Where does my counting argument fails (Counting of numbers with freshness $k$)?
$begingroup$
Doing some recreational math, i was trying to reproduce the combinatorial proof of the identity
$$
n^n=sum_{k=1}^nfrac{n!}{(n-k)!}cdot kcdot n^{n-k-1},,,,(*)
$$
mentioned by OP in the Question Non combinatorial proof of formula for nn
?
To be honest this shouldn't be too difficult:
the number of all sequences with length $n$ out of numbers $1,2,..,n$ is clearly $underbrace{ n cdot n...cdot n}_{n,,times}=n^n$ which establishs the l.h.s. . For the r.h.s we demand that the first $k$ numbers are distinct but that the rest of the sequence can be choosen freely as before (Denote this number by $a_k$). In the end we sum over all $a_k$. I get
$$
a_k=underbrace{ncdot(n-1)...cdot(n-k+1)}_{choose,,k,,numbers\distinct}timesunderbrace{n^{n-k}}_{choose,,n-k,,numbers\freely}\=
frac{n!}{(n-k)!}n^{n-k} ,,,,(**)
$$
But this is not the same then the terms under the summation sign in $(*)$ so what am i doing wrong?
Edit 1:
I think i got it, i have to discount for the sequences with $k'>k$ distinct numbers (they can still fall out of the random part of my product) .Is this correct?
Edit 2:
To finish this, the correct quantity is given by the difference of sequences with $k$ and $k+1$ distinct numbers:
$$
a_{k+1}-a_{k}=-frac{n!}{n^{n-k}}left(frac{1}{(n-k-1)!n}-frac{1}{(n-k)!}right)=-frac{k}{n(n-k)!}frac{n!}{n^{n-k}}
$$
Done!
Edit3:
To avoid algebra we can also reason as follows (credit to @darijgrinberg):
Add to $(**)$ the following condtion: the $k+1$ number is fixed to ne one of the numbers in the first part of the product so that we don't get a sequence with $k+1$ distinct numbers. There are $k$ possiblities. Afterwards we can choose all remaining $n-k-1$ numbers at our wish. This results in a replacement $n^{n-k}rightarrow ktimes n^{n-k-1}$
combinatorics summation
asked Jan 5 at 13:59
tiredtired
10.6k12043
$endgroup$
add a comment |
$begingroup$
Doing some recreational math, i was trying to reproduce the combinatorial proof of the identity
$$
n^n=sum_{k=1}^nfrac{n!}{(n-k)!}cdot kcdot n^{n-k-1},,,,(*)
$$
mentioned by OP in the Question Non combinatorial proof of formula for nn
?
To be honest this shouldn't be too difficult:
the number of all sequences with length $n$ out of numbers $1,2,..,n$ is clearly $underbrace{ n cdot n...cdot n}_{n,,times}=n^n$ which establishs the l.h.s. . For the r.h.s we demand that the first $k$ numbers are distinct but that the rest of the sequence can be choosen freely as before (Denote this number by $a_k$). In the end we sum over all $a_k$. I get
$$
a_k=underbrace{ncdot(n-1)...cdot(n-k+1)}_{choose,,k,,numbers\distinct}timesunderbrace{n^{n-k}}_{choose,,n-k,,numbers\freely}\=
frac{n!}{(n-k)!}n^{n-k} ,,,,(**)
$$
But this is not the same then the terms under the summation sign in $(*)$ so what am i doing wrong?
Edit 1:
I think i got it, i have to discount for the sequences with $k'>k$ distinct numbers (they can still fall out of the random part of my product) .Is this correct?
Edit 2:
To finish this, the correct quantity is given by the difference of sequences with $k$ and $k+1$ distinct numbers:
$$
a_{k+1}-a_{k}=-frac{n!}{n^{n-k}}left(frac{1}{(n-k-1)!n}-frac{1}{(n-k)!}right)=-frac{k}{n(n-k)!}frac{n!}{n^{n-k}}
$$
Done!
