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I am studding a research paper in winch author presented a analytical model for set traversal and different cases of time complexity.
I am not understanding the one point in the model that is related to summation formulation equation n# 18.
$fracm_1 + 12t + fracl_2 + 12t + .... + fracl_n + 12t$
$ = frac12 bigr(m_1 + n + sum_a=l_2^l_n abigr) t$
Why $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$
why its not $sum_a=2^n l_a$
algebra-precalculus discrete-mathematics summation
asked 3 hours ago
Ahmad Bilal
455
add a comment |Â
up vote
3
down vote
favorite
I am studding a research paper in winch author presented a analytical model for set traversal and different cases of time complexity.
I am not understanding the one point in the model that is related to summation formulation equation n# 18.
$fracm_1 + 12t + fracl_2 + 12t + .... + fracl_n + 12t$
$ = frac12 bigr(m_1 + n + sum_a=l_2^l_n abigr) t$
Why $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$
why its not $sum_a=2^n l_a$
algebra-precalculus discrete-mathematics summation
asked 3 hours ago
Ahmad Bilal
455
1
...because they wanted to write it like that?
– Sean Roberson
3 hours ago
2
Because the authors are innumerate?
– Lord Shark the Unknown
3 hours ago
1
Your final line should have $sum_a=2^n l_a$.
– Lord Shark the Unknown
3 hours ago
1
@LordSharktheUnknown Or, better, replace $a$ by $i$ rather than $i$ by $a$.
– Ethan Bolker
3 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am studding a research paper in winch author presented a analytical model for set traversal and different cases of time complexity.
I am not understanding the one point in the model that is related to summation formulation equation n# 18.
$fracm_1 + 12t + fracl_2 + 12t + .... + fracl_n + 12t$
$ = frac12 bigr(m_1 + n + sum_a=l_2^l_n abigr) t$
Why $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$
why its not $sum_a=2^n l_a$
algebra-precalculus discrete-mathematics summation
asked 3 hours ago
Ahmad Bilal
455
I am studding a research paper in winch author presented a analytical model for set traversal and different cases of time complexity.
I am not understanding the one point in the model that is related to summation formulation equation n# 18.
$fracm_1 + 12t + fracl_2 + 12t + .... + fracl_n + 12t$
$ = frac12 bigr(m_1 + n + sum_a=l_2^l_n abigr) t$
Why $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$
why its not $sum_a=2^n l_a$
algebra-precalculus discrete-mathematics summation
algebra-precalculus discrete-mathematics summation
asked 3 hours ago
Ahmad Bilal
455
asked 3 hours ago
Ahmad Bilal
455
edited 3 hours ago
asked 3 hours ago
Ahmad Bilal
455
asked 3 hours ago
Ahmad Bilal
455
asked 3 hours ago
Ahmad Bilal
455
455
1
...because they wanted to write it like that?
– Sean Roberson
3 hours ago
2
Because the authors are innumerate?
– Lord Shark the Unknown
3 hours ago
1
Your final line should have $sum_a=2^n l_a$.
– Lord Shark the Unknown
3 hours ago
1
@LordSharktheUnknown Or, better, replace $a$ by $i$ rather than $i$ by $a$.
– Ethan Bolker
3 hours ago
add a comment |Â
1
...because they wanted to write it like that?
– Sean Roberson
3 hours ago
2
Because the authors are innumerate?
– Lord Shark the Unknown
3 hours ago
1
Your final line should have $sum_a=2^n l_a$.
– Lord Shark the Unknown
3 hours ago
1
@LordSharktheUnknown Or, better, replace $a$ by $i$ rather than $i$ by $a$.
– Ethan Bolker
3 hours ago
1
1
...because they wanted to write it like that?
– Sean Roberson
3 hours ago
...because they wanted to write it like that?
– Sean Roberson
3 hours ago
2
2
Because the authors are innumerate?
– Lord Shark the Unknown
3 hours ago
Because the authors are innumerate?
– Lord Shark the Unknown
3 hours ago
1
1
Your final line should have $sum_a=2^n l_a$.
– Lord Shark the Unknown
3 hours ago
Your final line should have $sum_a=2^n l_a$.
– Lord Shark the Unknown
3 hours ago
1
1
@LordSharktheUnknown Or, better, replace $a$ by $i$ rather than $i$ by $a$.
– Ethan Bolker
3 hours ago
@LordSharktheUnknown Or, better, replace $a$ by $i$ rather than $i$ by $a$.
