Mixing
Given $f$ continuous in an interval, prove that there exists x such that $f(x+1/n) = f(x)$.
$begingroup$
Let $f$ be a continuous function in $[0,1]$ such that $f(1) = f(0)$. let $n$ be a natural number. I have to show that there exists $x in [0, 1-1/n]$ such that $f(x+1/n) = f(x)$.
My question is, can it be shown for all $n$? I could show that for $n=2$ the statement is true by looking at $g(x) = f(x+1/2) - f(x)$ and then choosing $g(0)$ and $g(1/2)$ and by using the intermediate value theorem. it's easy to show that $x=1/2$ works.
I'm struggling however to prove it for all $n$. I tried doing the same thing but with $x= 1-1/n$, it didn't get me anywhere and it seems like I'm quite stuck.
i got:
$g(1-1/n) = f(1) - f(1-1/n)$
$g(0) = f(1/n) - f(0) = f(1/n) - f(1)$
if i could show that $g(0)cdot g(1-1/n) le 0$ then I'll be able to use the intermediate value theorem but I can't seem to succeed.
Thank you for your time!
calculus continuity
edited Jan 7 at 20:47
Tito Eliatron
1,483622
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
$endgroup$
add a comment |
$begingroup$
Let $f$ be a continuous function in $[0,1]$ such that $f(1) = f(0)$. let $n$ be a natural number. I have to show that there exists $x in [0, 1-1/n]$ such that $f(x+1/n) = f(x)$.
My question is, can it be shown for all $n$? I could show that for $n=2$ the statement is true by looking at $g(x) = f(x+1/2) - f(x)$ and then choosing $g(0)$ and $g(1/2)$ and by using the intermediate value theorem. it's easy to show that $x=1/2$ works.
I'm struggling however to prove it for all $n$. I tried doing the same thing but with $x= 1-1/n$, it didn't get me anywhere and it seems like I'm quite stuck.
i got:
$g(1-1/n) = f(1) - f(1-1/n)$
$g(0) = f(1/n) - f(0) = f(1/n) - f(1)$
if i could show that $g(0)cdot g(1-1/n) le 0$ then I'll be able to use the intermediate value theorem but I can't seem to succeed.
Thank you for your time!
calculus continuity
edited Jan 7 at 20:47
Tito Eliatron
1,483622
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
$endgroup$
add a comment |
$begingroup$
Let $f$ be a continuous function in $[0,1]$ such that $f(1) = f(0)$. let $n$ be a natural number. I have to show that there exists $x in [0, 1-1/n]$ such that $f(x+1/n) = f(x)$.
My question is, can it be shown for all $n$? I could show that for $n=2$ the statement is true by looking at $g(x) = f(x+1/2) - f(x)$ and then choosing $g(0)$ and $g(1/2)$ and by using the intermediate value theorem. it's easy to show that $x=1/2$ works.
I'm struggling however to prove it for all $n$. I tried doing the same thing but with $x= 1-1/n$, it didn't get me anywhere and it seems like I'm quite stuck.
i got:
$g(1-1/n) = f(1) - f(1-1/n)$
$g(0) = f(1/n) - f(0) = f(1/n) - f(1)$
if i could show that $g(0)cdot g(1-1/n) le 0$ then I'll be able to use the intermediate value theorem but I can't seem to succeed.
Thank you for your time!
calculus continuity
edited Jan 7 at 20:47
Tito Eliatron
1,483622
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
$endgroup$
Let $f$ be a continuous function in $[0,1]$ such that $f(1) = f(0)$. let $n$ be a natural number. I have to show that there exists $x in [0, 1-1/n]$ such that $f(x+1/n) = f(x)$.
My question is, can it be shown for all $n$? I could show that for $n=2$ the statement is true by looking at $g(x) = f(x+1/2) - f(x)$ and then choosing $g(0)$ and $g(1/2)$ and by using the intermediate value theorem. it's easy to show that $x=1/2$ works.
I'm struggling however to prove it for all $n$. I tried doing the same thing but with $x= 1-1/n$, it didn't get me anywhere and it seems like I'm quite stuck.
i got:
$g(1-1/n) = f(1) - f(1-1/n)$
$g(0) = f(1/n) - f(0) = f(1/n) - f(1)$
if i could show that $g(0)cdot g(1-1/n) le 0$ then I'll be able to use the intermediate value theorem but I can't seem to succeed.
Thank you for your time!
calculus continuity
calculus continuity
edited Jan 7 at 20:47
Tito Eliatron
1,483622
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
edited Jan 7 at 20:47
Tito Eliatron
1,483622
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
edited Jan 7 at 20:47
Tito Eliatron
1,483622
edited Jan 7 at 20:47
Tito Eliatron
1,483622
edited Jan 7 at 20:47
Tito Eliatron
1,483622
1,483622
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
asked Jan 7 at 20:40
Buk LauBuk Lau
1878
1878
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?
answered Jan 7 at 20:45
MindlackMindlack
3,39217
$endgroup$
$begingroup$
What made you think this way? I can't even start to think of a way to apply it to my questions.. i don't see what it tells me. I'm sorry i'm still learning so i'm not that great
$endgroup$
– Buk Lau
Jan 7 at 20:50
$begingroup$
Show the sum is zero. Deduce that either: at least one of the $g(k/n)$ is zero, or that at least one is negative and at least one positive.
