Let $f:mathbb{I} to mathbb{R}$ continuous function such that $f(0)=f(1)$.
up vote
1
down vote
favorite
$mathbb{I} = [0,1]$
Let $f:mathbb{I} to mathbb{R}$ continuous function such that $f(0)=f(1)$. Prove that for all $n in mathbb{N} $ there $ x in mathbb{I}$ such that $ x + frac{1}{n} in mathbb{I}$ and $f( x + frac{1}{n})=f(x)$
Could you help me by giving me an idea of how to do it?
real-analysis general-topology functions continuity
asked Nov 16 at 1:45
Walys Herrera
63
|
show 1 more comment
up vote
1
down vote
favorite
$mathbb{I} = [0,1]$
Let $f:mathbb{I} to mathbb{R}$ continuous function such that $f(0)=f(1)$. Prove that for all $n in mathbb{N} $ there $ x in mathbb{I}$ such that $ x + frac{1}{n} in mathbb{I}$ and $f( x + frac{1}{n})=f(x)$
Could you help me by giving me an idea of how to do it?
real-analysis general-topology functions continuity
asked Nov 16 at 1:45
Walys Herrera
63
No derivatives, but maybe there's a theorem we can use to show that $f(x+1/n)-f(x)$ has at least one zero on $mathbb{I}$.
– Prototank
Nov 16 at 2:00
No, only this is questions and it is about of continuous function
– Walys Herrera
Nov 16 at 2:00
Any hunch? What was your 1st idea when see this? What have you learned?
– xbh
Nov 16 at 2:01
I think that i should work with succesiones for the $frac{1}{n} + x $
– Walys Herrera
Nov 16 at 2:03
Is that induction?
– Prototank
Nov 16 at 2:05
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$mathbb{I} = [0,1]$
Let $f:mathbb{I} to mathbb{R}$ continuous function such that $f(0)=f(1)$. Prove that for all $n in mathbb{N} $ there $ x in mathbb{I}$ such that $ x + frac{1}{n} in mathbb{I}$ and $f( x + frac{1}{n})=f(x)$
Could you help me by giving me an idea of how to do it?
real-analysis general-topology functions continuity
asked Nov 16 at 1:45
Walys Herrera
63
$mathbb{I} = [0,1]$
Let $f:mathbb{I} to mathbb{R}$ continuous function such that $f(0)=f(1)$. Prove that for all $n in mathbb{N} $ there $ x in mathbb{I}$ such that $ x + frac{1}{n} in mathbb{I}$ and $f( x + frac{1}{n})=f(x)$
Could you help me by giving me an idea of how to do it?
real-analysis general-topology functions continuity
real-analysis general-topology functions continuity
asked Nov 16 at 1:45
Walys Herrera
63
asked Nov 16 at 1:45
Walys Herrera
63
edited Nov 16 at 1:50
asked Nov 16 at 1:45
Walys Herrera
63
asked Nov 16 at 1:45
Walys Herrera
63
asked Nov 16 at 1:45
Walys Herrera
63
63
No derivatives, but maybe there's a theorem we can use to show that $f(x+1/n)-f(x)$ has at least one zero on $mathbb{I}$.
– Prototank
Nov 16 at 2:00
No, only this is questions and it is about of continuous function
– Walys Herrera
Nov 16 at 2:00
Any hunch? What was your 1st idea when see this? What have you learned?
– xbh
Nov 16 at 2:01
I think that i should work with succesiones for the $frac{1}{n} + x $
– Walys Herrera
Nov 16 at 2:03
Is that induction?
– Prototank
Nov 16 at 2:05
|
show 1 more comment
No derivatives, but maybe there's a theorem we can use to show that $f(x+1/n)-f(x)$ has at least one zero on $mathbb{I}$.
– Prototank
Nov 16 at 2:00
No, only this is questions and it is about of continuous function
– Walys Herrera
Nov 16 at 2:00
Any hunch? What was your 1st idea when see this? What have you learned?