– Ethan Bolker
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Sloppy notation.
The author intended $sumlimits_a in l_2, l_3, .... , l_n a$ which would be the same thing as $sum_a=2^n l_a$.
However I'd argue that $sumlimits_a=l_2^l_n a$ is not actually wrong per se.
The notation for $sum$ is $sumlimits_textsome condition term = $ "the sum of all the $terms$ for which that condition is true".
(A practical example of this would be $sumlimits_24;ain mathbb N a = $ the sum of all the divisors of $24=1 + 2 + 3 + 4+ 6 + 8 +12 + 24$.)
This means the familiar $sumlimits_a=2^n l_a$ actually means $sumlimits_ain 2,3,4,...n l_a$ and that $_a=2^n$ is just shorthand for $ain 2.....n$.
Except when were we ever actually TAUGHT that $_a=2^n$ is just shorthand for $a in 2.....n$? Well, maybe you were but I never was. I had to pick it up on the streets...
Anyway, if $_a=2^n$ is proper shorthand for $a in 2,.... n$ then isn't it okay that $_a=l_2^l_n$ proper shorthand for $a in l_2, .... l_n$?
Meh... maybe. I admit it looks weird and we have to think but... is it any more weird than $_a=2^n$?
(That was rhetorical.)
(Anyway, welcome to learning mathematics on the street.)
====ADDENDUM====
In a comment in Marcus Scheurer's answer, I realize that $sumlimits_a=l_2^l_n a$ could just as legitimately be interpretated as $l_2 + (l_2 + 1) + (l_2 + 3) + ..... +(l_2 + lfloor l_n - l_2rfloor)$.
It would take a bit bit of a bean-counter mind to interpret it that way if the individual $l_i$ values weren't specifically spelled out.
But that is a technical and correct interpretation.
So... was the author wrong or right? Arguments can be made both ways.
But it's what the author did. And the author clearly meant it to be interpreted as $sumlimits_i=2^n l_i$.
Those are the existential facts. What we do with them is up to us.
answered 1 hour ago
fleablood
62.3k22678
add a comment |Â
up vote
2
down vote
Your doubts are reasonable and your proposal is correct. The formula (18) in the referred paper and some more show all the same kind of typos (or miscalculations).
We have
beginalign*
l_2 + l_3 + l_4 + ..... + l_ncolorblue=sum_a=2^n l_a
endalign*
and not $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$, since $$sum_a=l_2^l_n a=sum_l_2leq aleq l_na$$
answered 1 hour ago
Markus Scheuer
57.2k452137
Is the notation $sumlimits_a=l_2^l_n a = sumlimits_ain l_2...l_n a$ really any more of an abuse of notation than $sumlimits_i=2^n l_i = sumlimits_i in 2....n l_i$?
– fleablood
59 mins ago
@fleablood: The author might have somewhat of this kind in mind, but as it is written we have $sum_i=l_2^nl_i=l_2+(l_2+1)+cdots+l_n$ which is not the same as $sum_iin2,ldots,nl_i$.
– Markus Scheuer
54 mins ago
That's not what we have written. We have $sumlimits_a=l_2^l_n a$ which can be interpreted either as $l_2 + (l_2 +1) + (l_2 + 3)+.... + l_n$ (all assuming that $l_n$ is a positive integer value larger than $l_2$). Or as $l_2 + l_3 + l_4 + .... + l_n$. Obviously in context the author meant the latter. And I agree he should have been clearer with the more common $sumlimits_i=2^n l_i$. But can we really claim that that is "wrong" when $sumlimits_i=2^n$ to mean $sumlimits_iin2,...,n$ is every bit as informal and relays just as much on "being clear in context"?
– fleablood
45 mins ago
@fleablood: We have $sum_a=l_2^l_na=sum_l_2leq a leq l_na$ which is not the same as $sum_ain2,ldots,nl_a$. You might want to check formula (2.2), (2.3) and (2.4) in Concrete Mathematics.
– Markus Scheuer
40 mins ago
true...writing $_a=j^k$ is taken to mean $jle a le k; a-jin mathbb Z$. But is that written in stone? I agree with you I'm inclined to call what the author did wrong and ambiguous at best. Is interpreting $_a=j^k$ to mean $j le a le k$ any more legitimate? What if $j$ and $k$ aren't integers? Why is the 'step' assumed to be $1$? What if $k < j$. etc. ... well, theres a lot to be said for convention and we do have answers to all of those. But there's also something to be said about interpretation through context. (A subtle aspect is $a$ is not usually taken to be an integral index.)