$endgroup$
– Umberto P.
Jan 7 at 20:58
$begingroup$
I still don't know what you mean. If i spread the sum i'll end up with it being equal to f(1) - f(0) which is 0. how can I conclude that one of the elements is a zero, or that one has to be negative and another has to be positive? Thank you for your time!
$endgroup$
– Buk Lau
Jan 7 at 21:13
$begingroup$
The sum is zero: so you cannot have every element being nonzero with the same sign: so there must be some positive number and some negative number: $g$ takes some positive and negative values, so by the intermediate value theorem it vanishes at some point.
$endgroup$
– Mindlack
Jan 7 at 21:21
1
$begingroup$
Keep working and you will be! You certainly are bright enough to figure out that kind of trick if you solve enough problems. Creativity (in math at least) is not some predestined aspect of your soul, you can make it grow (or let it atrophy)!
$endgroup$
– Mindlack
Jan 7 at 21:37
|
show 1 more comment
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?
answered Jan 7 at 20:45
MindlackMindlack
3,39217
$endgroup$
$begingroup$
What made you think this way? I can't even start to think of a way to apply it to my questions.. i don't see what it tells me. I'm sorry i'm still learning so i'm not that great
$endgroup$
– Buk Lau
Jan 7 at 20:50
$begingroup$
Show the sum is zero. Deduce that either: at least one of the $g(k/n)$ is zero, or that at least one is negative and at least one positive.
$endgroup$
– Umberto P.
Jan 7 at 20:58
$begingroup$
I still don't know what you mean. If i spread the sum i'll end up with it being equal to f(1) - f(0) which is 0. how can I conclude that one of the elements is a zero, or that one has to be negative and another has to be positive? Thank you for your time!
$endgroup$
– Buk Lau
Jan 7 at 21:13
$begingroup$
The sum is zero: so you cannot have every element being nonzero with the same sign: so there must be some positive number and some negative number: $g$ takes some positive and negative values, so by the intermediate value theorem it vanishes at some point.
$endgroup$
– Mindlack
Jan 7 at 21:21
1
$begingroup$
Keep working and you will be! You certainly are bright enough to figure out that kind of trick if you solve enough problems. Creativity (in math at least) is not some predestined aspect of your soul, you can make it grow (or let it atrophy)!
$endgroup$
– Mindlack
Jan 7 at 21:37
|
show 1 more comment
$begingroup$
Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?
answered Jan 7 at 20:45
MindlackMindlack
3,39217
$endgroup$
$begingroup$
What made you think this way? I can't even start to think of a way to apply it to my questions.. i don't see what it tells me. I'm sorry i'm still learning so i'm not that great
$endgroup$
– Buk Lau
Jan 7 at 20:50
$begingroup$
Show the sum is zero. Deduce that either: at least one of the $g(k/n)$ is zero, or that at least one is negative and at least one positive.
$endgroup$
– Umberto P.
Jan 7 at 20:58
$begingroup$
I still don't know what you mean. If i spread the sum i'll end up with it being equal to f(1) - f(0) which is 0. how can I conclude that one of the elements is a zero, or that one has to be negative and another has to be positive? Thank you for your time!
$endgroup$
– Buk Lau
Jan 7 at 21:13
$begingroup$
The sum is zero: so you cannot have every element being nonzero with the same sign: so there must be some positive number and some negative number: $g$ takes some positive and negative values, so by the intermediate value theorem it vanishes at some point.
$endgroup$
– Mindlack
Jan 7 at 21:21
1
$begingroup$
Keep working and you will be! You certainly are bright enough to figure out that kind of trick if you solve enough problems. Creativity (in math at least) is not some predestined aspect of your soul, you can make it grow (or let it atrophy)!
$endgroup$
– Mindlack
Jan 7 at 21:37
|
show 1 more comment
$begingroup$
Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?
answered Jan 7 at 20:45
MindlackMindlack
3,39217
$endgroup$
Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?
answered Jan 7 at 20:45
MindlackMindlack
3,39217
answered Jan 7 at 20:45
MindlackMindlack
3,39217
answered Jan 7 at 20:45
MindlackMindlack
3,39217
answered Jan 7 at 20:45
MindlackMindlack
3,39217
3,39217
$begingroup$
What made you think this way? I can't even start to think of a way to apply it to my questions.. i don't see what it tells me. I'm sorry i'm still learning so i'm not that great
$endgroup$
– Buk Lau
Jan 7 at 20:50
$begingroup$
Show the sum is zero. Deduce that either: at least one of the $g(k/n)$ is zero, or that at least one is negative and at least one positive.