– xbh
Nov 16 at 2:01
I think that i should work with succesiones for the $frac{1}{n} + x $
– Walys Herrera
Nov 16 at 2:03
Is that induction?
– Prototank
Nov 16 at 2:05
No derivatives, but maybe there's a theorem we can use to show that $f(x+1/n)-f(x)$ has at least one zero on $mathbb{I}$.
– Prototank
Nov 16 at 2:00
No derivatives, but maybe there's a theorem we can use to show that $f(x+1/n)-f(x)$ has at least one zero on $mathbb{I}$.
– Prototank
Nov 16 at 2:00
No, only this is questions and it is about of continuous function
– Walys Herrera
Nov 16 at 2:00
No, only this is questions and it is about of continuous function
– Walys Herrera
Nov 16 at 2:00
Any hunch? What was your 1st idea when see this? What have you learned?
– xbh
Nov 16 at 2:01
Any hunch? What was your 1st idea when see this? What have you learned?
– xbh
Nov 16 at 2:01
I think that i should work with succesiones for the $frac{1}{n} + x $
– Walys Herrera
Nov 16 at 2:03
I think that i should work with succesiones for the $frac{1}{n} + x $
– Walys Herrera
Nov 16 at 2:03
Is that induction?
– Prototank
Nov 16 at 2:05
Is that induction?
– Prototank
Nov 16 at 2:05
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
4
down vote
Suppose there is no such $x$. Then either $f(x+frac 1 n ) >f(x)$ for all $x$ or $f(x+frac 1 n ) <f(x)$ for all $x$ (by IVP applied to the continuous function $f(x+frac 1 n ) -f(x))$. Assume that $f(x+frac 1 n ) >f(x)$ for all $x$. (the proof is similar in the other case). Then $f(0)<f(frac 1 n) <f(frac 2 n)<cdots <f(1)$ which is a contradiction.
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
add a comment |
up vote
0
down vote
Extend $f$ to $mathbb{R}$ periodically (and so continuously by $f(0) = f(1)$). Let $g(x): = f(x+1/n) - f(x)$. Then we have
$$
g(x) + gleft(x+frac{1}{n}right) + cdots + gleft(x + frac{n-1}{n}right) = 0
$$
This implies that $g(x)$ can't have always same sign, and so $g$ has a zero by intermediate value theorem.
I'm sure that there's better way to show that $f(x+alpha) = f(x)$ has root for any $0<alpha < 1$, since this method only works for $alphain mathbb{Q}$. But I don't have any idea for this now.
answered Nov 16 at 2:06
Seewoo Lee
5,941826
Continuity implies what you conjecture, plus compactness of $mathbb{I}$, I think.
– Prototank
Nov 16 at 2:07
@Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number.
– Seewoo Lee
Nov 16 at 2:13
1
The sequence of $r_n$ converges to $alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:Ktomathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=lbrace (x,y):x,ygeq 0 x+yleq 1rbrace$. Choose a convergent subsequence in $K$. How does that look?
– Prototank
Nov 16 at 2:23
@Prototank That seems good for me. Thanks!
– Seewoo Lee
Nov 16 at 3:37
But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $alpha$ other than some $1/n$ the claim is false (even if $alpha$ is rational). Let p be a periodic function whose period is such an $alpha$, ($0 < alpha < 1$), and $p(0) = 0$, $p(1) = c neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + alpha) - f(x) = -calpha neq 0$ for all $x in [0, 1-alpha]$.
– David Hartley
Nov 17 at 2:14
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Suppose there is no such $x$. Then either $f(x+frac 1 n ) >f(x)$ for all $x$ or $f(x+frac 1 n ) <f(x)$ for all $x$ (by IVP applied to the continuous function $f(x+frac 1 n ) -f(x))$. Assume that $f(x+frac 1 n ) >f(x)$ for all $x$. (the proof is similar in the other case). Then $f(0)<f(frac 1 n) <f(frac 2 n)<cdots <f(1)$ which is a contradiction.