Edit3:
To avoid algebra we can also reason as follows (credit to @darijgrinberg):
Add to $(**)$ the following condtion: the $k+1$ number is fixed to ne one of the numbers in the first part of the product so that we don't get a sequence with $k+1$ distinct numbers. There are $k$ possiblities. Afterwards we can choose all remaining $n-k-1$ numbers at our wish. This results in a replacement $n^{n-k}rightarrow ktimes n^{n-k-1}$
combinatorics summation
asked Jan 5 at 13:59
tiredtired
10.6k12043
$endgroup$
2
$begingroup$
You don't choose the last $n-k$ numbers freely. You have to make sure that the first of them violates the distinctness condition, i.e., equals one of the preceding $k$ numbers. That's where the $k$ factor will come from.
$endgroup$
– darij grinberg
Jan 5 at 14:15
$begingroup$
@darijgrinberg thanks: i was already suspecting that. See my edit!
$endgroup$
– tired
Jan 5 at 14:17
$begingroup$
@darijgrinberg do you have any idea how to see the last Formula (after edit) directly without resorting to annoying Algebra? Thx!
$endgroup$
– tired
Jan 5 at 14:22
1
$begingroup$
You don't have to subtract anything. Just fix your argument as I suggested in the first comment.
$endgroup$
– darij grinberg
Jan 5 at 14:43
$begingroup$
apart from the sign, edit 2 is correct.
$endgroup$
– G Cab
Jan 5 at 14:57
add a comment |
$begingroup$
Doing some recreational math, i was trying to reproduce the combinatorial proof of the identity
$$
n^n=sum_{k=1}^nfrac{n!}{(n-k)!}cdot kcdot n^{n-k-1},,,,(*)
$$
mentioned by OP in the Question Non combinatorial proof of formula for nn
?
To be honest this shouldn't be too difficult:
the number of all sequences with length $n$ out of numbers $1,2,..,n$ is clearly $underbrace{ n cdot n...cdot n}_{n,,times}=n^n$ which establishs the l.h.s. . For the r.h.s we demand that the first $k$ numbers are distinct but that the rest of the sequence can be choosen freely as before (Denote this number by $a_k$). In the end we sum over all $a_k$. I get
$$
a_k=underbrace{ncdot(n-1)...cdot(n-k+1)}_{choose,,k,,numbers\distinct}timesunderbrace{n^{n-k}}_{choose,,n-k,,numbers\freely}\=
frac{n!}{(n-k)!}n^{n-k} ,,,,(**)
$$
But this is not the same then the terms under the summation sign in $(*)$ so what am i doing wrong?
Edit 1:
I think i got it, i have to discount for the sequences with $k'>k$ distinct numbers (they can still fall out of the random part of my product) .Is this correct?
Edit 2:
To finish this, the correct quantity is given by the difference of sequences with $k$ and $k+1$ distinct numbers:
$$
a_{k+1}-a_{k}=-frac{n!}{n^{n-k}}left(frac{1}{(n-k-1)!n}-frac{1}{(n-k)!}right)=-frac{k}{n(n-k)!}frac{n!}{n^{n-k}}
$$
Done!
Edit3:
To avoid algebra we can also reason as follows (credit to @darijgrinberg):
Add to $(**)$ the following condtion: the $k+1$ number is fixed to ne one of the numbers in the first part of the product so that we don't get a sequence with $k+1$ distinct numbers. There are $k$ possiblities. Afterwards we can choose all remaining $n-k-1$ numbers at our wish. This results in a replacement $n^{n-k}rightarrow ktimes n^{n-k-1}$
combinatorics summation
asked Jan 5 at 13:59
tiredtired
10.6k12043
$endgroup$
Doing some recreational math, i was trying to reproduce the combinatorial proof of the identity
$$
n^n=sum_{k=1}^nfrac{n!}{(n-k)!}cdot kcdot n^{n-k-1},,,,(*)
$$
mentioned by OP in the Question Non combinatorial proof of formula for nn
?
