Notation for repeated composition of functions












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$begingroup$


I have a repeated composition of functions ${T_n}(z) = {tau _0} circ {tau _1} circ {tau _2} circ cdots circ {tau _n}(z)$



By analogy with $sumlimits_{i = 1}^n {} ,prodlimits_{i = 1}^n {} ,bigcuplimits_{i = 1}^n {} ,bigcaplimits_{i = 1}^n {} ,$
I want to write ${T_n}(z) = left( {mathop circ limits_{i = 0}^n {tau _i}} right)(z)$ or even ${T_n}(z) = {mathop circ limits_{i = 0}^n {tau _i}} (z)$. Can I do this?










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    5












    $begingroup$


    I have a repeated composition of functions ${T_n}(z) = {tau _0} circ {tau _1} circ {tau _2} circ cdots circ {tau _n}(z)$



    By analogy with $sumlimits_{i = 1}^n {} ,prodlimits_{i = 1}^n {} ,bigcuplimits_{i = 1}^n {} ,bigcaplimits_{i = 1}^n {} ,$
    I want to write ${T_n}(z) = left( {mathop circ limits_{i = 0}^n {tau _i}} right)(z)$ or even ${T_n}(z) = {mathop circ limits_{i = 0}^n {tau _i}} (z)$. Can I do this?










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      5












      5








      5


      1



      $begingroup$


      I have a repeated composition of functions ${T_n}(z) = {tau _0} circ {tau _1} circ {tau _2} circ cdots circ {tau _n}(z)$



      By analogy with $sumlimits_{i = 1}^n {} ,prodlimits_{i = 1}^n {} ,bigcuplimits_{i = 1}^n {} ,bigcaplimits_{i = 1}^n {} ,$
      I want to write ${T_n}(z) = left( {mathop circ limits_{i = 0}^n {tau _i}} right)(z)$ or even ${T_n}(z) = {mathop circ limits_{i = 0}^n {tau _i}} (z)$. Can I do this?










      share|cite|improve this question









      $endgroup$




      I have a repeated composition of functions ${T_n}(z) = {tau _0} circ {tau _1} circ {tau _2} circ cdots circ {tau _n}(z)$



      By analogy with $sumlimits_{i = 1}^n {} ,prodlimits_{i = 1}^n {} ,bigcuplimits_{i = 1}^n {} ,bigcaplimits_{i = 1}^n {} ,$
      I want to write ${T_n}(z) = left( {mathop circ limits_{i = 0}^n {tau _i}} right)(z)$ or even ${T_n}(z) = {mathop circ limits_{i = 0}^n {tau _i}} (z)$. Can I do this?







      notation function-and-relation-composition






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      asked Sep 10 '14 at 13:07









      BudgieJaneBudgieJane

      667




      667






















          3 Answers
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          $begingroup$

          If I were writing something in which I had to do a large number of these, the following is probably not quite what I would do:
          $$
          overset{n}{underset{k=0}bigcirc} f_k quad text{ or } quad overset{0}{underset{k=n}bigcirc} f_k .
          $$

          Instead, I'd go over to tex.stackexchange.com and ask how to make this thing look respectable instead of like a workaround. I'd probably want it to be comparable in size and boldness to something like $displaystylebigcap$ in $displaystylebigcap_{k=0}^n A_k$ or to $displaystylebigoplus$. Before the begin{document} I'd put newcommand{Circ}{blah blah blah} (with a capital "C" distinguishing it from circ.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            Basically you can do anything. Notice, however, that the classical operators $sum$ and $prod$ do not coincide with the symbols they represent. And old books used to have $sum$ instead of $bigcup$ and $prod$ instead of $bigcap$.



            I personally understand the notation $mathop{circ}_{n=1}^N$, but it doesn't look appealing. By analogy, why don't you define $$mathop{rm C}limits_{n=1}^N $$ or $$mathop{rm K}limits_{n=1}^N ?$$
            Anyway, I don't like them either...






