Mixing
When Sigma-summation stops with a decimal value
$displaystyle sum_{n=1}^{log_2x}$...
When you have a sum like this, what is the result of $log2_x$ since it may return a decimal/fractional value? Is it truncated, ceiled or floored?
I know I could do:
$displaystyle sum_{n=1}^{lceillog_2xrceil}$ or
$displaystyle sum_{n=1}^{lfloorlog_2xrfloor}$
to make it a whole number(integer) without any fuzz, but I've seen similar notation before without truncation or ceiling. So I havent yet found out what what the value goes to (when this summation stops).
summation
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
add a comment |
$displaystyle sum_{n=1}^{log_2x}$...
When you have a sum like this, what is the result of $log2_x$ since it may return a decimal/fractional value? Is it truncated, ceiled or floored?
I know I could do:
$displaystyle sum_{n=1}^{lceillog_2xrceil}$ or
$displaystyle sum_{n=1}^{lfloorlog_2xrfloor}$
to make it a whole number(integer) without any fuzz, but I've seen similar notation before without truncation or ceiling. So I havent yet found out what what the value goes to (when this summation stops).
summation
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
1
Usually this kind of thing is written in asymptotic contexts where the precise rounding rule doesn't change the final outcome.
– Ian
Nov 20 '18 at 17:07
add a comment |
$displaystyle sum_{n=1}^{log_2x}$...
When you have a sum like this, what is the result of $log2_x$ since it may return a decimal/fractional value? Is it truncated, ceiled or floored?
I know I could do:
$displaystyle sum_{n=1}^{lceillog_2xrceil}$ or
$displaystyle sum_{n=1}^{lfloorlog_2xrfloor}$
to make it a whole number(integer) without any fuzz, but I've seen similar notation before without truncation or ceiling. So I havent yet found out what what the value goes to (when this summation stops).
summation
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
$displaystyle sum_{n=1}^{log_2x}$...
When you have a sum like this, what is the result of $log2_x$ since it may return a decimal/fractional value? Is it truncated, ceiled or floored?
I know I could do:
$displaystyle sum_{n=1}^{lceillog_2xrceil}$ or
$displaystyle sum_{n=1}^{lfloorlog_2xrfloor}$
to make it a whole number(integer) without any fuzz, but I've seen similar notation before without truncation or ceiling. So I havent yet found out what what the value goes to (when this summation stops).
summation
summation
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
asked Nov 20 '18 at 17:03
Natural Number Guy
408517
408517
1
Usually this kind of thing is written in asymptotic contexts where the precise rounding rule doesn't change the final outcome.
– Ian
Nov 20 '18 at 17:07
add a comment |
1
Usually this kind of thing is written in asymptotic contexts where the precise rounding rule doesn't change the final outcome.
– Ian
Nov 20 '18 at 17:07
1
1
Usually this kind of thing is written in asymptotic contexts where the precise rounding rule doesn't change the final outcome.
– Ian
Nov 20 '18 at 17:07
Usually this kind of thing is written in asymptotic contexts where the precise rounding rule doesn't change the final outcome.
– Ian
Nov 20 '18 at 17:07
add a comment |
1 Answer
1
active
oldest
votes
The usual convention is that
$$ sum_{n=1}^{N} a_n $$
is the sum of all $a_n$ with indices between the lower and upper limits of summation. Hence
$$ sum_{n=1}^{N} a_n = sum_{1le n le N} a_n = sum_{n=1}^{lfloor N rfloor} a_n = a_1 + a_2 + dotsb + a_{lfloor N rfloor}. $$
That being said, if you are interested in what happens as $N$ goes to infinity (as is often the case), it doesn't really matter which convention you choose.
The following is probably overkill, but the above can actually be justified in a meaningful way via a branch of mathematics called measure theory: a series is a special kind of integral. Indeed, if you know any measure theory, a series is an integral of the form
$$ sum_{n=1}^{N} a_n = int_{[0,N]} a(x) ,mathrm{d}mu(x), $$
where $a : mathbb{R} to mathbb{R}$ (or some other codomain) such that $$ a(x) = begin{cases} a_n & text{if $x = n in mathbb{N}$, and} \ 0 & text{otherwise}, end{cases} $$
and $mu$ is counting measure on $mathbb{N}$. If $N$ is not an integer, then
begin{align}
sum_{n=1}^{N} a_n &= int_{[0,N]} a(x),mathrm{d}mu(x) && text{(by definition)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) + int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) && text{(additivity of integrals)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) tag{1}\
&= sum_{n=1}^{lfloor N rfloor} a_n. && text{(definition)}
end{align}
The only mystery here is at (1). Here, we are using the fact that the measure of an interval which contains no integers must be zero. That is,
$$ mu((lfloor N rfloor, N]) = 0. $$
In a measure space, integration over a set of measure zero is always zero. Hence
$$ int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) = 0. $$
In short, the identity above is justified, that is
$$ sum_{n=1}^{N} a_n = sum_{n=1}^{lfloor N rfloor} a_n. $$
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
Ok. so if $displaystyle sum_{n=1}^{3.9}1$, would give the result = $1+1+1 = 3$ ?
