Mixing
Notation for repeated division
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For addition, there is $sum$
For multiplication, there is $prod$
What about for division?
Something like $/_i=1^n a_i = a_1 div a_2 div ... div a_n$?
notation
asked Jul 14 at 16:21
Nick
3,62163261
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up vote
1
down vote
favorite
For addition, there is $sum$
For multiplication, there is $prod$
What about for division?
Something like $/_i=1^n a_i = a_1 div a_2 div ... div a_n$?
notation
asked Jul 14 at 16:21
Nick
3,62163261
3
Like subtraction, division is not an associative operation. This is probably why these operators are not used, as their meaning might be ambiguous.
– Yves Daoust
Jul 14 at 16:36
@YvesDaoust: $sum_i=1^n a_0 - a_i$ is not ambiguous.
– Nick
Jul 14 at 16:38
2
Do you mean $sum_i=1^n(a_0-a_i)$ ? This has no connection with the definitions of repeated operators ! Btw, you bring a new interpretation I would never have thought of. Even more ambiguous.
– Yves Daoust
Jul 14 at 16:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For addition, there is $sum$
For multiplication, there is $prod$
What about for division?
Something like $/_i=1^n a_i = a_1 div a_2 div ... div a_n$?
notation
asked Jul 14 at 16:21
Nick
3,62163261
For addition, there is $sum$
For multiplication, there is $prod$
What about for division?
Something like $/_i=1^n a_i = a_1 div a_2 div ... div a_n$?
notation
asked Jul 14 at 16:21
Nick
3,62163261
asked Jul 14 at 16:21
Nick
3,62163261
asked Jul 14 at 16:21
Nick
3,62163261
asked Jul 14 at 16:21
Nick
3,62163261
3,62163261
3
Like subtraction, division is not an associative operation. This is probably why these operators are not used, as their meaning might be ambiguous.
– Yves Daoust
Jul 14 at 16:36
@YvesDaoust: $sum_i=1^n a_0 - a_i$ is not ambiguous.
– Nick
Jul 14 at 16:38
2
Do you mean $sum_i=1^n(a_0-a_i)$ ? This has no connection with the definitions of repeated operators ! Btw, you bring a new interpretation I would never have thought of. Even more ambiguous.
– Yves Daoust
Jul 14 at 16:58
add a comment |Â
3
Like subtraction, division is not an associative operation. This is probably why these operators are not used, as their meaning might be ambiguous.
– Yves Daoust
Jul 14 at 16:36
@YvesDaoust: $sum_i=1^n a_0 - a_i$ is not ambiguous.
– Nick
Jul 14 at 16:38
2
Do you mean $sum_i=1^n(a_0-a_i)$ ? This has no connection with the definitions of repeated operators ! Btw, you bring a new interpretation I would never have thought of. Even more ambiguous.
– Yves Daoust
Jul 14 at 16:58
3
3
Like subtraction, division is not an associative operation. This is probably why these operators are not used, as their meaning might be ambiguous.
– Yves Daoust
Jul 14 at 16:36
Like subtraction, division is not an associative operation. This is probably why these operators are not used, as their meaning might be ambiguous.
– Yves Daoust
Jul 14 at 16:36
@YvesDaoust: $sum_i=1^n a_0 - a_i$ is not ambiguous.
– Nick
Jul 14 at 16:38
@YvesDaoust: $sum_i=1^n a_0 - a_i$ is not ambiguous.
– Nick
Jul 14 at 16:38
2
2
Do you mean $sum_i=1^n(a_0-a_i)$ ? This has no connection with the definitions of repeated operators ! Btw, you bring a new interpretation I would never have thought of. Even more ambiguous.
– Yves Daoust
Jul 14 at 16:58
Do you mean $sum_i=1^n(a_0-a_i)$ ? This has no connection with the definitions of repeated operators ! Btw, you bring a new interpretation I would never have thought of. Even more ambiguous.
– Yves Daoust
Jul 14 at 16:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
I don't believe there is such a symbol, and you can likely get what you want pretty simply with $prod$ alone.
