Splitting up compound inequalities of multiple variables
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I'm reading through Concrete Mathematics 2nd ed. In the last example in section 2.4 Multiple Sums (p.40), they go from this:
$$ sum_{1le j < k+jle n} frac{1}{k} $$
to this, in one step:
$$ sum_{1le kle n} sum_{1le jle n-k} frac{1}{k} $$
The purpose is to sum first on $j$ which is desirable since $j$ does not appear in $frac1k$. ($n$ is constant while $j,k$ are index variables.)
I can understand this step graphically by plotting it, but I can't figure out how they did it purely symbolically. Is there a general non-graphical way to split up these compound inequalities, i.e. to factor out a certain variable like $j$?
sequences-and-series inequality summation
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
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add a comment |
$begingroup$
I'm reading through Concrete Mathematics 2nd ed. In the last example in section 2.4 Multiple Sums (p.40), they go from this:
$$ sum_{1le j < k+jle n} frac{1}{k} $$
to this, in one step:
$$ sum_{1le kle n} sum_{1le jle n-k} frac{1}{k} $$
The purpose is to sum first on $j$ which is desirable since $j$ does not appear in $frac1k$. ($n$ is constant while $j,k$ are index variables.)
I can understand this step graphically by plotting it, but I can't figure out how they did it purely symbolically. Is there a general non-graphical way to split up these compound inequalities, i.e. to factor out a certain variable like $j$?
sequences-and-series inequality summation
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
$endgroup$
$begingroup$
Generally when you want to evaluate a double summation, you can make a 2d grid and note that you can either sum by rows or by columns. Try that here
$endgroup$
– Sandeep Silwal
Dec 27 '18 at 20:41
$begingroup$
For people who might be interested in future, what part of the book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 28 '18 at 5:42
add a comment |
$begingroup$
I'm reading through Concrete Mathematics 2nd ed. In the last example in section 2.4 Multiple Sums (p.40), they go from this:
$$ sum_{1le j < k+jle n} frac{1}{k} $$
to this, in one step:
$$ sum_{1le kle n} sum_{1le jle n-k} frac{1}{k} $$
The purpose is to sum first on $j$ which is desirable since $j$ does not appear in $frac1k$. ($n$ is constant while $j,k$ are index variables.)
I can understand this step graphically by plotting it, but I can't figure out how they did it purely symbolically. Is there a general non-graphical way to split up these compound inequalities, i.e. to factor out a certain variable like $j$?
sequences-and-series inequality summation
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
$endgroup$
I'm reading through Concrete Mathematics 2nd ed. In the last example in section 2.4 Multiple Sums (p.40), they go from this:
$$ sum_{1le j < k+jle n} frac{1}{k} $$
to this, in one step:
$$ sum_{1le kle n} sum_{1le jle n-k} frac{1}{k} $$
The purpose is to sum first on $j$ which is desirable since $j$ does not appear in $frac1k$. ($n$ is constant while $j,k$ are index variables.)
I can understand this step graphically by plotting it, but I can't figure out how they did it purely symbolically. Is there a general non-graphical way to split up these compound inequalities, i.e. to factor out a certain variable like $j$?
