{m+1})=0$ for every $min mathbb N$



    I need to:



    Find the value of $f^{(n)}(0)$ for each $nin mathbb N$



    Determine the value of $f(x)$ for every $xin(-1,1)$



    I claim that $f^{(n)}(0)=0$ for all $nin mathbb N$ and I wanted to prove this by induction. When $n=1$, we note that by (2), $f'(0)=0$ by the sequential criterion for limits with $mrightarrow infty$. However, I am not sure on how to prove the inductive step.



    I note that from (1), we have $f(x)lefrac{1}{1-x}$ by Taylor's Theorem.



    Any help is greatly appreciated!










    share|cite|improve this question













    marked as duplicate by Martin R, Servaes, Community Nov 23 '18 at 11:38


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















      3












      3








      3


      1






      This question already has an answer here:




      • Let $f:(-1,1) rightarrow mathbb{R}$ be smooth on $(-1,1)$, $f(0) = 1$. Find $f(x)$.

        1 answer




      Let $f:(-1,1)rightarrowmathbb R$ be infinitely differentiable on $(-1,1)$ with $f(0)=1$ and has the following properties:



      (1) $vert f^{(n)}(x)vert le n!$ for every $xin (-1, 1)$ and for every $nin mathbb N$



      (2) $f'(frac{1}{m+1})=0$ for every $min mathbb N$



      I need to:



      Find the value of $f^{(n)}(0)$ for each $nin mathbb N$



      Determine the value of $f(x)$ for every $xin(-1,1)$



      I claim that $f^{(n)}(0)=0$ for all $nin mathbb N$ and I wanted to prove this by induction. When $n=1$, we note that by (2), $f'(0)=0$ by the sequential criterion for limits with $mrightarrow infty$. However, I am not sure on how to prove the inductive step.



      I note that from (1), we have $f(x)lefrac{1}{1-x}$ by Taylor's Theorem.



      Any help is greatly appreciated!










      share|cite|improve this question














      This question already has an answer here:




      • Let $f:(-1,1) rightarrow mathbb{R}$ be smooth on $(-1,1)$, $f(0) = 1$. Find $f(x)$.

        1 answer




      Let $f:(-1,1)rightarrowmathbb R$ be infinitely differentiable on $(-1,1)$ with $f(0)=1$ and has the following properties:



      (1) $vert f^{(n)}(x)vert le n!$ for every $xin (-1, 1)$ and for every $nin mathbb N$



      (2) $f'(frac{1}{m+1})=0$ for every $min mathbb N$



      I need to:



      Find the value of $f^{(n)}(0)$ for each $nin mathbb N$



      Determine the value of $f(x)$ for every $xin(-1,1)$



      I claim that $f^{(n)}(0)=0$ for all $nin mathbb N$ and I wanted to prove this by induction. When $n=1$, we note that by (2), $f'(0)=0$ by the sequential criterion for limits with $mrightarrow infty$. However, I am not sure on how to prove the inductive step.



      I note that from (1), we have $f(x)lefrac{1}{1-x}$ by Taylor's Theorem.



      Any help is greatly appreciated!





      This question already has an answer here:




      • Let $f:(-1,1) rightarrow mathbb{R}$ be smooth on $(-1,1)$, $f(0) = 1$. Find $f(x)$.

        1 answer








      calculus taylor-expansion






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      asked Nov 23 '18 at 9:08









      Derp

      228214




      228214




      marked as duplicate by Martin R, Servaes, Community Nov 23 '18 at 11:38


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Martin R, Servaes, Community Nov 23 '18 at 11:38


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          2 Answers
          2






          active

          oldest

          votes


















          4














          Rolle's theorem implies that $f''$ has a zero in the interval $(frac{1}{m+2}, frac{1}{m+1})$
          for each $m in Bbb N$, therefore $f''(0) = 0$.



          The same argument can be repeated to show that each derivative $f^{(k)}$ has
          a sequence of zeros converging to $0$.



          (The estimate $vert f^{(n)}(x)vert le n!$ is not needed for this
          conclusion.)






          share|cite|improve this answer























          • Nice answer (+1). I realized that also by my approach the estimate of $n$-th derivative is not needed
            – Robert Z
            Nov 23 '18 at 9:41












          • @RobertZ: Thanks. – As it turns out, the question has been asked and answered before, unfortunately I did not notice that before posting the answer.
            – Martin R
            Nov 23 '18 at 9:57










          • As far as I can see no answers there pointed out that the estimate on $f^{(n)}$ is superfluous.
            – Robert Z
            Nov 23 '18 at 12:51





















          2














          Hint. Let us consider the case $n=2$ (try by your own the case $n>2$)



