Notation for repeated composition of functions
$begingroup$
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?
notation function-and-relation-composition
$endgroup$
add a comment |
$begingroup$
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?
notation function-and-relation-composition
$endgroup$
add a comment |
$begingroup$
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?
notation function-and-relation-composition
$endgroup$
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
notation function-and-relation-composition
asked Sep 10 '14 at 13:07
BudgieJaneBudgieJane
667
667
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3 Answers
3
active
oldest
votes
$begingroup$
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
.
$endgroup$
add a comment |
$begingroup$
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...
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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., $G_2 = f_0 circ f_1 circ f_2$.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
.
$endgroup$
add a comment |
$begingroup$
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
.
$endgroup$
add a comment |
$begingroup$
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
.
$endgroup$
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
.
edited Dec 21 '18 at 10:41
Sik Feng Cheong
1579
1579
answered Jan 8 '15 at 23:28
Michael HardyMichael Hardy
1
1
add a comment |
add a comment |
$begingroup$
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...
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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...
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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...
$endgroup$
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...
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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., $G_2 = f_0 circ f_1 circ f_2$.
$endgroup$
add a comment |
$begingroup$
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., $G_2 = f_0 circ f_1 circ f_2$.
$endgroup$
add a comment |
$begingroup$
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., $G_2 = f_0 circ f_1 circ f_2$.
$endgroup$
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., $G_2 = f_0 circ f_1 circ f_2$.
answered Oct 24 '16 at 1:36
chharveychharvey
1,37131839
1,37131839
add a comment |
add a comment |
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