sequences-and-series inequality summation
sequences-and-series inequality summation
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
edited Dec 28 '18 at 22:55
EpicOrange
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
asked Dec 27 '18 at 20:21
EpicOrangeEpicOrange
1155
1155
$begingroup$
Generally when you want to evaluate a double summation, you can make a 2d grid and note that you can either sum by rows or by columns. Try that here
$endgroup$
– Sandeep Silwal
Dec 27 '18 at 20:41
$begingroup$
For people who might be interested in future, what part of the book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 28 '18 at 5:42
add a comment |
$begingroup$
Generally when you want to evaluate a double summation, you can make a 2d grid and note that you can either sum by rows or by columns. Try that here
$endgroup$
– Sandeep Silwal
Dec 27 '18 at 20:41
$begingroup$
For people who might be interested in future, what part of the book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 28 '18 at 5:42
$begingroup$
Generally when you want to evaluate a double summation, you can make a 2d grid and note that you can either sum by rows or by columns. Try that here
$endgroup$
– Sandeep Silwal
Dec 27 '18 at 20:41
$begingroup$
Generally when you want to evaluate a double summation, you can make a 2d grid and note that you can either sum by rows or by columns. Try that here
$endgroup$
– Sandeep Silwal
Dec 27 '18 at 20:41
$begingroup$
For people who might be interested in future, what part of the book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 28 '18 at 5:42
$begingroup$
For people who might be interested in future, what part of the book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 28 '18 at 5:42
add a comment |
1 Answer
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$begingroup$
We obtain
begin{align*}
sum_{color{blue}{1leq j<k+jleq n}}frac{1}{k}&=sum_{{1leq kleq n-1,1leq jleq n-1}atop{k+jleq n}}frac{1}{k}tag{1}\
&=sum_{1leq kleq n-1}sum_{1leq jleq n-k}frac{1}{k}tag{2}\
&=sum_{color{blue}{1leq kleq n}}sum_{color{blue}{1leq jleq n-k}}frac{1}{k}tag{3}
end{align*}
Comment:
In (1) we write the index region of the RHS somewhat more conveniently as preparation for the next step(s). We conclude thereby from $1leq j<k+j$ that $kgeq 1$ and from $k+jleq n$ that $kleq n-1$ resulting in $1leq kleq n-1$. Analogously we see that $jleq n-1$, since $k+jleq n$ and $kgeq 1$.
In (2) we use from (1) that $jleq n-k$, since $k+jleq n$. We observe from (1) that both $1leq jleq n-1$ and $jleq n-k$ has to be valid which can be written as $1leq jleq n-k$.
In (3) we write for convenience only $n$ instead of $n-1$ as upper limit of the index $k$ without changing anything, since the index region $1leq jleq 0$ of the inner sum is then the empty set, i.e. the inner sum $sum_{1leq jleq 0}frac{1}{k}=0$ when $k=n$.
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
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add a comment |
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$begingroup$
We obtain
begin{align*}
sum_{color{blue}{1leq j<k+jleq n}}frac{1}{k}&=sum_{{1leq kleq n-1,1leq jleq n-1}atop{k+jleq n}}frac{1}{k}tag{1}\
&=sum_{1leq kleq n-1}sum_{1leq jleq n-k}frac{1}{k}tag{2}\
&=sum_{color{blue}{1leq kleq n}}sum_{color{blue}{1leq jleq n-k}}frac{1}{k}tag{3}
end{align*}
Comment:
In (1) we write the index region of the RHS somewhat more conveniently as preparation for the next step(s). We conclude thereby from $1leq j<k+j$ that $kgeq 1$ and from $k+jleq n$ that $kleq n-1$ resulting in $1leq kleq n-1$. Analogously we see that $jleq n-1$, since $k+jleq n$ and $kgeq 1$.
In (2) we use from (1) that $jleq n-k$, since $k+jleq n$. We observe from (1) that both $1leq jleq n-1$ and $jleq n-k$ has to be valid which can be written as $1leq jleq n-k$.
In (3) we write for convenience only $n$ instead of $n-1$ as upper limit of the index $k$ without changing anything, since the index region $1leq jleq 0$ of the inner sum is then the empty set, i.e. the inner sum $sum_{1leq jleq 0}frac{1}{k}=0$ when $k=n$.
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
$endgroup$
add a comment |
$begingroup$
We obtain
begin{align*}
sum_{color{blue}{1leq j<k+jleq n}}frac{1}{k}&=sum_{{1leq kleq n-1,1leq jleq n-1}atop{k+jleq n}}frac{1}{k}tag{1}\
&=sum_{1leq kleq n-1}sum_{1leq jleq n-k}frac{1}{k}tag{2}\
&=sum_{color{blue}{1leq kleq n}}sum_{color{blue}{1leq jleq n-k}}frac{1}{k}tag{3}
end{align*}
Comment:
In (1) we write the index region of the RHS somewhat more conveniently as preparation for the next step(s). We conclude thereby from $1leq j<k+j$ that $kgeq 1$ and from $k+jleq n$ that $kleq n-1$ resulting in $1leq kleq n-1$. Analogously we see that $jleq n-1$, since $k+jleq n$ and $kgeq 1$.