To be honest this shouldn't be too difficult:
the number of all sequences with length $n$ out of numbers $1,2,..,n$ is clearly $underbrace{ n cdot n...cdot n}_{n,,times}=n^n$ which establishs the l.h.s. . For the r.h.s we demand that the first $k$ numbers are distinct but that the rest of the sequence can be choosen freely as before (Denote this number by $a_k$). In the end we sum over all $a_k$. I get
$$
a_k=underbrace{ncdot(n-1)...cdot(n-k+1)}_{choose,,k,,numbers\distinct}timesunderbrace{n^{n-k}}_{choose,,n-k,,numbers\freely}\=
frac{n!}{(n-k)!}n^{n-k} ,,,,(**)
$$
But this is not the same then the terms under the summation sign in $(*)$ so what am i doing wrong?
Edit 1:
I think i got it, i have to discount for the sequences with $k'>k$ distinct numbers (they can still fall out of the random part of my product) .Is this correct?
Edit 2:
To finish this, the correct quantity is given by the difference of sequences with $k$ and $k+1$ distinct numbers:
$$
a_{k+1}-a_{k}=-frac{n!}{n^{n-k}}left(frac{1}{(n-k-1)!n}-frac{1}{(n-k)!}right)=-frac{k}{n(n-k)!}frac{n!}{n^{n-k}}
$$
Done!
Edit3:
To avoid algebra we can also reason as follows (credit to @darijgrinberg):
Add to $(**)$ the following condtion: the $k+1$ number is fixed to ne one of the numbers in the first part of the product so that we don't get a sequence with $k+1$ distinct numbers. There are $k$ possiblities. Afterwards we can choose all remaining $n-k-1$ numbers at our wish. This results in a replacement $n^{n-k}rightarrow ktimes n^{n-k-1}$
combinatorics summation
combinatorics summation
asked Jan 5 at 13:59
tiredtired
10.6k12043
asked Jan 5 at 13:59
tiredtired
10.6k12043
edited Jan 5 at 15:52
tired
asked Jan 5 at 13:59
tiredtired
10.6k12043
asked Jan 5 at 13:59
tiredtired
10.6k12043
asked Jan 5 at 13:59
tiredtired
10.6k12043
10.6k12043
2
$begingroup$
You don't choose the last $n-k$ numbers freely. You have to make sure that the first of them violates the distinctness condition, i.e., equals one of the preceding $k$ numbers. That's where the $k$ factor will come from.
$endgroup$
– darij grinberg
Jan 5 at 14:15
$begingroup$
@darijgrinberg thanks: i was already suspecting that. See my edit!
$endgroup$
– tired
Jan 5 at 14:17
$begingroup$
@darijgrinberg do you have any idea how to see the last Formula (after edit) directly without resorting to annoying Algebra? Thx!
$endgroup$
– tired
Jan 5 at 14:22
1
$begingroup$
You don't have to subtract anything. Just fix your argument as I suggested in the first comment.
$endgroup$
– darij grinberg
Jan 5 at 14:43
$begingroup$
apart from the sign, edit 2 is correct.
$endgroup$
– G Cab
Jan 5 at 14:57
add a comment |
2
$begingroup$
You don't choose the last $n-k$ numbers freely. You have to make sure that the first of them violates the distinctness condition, i.e., equals one of the preceding $k$ numbers. That's where the $k$ factor will come from.
$endgroup$
– darij grinberg
Jan 5 at 14:15
$begingroup$
@darijgrinberg thanks: i was already suspecting that. See my edit!
$endgroup$
– tired
Jan 5 at 14:17
$begingroup$
@darijgrinberg do you have any idea how to see the last Formula (after edit) directly without resorting to annoying Algebra? Thx!
$endgroup$
– tired
Jan 5 at 14:22
1
$begingroup$
You don't have to subtract anything. Just fix your argument as I suggested in the first comment.
$endgroup$
– darij grinberg
Jan 5 at 14:43
$begingroup$
apart from the sign, edit 2 is correct.
$endgroup$
– G Cab
Jan 5 at 14:57
2
2
$begingroup$
You don't choose the last $n-k$ numbers freely. You have to make sure that the first of them violates the distinctness condition, i.e., equals one of the preceding $k$ numbers. That's where the $k$ factor will come from.