          For any $minmathbb{N}$, by using the Taylor expansion of $f'$ at $0$, we have that there exists $t_min (-1,1)$ such that
          $$f'left(frac{1}{m+1}right)=f'(0)+f''(0)cdot frac{1}{m+1}+frac{f'''(t_m)}{2}cdot frac{1}{(m+1)^2}.$$
          Since $f'left(frac{1}{m+1}right)=0$ and we already know that $f'(0)=0$, it follows
          $$f''(0)=-frac{f'''(t_m)}{2}cdot frac{1}{(m+1)}$$
          which implies that as $mto+infty$,
          $$|f''(0)|=frac{|f'''(t_m)|}{2}cdot frac{1} {(m+1)}leqfrac{M_3} {2(m+1)}to 0.$$
          where $M_3=max_{xin[-1/2,1/2]}| f^{(3)}(x)|$ (the assumption $vert f^{(n)}(x)vert le n!$ in $(-1,1)$ is not needed).






          share|cite|improve this answer






























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            Rolle's theorem implies that $f''$ has a zero in the interval $(frac{1}{m+2}, frac{1}{m+1})$
            for each $m in Bbb N$, therefore $f''(0) = 0$.



            The same argument can be repeated to show that each derivative $f^{(k)}$ has
            a sequence of zeros converging to $0$.



            (The estimate $vert f^{(n)}(x)vert le n!$ is not needed for this
            conclusion.)






            share|cite|improve this answer























            • Nice answer (+1). I realized that also by my approach the estimate of $n$-th derivative is not needed
              – Robert Z
              Nov 23 '18 at 9:41












            • @RobertZ: Thanks. – As it turns out, the question has been asked and answered before, unfortunately I did not notice that before posting the answer.
              – Martin R
              Nov 23 '18 at 9:57










            • As far as I can see no answers there pointed out that the estimate on $f^{(n)}$ is superfluous.
              – Robert Z
              Nov 23 '18 at 12:51


















            4














            Rolle's theorem implies that $f''$ has a zero in the interval $(frac{1}{m+2}, frac{1}{m+1})$
            for each $m in Bbb N$, therefore $f''(0) = 0$.



            The same argument can be repeated to show that each derivative $f^{(k)}$ has
            a sequence of zeros converging to $0$.



            (The estimate $vert f^{(n)}(x)vert le n!$ is not needed for this
            conclusion.)






            share|cite|improve this answer























            • Nice answer (+1). I realized that also by my approach the estimate of $n$-th derivative is not needed
              – Robert Z
              Nov 23 '18 at 9:41












            • @RobertZ: Thanks. – As it turns out, the question has been asked and answered before, unfortunately I did not notice that before posting the answer.
              – Martin R
              Nov 23 '18 at 9:57










            • As far as I can see no answers there pointed out that the estimate on $f^{(n)}$ is superfluous.
              – Robert Z
              Nov 23 '18 at 12:51
















            4












            4








            4






            Rolle's theorem implies that $f''$ has a zero in the interval $(frac{1}{m+2}, frac{1}{m+1})$
            for each $m in Bbb N$, therefore $f''(0) = 0$.



            The same argument can be repeated to show that each derivative $f^{(k)}$ has
            a sequence of zeros converging to $0$.



            (The estimate $vert f^{(n)}(x)vert le n!$ is not needed for this
            conclusion.)






            share|cite|improve this answer














            Rolle's theorem implies that $f''$ has a zero in the interval $(frac{1}{m+2}, frac{1}{m+1})$
            for each $m in Bbb N$, therefore $f''(0) = 0$.



            The same argument can be repeated to show that each derivative $f^{(k)}$ has
            a sequence of zeros converging to $0$.



            (The estimate $vert f^{(n)}(x)vert le n!$ is not needed for this
            conclusion.)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 23 '18 at 9:32

























            answered Nov 23 '18 at 9:21









            Martin R

            27k33252




            27k33252












            • Nice answer (+1). I realized that also by my approach the estimate of $n$-th derivative is not needed
              – Robert Z
              Nov 23 '18 at 9:41












            • @RobertZ: Thanks. – As it turns out, the question has been asked and answered before, unfortunately I did not notice that before posting the answer.
              – Martin R
              Nov 23 '18 at 9:57










            • As far as I can see no answers there pointed out that the estimate on $f^{(n)}$ is superfluous.
              – Robert Z
              Nov 23 '18 at 12:51




















            • Nice answer (+1). I realized that also by my approach the estimate of $n$-th derivative is not needed
              – Robert Z
              Nov 23 '18 at 9:41












            • @RobertZ: Thanks. – As it turns out, the question has been asked and answered before, unfortunately I did not notice that before posting the answer.
              – Martin R
              Nov 23 '18 at 9:57










            • As far as I can see no answers there pointed out that the estimate on $f^{(n)}$ is superfluous.
              – Robert Z
              Nov 23 '18 at 12:51


















            Nice answer (+1). I realized that also by my approach the estimate of $n$-th derivative is not needed
            – Robert Z
            Nov 23 '18 at 9:41






            Nice answer (+1). I realized that also by my approach the estimate of $n$-th derivative is not needed
            – Robert Z
            Nov 23 '18 at 9:41














            @RobertZ: Thanks. – As it turns out, the question has been asked and answered before, unfortunately I did not notice that before posting the answer.
            – Martin R
            Nov 23 '18 at 9:57