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
add a comment |
up vote
4
down vote
Suppose there is no such $x$. Then either $f(x+frac 1 n ) >f(x)$ for all $x$ or $f(x+frac 1 n ) <f(x)$ for all $x$ (by IVP applied to the continuous function $f(x+frac 1 n ) -f(x))$. Assume that $f(x+frac 1 n ) >f(x)$ for all $x$. (the proof is similar in the other case). Then $f(0)<f(frac 1 n) <f(frac 2 n)<cdots <f(1)$ which is a contradiction.
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
add a comment |
up vote
4
down vote
up vote
4
down vote
Suppose there is no such $x$. Then either $f(x+frac 1 n ) >f(x)$ for all $x$ or $f(x+frac 1 n ) <f(x)$ for all $x$ (by IVP applied to the continuous function $f(x+frac 1 n ) -f(x))$. Assume that $f(x+frac 1 n ) >f(x)$ for all $x$. (the proof is similar in the other case). Then $f(0)<f(frac 1 n) <f(frac 2 n)<cdots <f(1)$ which is a contradiction.
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
Suppose there is no such $x$. Then either $f(x+frac 1 n ) >f(x)$ for all $x$ or $f(x+frac 1 n ) <f(x)$ for all $x$ (by IVP applied to the continuous function $f(x+frac 1 n ) -f(x))$. Assume that $f(x+frac 1 n ) >f(x)$ for all $x$. (the proof is similar in the other case). Then $f(0)<f(frac 1 n) <f(frac 2 n)<cdots <f(1)$ which is a contradiction.
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
answered Nov 16 at 5:38
Kavi Rama Murthy
42.1k31751
42.1k31751
add a comment |
add a comment |
up vote
0
down vote
Extend $f$ to $mathbb{R}$ periodically (and so continuously by $f(0) = f(1)$). Let $g(x): = f(x+1/n) - f(x)$. Then we have
$$
g(x) + gleft(x+frac{1}{n}right) + cdots + gleft(x + frac{n-1}{n}right) = 0
$$
This implies that $g(x)$ can't have always same sign, and so $g$ has a zero by intermediate value theorem.
I'm sure that there's better way to show that $f(x+alpha) = f(x)$ has root for any $0<alpha < 1$, since this method only works for $alphain mathbb{Q}$. But I don't have any idea for this now.
answered Nov 16 at 2:06
Seewoo Lee
5,941826
Continuity implies what you conjecture, plus compactness of $mathbb{I}$, I think.
– Prototank
Nov 16 at 2:07
@Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number.
– Seewoo Lee
Nov 16 at 2:13
1
The sequence of $r_n$ converges to $alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:Ktomathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=lbrace (x,y):x,ygeq 0 x+yleq 1rbrace$. Choose a convergent subsequence in $K$. How does that look?
– Prototank
Nov 16 at 2:23
@Prototank That seems good for me. Thanks!
– Seewoo Lee
Nov 16 at 3:37
But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $alpha$ other than some $1/n$ the claim is false (even if $alpha$ is rational). Let p be a periodic function whose period is such an $alpha$, ($0 < alpha < 1$), and $p(0) = 0$, $p(1) = c neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + alpha) - f(x) = -calpha neq 0$ for all $x in [0, 1-alpha]$.
– David Hartley
Nov 17 at 2:14
|
show 1 more comment
up vote
0
down vote
Extend $f$ to $mathbb{R}$ periodically (and so continuously by $f(0) = f(1)$). Let $g(x): = f(x+1/n) - f(x)$. Then we have
$$
g(x) + gleft(x+frac{1}{n}right) + cdots + gleft(x + frac{n-1}{n}right) = 0
$$
This implies that $g(x)$ can't have always same sign, and so $g$ has a zero by intermediate value theorem.