$endgroup$
– darij grinberg
Jan 5 at 14:15
$begingroup$
You don't choose the last $n-k$ numbers freely. You have to make sure that the first of them violates the distinctness condition, i.e., equals one of the preceding $k$ numbers. That's where the $k$ factor will come from.
$endgroup$
– darij grinberg
Jan 5 at 14:15
$begingroup$
@darijgrinberg thanks: i was already suspecting that. See my edit!
$endgroup$
– tired
Jan 5 at 14:17
$begingroup$
@darijgrinberg thanks: i was already suspecting that. See my edit!
$endgroup$
– tired
Jan 5 at 14:17
$begingroup$
@darijgrinberg do you have any idea how to see the last Formula (after edit) directly without resorting to annoying Algebra? Thx!
$endgroup$
– tired
Jan 5 at 14:22
$begingroup$
@darijgrinberg do you have any idea how to see the last Formula (after edit) directly without resorting to annoying Algebra? Thx!
$endgroup$
– tired
Jan 5 at 14:22
1
1
$begingroup$
You don't have to subtract anything. Just fix your argument as I suggested in the first comment.
$endgroup$
– darij grinberg
Jan 5 at 14:43
$begingroup$
You don't have to subtract anything. Just fix your argument as I suggested in the first comment.
$endgroup$
– darij grinberg
Jan 5 at 14:43
$begingroup$
apart from the sign, edit 2 is correct.
$endgroup$
– G Cab
Jan 5 at 14:57
$begingroup$
apart from the sign, edit 2 is correct.
$endgroup$
– G Cab
Jan 5 at 14:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I cannot tell if you already figured it out, but in case you haven't:
$$
a_k = underbrace{ncdot (n-1)cdotldotscdot (n-k+1)}_text{choose first $k$ numbers distinct}timesunderbrace{k}_{begin{array}{c}text{choose $(k+1)^{st}$ number equal to }\text{one of the first $k$ numbers}end{array}}times underbrace{n^{n-k-1}}_text{choose remaining numbers freely}
$$
The middle term must be $k$ because you need the "freshness" to be exactly $k$, meaning the first $k$ numbers are distinct, but the first $k+1$ numbers are not. Therefore, the $(k+1)^{st}$ number needs to be equal to one of the previous $k$.
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
$endgroup$
$begingroup$
Thanks, this is exactely what i added to my Question a few seconds ago. So now i have two different Proofs relying purely on combinatorics, cool!!! :)
$endgroup$
– tired
Jan 5 at 15:54
1
$begingroup$
@tired Indeed! Since you are doing this for recreation, let me mention one other connection. Let $f_{n,k}=binom{n}k kn^{n-k-1}$, then the formula can be written $$n^n=sum_{k=1}^n k! f_{n,k}.$$ This is interesting because $f_{n,k}$ has a combinatorial interpretation; it is the number of "rooted forests" (forest with a root in each component) on a set of size $n$ with $k$ components. This shows that a function can be described as rooted forest together with a permutation $pi$ of the roots. The correspondence is: if $v$ is a not a root, $f(v)$ is its parent; if $v$ is a root, $f(v)=pi(v)$.
$endgroup$
– Mike Earnest
Jan 5 at 15:59
$begingroup$
this is cool indeed. I wish i would have done more of this stuff back at University. Another Thing that i found nice is to divide everything by $n^n$ and use Stirling. We then see that the probability to get sequences with freshness $k$ is given by $k/n$ for sufficently small $k/n$ ...
$endgroup$
– tired
Jan 5 at 16:09
1
$begingroup$
@tired That probability distribution is cool, too! Letting $X_n$ be the freshness, then $P(X_n/sqrt{n}>t)to e^{-t^2/2}$ as $ntoinfty$, so $X_n$ has an asymptotically Weibull distribution. Then $E[X_n]sim sqrt{npi/2}$, so in a random function, around $sqrt{npi/2}$ elements will be contained in a cycle, meaning $f^m(x)=x$ for some $m$.