– Natural Number Guy
Nov 20 '18 at 17:16
Yes, that is correct.
– Xander Henderson
Nov 20 '18 at 17:20
add a comment |
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1 Answer
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The usual convention is that
$$ sum_{n=1}^{N} a_n $$
is the sum of all $a_n$ with indices between the lower and upper limits of summation. Hence
$$ sum_{n=1}^{N} a_n = sum_{1le n le N} a_n = sum_{n=1}^{lfloor N rfloor} a_n = a_1 + a_2 + dotsb + a_{lfloor N rfloor}. $$
That being said, if you are interested in what happens as $N$ goes to infinity (as is often the case), it doesn't really matter which convention you choose.
The following is probably overkill, but the above can actually be justified in a meaningful way via a branch of mathematics called measure theory: a series is a special kind of integral. Indeed, if you know any measure theory, a series is an integral of the form
$$ sum_{n=1}^{N} a_n = int_{[0,N]} a(x) ,mathrm{d}mu(x), $$
where $a : mathbb{R} to mathbb{R}$ (or some other codomain) such that $$ a(x) = begin{cases} a_n & text{if $x = n in mathbb{N}$, and} \ 0 & text{otherwise}, end{cases} $$
and $mu$ is counting measure on $mathbb{N}$. If $N$ is not an integer, then
begin{align}
sum_{n=1}^{N} a_n &= int_{[0,N]} a(x),mathrm{d}mu(x) && text{(by definition)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) + int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) && text{(additivity of integrals)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) tag{1}\
&= sum_{n=1}^{lfloor N rfloor} a_n. && text{(definition)}
end{align}
The only mystery here is at (1). Here, we are using the fact that the measure of an interval which contains no integers must be zero. That is,
$$ mu((lfloor N rfloor, N]) = 0. $$
In a measure space, integration over a set of measure zero is always zero. Hence
$$ int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) = 0. $$
In short, the identity above is justified, that is
$$ sum_{n=1}^{N} a_n = sum_{n=1}^{lfloor N rfloor} a_n. $$
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
Ok. so if $displaystyle sum_{n=1}^{3.9}1$, would give the result = $1+1+1 = 3$ ?
– Natural Number Guy
Nov 20 '18 at 17:16
Yes, that is correct.
– Xander Henderson
Nov 20 '18 at 17:20
add a comment |
The usual convention is that
$$ sum_{n=1}^{N} a_n $$
is the sum of all $a_n$ with indices between the lower and upper limits of summation. Hence
$$ sum_{n=1}^{N} a_n = sum_{1le n le N} a_n = sum_{n=1}^{lfloor N rfloor} a_n = a_1 + a_2 + dotsb + a_{lfloor N rfloor}. $$
That being said, if you are interested in what happens as $N$ goes to infinity (as is often the case), it doesn't really matter which convention you choose.
The following is probably overkill, but the above can actually be justified in a meaningful way via a branch of mathematics called measure theory: a series is a special kind of integral. Indeed, if you know any measure theory, a series is an integral of the form
$$ sum_{n=1}^{N} a_n = int_{[0,N]} a(x) ,mathrm{d}mu(x), $$
where $a : mathbb{R} to mathbb{R}$ (or some other codomain) such that $$ a(x) = begin{cases} a_n & text{if $x = n in mathbb{N}$, and} \ 0 & text{otherwise}, end{cases} $$
and $mu$ is counting measure on $mathbb{N}$. If $N$ is not an integer, then
begin{align}
sum_{n=1}^{N} a_n &= int_{[0,N]} a(x),mathrm{d}mu(x) && text{(by definition)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) + int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) && text{(additivity of integrals)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) tag{1}\
&= sum_{n=1}^{lfloor N rfloor} a_n. && text{(definition)}
end{align}
The only mystery here is at (1). Here, we are using the fact that the measure of an interval which contains no integers must be zero. That is,
$$ mu((lfloor N rfloor, N]) = 0. $$
In a measure space, integration over a set of measure zero is always zero. Hence
$$ int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) = 0. $$
In short, the identity above is justified, that is
$$ sum_{n=1}^{N} a_n = sum_{n=1}^{lfloor N rfloor} a_n. $$
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
Ok. so if $displaystyle sum_{n=1}^{3.9}1$, would give the result = $1+1+1 = 3$ ?
– Natural Number Guy
Nov 20 '18 at 17:16
Yes, that is correct.
– Xander Henderson
Nov 20 '18 at 17:20
add a comment |
The usual convention is that
$$ sum_{n=1}^{N} a_n $$
is the sum of all $a_n$ with indices between the lower and upper limits of summation. Hence
$$ sum_{n=1}^{N} a_n = sum_{1le n le N} a_n = sum_{n=1}^{lfloor N rfloor} a_n = a_1 + a_2 + dotsb + a_{lfloor N rfloor}. $$
That being said, if you are interested in what happens as $N$ goes to infinity (as is often the case), it doesn't really matter which convention you choose.