For instance, if $N=4$, you probably intended one of:
$$a_1/a_2/a_3/a_4=dfraca_1 displaystyleprod_i=2^4 a_i$$
or
$$a_1/(a_2/(a_3/a_4))=dfracdisplaystyleprod_i=1^2a_2i-1displaystyleprod_i=1^2a_2itext.$$
answered Jul 14 at 16:28
Mark S.
10.8k22363
add a comment |Â
up vote
2
down vote
There can be a justification for a repeated subtraction, provided it is defined with right to left associativity.
Because $a_0-(a_1-(a_2-(a_3-a_4)))=a_0-a_1+a_2-a_3+a_4$ is an alternating series, which is a very frequent pattern, whereas $(((a_0-a_1)-a_2)-a_3)-a_4=a_0-a_1-a_1-a_2-a_2$ is a mere summation in disguise, and we don't need a specific notation for this very rare case.
The case of repeated division is clearer:
$$fraca_0a_1cdot a_2cdot a_3cdot a_4cdots$$
is well replaced by a product and
$$fraca_0cdot a_2cdot a_4cdotsa_1cdot a_3cdots$$ is of very limited use because products with the same factor expressions at the numerator and denominator are exceptional.
By natural selection, these operators have died out.
answered Jul 14 at 16:52
Yves Daoust
111k665205
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I don't believe there is such a symbol, and you can likely get what you want pretty simply with $prod$ alone.
For instance, if $N=4$, you probably intended one of:
$$a_1/a_2/a_3/a_4=dfraca_1 displaystyleprod_i=2^4 a_i$$
or
$$a_1/(a_2/(a_3/a_4))=dfracdisplaystyleprod_i=1^2a_2i-1displaystyleprod_i=1^2a_2itext.$$
answered Jul 14 at 16:28
Mark S.
10.8k22363
add a comment |Â
up vote
3
down vote
I don't believe there is such a symbol, and you can likely get what you want pretty simply with $prod$ alone.
For instance, if $N=4$, you probably intended one of:
$$a_1/a_2/a_3/a_4=dfraca_1 displaystyleprod_i=2^4 a_i$$
or
$$a_1/(a_2/(a_3/a_4))=dfracdisplaystyleprod_i=1^2a_2i-1displaystyleprod_i=1^2a_2itext.$$
answered Jul 14 at 16:28
Mark S.
10.8k22363
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I don't believe there is such a symbol, and you can likely get what you want pretty simply with $prod$ alone.
For instance, if $N=4$, you probably intended one of:
$$a_1/a_2/a_3/a_4=dfraca_1 displaystyleprod_i=2^4 a_i$$
or
$$a_1/(a_2/(a_3/a_4))=dfracdisplaystyleprod_i=1^2a_2i-1displaystyleprod_i=1^2a_2itext.$$
answered Jul 14 at 16:28
Mark S.
10.8k22363
I don't believe there is such a symbol, and you can likely get what you want pretty simply with $prod$ alone.
For instance, if $N=4$, you probably intended one of:
$$a_1/a_2/a_3/a_4=dfraca_1 displaystyleprod_i=2^4 a_i$$
or
$$a_1/(a_2/(a_3/a_4))=dfracdisplaystyleprod_i=1^2a_2i-1displaystyleprod_i=1^2a_2itext.$$
answered Jul 14 at 16:28
Mark S.
10.8k22363
answered Jul 14 at 16:28
Mark S.
10.8k22363
answered Jul 14 at 16:28
Mark S.
10.8k22363
answered Jul 14 at 16:28
Mark S.
10.8k22363
10.8k22363
add a comment |Â
add a comment |Â
up vote
2
down vote
There can be a justification for a repeated subtraction, provided it is defined with right to left associativity.
Because $a_0-(a_1-(a_2-(a_3-a_4)))=a_0-a_1+a_2-a_3+a_4$ is an alternating series, which is a very frequent pattern, whereas $(((a_0-a_1)-a_2)-a_3)-a_4=a_0-a_1-a_1-a_2-a_2$ is a mere summation in disguise, and we don't need a specific notation for this very rare case.