$endgroup$
– Mike Earnest
Jan 5 at 16:22
add a comment |
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$begingroup$
I cannot tell if you already figured it out, but in case you haven't:
$$
a_k = underbrace{ncdot (n-1)cdotldotscdot (n-k+1)}_text{choose first $k$ numbers distinct}timesunderbrace{k}_{begin{array}{c}text{choose $(k+1)^{st}$ number equal to }\text{one of the first $k$ numbers}end{array}}times underbrace{n^{n-k-1}}_text{choose remaining numbers freely}
$$
The middle term must be $k$ because you need the "freshness" to be exactly $k$, meaning the first $k$ numbers are distinct, but the first $k+1$ numbers are not. Therefore, the $(k+1)^{st}$ number needs to be equal to one of the previous $k$.
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
$endgroup$
$begingroup$
Thanks, this is exactely what i added to my Question a few seconds ago. So now i have two different Proofs relying purely on combinatorics, cool!!! :)
$endgroup$
– tired
Jan 5 at 15:54
1
$begingroup$
@tired Indeed! Since you are doing this for recreation, let me mention one other connection. Let $f_{n,k}=binom{n}k kn^{n-k-1}$, then the formula can be written $$n^n=sum_{k=1}^n k! f_{n,k}.$$ This is interesting because $f_{n,k}$ has a combinatorial interpretation; it is the number of "rooted forests" (forest with a root in each component) on a set of size $n$ with $k$ components. This shows that a function can be described as rooted forest together with a permutation $pi$ of the roots. The correspondence is: if $v$ is a not a root, $f(v)$ is its parent; if $v$ is a root, $f(v)=pi(v)$.
$endgroup$
– Mike Earnest
Jan 5 at 15:59
$begingroup$
this is cool indeed. I wish i would have done more of this stuff back at University. Another Thing that i found nice is to divide everything by $n^n$ and use Stirling. We then see that the probability to get sequences with freshness $k$ is given by $k/n$ for sufficently small $k/n$ ...
$endgroup$
– tired
Jan 5 at 16:09
1
$begingroup$
@tired That probability distribution is cool, too! Letting $X_n$ be the freshness, then $P(X_n/sqrt{n}>t)to e^{-t^2/2}$ as $ntoinfty$, so $X_n$ has an asymptotically Weibull distribution. Then $E[X_n]sim sqrt{npi/2}$, so in a random function, around $sqrt{npi/2}$ elements will be contained in a cycle, meaning $f^m(x)=x$ for some $m$.
$endgroup$
– Mike Earnest
Jan 5 at 16:22
add a comment |
$begingroup$
I cannot tell if you already figured it out, but in case you haven't:
$$
a_k = underbrace{ncdot (n-1)cdotldotscdot (n-k+1)}_text{choose first $k$ numbers distinct}timesunderbrace{k}_{begin{array}{c}text{choose $(k+1)^{st}$ number equal to }\text{one of the first $k$ numbers}end{array}}times underbrace{n^{n-k-1}}_text{choose remaining numbers freely}
$$
The middle term must be $k$ because you need the "freshness" to be exactly $k$, meaning the first $k$ numbers are distinct, but the first $k+1$ numbers are not. Therefore, the $(k+1)^{st}$ number needs to be equal to one of the previous $k$.
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
$endgroup$
$begingroup$
Thanks, this is exactely what i added to my Question a few seconds ago. So now i have two different Proofs relying purely on combinatorics, cool!!! :)
$endgroup$
– tired
Jan 5 at 15:54
1
$begingroup$
@tired Indeed! Since you are doing this for recreation, let me mention one other connection. Let $f_{n,k}=binom{n}k kn^{n-k-1}$, then the formula can be written $$n^n=sum_{k=1}^n k! f_{n,k}.$$ This is interesting because $f_{n,k}$ has a combinatorial interpretation; it is the number of "rooted forests" (forest with a root in each component) on a set of size $n$ with $k$ components. This shows that a function can be described as rooted forest together with a permutation $pi$ of the roots. The correspondence is: if $v$ is a not a root, $f(v)$ is its parent; if $v$ is a root, $f(v)=pi(v)$.