– fleablood
22 mins ago
 |Â
show 1 more comment
up vote
0
down vote
He actually does it in all the other equations as well. For example, look at equation (13):
$$
frac1l_2 sum_i = 1^l_2t_i = fracl_2 + 12t
$$
His l's are already decided in the definition of set $A_i$.
answered 2 hours ago
Abhi Devathi
1
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Who is "He" to whom you refer about "His" work?
– amWhy
2 hours ago
@amWhy presumably the author(s) of the linked paper.
– Brahadeesh
1 hour ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Brahadeesh
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Sloppy notation.
The author intended $sumlimits_a in l_2, l_3, .... , l_n a$ which would be the same thing as $sum_a=2^n l_a$.
However I'd argue that $sumlimits_a=l_2^l_n a$ is not actually wrong per se.
The notation for $sum$ is $sumlimits_textsome condition term = $ "the sum of all the $terms$ for which that condition is true".
(A practical example of this would be $sumlimits_24;ain mathbb N a = $ the sum of all the divisors of $24=1 + 2 + 3 + 4+ 6 + 8 +12 + 24$.)
This means the familiar $sumlimits_a=2^n l_a$ actually means $sumlimits_ain 2,3,4,...n l_a$ and that $_a=2^n$ is just shorthand for $ain 2.....n$.
Except when were we ever actually TAUGHT that $_a=2^n$ is just shorthand for $a in 2.....n$? Well, maybe you were but I never was. I had to pick it up on the streets...
Anyway, if $_a=2^n$ is proper shorthand for $a in 2,.... n$ then isn't it okay that $_a=l_2^l_n$ proper shorthand for $a in l_2, .... l_n$?
Meh... maybe. I admit it looks weird and we have to think but... is it any more weird than $_a=2^n$?
(That was rhetorical.)
(Anyway, welcome to learning mathematics on the street.)
====ADDENDUM====
In a comment in Marcus Scheurer's answer, I realize that $sumlimits_a=l_2^l_n a$ could just as legitimately be interpretated as $l_2 + (l_2 + 1) + (l_2 + 3) + ..... +(l_2 + lfloor l_n - l_2rfloor)$.
It would take a bit bit of a bean-counter mind to interpret it that way if the individual $l_i$ values weren't specifically spelled out.
But that is a technical and correct interpretation.
So... was the author wrong or right? Arguments can be made both ways.
But it's what the author did. And the author clearly meant it to be interpreted as $sumlimits_i=2^n l_i$.
Those are the existential facts. What we do with them is up to us.
answered 1 hour ago
fleablood
62.3k22678
add a comment |Â
up vote
2
down vote
accepted
Sloppy notation.
The author intended $sumlimits_a in l_2, l_3, .... , l_n a$ which would be the same thing as $sum_a=2^n l_a$.
However I'd argue that $sumlimits_a=l_2^l_n a$ is not actually wrong per se.
The notation for $sum$ is $sumlimits_textsome condition term = $ "the sum of all the $terms$ for which that condition is true".
(A practical example of this would be $sumlimits_24;ain mathbb N a = $ the sum of all the divisors of $24=1 + 2 + 3 + 4+ 6 + 8 +12 + 24$.)
This means the familiar $sumlimits_a=2^n l_a$ actually means $sumlimits_ain 2,3,4,...n l_a$ and that $_a=2^n$ is just shorthand for $ain 2.....n$.
Except when were we ever actually TAUGHT that $_a=2^n$ is just shorthand for $a in 2.....n$? Well, maybe you were but I never was. I had to pick it up on the streets...
Anyway, if $_a=2^n$ is proper shorthand for $a in 2,.... n$ then isn't it okay that $_a=l_2^l_n$ proper shorthand for $a in l_2, .... l_n$?
Meh... maybe. I admit it looks weird and we have to think but... is it any more weird than $_a=2^n$?
(That was rhetorical.)
(Anyway, welcome to learning mathematics on the street.)
====ADDENDUM====
In a comment in Marcus Scheurer's answer, I realize that $sumlimits_a=l_2^l_n a$ could just as legitimately be interpretated as $l_2 + (l_2 + 1) + (l_2 + 3) + ..... +(l_2 + lfloor l_n - l_2rfloor)$.
It would take a bit bit of a bean-counter mind to interpret it that way if the individual $l_i$ values weren't specifically spelled out.
