Given $f$ continuous in an interval, prove that there exists x such that $f(x+1/n) = f(x)$.












3












$begingroup$


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







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$endgroup$

















    3












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 20:47









      Tito Eliatron

      1,601622




      1,601622










      asked Jan 7 at 20:40









      Buk LauBuk Lau

      1878




      1878






















          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?






          share|cite|improve this answer









          $endgroup$



          Hint: what is $sum_{k=0}^{n-1}{gleft(frac{k}{n}right)}$? Thus, can $g$ not vanish?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 20:45









          MindlackMindlack

          4,585210




          4,585210












          • $begingroup$
            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












          • $begingroup$
            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


















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