function not having same value at every point [closed]












2












$begingroup$


Let $f : (−1,1) toBbb R$ be a twice differentiable function such that $f''(0) > 0$. Show that there exists $n in Bbb N$ such that $f(frac 1n)$ is not equals $1$.



I know that $f$ has a minima at $0$ and $f'(0)$ is $0$ but how to proceed from here?






real-analysis







share|cite|improve this question











$endgroup$



closed as off-topic by RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin Dec 18 '18 at 18:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    Hint: Consider the sequence $(1/n)$ and $(f(1/n))$ as $ntoinfty$.
    $endgroup$
    – YiFan
    Dec 18 '18 at 4:42






  • 1




    $begingroup$
    Yes but why is f (0) not 1 I don't underztand
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 4:45






  • 1




    $begingroup$
    Why does $f$ have a minima at 0? (Presumably you mean minimiser?)
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:13








  • 1




    $begingroup$
    f"(0) is positive so f has a local minima at 0
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 5:18






  • 1




    $begingroup$
    @DebarghyaMukherjee: $x mapsto (x-1)^2$ has a positive second derivative everywhere but does not have a local minimum at $x=0$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:23
















2












$begingroup$


Let $f : (−1,1) toBbb R$ be a twice differentiable function such that $f''(0) > 0$. Show that there exists $n in Bbb N$ such that $f(frac 1n)$ is not equals $1$.



I know that $f$ has a minima at $0$ and $f'(0)$ is $0$ but how to proceed from here?






real-analysis







share|cite|improve this question











$endgroup$



closed as off-topic by RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin Dec 18 '18 at 18:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    Hint: Consider the sequence $(1/n)$ and $(f(1/n))$ as $ntoinfty$.
    $endgroup$
    – YiFan
    Dec 18 '18 at 4:42






  • 1




    $begingroup$
    Yes but why is f (0) not 1 I don't underztand
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 4:45






  • 1




    $begingroup$
    Why does $f$ have a minima at 0? (Presumably you mean minimiser?)
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:13








  • 1




    $begingroup$
    f"(0) is positive so f has a local minima at 0
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 5:18






  • 1




    $begingroup$
    @DebarghyaMukherjee: $x mapsto (x-1)^2$ has a positive second derivative everywhere but does not have a local minimum at $x=0$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:23














2












2








2


1



$begingroup$


Let $f : (−1,1) toBbb R$ be a twice differentiable function such that $f''(0) > 0$. Show that there exists $n in Bbb N$ such that $f(frac 1n)$ is not equals $1$.



I know that $f$ has a minima at $0$ and $f'(0)$ is $0$ but how to proceed from here?






real-analysis







share|cite|improve this question











$endgroup$




Let $f : (−1,1) toBbb R$ be a twice differentiable function such that $f''(0) > 0$. Show that there exists $n in Bbb N$ such that $f(frac 1n)$ is not equals $1$.



I know that $f$ has a minima at $0$ and $f'(0)$ is $0$ but how to proceed from here?






real-analysis




real-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 5:11









BigbearZzz

8,51721652




8,51721652










asked Dec 18 '18 at 4:36









Debarghya MukherjeeDebarghya Mukherjee

352




352




closed as off-topic by RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin Dec 18 '18 at 18:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin Dec 18 '18 at 18:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Brahadeesh, user10354138, José Carlos Santos, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Hint: Consider the sequence $(1/n)$ and $(f(1/n))$ as $ntoinfty$.
    $endgroup$
    – YiFan
    Dec 18 '18 at 4:42






  • 1




    $begingroup$
    Yes but why is f (0) not 1 I don't underztand
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 4:45






  • 1




    $begingroup$
    Why does $f$ have a minima at 0? (Presumably you mean minimiser?)
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:13








  • 1




    $begingroup$
    f"(0) is positive so f has a local minima at 0
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 5:18