But that is a technical and correct interpretation.
So... was the author wrong or right? Arguments can be made both ways.
But it's what the author did. And the author clearly meant it to be interpreted as $sumlimits_i=2^n l_i$.
Those are the existential facts. What we do with them is up to us.
answered 1 hour ago
fleablood
62.3k22678
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Sloppy notation.
The author intended $sumlimits_a in l_2, l_3, .... , l_n a$ which would be the same thing as $sum_a=2^n l_a$.
However I'd argue that $sumlimits_a=l_2^l_n a$ is not actually wrong per se.
The notation for $sum$ is $sumlimits_textsome condition term = $ "the sum of all the $terms$ for which that condition is true".
(A practical example of this would be $sumlimits_24;ain mathbb N a = $ the sum of all the divisors of $24=1 + 2 + 3 + 4+ 6 + 8 +12 + 24$.)
This means the familiar $sumlimits_a=2^n l_a$ actually means $sumlimits_ain 2,3,4,...n l_a$ and that $_a=2^n$ is just shorthand for $ain 2.....n$.
Except when were we ever actually TAUGHT that $_a=2^n$ is just shorthand for $a in 2.....n$? Well, maybe you were but I never was. I had to pick it up on the streets...
Anyway, if $_a=2^n$ is proper shorthand for $a in 2,.... n$ then isn't it okay that $_a=l_2^l_n$ proper shorthand for $a in l_2, .... l_n$?
Meh... maybe. I admit it looks weird and we have to think but... is it any more weird than $_a=2^n$?
(That was rhetorical.)
(Anyway, welcome to learning mathematics on the street.)
====ADDENDUM====
In a comment in Marcus Scheurer's answer, I realize that $sumlimits_a=l_2^l_n a$ could just as legitimately be interpretated as $l_2 + (l_2 + 1) + (l_2 + 3) + ..... +(l_2 + lfloor l_n - l_2rfloor)$.
It would take a bit bit of a bean-counter mind to interpret it that way if the individual $l_i$ values weren't specifically spelled out.
But that is a technical and correct interpretation.
So... was the author wrong or right? Arguments can be made both ways.
But it's what the author did. And the author clearly meant it to be interpreted as $sumlimits_i=2^n l_i$.
Those are the existential facts. What we do with them is up to us.
answered 1 hour ago
fleablood
62.3k22678
Sloppy notation.
The author intended $sumlimits_a in l_2, l_3, .... , l_n a$ which would be the same thing as $sum_a=2^n l_a$.
However I'd argue that $sumlimits_a=l_2^l_n a$ is not actually wrong per se.
The notation for $sum$ is $sumlimits_textsome condition term = $ "the sum of all the $terms$ for which that condition is true".
(A practical example of this would be $sumlimits_24;ain mathbb N a = $ the sum of all the divisors of $24=1 + 2 + 3 + 4+ 6 + 8 +12 + 24$.)
This means the familiar $sumlimits_a=2^n l_a$ actually means $sumlimits_ain 2,3,4,...n l_a$ and that $_a=2^n$ is just shorthand for $ain 2.....n$.
Except when were we ever actually TAUGHT that $_a=2^n$ is just shorthand for $a in 2.....n$? Well, maybe you were but I never was. I had to pick it up on the streets...
Anyway, if $_a=2^n$ is proper shorthand for $a in 2,.... n$ then isn't it okay that $_a=l_2^l_n$ proper shorthand for $a in l_2, .... l_n$?
Meh... maybe. I admit it looks weird and we have to think but... is it any more weird than $_a=2^n$?
(That was rhetorical.)
(Anyway, welcome to learning mathematics on the street.)
====ADDENDUM====
In a comment in Marcus Scheurer's answer, I realize that $sumlimits_a=l_2^l_n a$ could just as legitimately be interpretated as $l_2 + (l_2 + 1) + (l_2 + 3) + ..... +(l_2 + lfloor l_n - l_2rfloor)$.
It would take a bit bit of a bean-counter mind to interpret it that way if the individual $l_i$ values weren't specifically spelled out.
But that is a technical and correct interpretation.
So... was the author wrong or right? Arguments can be made both ways.
But it's what the author did. And the author clearly meant it to be interpreted as $sumlimits_i=2^n l_i$.
