Proof Stirling equality
Proof that
$k^n = sum_{j=1}^{k}binom{k}{j}j!S(n,j)$
To justify this last equality you have to consider, firstly, that if $X$ and $Y$ are two sets with $n$ and $k$ elements, respectively, then there are $k^{n}$ applications of $X$ and $Y$. To reach the equality it have to use that these applications can be constructed in the following way: the images of the elements of X are a series of elements of Y, say j of them, where $1leq jleq k$ (the rest will not have preimage). Then, first, we decided that elements of Y have preimagenes and, once decided that the elements "arrives" the application, it only remains to build a suprayective application that has only these elements as an image. This is the process that should lead to the desired formula.
discrete-mathematics proof-verification stirling-numbers
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
add a comment |
Proof that
$k^n = sum_{j=1}^{k}binom{k}{j}j!S(n,j)$
To justify this last equality you have to consider, firstly, that if $X$ and $Y$ are two sets with $n$ and $k$ elements, respectively, then there are $k^{n}$ applications of $X$ and $Y$. To reach the equality it have to use that these applications can be constructed in the following way: the images of the elements of X are a series of elements of Y, say j of them, where $1leq jleq k$ (the rest will not have preimage). Then, first, we decided that elements of Y have preimagenes and, once decided that the elements "arrives" the application, it only remains to build a suprayective application that has only these elements as an image. This is the process that should lead to the desired formula.
discrete-mathematics proof-verification stirling-numbers
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
"applications"? Do you means maps? There is a typo in the header. And you can add the tag proof-verification :)
– Stockfish
Nov 29 '18 at 9:09
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Nov 29 '18 at 9:11
Your proof is correct modulo small language issues.
– Michal Adamaszek
Nov 29 '18 at 11:06
add a comment |
Proof that
$k^n = sum_{j=1}^{k}binom{k}{j}j!S(n,j)$
To justify this last equality you have to consider, firstly, that if $X$ and $Y$ are two sets with $n$ and $k$ elements, respectively, then there are $k^{n}$ applications of $X$ and $Y$. To reach the equality it have to use that these applications can be constructed in the following way: the images of the elements of X are a series of elements of Y, say j of them, where $1leq jleq k$ (the rest will not have preimage). Then, first, we decided that elements of Y have preimagenes and, once decided that the elements "arrives" the application, it only remains to build a suprayective application that has only these elements as an image. This is the process that should lead to the desired formula.
discrete-mathematics proof-verification stirling-numbers
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
Proof that
$k^n = sum_{j=1}^{k}binom{k}{j}j!S(n,j)$
To justify this last equality you have to consider, firstly, that if $X$ and $Y$ are two sets with $n$ and $k$ elements, respectively, then there are $k^{n}$ applications of $X$ and $Y$. To reach the equality it have to use that these applications can be constructed in the following way: the images of the elements of X are a series of elements of Y, say j of them, where $1leq jleq k$ (the rest will not have preimage). Then, first, we decided that elements of Y have preimagenes and, once decided that the elements "arrives" the application, it only remains to build a suprayective application that has only these elements as an image. This is the process that should lead to the desired formula.
discrete-mathematics proof-verification stirling-numbers
discrete-mathematics proof-verification stirling-numbers
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
edited Nov 29 '18 at 9:12
Str0nger
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
asked Nov 29 '18 at 9:07
Str0ngerStr0nger
12
12
"applications"? Do you means maps? There is a typo in the header. And you can add the tag proof-verification :)
– Stockfish
Nov 29 '18 at 9:09
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Nov 29 '18 at 9:11
Your proof is correct modulo small language issues.
– Michal Adamaszek
Nov 29 '18 at 11:06
add a comment |
"applications"? Do you means maps? There is a typo in the header. And you can add the tag proof-verification :)
– Stockfish
Nov 29 '18 at 9:09
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Nov 29 '18 at 9:11
Your proof is correct modulo small language issues.
– Michal Adamaszek
Nov 29 '18 at 11:06
"applications"? Do you means maps? There is a typo in the header. And you can add the tag proof-verification :)
– Stockfish
Nov 29 '18 at 9:09
"applications"? Do you means maps? There is a typo in the header. And you can add the tag proof-verification :)
– Stockfish
Nov 29 '18 at 9:09
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Nov 29 '18 at 9:11
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Nov 29 '18 at 9:11
Your proof is correct modulo small language issues.
– Michal Adamaszek
Nov 29 '18 at 11:06
Your proof is correct modulo small language issues.
– Michal Adamaszek
Nov 29 '18 at 11:06
add a comment |
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"applications"? Do you means maps? There is a typo in the header. And you can add the tag proof-verification :)
– Stockfish
Nov 29 '18 at 9:09
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
Nov 29 '18 at 9:11
Your proof is correct modulo small language issues.
– Michal Adamaszek
Nov 29 '18 at 11:06