$endgroup$
– Umberto P.
Jan 7 at 20:58
$begingroup$
I still don't know what you mean. If i spread the sum i'll end up with it being equal to f(1) - f(0) which is 0. how can I conclude that one of the elements is a zero, or that one has to be negative and another has to be positive? Thank you for your time!
$endgroup$
– Buk Lau
Jan 7 at 21:13
$begingroup$
The sum is zero: so you cannot have every element being nonzero with the same sign: so there must be some positive number and some negative number: $g$ takes some positive and negative values, so by the intermediate value theorem it vanishes at some point.
$endgroup$
– Mindlack
Jan 7 at 21:21
1
$begingroup$
Keep working and you will be! You certainly are bright enough to figure out that kind of trick if you solve enough problems. Creativity (in math at least) is not some predestined aspect of your soul, you can make it grow (or let it atrophy)!
$endgroup$
– Mindlack
Jan 7 at 21:37
|
show 1 more comment
$begingroup$
What made you think this way? I can't even start to think of a way to apply it to my questions.. i don't see what it tells me. I'm sorry i'm still learning so i'm not that great
$endgroup$
– Buk Lau
Jan 7 at 20:50
$begingroup$
Show the sum is zero. Deduce that either: at least one of the $g(k/n)$ is zero, or that at least one is negative and at least one positive.
$endgroup$
– Umberto P.
Jan 7 at 20:58
$begingroup$
I still don't know what you mean. If i spread the sum i'll end up with it being equal to f(1) - f(0) which is 0. how can I conclude that one of the elements is a zero, or that one has to be negative and another has to be positive? Thank you for your time!
$endgroup$
– Buk Lau
Jan 7 at 21:13
$begingroup$
The sum is zero: so you cannot have every element being nonzero with the same sign: so there must be some positive number and some negative number: $g$ takes some positive and negative values, so by the intermediate value theorem it vanishes at some point.
$endgroup$
– Mindlack
Jan 7 at 21:21
1
$begingroup$
Keep working and you will be! You certainly are bright enough to figure out that kind of trick if you solve enough problems. Creativity (in math at least) is not some predestined aspect of your soul, you can make it grow (or let it atrophy)!
$endgroup$
– Mindlack
Jan 7 at 21:37
$begingroup$
What made you think this way? I can't even start to think of a way to apply it to my questions.. i don't see what it tells me. I'm sorry i'm still learning so i'm not that great
$endgroup$
– Buk Lau
Jan 7 at 20:50
$begingroup$
What made you think this way? I can't even start to think of a way to apply it to my questions.. i don't see what it tells me. I'm sorry i'm still learning so i'm not that great
$endgroup$
– Buk Lau
Jan 7 at 20:50
$begingroup$
Show the sum is zero. Deduce that either: at least one of the $g(k/n)$ is zero, or that at least one is negative and at least one positive.
$endgroup$
– Umberto P.
Jan 7 at 20:58
$begingroup$
Show the sum is zero. Deduce that either: at least one of the $g(k/n)$ is zero, or that at least one is negative and at least one positive.
$endgroup$
– Umberto P.
Jan 7 at 20:58
$begingroup$
I still don't know what you mean. If i spread the sum i'll end up with it being equal to f(1) - f(0) which is 0. how can I conclude that one of the elements is a zero, or that one has to be negative and another has to be positive? Thank you for your time!
$endgroup$
– Buk Lau
Jan 7 at 21:13
$begingroup$
I still don't know what you mean. If i spread the sum i'll end up with it being equal to f(1) - f(0) which is 0. how can I conclude that one of the elements is a zero, or that one has to be negative and another has to be positive? Thank you for your time!
$endgroup$
– Buk Lau
Jan 7 at 21:13
$begingroup$
The sum is zero: so you cannot have every element being nonzero with the same sign: so there must be some positive number and some negative number: $g$ takes some positive and negative values, so by the intermediate value theorem it vanishes at some point.
$endgroup$
– Mindlack
Jan 7 at 21:21
$begingroup$
The sum is zero: so you cannot have every element being nonzero with the same sign: so there must be some positive number and some negative number: $g$ takes some positive and negative values, so by the intermediate value theorem it vanishes at some point.
$endgroup$
– Mindlack
Jan 7 at 21:21
1
1
$begingroup$
Keep working and you will be! You certainly are bright enough to figure out that kind of trick if you solve enough problems. Creativity (in math at least) is not some predestined aspect of your soul, you can make it grow (or let it atrophy)!
$endgroup$
– Mindlack
Jan 7 at 21:37
$begingroup$
Keep working and you will be! You certainly are bright enough to figure out that kind of trick if you solve enough problems. Creativity (in math at least) is not some predestined aspect of your soul, you can make it grow (or let it atrophy)!
$endgroup$
– Mindlack
Jan 7 at 21:37
|
show 1 more comment
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