I'm sure that there's better way to show that $f(x+alpha) = f(x)$ has root for any $0<alpha < 1$, since this method only works for $alphain mathbb{Q}$. But I don't have any idea for this now.
answered Nov 16 at 2:06
Seewoo Lee
5,941826
Continuity implies what you conjecture, plus compactness of $mathbb{I}$, I think.
– Prototank
Nov 16 at 2:07
@Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number.
– Seewoo Lee
Nov 16 at 2:13
1
The sequence of $r_n$ converges to $alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:Ktomathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=lbrace (x,y):x,ygeq 0 x+yleq 1rbrace$. Choose a convergent subsequence in $K$. How does that look?
– Prototank
Nov 16 at 2:23
@Prototank That seems good for me. Thanks!
– Seewoo Lee
Nov 16 at 3:37
But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $alpha$ other than some $1/n$ the claim is false (even if $alpha$ is rational). Let p be a periodic function whose period is such an $alpha$, ($0 < alpha < 1$), and $p(0) = 0$, $p(1) = c neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + alpha) - f(x) = -calpha neq 0$ for all $x in [0, 1-alpha]$.
– David Hartley
Nov 17 at 2:14
|
show 1 more comment
up vote
0
down vote
up vote
0
down vote
Extend $f$ to $mathbb{R}$ periodically (and so continuously by $f(0) = f(1)$). Let $g(x): = f(x+1/n) - f(x)$. Then we have
$$
g(x) + gleft(x+frac{1}{n}right) + cdots + gleft(x + frac{n-1}{n}right) = 0
$$
This implies that $g(x)$ can't have always same sign, and so $g$ has a zero by intermediate value theorem.
I'm sure that there's better way to show that $f(x+alpha) = f(x)$ has root for any $0<alpha < 1$, since this method only works for $alphain mathbb{Q}$. But I don't have any idea for this now.
answered Nov 16 at 2:06
Seewoo Lee
5,941826
Extend $f$ to $mathbb{R}$ periodically (and so continuously by $f(0) = f(1)$). Let $g(x): = f(x+1/n) - f(x)$. Then we have
$$
g(x) + gleft(x+frac{1}{n}right) + cdots + gleft(x + frac{n-1}{n}right) = 0
$$
This implies that $g(x)$ can't have always same sign, and so $g$ has a zero by intermediate value theorem.
I'm sure that there's better way to show that $f(x+alpha) = f(x)$ has root for any $0<alpha < 1$, since this method only works for $alphain mathbb{Q}$. But I don't have any idea for this now.
answered Nov 16 at 2:06
Seewoo Lee
5,941826
answered Nov 16 at 2:06
Seewoo Lee
5,941826
answered Nov 16 at 2:06
Seewoo Lee
5,941826
answered Nov 16 at 2:06
Seewoo Lee
5,941826
5,941826
Continuity implies what you conjecture, plus compactness of $mathbb{I}$, I think.
– Prototank
Nov 16 at 2:07
@Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number.
– Seewoo Lee
Nov 16 at 2:13
1
The sequence of $r_n$ converges to $alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:Ktomathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=lbrace (x,y):x,ygeq 0 x+yleq 1rbrace$. Choose a convergent subsequence in $K$. How does that look?
– Prototank
Nov 16 at 2:23
@Prototank That seems good for me. Thanks!
– Seewoo Lee
Nov 16 at 3:37
But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $alpha$ other than some $1/n$ the claim is false (even if $alpha$ is rational). Let p be a periodic function whose period is such an $alpha$, ($0 < alpha < 1$), and $p(0) = 0$, $p(1) = c neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + alpha) - f(x) = -calpha neq 0$ for all $x in [0, 1-alpha]$.
– David Hartley
Nov 17 at 2:14
|
show 1 more comment
Continuity implies what you conjecture, plus compactness of $mathbb{I}$, I think.
– Prototank
Nov 16 at 2:07
@Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number.