Those are the existential facts. What we do with them is up to us.
answered 1 hour ago
fleablood
62.3k22678
edited 32 mins ago
answered 1 hour ago
fleablood
62.3k22678
answered 1 hour ago
fleablood
62.3k22678
answered 1 hour ago
fleablood
62.3k22678
62.3k22678
add a comment |Â
add a comment |Â
up vote
2
down vote
Your doubts are reasonable and your proposal is correct. The formula (18) in the referred paper and some more show all the same kind of typos (or miscalculations).
We have
beginalign*
l_2 + l_3 + l_4 + ..... + l_ncolorblue=sum_a=2^n l_a
endalign*
and not $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$, since $$sum_a=l_2^l_n a=sum_l_2leq aleq l_na$$
answered 1 hour ago
Markus Scheuer
57.2k452137
Is the notation $sumlimits_a=l_2^l_n a = sumlimits_ain l_2...l_n a$ really any more of an abuse of notation than $sumlimits_i=2^n l_i = sumlimits_i in 2....n l_i$?
– fleablood
59 mins ago
@fleablood: The author might have somewhat of this kind in mind, but as it is written we have $sum_i=l_2^nl_i=l_2+(l_2+1)+cdots+l_n$ which is not the same as $sum_iin2,ldots,nl_i$.
– Markus Scheuer
54 mins ago
That's not what we have written. We have $sumlimits_a=l_2^l_n a$ which can be interpreted either as $l_2 + (l_2 +1) + (l_2 + 3)+.... + l_n$ (all assuming that $l_n$ is a positive integer value larger than $l_2$). Or as $l_2 + l_3 + l_4 + .... + l_n$. Obviously in context the author meant the latter. And I agree he should have been clearer with the more common $sumlimits_i=2^n l_i$. But can we really claim that that is "wrong" when $sumlimits_i=2^n$ to mean $sumlimits_iin2,...,n$ is every bit as informal and relays just as much on "being clear in context"?
– fleablood
45 mins ago
@fleablood: We have $sum_a=l_2^l_na=sum_l_2leq a leq l_na$ which is not the same as $sum_ain2,ldots,nl_a$. You might want to check formula (2.2), (2.3) and (2.4) in Concrete Mathematics.
– Markus Scheuer
40 mins ago
true...writing $_a=j^k$ is taken to mean $jle a le k; a-jin mathbb Z$. But is that written in stone? I agree with you I'm inclined to call what the author did wrong and ambiguous at best. Is interpreting $_a=j^k$ to mean $j le a le k$ any more legitimate? What if $j$ and $k$ aren't integers? Why is the 'step' assumed to be $1$? What if $k < j$. etc. ... well, theres a lot to be said for convention and we do have answers to all of those. But there's also something to be said about interpretation through context. (A subtle aspect is $a$ is not usually taken to be an integral index.)
– fleablood
22 mins ago
 |Â
show 1 more comment
up vote
2
down vote
Your doubts are reasonable and your proposal is correct. The formula (18) in the referred paper and some more show all the same kind of typos (or miscalculations).
We have
beginalign*
l_2 + l_3 + l_4 + ..... + l_ncolorblue=sum_a=2^n l_a
endalign*
and not $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$, since $$sum_a=l_2^l_n a=sum_l_2leq aleq l_na$$
answered 1 hour ago
Markus Scheuer
57.2k452137
Is the notation $sumlimits_a=l_2^l_n a = sumlimits_ain l_2...l_n a$ really any more of an abuse of notation than $sumlimits_i=2^n l_i = sumlimits_i in 2....n l_i$?
– fleablood
59 mins ago
@fleablood: The author might have somewhat of this kind in mind, but as it is written we have $sum_i=l_2^nl_i=l_2+(l_2+1)+cdots+l_n$ which is not the same as $sum_iin2,ldots,nl_i$.
– Markus Scheuer
54 mins ago
That's not what we have written. We have $sumlimits_a=l_2^l_n a$ which can be interpreted either as $l_2 + (l_2 +1) + (l_2 + 3)+.... + l_n$ (all assuming that $l_n$ is a positive integer value larger than $l_2$). Or as $l_2 + l_3 + l_4 + .... + l_n$. Obviously in context the author meant the latter. And I agree he should have been clearer with the more common $sumlimits_i=2^n l_i$. But can we really claim that that is "wrong" when $sumlimits_i=2^n$ to mean $sumlimits_iin2,...,n$ is every bit as informal and relays just as much on "being clear in context"?