The case of repeated division is clearer:
$$fraca_0a_1cdot a_2cdot a_3cdot a_4cdots$$
is well replaced by a product and
$$fraca_0cdot a_2cdot a_4cdotsa_1cdot a_3cdots$$ is of very limited use because products with the same factor expressions at the numerator and denominator are exceptional.
By natural selection, these operators have died out.
answered Jul 14 at 16:52
Yves Daoust
111k665205
add a comment |Â
up vote
2
down vote
There can be a justification for a repeated subtraction, provided it is defined with right to left associativity.
Because $a_0-(a_1-(a_2-(a_3-a_4)))=a_0-a_1+a_2-a_3+a_4$ is an alternating series, which is a very frequent pattern, whereas $(((a_0-a_1)-a_2)-a_3)-a_4=a_0-a_1-a_1-a_2-a_2$ is a mere summation in disguise, and we don't need a specific notation for this very rare case.
The case of repeated division is clearer:
$$fraca_0a_1cdot a_2cdot a_3cdot a_4cdots$$
is well replaced by a product and
$$fraca_0cdot a_2cdot a_4cdotsa_1cdot a_3cdots$$ is of very limited use because products with the same factor expressions at the numerator and denominator are exceptional.
By natural selection, these operators have died out.
answered Jul 14 at 16:52
Yves Daoust
111k665205
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There can be a justification for a repeated subtraction, provided it is defined with right to left associativity.
Because $a_0-(a_1-(a_2-(a_3-a_4)))=a_0-a_1+a_2-a_3+a_4$ is an alternating series, which is a very frequent pattern, whereas $(((a_0-a_1)-a_2)-a_3)-a_4=a_0-a_1-a_1-a_2-a_2$ is a mere summation in disguise, and we don't need a specific notation for this very rare case.
The case of repeated division is clearer:
$$fraca_0a_1cdot a_2cdot a_3cdot a_4cdots$$
is well replaced by a product and
$$fraca_0cdot a_2cdot a_4cdotsa_1cdot a_3cdots$$ is of very limited use because products with the same factor expressions at the numerator and denominator are exceptional.
By natural selection, these operators have died out.
answered Jul 14 at 16:52
Yves Daoust
111k665205
There can be a justification for a repeated subtraction, provided it is defined with right to left associativity.
Because $a_0-(a_1-(a_2-(a_3-a_4)))=a_0-a_1+a_2-a_3+a_4$ is an alternating series, which is a very frequent pattern, whereas $(((a_0-a_1)-a_2)-a_3)-a_4=a_0-a_1-a_1-a_2-a_2$ is a mere summation in disguise, and we don't need a specific notation for this very rare case.
The case of repeated division is clearer:
$$fraca_0a_1cdot a_2cdot a_3cdot a_4cdots$$
is well replaced by a product and
$$fraca_0cdot a_2cdot a_4cdotsa_1cdot a_3cdots$$ is of very limited use because products with the same factor expressions at the numerator and denominator are exceptional.
By natural selection, these operators have died out.
answered Jul 14 at 16:52
Yves Daoust
111k665205
edited Jul 14 at 17:00
answered Jul 14 at 16:52
Yves Daoust
111k665205
answered Jul 14 at 16:52
Yves Daoust
111k665205
answered Jul 14 at 16:52
Yves Daoust
111k665205
111k665205
add a comment |Â
add a comment |Â
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3
Like subtraction, division is not an associative operation. This is probably why these operators are not used, as their meaning might be ambiguous.
– Yves Daoust
Jul 14 at 16:36
@YvesDaoust: $sum_i=1^n a_0 - a_i$ is not ambiguous.
– Nick
Jul 14 at 16:38
2
Do you mean $sum_i=1^n(a_0-a_i)$ ? This has no connection with the definitions of repeated operators ! Btw, you bring a new interpretation I would never have thought of. Even more ambiguous.
– Yves Daoust
Jul 14 at 16:58