combinatorial argument for Exponential Generating Function [closed]
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Could you tell me how can I use a combinatorial argument to prove that $(e^x)^n = e^nx?$
combinatorics exponentiation
edited Sep 2 at 13:09
amWhy
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asked Sep 2 at 12:47
Saleh Verdi
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closed as off-topic by amWhy, Shailesh, José Carlos Santos, m_t_, Isaac Browne Sep 2 at 14:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Shailesh, Jos̩ Carlos Santos, m_t_, Isaac Browne
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up vote
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favorite
Could you tell me how can I use a combinatorial argument to prove that $(e^x)^n = e^nx?$
combinatorics exponentiation
edited Sep 2 at 13:09
amWhy
190k26221433
asked Sep 2 at 12:47
Saleh Verdi
1
closed as off-topic by amWhy, Shailesh, José Carlos Santos, m_t_, Isaac Browne Sep 2 at 14:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Shailesh, Jos̩ Carlos Santos, m_t_, Isaac Browne
The sum of all multinomial coefficients of the form $dfracr!r_1! r_2!cdots r_m!$ where we keep $r$ fixed and sum over the $r_j$ while keeping $sum_j r_j = r$ is $m^r$. This is because the multinomial coefficient gives you the number of ways you can choose the $r_j$ from the $r$ to, say, put them into box nr. $j$, and if you sum over all such partitions, you'll get the total number of possibilities of putting the $r$ objects in $m$ boxes. For each object there are $m$ choices, so the total is $m^r$.
– Count Iblis
Sep 2 at 13:13
Please show your work.
– Marko Riedel
Sep 2 at 14:02
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Could you tell me how can I use a combinatorial argument to prove that $(e^x)^n = e^nx?$
combinatorics exponentiation
edited Sep 2 at 13:09
amWhy
190k26221433
asked Sep 2 at 12:47
Saleh Verdi
1
Could you tell me how can I use a combinatorial argument to prove that $(e^x)^n = e^nx?$
combinatorics exponentiation
combinatorics exponentiation
edited Sep 2 at 13:09
amWhy
190k26221433
asked Sep 2 at 12:47
Saleh Verdi
1
edited Sep 2 at 13:09
amWhy
190k26221433
asked Sep 2 at 12:47
Saleh Verdi
1
edited Sep 2 at 13:09
amWhy
190k26221433
edited Sep 2 at 13:09
amWhy
190k26221433
edited Sep 2 at 13:09
amWhy
190k26221433
190k26221433
asked Sep 2 at 12:47
Saleh Verdi
1
asked Sep 2 at 12:47
Saleh Verdi
1
asked Sep 2 at 12:47
Saleh Verdi
1
1
closed as off-topic by amWhy, Shailesh, José Carlos Santos, m_t_, Isaac Browne Sep 2 at 14:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Shailesh, Jos̩ Carlos Santos, m_t_, Isaac Browne
closed as off-topic by amWhy, Shailesh, José Carlos Santos, m_t_, Isaac Browne Sep 2 at 14:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Shailesh, Jos̩ Carlos Santos, m_t_, Isaac Browne
The sum of all multinomial coefficients of the form $dfracr!r_1! r_2!cdots r_m!$ where we keep $r$ fixed and sum over the $r_j$ while keeping $sum_j r_j = r$ is $m^r$. This is because the multinomial coefficient gives you the number of ways you can choose the $r_j$ from the $r$ to, say, put them into box nr. $j$, and if you sum over all such partitions, you'll get the total number of possibilities of putting the $r$ objects in $m$ boxes. For each object there are $m$ choices, so the total is $m^r$.
– Count Iblis
Sep 2 at 13:13
Please show your work.