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
              $endgroup$
              – BudgieJane
              Sep 10 '14 at 17:55



















            2












            $begingroup$

            Let $f_n(x)$ be a sequence of functions indexed by $n$.



            Define a new sequence of functions $F_k(x)$ indexed by $k$:



            $$
            F_k(x) =
            begin{cases}
            f_0(x) &: k=0\
            (f_kcirc F_{k-1})(x)&: kgt 0
            end{cases}
            $$



            So you have, for example
            $$
            begin{align}
            F_0(x) &= f_0(x)\
            F_1(x) &= (f_1 circ f_0)(x)\
            F_2(x) &= (f_2 circ F_1)(x) = (f_2 circ f_1 circ f_0)(x)\
            &dots\
            F_{n}(x) &= (f_ncirc f_{n-1} circ cdots circ f_0)(x)\
            end{align}
            $$



            Then for your composition $f_ncirc f_{n-1} circ cdots circ f_0$ of $n$ terms you can simply write $F_n$.



            Of course you can define a new sequence $G_n$ for the ascending direction, i.e., $G_2 = f_0 circ f_1 circ f_2$.






            share|cite|improve this answer









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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






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              3












              $begingroup$

              If I were writing something in which I had to do a large number of these, the following is probably not quite what I would do:
              $$
              overset{n}{underset{k=0}bigcirc} f_k quad text{ or } quad overset{0}{underset{k=n}bigcirc} f_k .
              $$

              Instead, I'd go over to tex.stackexchange.com and ask how to make this thing look respectable instead of like a workaround. I'd probably want it to be comparable in size and boldness to something like $displaystylebigcap$ in $displaystylebigcap_{k=0}^n A_k$ or to $displaystylebigoplus$. Before the begin{document} I'd put newcommand{Circ}{blah blah blah} (with a capital "C" distinguishing it from circ.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                If I were writing something in which I had to do a large number of these, the following is probably not quite what I would do:
                $$
                overset{n}{underset{k=0}bigcirc} f_k quad text{ or } quad overset{0}{underset{k=n}bigcirc} f_k .
                $$

                Instead, I'd go over to tex.stackexchange.com and ask how to make this thing look respectable instead of like a workaround. I'd probably want it to be comparable in size and boldness to something like $displaystylebigcap$ in $displaystylebigcap_{k=0}^n A_k$ or to $displaystylebigoplus$. Before the begin{document} I'd put newcommand{Circ}{blah blah blah} (with a capital "C" distinguishing it from circ.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  If I were writing something in which I had to do a large number of these, the following is probably not quite what I would do:
                  $$
                  overset{n}{underset{k=0}bigcirc} f_k quad text{ or } quad overset{0}{underset{k=n}bigcirc} f_k .
                  $$

                  Instead, I'd go over to tex.stackexchange.com and ask how to make this thing look respectable instead of like a workaround. I'd probably want it to be comparable in size and boldness to something like $displaystylebigcap$ in $displaystylebigcap_{k=0}^n A_k$ or to $displaystylebigoplus$. Before the begin{document} I'd put newcommand{Circ}{blah blah blah} (with a capital "C" distinguishing it from circ.






                  share|cite|improve this answer











                  $endgroup$



                  If I were writing something in which I had to do a large number of these, the following is probably not quite what I would do:
                  $$
                  overset{n}{underset{k=0}bigcirc} f_k quad text{ or } quad overset{0}{underset{k=n}bigcirc} f_k .
                  $$

                  Instead, I'd go over to tex.stackexchange.com and ask how to make this thing look respectable instead of like a workaround. I'd probably want it to be comparable in size and boldness to something like $displaystylebigcap$ in $displaystylebigcap_{k=0}^n A_k$ or to $displaystylebigoplus$. Before the begin{document} I'd put newcommand{Circ}{blah blah blah} (with a capital "C" distinguishing it from circ.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 21 '18 at 10:41









                  Sik Feng Cheong

                  1579




                  1579










                  answered Jan 8 '15 at 23:28









                  Michael HardyMichael Hardy

                  1




                  1























                      2












                      $begingroup$

                      Basically you can do anything. Notice, however, that the classical operators $sum$ and $prod$ do not coincide with the symbols they represent. And old books used to have $sum$ instead of $bigcup$ and $prod$ instead of $bigcap$.