– fleablood
45 mins ago
@fleablood: We have $sum_a=l_2^l_na=sum_l_2leq a leq l_na$ which is not the same as $sum_ain2,ldots,nl_a$. You might want to check formula (2.2), (2.3) and (2.4) in Concrete Mathematics.
– Markus Scheuer
40 mins ago
true...writing $_a=j^k$ is taken to mean $jle a le k; a-jin mathbb Z$. But is that written in stone? I agree with you I'm inclined to call what the author did wrong and ambiguous at best. Is interpreting $_a=j^k$ to mean $j le a le k$ any more legitimate? What if $j$ and $k$ aren't integers? Why is the 'step' assumed to be $1$? What if $k < j$. etc. ... well, theres a lot to be said for convention and we do have answers to all of those. But there's also something to be said about interpretation through context. (A subtle aspect is $a$ is not usually taken to be an integral index.)
– fleablood
22 mins ago
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
Your doubts are reasonable and your proposal is correct. The formula (18) in the referred paper and some more show all the same kind of typos (or miscalculations).
We have
beginalign*
l_2 + l_3 + l_4 + ..... + l_ncolorblue=sum_a=2^n l_a
endalign*
and not $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$, since $$sum_a=l_2^l_n a=sum_l_2leq aleq l_na$$
answered 1 hour ago
Markus Scheuer
57.2k452137
Your doubts are reasonable and your proposal is correct. The formula (18) in the referred paper and some more show all the same kind of typos (or miscalculations).
We have
beginalign*
l_2 + l_3 + l_4 + ..... + l_ncolorblue=sum_a=2^n l_a
endalign*
and not $l_2 + l_3 + l_4 + ..... + l_n = sum_a=l_2^l_n a$, since $$sum_a=l_2^l_n a=sum_l_2leq aleq l_na$$
answered 1 hour ago
Markus Scheuer
57.2k452137
edited 36 mins ago
answered 1 hour ago
Markus Scheuer
57.2k452137
answered 1 hour ago
Markus Scheuer
57.2k452137
answered 1 hour ago
Markus Scheuer
57.2k452137
57.2k452137
Is the notation $sumlimits_a=l_2^l_n a = sumlimits_ain l_2...l_n a$ really any more of an abuse of notation than $sumlimits_i=2^n l_i = sumlimits_i in 2....n l_i$?
– fleablood
59 mins ago
@fleablood: The author might have somewhat of this kind in mind, but as it is written we have $sum_i=l_2^nl_i=l_2+(l_2+1)+cdots+l_n$ which is not the same as $sum_iin2,ldots,nl_i$.
– Markus Scheuer
54 mins ago
That's not what we have written. We have $sumlimits_a=l_2^l_n a$ which can be interpreted either as $l_2 + (l_2 +1) + (l_2 + 3)+.... + l_n$ (all assuming that $l_n$ is a positive integer value larger than $l_2$). Or as $l_2 + l_3 + l_4 + .... + l_n$. Obviously in context the author meant the latter. And I agree he should have been clearer with the more common $sumlimits_i=2^n l_i$. But can we really claim that that is "wrong" when $sumlimits_i=2^n$ to mean $sumlimits_iin2,...,n$ is every bit as informal and relays just as much on "being clear in context"?
– fleablood
45 mins ago
@fleablood: We have $sum_a=l_2^l_na=sum_l_2leq a leq l_na$ which is not the same as $sum_ain2,ldots,nl_a$. You might want to check formula (2.2), (2.3) and (2.4) in Concrete Mathematics.
– Markus Scheuer
40 mins ago
true...writing $_a=j^k$ is taken to mean $jle a le k; a-jin mathbb Z$. But is that written in stone? I agree with you I'm inclined to call what the author did wrong and ambiguous at best. Is interpreting $_a=j^k$ to mean $j le a le k$ any more legitimate? What if $j$ and $k$ aren't integers? Why is the 'step' assumed to be $1$? What if $k < j$. etc. ... well, theres a lot to be said for convention and we do have answers to all of those. But there's also something to be said about interpretation through context. (A subtle aspect is $a$ is not usually taken to be an integral index.)
– fleablood
22 mins ago
 |Â
show 1 more comment
Is the notation $sumlimits_a=l_2^l_n a = sumlimits_ain l_2...l_n a$ really any more of an abuse of notation than $sumlimits_i=2^n l_i = sumlimits_i in 2....n l_i$?