            @RobertZ: Thanks. – As it turns out, the question has been asked and answered before, unfortunately I did not notice that before posting the answer.
            – Martin R
            Nov 23 '18 at 9:57












            As far as I can see no answers there pointed out that the estimate on $f^{(n)}$ is superfluous.
            – Robert Z
            Nov 23 '18 at 12:51






            As far as I can see no answers there pointed out that the estimate on $f^{(n)}$ is superfluous.
            – Robert Z
            Nov 23 '18 at 12:51













            2














            Hint. Let us consider the case $n=2$ (try by your own the case $n>2$)



            For any $minmathbb{N}$, by using the Taylor expansion of $f'$ at $0$, we have that there exists $t_min (-1,1)$ such that
            $$f'left(frac{1}{m+1}right)=f'(0)+f''(0)cdot frac{1}{m+1}+frac{f'''(t_m)}{2}cdot frac{1}{(m+1)^2}.$$
            Since $f'left(frac{1}{m+1}right)=0$ and we already know that $f'(0)=0$, it follows
            $$f''(0)=-frac{f'''(t_m)}{2}cdot frac{1}{(m+1)}$$
            which implies that as $mto+infty$,
            $$|f''(0)|=frac{|f'''(t_m)|}{2}cdot frac{1} {(m+1)}leqfrac{M_3} {2(m+1)}to 0.$$
            where $M_3=max_{xin[-1/2,1/2]}| f^{(3)}(x)|$ (the assumption $vert f^{(n)}(x)vert le n!$ in $(-1,1)$ is not needed).






            share|cite|improve this answer




























              2














              Hint. Let us consider the case $n=2$ (try by your own the case $n>2$)



              For any $minmathbb{N}$, by using the Taylor expansion of $f'$ at $0$, we have that there exists $t_min (-1,1)$ such that
              $$f'left(frac{1}{m+1}right)=f'(0)+f''(0)cdot frac{1}{m+1}+frac{f'''(t_m)}{2}cdot frac{1}{(m+1)^2}.$$
              Since $f'left(frac{1}{m+1}right)=0$ and we already know that $f'(0)=0$, it follows
              $$f''(0)=-frac{f'''(t_m)}{2}cdot frac{1}{(m+1)}$$
              which implies that as $mto+infty$,
              $$|f''(0)|=frac{|f'''(t_m)|}{2}cdot frac{1} {(m+1)}leqfrac{M_3} {2(m+1)}to 0.$$
              where $M_3=max_{xin[-1/2,1/2]}| f^{(3)}(x)|$ (the assumption $vert f^{(n)}(x)vert le n!$ in $(-1,1)$ is not needed).






              share|cite|improve this answer


























                2












                2








                2






                Hint. Let us consider the case $n=2$ (try by your own the case $n>2$)



                For any $minmathbb{N}$, by using the Taylor expansion of $f'$ at $0$, we have that there exists $t_min (-1,1)$ such that
                $$f'left(frac{1}{m+1}right)=f'(0)+f''(0)cdot frac{1}{m+1}+frac{f'''(t_m)}{2}cdot frac{1}{(m+1)^2}.$$
                Since $f'left(frac{1}{m+1}right)=0$ and we already know that $f'(0)=0$, it follows
                $$f''(0)=-frac{f'''(t_m)}{2}cdot frac{1}{(m+1)}$$
                which implies that as $mto+infty$,
                $$|f''(0)|=frac{|f'''(t_m)|}{2}cdot frac{1} {(m+1)}leqfrac{M_3} {2(m+1)}to 0.$$
                where $M_3=max_{xin[-1/2,1/2]}| f^{(3)}(x)|$ (the assumption $vert f^{(n)}(x)vert le n!$ in $(-1,1)$ is not needed).






                share|cite|improve this answer














                Hint. Let us consider the case $n=2$ (try by your own the case $n>2$)



                For any $minmathbb{N}$, by using the Taylor expansion of $f'$ at $0$, we have that there exists $t_min (-1,1)$ such that
                $$f'left(frac{1}{m+1}right)=f'(0)+f''(0)cdot frac{1}{m+1}+frac{f'''(t_m)}{2}cdot frac{1}{(m+1)^2}.$$
                Since $f'left(frac{1}{m+1}right)=0$ and we already know that $f'(0)=0$, it follows
                $$f''(0)=-frac{f'''(t_m)}{2}cdot frac{1}{(m+1)}$$
                which implies that as $mto+infty$,
                $$|f''(0)|=frac{|f'''(t_m)|}{2}cdot frac{1} {(m+1)}leqfrac{M_3} {2(m+1)}to 0.$$
                where $M_3=max_{xin[-1/2,1/2]}| f^{(3)}(x)|$ (the assumption $vert f^{(n)}(x)vert le n!$ in $(-1,1)$ is not needed).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 23 '18 at 9:39

























                answered Nov 23 '18 at 9:18









                Robert Z

                93.3k1061132




                93.3k1061132















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