– Marko Riedel
Sep 2 at 14:02
add a comment |Â
The sum of all multinomial coefficients of the form $dfracr!r_1! r_2!cdots r_m!$ where we keep $r$ fixed and sum over the $r_j$ while keeping $sum_j r_j = r$ is $m^r$. This is because the multinomial coefficient gives you the number of ways you can choose the $r_j$ from the $r$ to, say, put them into box nr. $j$, and if you sum over all such partitions, you'll get the total number of possibilities of putting the $r$ objects in $m$ boxes. For each object there are $m$ choices, so the total is $m^r$.
– Count Iblis
Sep 2 at 13:13
Please show your work.
– Marko Riedel
Sep 2 at 14:02
The sum of all multinomial coefficients of the form $dfracr!r_1! r_2!cdots r_m!$ where we keep $r$ fixed and sum over the $r_j$ while keeping $sum_j r_j = r$ is $m^r$. This is because the multinomial coefficient gives you the number of ways you can choose the $r_j$ from the $r$ to, say, put them into box nr. $j$, and if you sum over all such partitions, you'll get the total number of possibilities of putting the $r$ objects in $m$ boxes. For each object there are $m$ choices, so the total is $m^r$.
– Count Iblis
Sep 2 at 13:13
The sum of all multinomial coefficients of the form $dfracr!r_1! r_2!cdots r_m!$ where we keep $r$ fixed and sum over the $r_j$ while keeping $sum_j r_j = r$ is $m^r$. This is because the multinomial coefficient gives you the number of ways you can choose the $r_j$ from the $r$ to, say, put them into box nr. $j$, and if you sum over all such partitions, you'll get the total number of possibilities of putting the $r$ objects in $m$ boxes. For each object there are $m$ choices, so the total is $m^r$.
– Count Iblis
Sep 2 at 13:13
Please show your work.
– Marko Riedel
Sep 2 at 14:02
Please show your work.
– Marko Riedel
Sep 2 at 14:02
add a comment |Â
1 Answer
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The left side is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSEQ_=n(textscSET(mathcalZ)).$$
The right side is
$$left.textscSET(mathcalZ_1+cdots+mathcalZ_n)
right|_mathcalZ_k=mathcalZ =
textscSET(nmathcalZ).$$
The coefficient on $m![z^m]$ of the EGF $exp(z)^n$ for the first
counts the number of partitions of a row of $m$ slots into $n$ sets,
i.e. we assign an identifier $k$ from $[n]$ to every slot. The second
coefficient again on $m! [z^m]$ , this time of $exp(nz)$, extracts
all multisets on the symbols $mathcalZ_k$ with a total of $m$
symbols. Upon labeling these with labels from $[m]$ we once more
obtain an ordered partition of $[m]$ into $n$ sets, some possibly
empty. (A term $mathcalZ_k^q$ in the multiset means that the $q$
labels represent slots assigned the value $k.$) We may substitute
$mathcalZ$ for $mathcalZ_k$ because we only require the size of
the multiset, as opposed to its factorization into constituents.
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The left side is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSEQ_=n(textscSET(mathcalZ)).$$
The right side is
$$left.textscSET(mathcalZ_1+cdots+mathcalZ_n)
right|_mathcalZ_k=mathcalZ =
textscSET(nmathcalZ).$$
The coefficient on $m![z^m]$ of the EGF $exp(z)^n$ for the first
counts the number of partitions of a row of $m$ slots into $n$ sets,
i.e. we assign an identifier $k$ from $[n]$ to every slot. The second
coefficient again on $m! [z^m]$ , this time of $exp(nz)$, extracts
all multisets on the symbols $mathcalZ_k$ with a total of $m$
symbols. Upon labeling these with labels from $[m]$ we once more
obtain an ordered partition of $[m]$ into $n$ sets, some possibly
empty. (A term $mathcalZ_k^q$ in the multiset means that the $q$
labels represent slots assigned the value $k.$) We may substitute
$mathcalZ$ for $mathcalZ_k$ because we only require the size of
the multiset, as opposed to its factorization into constituents.