– fleablood
59 mins ago
@fleablood: The author might have somewhat of this kind in mind, but as it is written we have $sum_i=l_2^nl_i=l_2+(l_2+1)+cdots+l_n$ which is not the same as $sum_iin2,ldots,nl_i$.
– Markus Scheuer
54 mins ago
That's not what we have written. We have $sumlimits_a=l_2^l_n a$ which can be interpreted either as $l_2 + (l_2 +1) + (l_2 + 3)+.... + l_n$ (all assuming that $l_n$ is a positive integer value larger than $l_2$). Or as $l_2 + l_3 + l_4 + .... + l_n$. Obviously in context the author meant the latter. And I agree he should have been clearer with the more common $sumlimits_i=2^n l_i$. But can we really claim that that is "wrong" when $sumlimits_i=2^n$ to mean $sumlimits_iin2,...,n$ is every bit as informal and relays just as much on "being clear in context"?
– fleablood
45 mins ago
@fleablood: We have $sum_a=l_2^l_na=sum_l_2leq a leq l_na$ which is not the same as $sum_ain2,ldots,nl_a$. You might want to check formula (2.2), (2.3) and (2.4) in Concrete Mathematics.
– Markus Scheuer
40 mins ago
true...writing $_a=j^k$ is taken to mean $jle a le k; a-jin mathbb Z$. But is that written in stone? I agree with you I'm inclined to call what the author did wrong and ambiguous at best. Is interpreting $_a=j^k$ to mean $j le a le k$ any more legitimate? What if $j$ and $k$ aren't integers? Why is the 'step' assumed to be $1$? What if $k < j$. etc. ... well, theres a lot to be said for convention and we do have answers to all of those. But there's also something to be said about interpretation through context. (A subtle aspect is $a$ is not usually taken to be an integral index.)
– fleablood
22 mins ago
Is the notation $sumlimits_a=l_2^l_n a = sumlimits_ain l_2...l_n a$ really any more of an abuse of notation than $sumlimits_i=2^n l_i = sumlimits_i in 2....n l_i$?
– fleablood
59 mins ago
Is the notation $sumlimits_a=l_2^l_n a = sumlimits_ain l_2...l_n a$ really any more of an abuse of notation than $sumlimits_i=2^n l_i = sumlimits_i in 2....n l_i$?
– fleablood
59 mins ago
@fleablood: The author might have somewhat of this kind in mind, but as it is written we have $sum_i=l_2^nl_i=l_2+(l_2+1)+cdots+l_n$ which is not the same as $sum_iin2,ldots,nl_i$.
– Markus Scheuer
54 mins ago
@fleablood: The author might have somewhat of this kind in mind, but as it is written we have $sum_i=l_2^nl_i=l_2+(l_2+1)+cdots+l_n$ which is not the same as $sum_iin2,ldots,nl_i$.
– Markus Scheuer
54 mins ago
That's not what we have written. We have $sumlimits_a=l_2^l_n a$ which can be interpreted either as $l_2 + (l_2 +1) + (l_2 + 3)+.... + l_n$ (all assuming that $l_n$ is a positive integer value larger than $l_2$). Or as $l_2 + l_3 + l_4 + .... + l_n$. Obviously in context the author meant the latter. And I agree he should have been clearer with the more common $sumlimits_i=2^n l_i$. But can we really claim that that is "wrong" when $sumlimits_i=2^n$ to mean $sumlimits_iin2,...,n$ is every bit as informal and relays just as much on "being clear in context"?
– fleablood
45 mins ago
That's not what we have written. We have $sumlimits_a=l_2^l_n a$ which can be interpreted either as $l_2 + (l_2 +1) + (l_2 + 3)+.... + l_n$ (all assuming that $l_n$ is a positive integer value larger than $l_2$). Or as $l_2 + l_3 + l_4 + .... + l_n$. Obviously in context the author meant the latter. And I agree he should have been clearer with the more common $sumlimits_i=2^n l_i$. But can we really claim that that is "wrong" when $sumlimits_i=2^n$ to mean $sumlimits_iin2,...,n$ is every bit as informal and relays just as much on "being clear in context"?
– fleablood
45 mins ago
@fleablood: We have $sum_a=l_2^l_na=sum_l_2leq a leq l_na$ which is not the same as $sum_ain2,ldots,nl_a$. You might want to check formula (2.2), (2.3) and (2.4) in Concrete Mathematics.
