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Sum operator precedence
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$begingroup$
I'm trying to read some simple equations and in order to interpret them in the right way I need to know $sum$ and $prod $ operator range/precedence.
$$ sum p(s, a) +gamma $$
is equal to $sum(p(s,a) + gamma)$ or $sum(p(s,a)) + gamma$.
The same question is for product operator.
Also, for UCB1 formula
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}$$
should I treat it like this
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) + sqrt{frac{2log t}{N_t(a)}} Bigr) $$
or like this?
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) Bigr) + sqrt{frac{2log t}{N_t(a)}} $$
Could you please clarify those for me?
summation operator-theory products
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
$endgroup$
add a comment |
$begingroup$
I'm trying to read some simple equations and in order to interpret them in the right way I need to know $sum$ and $prod $ operator range/precedence.
$$ sum p(s, a) +gamma $$
is equal to $sum(p(s,a) + gamma)$ or $sum(p(s,a)) + gamma$.
The same question is for product operator.
Also, for UCB1 formula
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}$$
should I treat it like this
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) + sqrt{frac{2log t}{N_t(a)}} Bigr) $$
or like this?
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) Bigr) + sqrt{frac{2log t}{N_t(a)}} $$
Could you please clarify those for me?
summation operator-theory products
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
$endgroup$
$begingroup$
For $A_t$ it is the first option since the root term depends on $a$ which is iterated by $text{argmax}$. In other words $a$ cannot appear outside $text{argmax}$ like it does in the second option. For the sum I think it depends on the context.
$endgroup$
– M. Nestor
Mar 22 at 12:50
add a comment |
$begingroup$
I'm trying to read some simple equations and in order to interpret them in the right way I need to know $sum$ and $prod $ operator range/precedence.
$$ sum p(s, a) +gamma $$
is equal to $sum(p(s,a) + gamma)$ or $sum(p(s,a)) + gamma$.
The same question is for product operator.
Also, for UCB1 formula
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}$$
should I treat it like this
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) + sqrt{frac{2log t}{N_t(a)}} Bigr) $$
or like this?
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) Bigr) + sqrt{frac{2log t}{N_t(a)}} $$
Could you please clarify those for me?
summation operator-theory products
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
$endgroup$
I'm trying to read some simple equations and in order to interpret them in the right way I need to know $sum$ and $prod $ operator range/precedence.
$$ sum p(s, a) +gamma $$
is equal to $sum(p(s,a) + gamma)$ or $sum(p(s,a)) + gamma$.
The same question is for product operator.
Also, for UCB1 formula
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}$$
should I treat it like this
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) + sqrt{frac{2log t}{N_t(a)}} Bigr) $$
or like this?
$$ A_t = underset{ainmathcal{A}}{operatorname{argmax}}Bigl( Q_t(a) Bigr) + sqrt{frac{2log t}{N_t(a)}} $$
Could you please clarify those for me?
summation operator-theory products
summation operator-theory products
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
asked Mar 22 at 12:38
Most WantedMost Wanted
1326
1326
$begingroup$
For $A_t$ it is the first option since the root term depends on $a$ which is iterated by $text{argmax}$. In other words $a$ cannot appear outside $text{argmax}$ like it does in the second option. For the sum I think it depends on the context.
$endgroup$
– M. Nestor
Mar 22 at 12:50
add a comment |
$begingroup$
For $A_t$ it is the first option since the root term depends on $a$ which is iterated by $text{argmax}$. In other words $a$ cannot appear outside $text{argmax}$ like it does in the second option. For the sum I think it depends on the context.
$endgroup$
– M. Nestor
Mar 22 at 12:50
$begingroup$
For $A_t$ it is the first option since the root term depends on $a$ which is iterated by $text{argmax}$. In other words $a$ cannot appear outside $text{argmax}$ like it does in the second option. For the sum I think it depends on the context.
$endgroup$
– M. Nestor
Mar 22 at 12:50
$begingroup$
For $A_t$ it is the first option since the root term depends on $a$ which is iterated by $text{argmax}$. In other words $a$ cannot appear outside $text{argmax}$ like it does in the second option. For the sum I think it depends on the context.
$endgroup$
– M. Nestor
Mar 22 at 12:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The $sum$ operator and $+$ have the same precedence level, so
begin{align*}
sum p(s, a) +gamma &=left(sum p(s,a)right)+gamma
end{align*}
contrary to
begin{align*}
sum p(s, a) cdotgamma &=sum left(p(s,a)cdotgammaright)
end{align*}
The $max$ operator binds stronger than the $+$ operator, so
begin{align*}
underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}
&=left(underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a)right) + sqrt{frac{2log t}{N_t(a)}}
end{align*}
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
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$begingroup$
The $sum$ operator and $+$ have the same precedence level, so
begin{align*}
sum p(s, a) +gamma &=left(sum p(s,a)right)+gamma
end{align*}
contrary to
begin{align*}
sum p(s, a) cdotgamma &=sum left(p(s,a)cdotgammaright)
end{align*}
The $max$ operator binds stronger than the $+$ operator, so
begin{align*}
underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}
&=left(underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a)right) + sqrt{frac{2log t}{N_t(a)}}
end{align*}
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
$begingroup$
The $sum$ operator and $+$ have the same precedence level, so
begin{align*}
sum p(s, a) +gamma &=left(sum p(s,a)right)+gamma
end{align*}
contrary to
begin{align*}
sum p(s, a) cdotgamma &=sum left(p(s,a)cdotgammaright)
end{align*}
The $max$ operator binds stronger than the $+$ operator, so
begin{align*}
underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}
&=left(underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a)right) + sqrt{frac{2log t}{N_t(a)}}
end{align*}
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
$begingroup$
The $sum$ operator and $+$ have the same precedence level, so
begin{align*}
sum p(s, a) +gamma &=left(sum p(s,a)right)+gamma
end{align*}
contrary to
begin{align*}
sum p(s, a) cdotgamma &=sum left(p(s,a)cdotgammaright)
end{align*}
The $max$ operator binds stronger than the $+$ operator, so
begin{align*}
underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}
&=left(underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a)right) + sqrt{frac{2log t}{N_t(a)}}
end{align*}
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
The $sum$ operator and $+$ have the same precedence level, so
begin{align*}
sum p(s, a) +gamma &=left(sum p(s,a)right)+gamma
end{align*}
contrary to
begin{align*}
sum p(s, a) cdotgamma &=sum left(p(s,a)cdotgammaright)
end{align*}
The $max$ operator binds stronger than the $+$ operator, so
begin{align*}
underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a) + sqrt{frac{2log t}{N_t(a)}}
&=left(underset{ainmathcal{A}}{operatorname{argmax}} Q_t(a)right) + sqrt{frac{2log t}{N_t(a)}}
end{align*}
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
edited Mar 22 at 19:24
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
answered Mar 22 at 19:17
Markus ScheuerMarkus Scheuer
64.1k460152
64.1k460152
add a comment |
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$begingroup$
For $A_t$ it is the first option since the root term depends on $a$ which is iterated by $text{argmax}$. In other words $a$ cannot appear outside $text{argmax}$ like it does in the second option. For the sum I think it depends on the context.
$endgroup$
– M. Nestor
Mar 22 at 12:50