                      I personally understand the notation $mathop{circ}_{n=1}^N$, but it doesn't look appealing. By analogy, why don't you define $$mathop{rm C}limits_{n=1}^N $$ or $$mathop{rm K}limits_{n=1}^N ?$$
                      Anyway, I don't like them either...






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
                        $endgroup$
                        – BudgieJane
                        Sep 10 '14 at 17:55
















                      2












                      $begingroup$

                      Basically you can do anything. Notice, however, that the classical operators $sum$ and $prod$ do not coincide with the symbols they represent. And old books used to have $sum$ instead of $bigcup$ and $prod$ instead of $bigcap$.



                      I personally understand the notation $mathop{circ}_{n=1}^N$, but it doesn't look appealing. By analogy, why don't you define $$mathop{rm C}limits_{n=1}^N $$ or $$mathop{rm K}limits_{n=1}^N ?$$
                      Anyway, I don't like them either...






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
                        $endgroup$
                        – BudgieJane
                        Sep 10 '14 at 17:55














                      2












                      2








                      2





                      $begingroup$

                      Basically you can do anything. Notice, however, that the classical operators $sum$ and $prod$ do not coincide with the symbols they represent. And old books used to have $sum$ instead of $bigcup$ and $prod$ instead of $bigcap$.



                      I personally understand the notation $mathop{circ}_{n=1}^N$, but it doesn't look appealing. By analogy, why don't you define $$mathop{rm C}limits_{n=1}^N $$ or $$mathop{rm K}limits_{n=1}^N ?$$
                      Anyway, I don't like them either...






                      share|cite|improve this answer









                      $endgroup$



                      Basically you can do anything. Notice, however, that the classical operators $sum$ and $prod$ do not coincide with the symbols they represent. And old books used to have $sum$ instead of $bigcup$ and $prod$ instead of $bigcap$.



                      I personally understand the notation $mathop{circ}_{n=1}^N$, but it doesn't look appealing. By analogy, why don't you define $$mathop{rm C}limits_{n=1}^N $$ or $$mathop{rm K}limits_{n=1}^N ?$$
                      Anyway, I don't like them either...







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Sep 10 '14 at 13:14









                      SiminoreSiminore

                      30.4k33368




                      30.4k33368












                      • $begingroup$
                        I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
                        $endgroup$
                        – BudgieJane
                        Sep 10 '14 at 17:55


















                      • $begingroup$
                        I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
                        $endgroup$
                        – BudgieJane
                        Sep 10 '14 at 17:55
















                      $begingroup$
                      I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
                      $endgroup$
                      – BudgieJane
                      Sep 10 '14 at 17:55




                      $begingroup$
                      I can't use K as that's already in use for continued fractions, because K = Kettenbruch, the German for Continued Fraction.
                      $endgroup$
                      – BudgieJane
                      Sep 10 '14 at 17:55











                      2












                      $begingroup$

                      Let $f_n(x)$ be a sequence of functions indexed by $n$.



                      Define a new sequence of functions $F_k(x)$ indexed by $k$:



                      $$
                      F_k(x) =
                      begin{cases}
                      f_0(x) &: k=0\
                      (f_kcirc F_{k-1})(x)&: kgt 0
                      end{cases}
                      $$



                      So you have, for example
                      $$
                      begin{align}
                      F_0(x) &= f_0(x)\
                      F_1(x) &= (f_1 circ f_0)(x)\
                      F_2(x) &= (f_2 circ F_1)(x) = (f_2 circ f_1 circ f_0)(x)\
                      &dots\
                      F_{n}(x) &= (f_ncirc f_{n-1} circ cdots circ f_0)(x)\
                      end{align}
                      $$



                      Then for your composition $f_ncirc f_{n-1} circ cdots circ f_0$ of $n$ terms you can simply write $F_n$.