  • 1




    $begingroup$
    @DebarghyaMukherjee: $x mapsto (x-1)^2$ has a positive second derivative everywhere but does not have a local minimum at $x=0$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:23














  • 1




    $begingroup$
    Hint: Consider the sequence $(1/n)$ and $(f(1/n))$ as $ntoinfty$.
    $endgroup$
    – YiFan
    Dec 18 '18 at 4:42






  • 1




    $begingroup$
    Yes but why is f (0) not 1 I don't underztand
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 4:45






  • 1




    $begingroup$
    Why does $f$ have a minima at 0? (Presumably you mean minimiser?)
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:13








  • 1




    $begingroup$
    f"(0) is positive so f has a local minima at 0
    $endgroup$
    – Debarghya Mukherjee
    Dec 18 '18 at 5:18






  • 1




    $begingroup$
    @DebarghyaMukherjee: $x mapsto (x-1)^2$ has a positive second derivative everywhere but does not have a local minimum at $x=0$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 5:23








1




1




$begingroup$
Hint: Consider the sequence $(1/n)$ and $(f(1/n))$ as $ntoinfty$.
$endgroup$
– YiFan
Dec 18 '18 at 4:42




$begingroup$
Hint: Consider the sequence $(1/n)$ and $(f(1/n))$ as $ntoinfty$.
$endgroup$
– YiFan
Dec 18 '18 at 4:42




1




1




$begingroup$
Yes but why is f (0) not 1 I don't underztand
$endgroup$
– Debarghya Mukherjee
Dec 18 '18 at 4:45




$begingroup$
Yes but why is f (0) not 1 I don't underztand
$endgroup$
– Debarghya Mukherjee
Dec 18 '18 at 4:45




1




1




$begingroup$
Why does $f$ have a minima at 0? (Presumably you mean minimiser?)
$endgroup$
– copper.hat
Dec 18 '18 at 5:13






$begingroup$
Why does $f$ have a minima at 0? (Presumably you mean minimiser?)
$endgroup$
– copper.hat
Dec 18 '18 at 5:13






1




1




$begingroup$
f"(0) is positive so f has a local minima at 0
$endgroup$
– Debarghya Mukherjee
Dec 18 '18 at 5:18




$begingroup$
f"(0) is positive so f has a local minima at 0
$endgroup$
– Debarghya Mukherjee
Dec 18 '18 at 5:18




1




1




$begingroup$
@DebarghyaMukherjee: $x mapsto (x-1)^2$ has a positive second derivative everywhere but does not have a local minimum at $x=0$.
$endgroup$
– copper.hat
Dec 18 '18 at 5:23




$begingroup$
@DebarghyaMukherjee: $x mapsto (x-1)^2$ has a positive second derivative everywhere but does not have a local minimum at $x=0$.
$endgroup$
– copper.hat
Dec 18 '18 at 5:23










2 Answers
2






active

oldest

votes


















1












$begingroup$

By definition of derivative you get $f'(0)=0$. Since $f''(0) >0$ there exits $delta >0$ such that $f'(x)>f'(0)=0$ for $0<x<delta$. This makes $f$ strictly increasing in $(0,delta)$ making $f(frac 1 {n+1}) <f(frac 1 n)$ or $1<1$!






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Sir how can $f'(0)=0$. I guess given $f''(0)>0impliesdisplaystyle lim_{hto0}dfrac{f'(0+h)-f'(0)}{h}>0implies f'(0+h)>f'(0)$. Am I missing something here?
    $endgroup$
    – Yadati Kiran
    Dec 18 '18 at 6:00








  • 1




    $begingroup$
    $f'(0)=lim_{nto infty} frac {f(frac 1 n) -f(0)} {frac 1 n-0}$ By continuity $f(0)=1$ so numerator is $0$ for each $n$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 6:01












  • $begingroup$
    I get it now Sir. Thank you
    $endgroup$
    – Yadati Kiran
    Dec 18 '18 at 6:13



















