Notation of $d$-dimensional cuboids: $prod_{i = 1}^{d} [a_i, b_i] = times_{i = 1}^{d} [a_i, b_i]$?
When writing about axially parallel cuboids
begin{equation*}
Q := [a_i, b_i] times ldots times [a_d, b_d],
end{equation*}
where $a_i, b_i in mathbb{R}$ for all $i$ I often see those two notations and wonder why the first one is often used instead of the second:
begin{equation*}
prod_{i = 1}^{d} [a_i, b_i]
qquad text{and} qquad
times_{i = 1}^{d} [a_i, b_i]
end{equation*}
notation
asked Dec 26 at 16:45
Viktor Glombik
574425
add a comment |
When writing about axially parallel cuboids
begin{equation*}
Q := [a_i, b_i] times ldots times [a_d, b_d],
end{equation*}
where $a_i, b_i in mathbb{R}$ for all $i$ I often see those two notations and wonder why the first one is often used instead of the second:
begin{equation*}
prod_{i = 1}^{d} [a_i, b_i]
qquad text{and} qquad
times_{i = 1}^{d} [a_i, b_i]
end{equation*}
notation
asked Dec 26 at 16:45
Viktor Glombik
574425
I guess for the same reason $n!=prod_{i=1}^n i$ is preferred to $n!=times_{i=1}^n i$ and $frac{n(n+1)}2=sum_{i=0}^n i$ is preferred to $frac{n(n+1)}2=+_{i=0}^n i$ or to $bigoplus_{i=0}^n i$.
– Saucy O'Path
Dec 26 at 16:48
I'd prefer to use the second way of writing a Cartesian product, because the first way is sometimes used for set intersection instead of $bigcap$. I think there's no real reason behind it, other than the word "product", which implies some connection to multiplication
– Jakobian
Dec 26 at 17:04
@SaucyO'Path But I thought $$times_{j = 1}^{n} A_i = { (a_1, ldots, a_n): a_i in A_i forall i {1, ldots, n}}.$$How is that consistent with $n! = times_{j = 1}^{n} j$?
– Viktor Glombik
Dec 26 at 17:13
add a comment |
When writing about axially parallel cuboids
begin{equation*}
Q := [a_i, b_i] times ldots times [a_d, b_d],
end{equation*}
where $a_i, b_i in mathbb{R}$ for all $i$ I often see those two notations and wonder why the first one is often used instead of the second:
begin{equation*}
prod_{i = 1}^{d} [a_i, b_i]
qquad text{and} qquad
times_{i = 1}^{d} [a_i, b_i]
end{equation*}
notation
asked Dec 26 at 16:45
Viktor Glombik
574425
When writing about axially parallel cuboids
begin{equation*}
Q := [a_i, b_i] times ldots times [a_d, b_d],
end{equation*}
where $a_i, b_i in mathbb{R}$ for all $i$ I often see those two notations and wonder why the first one is often used instead of the second:
begin{equation*}
prod_{i = 1}^{d} [a_i, b_i]
qquad text{and} qquad
times_{i = 1}^{d} [a_i, b_i]
end{equation*}
notation
notation
asked Dec 26 at 16:45
Viktor Glombik
574425
asked Dec 26 at 16:45
Viktor Glombik
574425
asked Dec 26 at 16:45
Viktor Glombik
574425
asked Dec 26 at 16:45
Viktor Glombik
574425
asked Dec 26 at 16:45
Viktor Glombik
574425
574425
I guess for the same reason $n!=prod_{i=1}^n i$ is preferred to $n!=times_{i=1}^n i$ and $frac{n(n+1)}2=sum_{i=0}^n i$ is preferred to $frac{n(n+1)}2=+_{i=0}^n i$ or to $bigoplus_{i=0}^n i$.