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
add a comment |Â
up vote
1
down vote
The left side is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSEQ_=n(textscSET(mathcalZ)).$$
The right side is
$$left.textscSET(mathcalZ_1+cdots+mathcalZ_n)
right|_mathcalZ_k=mathcalZ =
textscSET(nmathcalZ).$$
The coefficient on $m![z^m]$ of the EGF $exp(z)^n$ for the first
counts the number of partitions of a row of $m$ slots into $n$ sets,
i.e. we assign an identifier $k$ from $[n]$ to every slot. The second
coefficient again on $m! [z^m]$ , this time of $exp(nz)$, extracts
all multisets on the symbols $mathcalZ_k$ with a total of $m$
symbols. Upon labeling these with labels from $[m]$ we once more
obtain an ordered partition of $[m]$ into $n$ sets, some possibly
empty. (A term $mathcalZ_k^q$ in the multiset means that the $q$
labels represent slots assigned the value $k.$) We may substitute
$mathcalZ$ for $mathcalZ_k$ because we only require the size of
the multiset, as opposed to its factorization into constituents.
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The left side is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSEQ_=n(textscSET(mathcalZ)).$$
The right side is
$$left.textscSET(mathcalZ_1+cdots+mathcalZ_n)
right|_mathcalZ_k=mathcalZ =
textscSET(nmathcalZ).$$
The coefficient on $m![z^m]$ of the EGF $exp(z)^n$ for the first
counts the number of partitions of a row of $m$ slots into $n$ sets,
i.e. we assign an identifier $k$ from $[n]$ to every slot. The second
coefficient again on $m! [z^m]$ , this time of $exp(nz)$, extracts
all multisets on the symbols $mathcalZ_k$ with a total of $m$
symbols. Upon labeling these with labels from $[m]$ we once more
obtain an ordered partition of $[m]$ into $n$ sets, some possibly
empty. (A term $mathcalZ_k^q$ in the multiset means that the $q$
labels represent slots assigned the value $k.$) We may substitute
$mathcalZ$ for $mathcalZ_k$ because we only require the size of
the multiset, as opposed to its factorization into constituents.
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
The left side is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSEQ_=n(textscSET(mathcalZ)).$$
The right side is
$$left.textscSET(mathcalZ_1+cdots+mathcalZ_n)
right|_mathcalZ_k=mathcalZ =
textscSET(nmathcalZ).$$
The coefficient on $m![z^m]$ of the EGF $exp(z)^n$ for the first
counts the number of partitions of a row of $m$ slots into $n$ sets,
i.e. we assign an identifier $k$ from $[n]$ to every slot. The second
coefficient again on $m! [z^m]$ , this time of $exp(nz)$, extracts
all multisets on the symbols $mathcalZ_k$ with a total of $m$
symbols. Upon labeling these with labels from $[m]$ we once more
obtain an ordered partition of $[m]$ into $n$ sets, some possibly
empty. (A term $mathcalZ_k^q$ in the multiset means that the $q$
labels represent slots assigned the value $k.$) We may substitute
$mathcalZ$ for $mathcalZ_k$ because we only require the size of
the multiset, as opposed to its factorization into constituents.
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
edited Sep 2 at 14:20
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
answered Sep 2 at 14:01
Marko Riedel
37.1k335107
37.1k335107
add a comment |Â
add a comment |Â
The sum of all multinomial coefficients of the form $dfracr!r_1! r_2!cdots r_m!$ where we keep $r$ fixed and sum over the $r_j$ while keeping $sum_j r_j = r$ is $m^r$. This is because the multinomial coefficient gives you the number of ways you can choose the $r_j$ from the $r$ to, say, put them into box nr. $j$, and if you sum over all such partitions, you'll get the total number of possibilities of putting the $r$ objects in $m$ boxes. For each object there are $m$ choices, so the total is $m^r$.
– Count Iblis
Sep 2 at 13:13
Please show your work.
– Marko Riedel
Sep 2 at 14:02