1












$begingroup$

Assume that



$forall n in Bbb N, ; f left ( dfrac{1}{n} right ) = 1; tag 1$



then by continuity,



$f(0) = displaystyle lim_{n to infty} f left ( dfrac{1}{n} right ) = 1; tag 2$



also,



$f'(0) = displaystyle lim_{n to infty} dfrac{f(1/n) - 1}{1/n} = 0; tag 3$



since



$f''(0) > 0, tag 4$



we find that $f(x)$ has a local minimum of value $f(0) = 1$ at $0$; now I claim that there is a fixed $epsilon > 0$ such that $f'(alpha) > 0$ for every $alpha in (0, epsilon)$; otherwise, we could find sequences $epsilon_n to 0$ and $alpha_n in (0, epsilon_n)$ with $f'(alpha_n) le 0$, and then



$f''(0) = displaystyle lim_{n to infty} dfrac{f'(alpha_n) - f'(0)}{alpha_n} = lim_{alpha_n to infty} dfrac{f'(alpha_n)}{alpha_n} le 0 Rightarrow Leftarrow f''(0) > 0; tag 5$



now since $f'(x)$ is differentiable, it is also continuous and thus we may apply the mean value theorem to any $beta in (0, epsilon)$ and obtain



$f(beta) - 1 = f(beta) - f(0) = f'(gamma) (beta - 0) = f'(gamma) beta, ; gamma in (0, beta); tag 6$



since $f'(gamma) > 0$ this yields



$f(beta) = 1 + f'(gamma) beta; tag 7$



we see that every $beta in (0, epsilon)$ satisfies



$f(beta) > 1, tag 8$



but this contradicts our assumption (1), and thus



$exists n in Bbb N, ; f left ( dfrac{1}{n} right ) ne 1. tag 9$



Note that we have in fact proved that $n$ in (9) may be taken arbitrarily large, and thus there are an infinite number of such $n$, which also follows from (7)-(8). In fact, we have proved the stronger statement than (1),



$exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$






share|cite|improve this answer











$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    By definition of derivative you get $f'(0)=0$. Since $f''(0) >0$ there exits $delta >0$ such that $f'(x)>f'(0)=0$ for $0<x<delta$. This makes $f$ strictly increasing in $(0,delta)$ making $f(frac 1 {n+1}) <f(frac 1 n)$ or $1<1$!






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Sir how can $f'(0)=0$. I guess given $f''(0)>0impliesdisplaystyle lim_{hto0}dfrac{f'(0+h)-f'(0)}{h}>0implies f'(0+h)>f'(0)$. Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:00








    • 1




      $begingroup$
      $f'(0)=lim_{nto infty} frac {f(frac 1 n) -f(0)} {frac 1 n-0}$ By continuity $f(0)=1$ so numerator is $0$ for each $n$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 18 '18 at 6:01












    • $begingroup$
      I get it now Sir. Thank you
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:13
















    1












    $begingroup$

    By definition of derivative you get $f'(0)=0$. Since $f''(0) >0$ there exits $delta >0$ such that $f'(x)>f'(0)=0$ for $0<x<delta$. This makes $f$ strictly increasing in $(0,delta)$ making $f(frac 1 {n+1}) <f(frac 1 n)$ or $1<1$!






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Sir how can $f'(0)=0$. I guess given $f''(0)>0impliesdisplaystyle lim_{hto0}dfrac{f'(0+h)-f'(0)}{h}>0implies f'(0+h)>f'(0)$. Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:00








    • 1




      $begingroup$
      $f'(0)=lim_{nto infty} frac {f(frac 1 n) -f(0)} {frac 1 n-0}$ By continuity $f(0)=1$ so numerator is $0$ for each $n$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 18 '18 at 6:01












    • $begingroup$
      I get it now Sir. Thank you
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:13














    1












    1








    1





    $begingroup$

    By definition of derivative you get $f'(0)=0$. Since $f''(0) >0$ there exits $delta >0$ such that $f'(x)>f'(0)=0$ for $0<x<delta$. This makes $f$ strictly increasing in $(0,delta)$ making $f(frac 1 {n+1}) <f(frac 1 n)$ or $1<1$!






    share|cite|improve this answer









    $endgroup$



    By definition of derivative you get $f'(0)=0$. Since $f''(0) >0$ there exits $delta >0$ such that $f'(x)>f'(0)=0$ for $0<x<delta$. This makes $f$ strictly increasing in $(0,delta)$ making $f(frac 1 {n+1}) <f(frac 1 n)$ or $1<1$!