                      Of course you can define a new sequence $G_n$ for the ascending direction, i.e., $G_2 = f_0 circ f_1 circ f_2$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Let $f_n(x)$ be a sequence of functions indexed by $n$.



                        Define a new sequence of functions $F_k(x)$ indexed by $k$:



                        $$
                        F_k(x) =
                        begin{cases}
                        f_0(x) &: k=0\
                        (f_kcirc F_{k-1})(x)&: kgt 0
                        end{cases}
                        $$



                        So you have, for example
                        $$
                        begin{align}
                        F_0(x) &= f_0(x)\
                        F_1(x) &= (f_1 circ f_0)(x)\
                        F_2(x) &= (f_2 circ F_1)(x) = (f_2 circ f_1 circ f_0)(x)\
                        &dots\
                        F_{n}(x) &= (f_ncirc f_{n-1} circ cdots circ f_0)(x)\
                        end{align}
                        $$



                        Then for your composition $f_ncirc f_{n-1} circ cdots circ f_0$ of $n$ terms you can simply write $F_n$.



                        Of course you can define a new sequence $G_n$ for the ascending direction, i.e., $G_2 = f_0 circ f_1 circ f_2$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Let $f_n(x)$ be a sequence of functions indexed by $n$.



                          Define a new sequence of functions $F_k(x)$ indexed by $k$:



                          $$
                          F_k(x) =
                          begin{cases}
                          f_0(x) &: k=0\
                          (f_kcirc F_{k-1})(x)&: kgt 0
                          end{cases}
                          $$



                          So you have, for example
                          $$
                          begin{align}
                          F_0(x) &= f_0(x)\
                          F_1(x) &= (f_1 circ f_0)(x)\
                          F_2(x) &= (f_2 circ F_1)(x) = (f_2 circ f_1 circ f_0)(x)\
                          &dots\
                          F_{n}(x) &= (f_ncirc f_{n-1} circ cdots circ f_0)(x)\
                          end{align}
                          $$



                          Then for your composition $f_ncirc f_{n-1} circ cdots circ f_0$ of $n$ terms you can simply write $F_n$.



                          Of course you can define a new sequence $G_n$ for the ascending direction, i.e., $G_2 = f_0 circ f_1 circ f_2$.






                          share|cite|improve this answer









                          $endgroup$



                          Let $f_n(x)$ be a sequence of functions indexed by $n$.



                          Define a new sequence of functions $F_k(x)$ indexed by $k$:



                          $$
                          F_k(x) =
                          begin{cases}
                          f_0(x) &: k=0\
                          (f_kcirc F_{k-1})(x)&: kgt 0
                          end{cases}
                          $$



                          So you have, for example
                          $$
                          begin{align}
                          F_0(x) &= f_0(x)\
                          F_1(x) &= (f_1 circ f_0)(x)\
                          F_2(x) &= (f_2 circ F_1)(x) = (f_2 circ f_1 circ f_0)(x)\
                          &dots\
                          F_{n}(x) &= (f_ncirc f_{n-1} circ cdots circ f_0)(x)\
                          end{align}
                          $$



                          Then for your composition $f_ncirc f_{n-1} circ cdots circ f_0$ of $n$ terms you can simply write $F_n$.



                          Of course you can define a new sequence $G_n$ for the ascending direction, i.e., $G_2 = f_0 circ f_1 circ f_2$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Oct 24 '16 at 1:36









                          chharveychharvey

                          1,37131839




                          1,37131839






























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