– Markus Scheuer
40 mins ago
@fleablood: We have $sum_a=l_2^l_na=sum_l_2leq a leq l_na$ which is not the same as $sum_ain2,ldots,nl_a$. You might want to check formula (2.2), (2.3) and (2.4) in Concrete Mathematics.
– Markus Scheuer
40 mins ago
true...writing $_a=j^k$ is taken to mean $jle a le k; a-jin mathbb Z$. But is that written in stone? I agree with you I'm inclined to call what the author did wrong and ambiguous at best. Is interpreting $_a=j^k$ to mean $j le a le k$ any more legitimate? What if $j$ and $k$ aren't integers? Why is the 'step' assumed to be $1$? What if $k < j$. etc. ... well, theres a lot to be said for convention and we do have answers to all of those. But there's also something to be said about interpretation through context. (A subtle aspect is $a$ is not usually taken to be an integral index.)
– fleablood
22 mins ago
true...writing $_a=j^k$ is taken to mean $jle a le k; a-jin mathbb Z$. But is that written in stone? I agree with you I'm inclined to call what the author did wrong and ambiguous at best. Is interpreting $_a=j^k$ to mean $j le a le k$ any more legitimate? What if $j$ and $k$ aren't integers? Why is the 'step' assumed to be $1$? What if $k < j$. etc. ... well, theres a lot to be said for convention and we do have answers to all of those. But there's also something to be said about interpretation through context. (A subtle aspect is $a$ is not usually taken to be an integral index.)
– fleablood
22 mins ago
 |Â
show 1 more comment
up vote
0
down vote
He actually does it in all the other equations as well. For example, look at equation (13):
$$
frac1l_2 sum_i = 1^l_2t_i = fracl_2 + 12t
$$
His l's are already decided in the definition of set $A_i$.
answered 2 hours ago
Abhi Devathi
1
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Who is "He" to whom you refer about "His" work?
– amWhy
2 hours ago
@amWhy presumably the author(s) of the linked paper.
– Brahadeesh
1 hour ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Brahadeesh
1 hour ago
add a comment |Â
up vote
0
down vote
He actually does it in all the other equations as well. For example, look at equation (13):
$$
frac1l_2 sum_i = 1^l_2t_i = fracl_2 + 12t
$$
His l's are already decided in the definition of set $A_i$.
answered 2 hours ago
Abhi Devathi
1
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Who is "He" to whom you refer about "His" work?
– amWhy
2 hours ago
@amWhy presumably the author(s) of the linked paper.
– Brahadeesh
1 hour ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Brahadeesh
1 hour ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
He actually does it in all the other equations as well. For example, look at equation (13):
$$
frac1l_2 sum_i = 1^l_2t_i = fracl_2 + 12t
$$
His l's are already decided in the definition of set $A_i$.
answered 2 hours ago
Abhi Devathi
1
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
He actually does it in all the other equations as well. For example, look at equation (13):
$$
frac1l_2 sum_i = 1^l_2t_i = fracl_2 + 12t
$$
His l's are already decided in the definition of set $A_i$.
answered 2 hours ago
Abhi Devathi
1
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Abhi Devathi
1
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Abhi Devathi
1
answered 2 hours ago
Abhi Devathi
1
1
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Abhi Devathi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Who is "He" to whom you refer about "His" work?
– amWhy
2 hours ago
@amWhy presumably the author(s) of the linked paper.
– Brahadeesh
1 hour ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Brahadeesh
1 hour ago
add a comment |Â
Who is "He" to whom you refer about "His" work?
– amWhy
2 hours ago
@amWhy presumably the author(s) of the linked paper.
– Brahadeesh
1 hour ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Brahadeesh
1 hour ago
Who is "He" to whom you refer about "His" work?
– amWhy
2 hours ago
Who is "He" to whom you refer about "His" work?
– amWhy
2 hours ago
@amWhy presumably the author(s) of the linked paper.
– Brahadeesh
1 hour ago
@amWhy presumably the author(s) of the linked paper.
– Brahadeesh
1 hour ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Brahadeesh
1 hour ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– Brahadeesh
1 hour ago
add a comment |Â
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1
...because they wanted to write it like that?
– Sean Roberson
3 hours ago
2
Because the authors are innumerate?
– Lord Shark the Unknown
3 hours ago
1
Your final line should have $sum_a=2^n l_a$.
– Lord Shark the Unknown
3 hours ago
1
@LordSharktheUnknown Or, better, replace $a$ by $i$ rather than $i$ by $a$.
– Ethan Bolker
3 hours ago