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 18 '18 at 5:44









    Kavi Rama MurthyKavi Rama Murthy

    55.4k42057




    55.4k42057








    • 1




      $begingroup$
      Sir how can $f'(0)=0$. I guess given $f''(0)>0impliesdisplaystyle lim_{hto0}dfrac{f'(0+h)-f'(0)}{h}>0implies f'(0+h)>f'(0)$. Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:00








    • 1




      $begingroup$
      $f'(0)=lim_{nto infty} frac {f(frac 1 n) -f(0)} {frac 1 n-0}$ By continuity $f(0)=1$ so numerator is $0$ for each $n$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 18 '18 at 6:01












    • $begingroup$
      I get it now Sir. Thank you
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:13














    • 1




      $begingroup$
      Sir how can $f'(0)=0$. I guess given $f''(0)>0impliesdisplaystyle lim_{hto0}dfrac{f'(0+h)-f'(0)}{h}>0implies f'(0+h)>f'(0)$. Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:00








    • 1




      $begingroup$
      $f'(0)=lim_{nto infty} frac {f(frac 1 n) -f(0)} {frac 1 n-0}$ By continuity $f(0)=1$ so numerator is $0$ for each $n$.
      $endgroup$
      – Kavi Rama Murthy
      Dec 18 '18 at 6:01












    • $begingroup$
      I get it now Sir. Thank you
      $endgroup$
      – Yadati Kiran
      Dec 18 '18 at 6:13








    1




    1




    $begingroup$
    Sir how can $f'(0)=0$. I guess given $f''(0)>0impliesdisplaystyle lim_{hto0}dfrac{f'(0+h)-f'(0)}{h}>0implies f'(0+h)>f'(0)$. Am I missing something here?
    $endgroup$
    – Yadati Kiran
    Dec 18 '18 at 6:00






    $begingroup$
    Sir how can $f'(0)=0$. I guess given $f''(0)>0impliesdisplaystyle lim_{hto0}dfrac{f'(0+h)-f'(0)}{h}>0implies f'(0+h)>f'(0)$. Am I missing something here?
    $endgroup$
    – Yadati Kiran
    Dec 18 '18 at 6:00






    1




    1




    $begingroup$
    $f'(0)=lim_{nto infty} frac {f(frac 1 n) -f(0)} {frac 1 n-0}$ By continuity $f(0)=1$ so numerator is $0$ for each $n$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 6:01






    $begingroup$
    $f'(0)=lim_{nto infty} frac {f(frac 1 n) -f(0)} {frac 1 n-0}$ By continuity $f(0)=1$ so numerator is $0$ for each $n$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 6:01














    $begingroup$
    I get it now Sir. Thank you
    $endgroup$
    – Yadati Kiran
    Dec 18 '18 at 6:13




    $begingroup$
    I get it now Sir. Thank you
    $endgroup$
    – Yadati Kiran
    Dec 18 '18 at 6:13











    1












    $begingroup$

    Assume that



    $forall n in Bbb N, ; f left ( dfrac{1}{n} right ) = 1; tag 1$



    then by continuity,



    $f(0) = displaystyle lim_{n to infty} f left ( dfrac{1}{n} right ) = 1; tag 2$



    also,



    $f'(0) = displaystyle lim_{n to infty} dfrac{f(1/n) - 1}{1/n} = 0; tag 3$



    since



    $f''(0) > 0, tag 4$



    we find that $f(x)$ has a local minimum of value $f(0) = 1$ at $0$; now I claim that there is a fixed $epsilon > 0$ such that $f'(alpha) > 0$ for every $alpha in (0, epsilon)$; otherwise, we could find sequences $epsilon_n to 0$ and $alpha_n in (0, epsilon_n)$ with $f'(alpha_n) le 0$, and then