In (2) we use from (1) that $jleq n-k$, since $k+jleq n$. We observe from (1) that both $1leq jleq n-1$ and $jleq n-k$ has to be valid which can be written as $1leq jleq n-k$.
In (3) we write for convenience only $n$ instead of $n-1$ as upper limit of the index $k$ without changing anything, since the index region $1leq jleq 0$ of the inner sum is then the empty set, i.e. the inner sum $sum_{1leq jleq 0}frac{1}{k}=0$ when $k=n$.
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
$endgroup$
add a comment |
$begingroup$
We obtain
begin{align*}
sum_{color{blue}{1leq j<k+jleq n}}frac{1}{k}&=sum_{{1leq kleq n-1,1leq jleq n-1}atop{k+jleq n}}frac{1}{k}tag{1}\
&=sum_{1leq kleq n-1}sum_{1leq jleq n-k}frac{1}{k}tag{2}\
&=sum_{color{blue}{1leq kleq n}}sum_{color{blue}{1leq jleq n-k}}frac{1}{k}tag{3}
end{align*}
Comment:
In (1) we write the index region of the RHS somewhat more conveniently as preparation for the next step(s). We conclude thereby from $1leq j<k+j$ that $kgeq 1$ and from $k+jleq n$ that $kleq n-1$ resulting in $1leq kleq n-1$. Analogously we see that $jleq n-1$, since $k+jleq n$ and $kgeq 1$.
In (2) we use from (1) that $jleq n-k$, since $k+jleq n$. We observe from (1) that both $1leq jleq n-1$ and $jleq n-k$ has to be valid which can be written as $1leq jleq n-k$.
In (3) we write for convenience only $n$ instead of $n-1$ as upper limit of the index $k$ without changing anything, since the index region $1leq jleq 0$ of the inner sum is then the empty set, i.e. the inner sum $sum_{1leq jleq 0}frac{1}{k}=0$ when $k=n$.
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
$endgroup$
We obtain
begin{align*}
sum_{color{blue}{1leq j<k+jleq n}}frac{1}{k}&=sum_{{1leq kleq n-1,1leq jleq n-1}atop{k+jleq n}}frac{1}{k}tag{1}\
&=sum_{1leq kleq n-1}sum_{1leq jleq n-k}frac{1}{k}tag{2}\
&=sum_{color{blue}{1leq kleq n}}sum_{color{blue}{1leq jleq n-k}}frac{1}{k}tag{3}
end{align*}
Comment:
In (1) we write the index region of the RHS somewhat more conveniently as preparation for the next step(s). We conclude thereby from $1leq j<k+j$ that $kgeq 1$ and from $k+jleq n$ that $kleq n-1$ resulting in $1leq kleq n-1$. Analogously we see that $jleq n-1$, since $k+jleq n$ and $kgeq 1$.
In (2) we use from (1) that $jleq n-k$, since $k+jleq n$. We observe from (1) that both $1leq jleq n-1$ and $jleq n-k$ has to be valid which can be written as $1leq jleq n-k$.
In (3) we write for convenience only $n$ instead of $n-1$ as upper limit of the index $k$ without changing anything, since the index region $1leq jleq 0$ of the inner sum is then the empty set, i.e. the inner sum $sum_{1leq jleq 0}frac{1}{k}=0$ when $k=n$.
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
answered Dec 28 '18 at 18:58
Markus ScheuerMarkus Scheuer
62.6k460150
62.6k460150
add a comment |
add a comment |
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$begingroup$
Generally when you want to evaluate a double summation, you can make a 2d grid and note that you can either sum by rows or by columns. Try that here
$endgroup$
– Sandeep Silwal
Dec 27 '18 at 20:41
$begingroup$
For people who might be interested in future, what part of the book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 28 '18 at 5:42