$endgroup$
– Mike Earnest
Jan 5 at 15:59
$begingroup$
this is cool indeed. I wish i would have done more of this stuff back at University. Another Thing that i found nice is to divide everything by $n^n$ and use Stirling. We then see that the probability to get sequences with freshness $k$ is given by $k/n$ for sufficently small $k/n$ ...
$endgroup$
– tired
Jan 5 at 16:09
1
$begingroup$
@tired That probability distribution is cool, too! Letting $X_n$ be the freshness, then $P(X_n/sqrt{n}>t)to e^{-t^2/2}$ as $ntoinfty$, so $X_n$ has an asymptotically Weibull distribution. Then $E[X_n]sim sqrt{npi/2}$, so in a random function, around $sqrt{npi/2}$ elements will be contained in a cycle, meaning $f^m(x)=x$ for some $m$.
$endgroup$
– Mike Earnest
Jan 5 at 16:22
add a comment |
$begingroup$
I cannot tell if you already figured it out, but in case you haven't:
$$
a_k = underbrace{ncdot (n-1)cdotldotscdot (n-k+1)}_text{choose first $k$ numbers distinct}timesunderbrace{k}_{begin{array}{c}text{choose $(k+1)^{st}$ number equal to }\text{one of the first $k$ numbers}end{array}}times underbrace{n^{n-k-1}}_text{choose remaining numbers freely}
$$
The middle term must be $k$ because you need the "freshness" to be exactly $k$, meaning the first $k$ numbers are distinct, but the first $k+1$ numbers are not. Therefore, the $(k+1)^{st}$ number needs to be equal to one of the previous $k$.
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
$endgroup$
I cannot tell if you already figured it out, but in case you haven't:
$$
a_k = underbrace{ncdot (n-1)cdotldotscdot (n-k+1)}_text{choose first $k$ numbers distinct}timesunderbrace{k}_{begin{array}{c}text{choose $(k+1)^{st}$ number equal to }\text{one of the first $k$ numbers}end{array}}times underbrace{n^{n-k-1}}_text{choose remaining numbers freely}
$$
The middle term must be $k$ because you need the "freshness" to be exactly $k$, meaning the first $k$ numbers are distinct, but the first $k+1$ numbers are not. Therefore, the $(k+1)^{st}$ number needs to be equal to one of the previous $k$.
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
answered Jan 5 at 15:52
Mike EarnestMike Earnest
20.9k11951
20.9k11951
$begingroup$
Thanks, this is exactely what i added to my Question a few seconds ago. So now i have two different Proofs relying purely on combinatorics, cool!!! :)
$endgroup$
– tired
Jan 5 at 15:54
1
$begingroup$
@tired Indeed! Since you are doing this for recreation, let me mention one other connection. Let $f_{n,k}=binom{n}k kn^{n-k-1}$, then the formula can be written $$n^n=sum_{k=1}^n k! f_{n,k}.$$ This is interesting because $f_{n,k}$ has a combinatorial interpretation; it is the number of "rooted forests" (forest with a root in each component) on a set of size $n$ with $k$ components. This shows that a function can be described as rooted forest together with a permutation $pi$ of the roots. The correspondence is: if $v$ is a not a root, $f(v)$ is its parent; if $v$ is a root, $f(v)=pi(v)$.
$endgroup$
– Mike Earnest
Jan 5 at 15:59
$begingroup$
this is cool indeed. I wish i would have done more of this stuff back at University. Another Thing that i found nice is to divide everything by $n^n$ and use Stirling. We then see that the probability to get sequences with freshness $k$ is given by $k/n$ for sufficently small $k/n$ ...
$endgroup$
– tired
Jan 5 at 16:09
1
$begingroup$
@tired That probability distribution is cool, too! Letting $X_n$ be the freshness, then $P(X_n/sqrt{n}>t)to e^{-t^2/2}$ as $ntoinfty$, so $X_n$ has an asymptotically Weibull distribution. Then $E[X_n]sim sqrt{npi/2}$, so in a random function, around $sqrt{npi/2}$ elements will be contained in a cycle, meaning $f^m(x)=x$ for some $m$.