    $f''(0) = displaystyle lim_{n to infty} dfrac{f'(alpha_n) - f'(0)}{alpha_n} = lim_{alpha_n to infty} dfrac{f'(alpha_n)}{alpha_n} le 0 Rightarrow Leftarrow f''(0) > 0; tag 5$



    now since $f'(x)$ is differentiable, it is also continuous and thus we may apply the mean value theorem to any $beta in (0, epsilon)$ and obtain



    $f(beta) - 1 = f(beta) - f(0) = f'(gamma) (beta - 0) = f'(gamma) beta, ; gamma in (0, beta); tag 6$



    since $f'(gamma) > 0$ this yields



    $f(beta) = 1 + f'(gamma) beta; tag 7$



    we see that every $beta in (0, epsilon)$ satisfies



    $f(beta) > 1, tag 8$



    but this contradicts our assumption (1), and thus



    $exists n in Bbb N, ; f left ( dfrac{1}{n} right ) ne 1. tag 9$



    Note that we have in fact proved that $n$ in (9) may be taken arbitrarily large, and thus there are an infinite number of such $n$, which also follows from (7)-(8). In fact, we have proved the stronger statement than (1),



    $exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Assume that



      $forall n in Bbb N, ; f left ( dfrac{1}{n} right ) = 1; tag 1$



      then by continuity,



      $f(0) = displaystyle lim_{n to infty} f left ( dfrac{1}{n} right ) = 1; tag 2$



      also,



      $f'(0) = displaystyle lim_{n to infty} dfrac{f(1/n) - 1}{1/n} = 0; tag 3$



      since



      $f''(0) > 0, tag 4$



      we find that $f(x)$ has a local minimum of value $f(0) = 1$ at $0$; now I claim that there is a fixed $epsilon > 0$ such that $f'(alpha) > 0$ for every $alpha in (0, epsilon)$; otherwise, we could find sequences $epsilon_n to 0$ and $alpha_n in (0, epsilon_n)$ with $f'(alpha_n) le 0$, and then



      $f''(0) = displaystyle lim_{n to infty} dfrac{f'(alpha_n) - f'(0)}{alpha_n} = lim_{alpha_n to infty} dfrac{f'(alpha_n)}{alpha_n} le 0 Rightarrow Leftarrow f''(0) > 0; tag 5$



      now since $f'(x)$ is differentiable, it is also continuous and thus we may apply the mean value theorem to any $beta in (0, epsilon)$ and obtain



      $f(beta) - 1 = f(beta) - f(0) = f'(gamma) (beta - 0) = f'(gamma) beta, ; gamma in (0, beta); tag 6$



      since $f'(gamma) > 0$ this yields



      $f(beta) = 1 + f'(gamma) beta; tag 7$



      we see that every $beta in (0, epsilon)$ satisfies



      $f(beta) > 1, tag 8$



      but this contradicts our assumption (1), and thus



      $exists n in Bbb N, ; f left ( dfrac{1}{n} right ) ne 1. tag 9$



      Note that we have in fact proved that $n$ in (9) may be taken arbitrarily large, and thus there are an infinite number of such $n$, which also follows from (7)-(8). In fact, we have proved the stronger statement than (1),