– Seewoo Lee
Nov 16 at 2:13
1
The sequence of $r_n$ converges to $alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:Ktomathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=lbrace (x,y):x,ygeq 0 x+yleq 1rbrace$. Choose a convergent subsequence in $K$. How does that look?
– Prototank
Nov 16 at 2:23
@Prototank That seems good for me. Thanks!
– Seewoo Lee
Nov 16 at 3:37
But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $alpha$ other than some $1/n$ the claim is false (even if $alpha$ is rational). Let p be a periodic function whose period is such an $alpha$, ($0 < alpha < 1$), and $p(0) = 0$, $p(1) = c neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + alpha) - f(x) = -calpha neq 0$ for all $x in [0, 1-alpha]$.
– David Hartley
Nov 17 at 2:14
Continuity implies what you conjecture, plus compactness of $mathbb{I}$, I think.
– Prototank
Nov 16 at 2:07
Continuity implies what you conjecture, plus compactness of $mathbb{I}$, I think.
– Prototank
Nov 16 at 2:07
@Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number.
– Seewoo Lee
Nov 16 at 2:13
@Prototank I think we have to be careful. If $r_{n}$ is a sequence of rational numbers that converges to $alpha$, then $g_{n}(x) = f(x+r_{n}) -f(x)$ has a solution $c_{n}$ for all $n$, but this doesn't imply that $c_{n}$ converges for some number.
– Seewoo Lee
Nov 16 at 2:13
1
1
The sequence of $r_n$ converges to $alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:Ktomathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=lbrace (x,y):x,ygeq 0 x+yleq 1rbrace$. Choose a convergent subsequence in $K$. How does that look?
– Prototank
Nov 16 at 2:23
The sequence of $r_n$ converges to $alpha$. Each $r_n$, by your argument, gives an $x_n$ where $f(x_n+r_n)-f(x_n)=0$. The function $G:Ktomathbb{R}$ by $G(x,y)=f(x+y)-f(x)$ is continuous and zero on the sequence $(x_n,r_n)$. I think $K$ needs to be the 2-simplex, the region $K=lbrace (x,y):x,ygeq 0 x+yleq 1rbrace$. Choose a convergent subsequence in $K$. How does that look?
– Prototank
Nov 16 at 2:23
@Prototank That seems good for me. Thanks!
– Seewoo Lee
Nov 16 at 3:37
@Prototank That seems good for me. Thanks!
– Seewoo Lee
Nov 16 at 3:37
But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $alpha$ other than some $1/n$ the claim is false (even if $alpha$ is rational). Let p be a periodic function whose period is such an $alpha$, ($0 < alpha < 1$), and $p(0) = 0$, $p(1) = c neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + alpha) - f(x) = -calpha neq 0$ for all $x in [0, 1-alpha]$.
– David Hartley
Nov 17 at 2:14
But f is only defined for $[0,1]$, your proof is incomplete until you show that the zero of g is in $[0, 1-1/n]$. Moreover, if you use an $alpha$ other than some $1/n$ the claim is false (even if $alpha$ is rational). Let p be a periodic function whose period is such an $alpha$, ($0 < alpha < 1$), and $p(0) = 0$, $p(1) = c neq 0$. Let $f(x) = p(x) - cx$. Then $f(0) = f(1) = 0$ and $f(x + alpha) - f(x) = -calpha neq 0$ for all $x in [0, 1-alpha]$.
– David Hartley
Nov 17 at 2:14
|
show 1 more comment
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No derivatives, but maybe there's a theorem we can use to show that $f(x+1/n)-f(x)$ has at least one zero on $mathbb{I}$.
– Prototank
Nov 16 at 2:00
No, only this is questions and it is about of continuous function
– Walys Herrera
Nov 16 at 2:00
Any hunch? What was your 1st idea when see this? What have you learned?
– xbh
Nov 16 at 2:01
I think that i should work with succesiones for the $frac{1}{n} + x $
– Walys Herrera
Nov 16 at 2:03
Is that induction?
– Prototank
Nov 16 at 2:05