The following is probably overkill, but the above can actually be justified in a meaningful way via a branch of mathematics called measure theory: a series is a special kind of integral. Indeed, if you know any measure theory, a series is an integral of the form
$$ sum_{n=1}^{N} a_n = int_{[0,N]} a(x) ,mathrm{d}mu(x), $$
where $a : mathbb{R} to mathbb{R}$ (or some other codomain) such that $$ a(x) = begin{cases} a_n & text{if $x = n in mathbb{N}$, and} \ 0 & text{otherwise}, end{cases} $$
and $mu$ is counting measure on $mathbb{N}$. If $N$ is not an integer, then
begin{align}
sum_{n=1}^{N} a_n &= int_{[0,N]} a(x),mathrm{d}mu(x) && text{(by definition)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) + int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) && text{(additivity of integrals)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) tag{1}\
&= sum_{n=1}^{lfloor N rfloor} a_n. && text{(definition)}
end{align}
The only mystery here is at (1). Here, we are using the fact that the measure of an interval which contains no integers must be zero. That is,
$$ mu((lfloor N rfloor, N]) = 0. $$
In a measure space, integration over a set of measure zero is always zero. Hence
$$ int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) = 0. $$
In short, the identity above is justified, that is
$$ sum_{n=1}^{N} a_n = sum_{n=1}^{lfloor N rfloor} a_n. $$
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
The usual convention is that
$$ sum_{n=1}^{N} a_n $$
is the sum of all $a_n$ with indices between the lower and upper limits of summation. Hence
$$ sum_{n=1}^{N} a_n = sum_{1le n le N} a_n = sum_{n=1}^{lfloor N rfloor} a_n = a_1 + a_2 + dotsb + a_{lfloor N rfloor}. $$
That being said, if you are interested in what happens as $N$ goes to infinity (as is often the case), it doesn't really matter which convention you choose.
The following is probably overkill, but the above can actually be justified in a meaningful way via a branch of mathematics called measure theory: a series is a special kind of integral. Indeed, if you know any measure theory, a series is an integral of the form
$$ sum_{n=1}^{N} a_n = int_{[0,N]} a(x) ,mathrm{d}mu(x), $$
where $a : mathbb{R} to mathbb{R}$ (or some other codomain) such that $$ a(x) = begin{cases} a_n & text{if $x = n in mathbb{N}$, and} \ 0 & text{otherwise}, end{cases} $$
and $mu$ is counting measure on $mathbb{N}$. If $N$ is not an integer, then
begin{align}
sum_{n=1}^{N} a_n &= int_{[0,N]} a(x),mathrm{d}mu(x) && text{(by definition)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) + int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) && text{(additivity of integrals)} \
&= int_{[0,lfloor N rfloor]} a(x) , mathrm{d}mu(x) tag{1}\
&= sum_{n=1}^{lfloor N rfloor} a_n. && text{(definition)}
end{align}
The only mystery here is at (1). Here, we are using the fact that the measure of an interval which contains no integers must be zero. That is,
$$ mu((lfloor N rfloor, N]) = 0. $$
In a measure space, integration over a set of measure zero is always zero. Hence
$$ int_{(lfloor Nrfloor, N]} a(x) ,mathrm{d}mu(x) = 0. $$
In short, the identity above is justified, that is
$$ sum_{n=1}^{N} a_n = sum_{n=1}^{lfloor N rfloor} a_n. $$
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
edited Nov 20 '18 at 17:19
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
answered Nov 20 '18 at 17:07
Xander Henderson
14.1k103554
14.1k103554
Ok. so if $displaystyle sum_{n=1}^{3.9}1$, would give the result = $1+1+1 = 3$ ?
– Natural Number Guy
Nov 20 '18 at 17:16
Yes, that is correct.
– Xander Henderson
Nov 20 '18 at 17:20
add a comment |
Ok. so if $displaystyle sum_{n=1}^{3.9}1$, would give the result = $1+1+1 = 3$ ?
– Natural Number Guy
Nov 20 '18 at 17:16
Yes, that is correct.
– Xander Henderson
Nov 20 '18 at 17:20
Ok. so if $displaystyle sum_{n=1}^{3.9}1$, would give the result = $1+1+1 = 3$ ?
– Natural Number Guy
Nov 20 '18 at 17:16
Ok. so if $displaystyle sum_{n=1}^{3.9}1$, would give the result = $1+1+1 = 3$ ?
– Natural Number Guy
Nov 20 '18 at 17:16
Yes, that is correct.
– Xander Henderson
Nov 20 '18 at 17:20
Yes, that is correct.
– Xander Henderson
Nov 20 '18 at 17:20
add a comment |
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1
Usually this kind of thing is written in asymptotic contexts where the precise rounding rule doesn't change the final outcome.
– Ian
Nov 20 '18 at 17:07