      $exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Assume that



        $forall n in Bbb N, ; f left ( dfrac{1}{n} right ) = 1; tag 1$



        then by continuity,



        $f(0) = displaystyle lim_{n to infty} f left ( dfrac{1}{n} right ) = 1; tag 2$



        also,



        $f'(0) = displaystyle lim_{n to infty} dfrac{f(1/n) - 1}{1/n} = 0; tag 3$



        since



        $f''(0) > 0, tag 4$



        we find that $f(x)$ has a local minimum of value $f(0) = 1$ at $0$; now I claim that there is a fixed $epsilon > 0$ such that $f'(alpha) > 0$ for every $alpha in (0, epsilon)$; otherwise, we could find sequences $epsilon_n to 0$ and $alpha_n in (0, epsilon_n)$ with $f'(alpha_n) le 0$, and then



        $f''(0) = displaystyle lim_{n to infty} dfrac{f'(alpha_n) - f'(0)}{alpha_n} = lim_{alpha_n to infty} dfrac{f'(alpha_n)}{alpha_n} le 0 Rightarrow Leftarrow f''(0) > 0; tag 5$



        now since $f'(x)$ is differentiable, it is also continuous and thus we may apply the mean value theorem to any $beta in (0, epsilon)$ and obtain



        $f(beta) - 1 = f(beta) - f(0) = f'(gamma) (beta - 0) = f'(gamma) beta, ; gamma in (0, beta); tag 6$



        since $f'(gamma) > 0$ this yields



        $f(beta) = 1 + f'(gamma) beta; tag 7$



        we see that every $beta in (0, epsilon)$ satisfies



        $f(beta) > 1, tag 8$



        but this contradicts our assumption (1), and thus



        $exists n in Bbb N, ; f left ( dfrac{1}{n} right ) ne 1. tag 9$



        Note that we have in fact proved that $n$ in (9) may be taken arbitrarily large, and thus there are an infinite number of such $n$, which also follows from (7)-(8). In fact, we have proved the stronger statement than (1),



        $exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$






        share|cite|improve this answer











        $endgroup$



        Assume that



        $forall n in Bbb N, ; f left ( dfrac{1}{n} right ) = 1; tag 1$



        then by continuity,



        $f(0) = displaystyle lim_{n to infty} f left ( dfrac{1}{n} right ) = 1; tag 2$



        also,



        $f'(0) = displaystyle lim_{n to infty} dfrac{f(1/n) - 1}{1/n} = 0; tag 3$



        since



        $f''(0) > 0, tag 4$



        we find that $f(x)$ has a local minimum of value $f(0) = 1$ at $0$; now I claim that there is a fixed $epsilon > 0$ such that $f'(alpha) > 0$ for every $alpha in (0, epsilon)$; otherwise, we could find sequences $epsilon_n to 0$ and $alpha_n in (0, epsilon_n)$ with $f'(alpha_n) le 0$, and then



        $f''(0) = displaystyle lim_{n to infty} dfrac{f'(alpha_n) - f'(0)}{alpha_n} = lim_{alpha_n to infty} dfrac{f'(alpha_n)}{alpha_n} le 0 Rightarrow Leftarrow f''(0) > 0; tag 5$



        now since $f'(x)$ is differentiable, it is also continuous and thus we may apply the mean value theorem to any $beta in (0, epsilon)$ and obtain



        $f(beta) - 1 = f(beta) - f(0) = f'(gamma) (beta - 0) = f'(gamma) beta, ; gamma in (0, beta); tag 6$



        since $f'(gamma) > 0$ this yields



        $f(beta) = 1 + f'(gamma) beta; tag 7$



        we see that every $beta in (0, epsilon)$ satisfies



        $f(beta) > 1, tag 8$



        but this contradicts our assumption (1), and thus



        $exists n in Bbb N, ; f left ( dfrac{1}{n} right ) ne 1. tag 9$



        Note that we have in fact proved that $n$ in (9) may be taken arbitrarily large, and thus there are an infinite number of such $n$, which also follows from (7)-(8). In fact, we have proved the stronger statement than (1),



        $exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 6:30

























        answered Dec 18 '18 at 6:04









        Robert LewisRobert Lewis

        45k22964




        45k22964















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