function not having same value at every point [closed]
$begingroup$
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
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
asked Dec 18 '18 at 4:36
Debarghya MukherjeeDebarghya Mukherjee
352
$endgroup$
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.
add a comment |
$begingroup$
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
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
asked Dec 18 '18 at 4:36
Debarghya MukherjeeDebarghya Mukherjee
352
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
asked Dec 18 '18 at 4:36
Debarghya MukherjeeDebarghya Mukherjee
352
$endgroup$
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
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
asked Dec 18 '18 at 4:36
Debarghya MukherjeeDebarghya Mukherjee
352
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
asked Dec 18 '18 at 4:36
Debarghya MukherjeeDebarghya Mukherjee
352
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
edited Dec 18 '18 at 5:11
BigbearZzz
8,51721652
8,51721652
asked Dec 18 '18 at 4:36
Debarghya MukherjeeDebarghya Mukherjee
352
asked Dec 18 '18 at 4:36
Debarghya MukherjeeDebarghya Mukherjee
352
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $1<1$!
answered Dec 18 '18 at 5:44
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
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
add a comment |
$begingroup$
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),
$exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $1<1$!
answered Dec 18 '18 at 5:44
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
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
add a comment |
$begingroup$
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 $1<1$!
answered Dec 18 '18 at 5:44
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
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
add a comment |
$begingroup$
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 $1<1$!
answered Dec 18 '18 at 5:44
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
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 $1<1$!
answered Dec 18 '18 at 5:44
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 18 '18 at 5:44
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 18 '18 at 5:44
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
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
add a comment |
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
add a comment |
$begingroup$
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),
$exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
$endgroup$
add a comment |
$begingroup$
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),
$exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
$endgroup$
add a comment |
$begingroup$
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),
$exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
$endgroup$
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),
$exists M in Bbb N, ; n > M Longrightarrow f left ( dfrac{1}{n} right ) ne 1. tag{10}$
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
edited Dec 18 '18 at 6:30
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
answered Dec 18 '18 at 6:04
Robert LewisRobert Lewis
45k22964
45k22964
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Hint: Consider the sequence $(1/n)$ and $(f(1/n))$ as $ntoinfty$.
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– YiFan
Dec 18 '18 at 4:42
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Yes but why is f (0) not 1 I don't underztand
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– Debarghya Mukherjee
Dec 18 '18 at 4:45
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Why does $f$ have a minima at 0? (Presumably you mean minimiser?)
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– copper.hat
Dec 18 '18 at 5:13
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f"(0) is positive so f has a local minima at 0
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– Debarghya Mukherjee
Dec 18 '18 at 5:18
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@DebarghyaMukherjee: $x mapsto (x-1)^2$ has a positive second derivative everywhere but does not have a local minimum at $x=0$.
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– copper.hat
Dec 18 '18 at 5:23