– Saucy O'Path
Dec 26 at 16:48
I'd prefer to use the second way of writing a Cartesian product, because the first way is sometimes used for set intersection instead of $bigcap$. I think there's no real reason behind it, other than the word "product", which implies some connection to multiplication
– Jakobian
Dec 26 at 17:04
@SaucyO'Path But I thought $$times_{j = 1}^{n} A_i = { (a_1, ldots, a_n): a_i in A_i forall i {1, ldots, n}}.$$How is that consistent with $n! = times_{j = 1}^{n} j$?
– Viktor Glombik
Dec 26 at 17:13
add a comment |
I guess for the same reason $n!=prod_{i=1}^n i$ is preferred to $n!=times_{i=1}^n i$ and $frac{n(n+1)}2=sum_{i=0}^n i$ is preferred to $frac{n(n+1)}2=+_{i=0}^n i$ or to $bigoplus_{i=0}^n i$.
– Saucy O'Path
Dec 26 at 16:48
I'd prefer to use the second way of writing a Cartesian product, because the first way is sometimes used for set intersection instead of $bigcap$. I think there's no real reason behind it, other than the word "product", which implies some connection to multiplication
– Jakobian
Dec 26 at 17:04
@SaucyO'Path But I thought $$times_{j = 1}^{n} A_i = { (a_1, ldots, a_n): a_i in A_i forall i {1, ldots, n}}.$$How is that consistent with $n! = times_{j = 1}^{n} j$?
– Viktor Glombik
Dec 26 at 17:13
I guess for the same reason $n!=prod_{i=1}^n i$ is preferred to $n!=times_{i=1}^n i$ and $frac{n(n+1)}2=sum_{i=0}^n i$ is preferred to $frac{n(n+1)}2=+_{i=0}^n i$ or to $bigoplus_{i=0}^n i$.
– Saucy O'Path
Dec 26 at 16:48
I guess for the same reason $n!=prod_{i=1}^n i$ is preferred to $n!=times_{i=1}^n i$ and $frac{n(n+1)}2=sum_{i=0}^n i$ is preferred to $frac{n(n+1)}2=+_{i=0}^n i$ or to $bigoplus_{i=0}^n i$.
– Saucy O'Path
Dec 26 at 16:48
I'd prefer to use the second way of writing a Cartesian product, because the first way is sometimes used for set intersection instead of $bigcap$. I think there's no real reason behind it, other than the word "product", which implies some connection to multiplication
– Jakobian
Dec 26 at 17:04
I'd prefer to use the second way of writing a Cartesian product, because the first way is sometimes used for set intersection instead of $bigcap$. I think there's no real reason behind it, other than the word "product", which implies some connection to multiplication
– Jakobian
Dec 26 at 17:04
@SaucyO'Path But I thought $$times_{j = 1}^{n} A_i = { (a_1, ldots, a_n): a_i in A_i forall i {1, ldots, n}}.$$How is that consistent with $n! = times_{j = 1}^{n} j$?
– Viktor Glombik
Dec 26 at 17:13
@SaucyO'Path But I thought $$times_{j = 1}^{n} A_i = { (a_1, ldots, a_n): a_i in A_i forall i {1, ldots, n}}.$$How is that consistent with $n! = times_{j = 1}^{n} j$?
– Viktor Glombik
Dec 26 at 17:13
add a comment |
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I guess for the same reason $n!=prod_{i=1}^n i$ is preferred to $n!=times_{i=1}^n i$ and $frac{n(n+1)}2=sum_{i=0}^n i$ is preferred to $frac{n(n+1)}2=+_{i=0}^n i$ or to $bigoplus_{i=0}^n i$.
– Saucy O'Path
Dec 26 at 16:48
I'd prefer to use the second way of writing a Cartesian product, because the first way is sometimes used for set intersection instead of $bigcap$. I think there's no real reason behind it, other than the word "product", which implies some connection to multiplication
– Jakobian
Dec 26 at 17:04
@SaucyO'Path But I thought $$times_{j = 1}^{n} A_i = { (a_1, ldots, a_n): a_i in A_i forall i {1, ldots, n}}.$$How is that consistent with $n! = times_{j = 1}^{n} j$?
– Viktor Glombik
Dec 26 at 17:13