$endgroup$
– Mike Earnest
Jan 5 at 16:22
add a comment |
$begingroup$
Thanks, this is exactely what i added to my Question a few seconds ago. So now i have two different Proofs relying purely on combinatorics, cool!!! :)
$endgroup$
– tired
Jan 5 at 15:54
1
$begingroup$
@tired Indeed! Since you are doing this for recreation, let me mention one other connection. Let $f_{n,k}=binom{n}k kn^{n-k-1}$, then the formula can be written $$n^n=sum_{k=1}^n k! f_{n,k}.$$ This is interesting because $f_{n,k}$ has a combinatorial interpretation; it is the number of "rooted forests" (forest with a root in each component) on a set of size $n$ with $k$ components. This shows that a function can be described as rooted forest together with a permutation $pi$ of the roots. The correspondence is: if $v$ is a not a root, $f(v)$ is its parent; if $v$ is a root, $f(v)=pi(v)$.
$endgroup$
– Mike Earnest
Jan 5 at 15:59
$begingroup$
this is cool indeed. I wish i would have done more of this stuff back at University. Another Thing that i found nice is to divide everything by $n^n$ and use Stirling. We then see that the probability to get sequences with freshness $k$ is given by $k/n$ for sufficently small $k/n$ ...
$endgroup$
– tired
Jan 5 at 16:09
1
$begingroup$
@tired That probability distribution is cool, too! Letting $X_n$ be the freshness, then $P(X_n/sqrt{n}>t)to e^{-t^2/2}$ as $ntoinfty$, so $X_n$ has an asymptotically Weibull distribution. Then $E[X_n]sim sqrt{npi/2}$, so in a random function, around $sqrt{npi/2}$ elements will be contained in a cycle, meaning $f^m(x)=x$ for some $m$.
$endgroup$
– Mike Earnest
Jan 5 at 16:22
$begingroup$
Thanks, this is exactely what i added to my Question a few seconds ago. So now i have two different Proofs relying purely on combinatorics, cool!!! :)
$endgroup$
– tired
Jan 5 at 15:54
$begingroup$
Thanks, this is exactely what i added to my Question a few seconds ago. So now i have two different Proofs relying purely on combinatorics, cool!!! :)
$endgroup$
– tired
Jan 5 at 15:54
1
1
$begingroup$
@tired Indeed! Since you are doing this for recreation, let me mention one other connection. Let $f_{n,k}=binom{n}k kn^{n-k-1}$, then the formula can be written $$n^n=sum_{k=1}^n k! f_{n,k}.$$ This is interesting because $f_{n,k}$ has a combinatorial interpretation; it is the number of "rooted forests" (forest with a root in each component) on a set of size $n$ with $k$ components. This shows that a function can be described as rooted forest together with a permutation $pi$ of the roots. The correspondence is: if $v$ is a not a root, $f(v)$ is its parent; if $v$ is a root, $f(v)=pi(v)$.
$endgroup$
– Mike Earnest
Jan 5 at 15:59
$begingroup$
@tired Indeed! Since you are doing this for recreation, let me mention one other connection. Let $f_{n,k}=binom{n}k kn^{n-k-1}$, then the formula can be written $$n^n=sum_{k=1}^n k! f_{n,k}.$$ This is interesting because $f_{n,k}$ has a combinatorial interpretation; it is the number of "rooted forests" (forest with a root in each component) on a set of size $n$ with $k$ components. This shows that a function can be described as rooted forest together with a permutation $pi$ of the roots. The correspondence is: if $v$ is a not a root, $f(v)$ is its parent; if $v$ is a root, $f(v)=pi(v)$.
$endgroup$
– Mike Earnest
Jan 5 at 15:59
$begingroup$
this is cool indeed. I wish i would have done more of this stuff back at University. Another Thing that i found nice is to divide everything by $n^n$ and use Stirling. We then see that the probability to get sequences with freshness $k$ is given by $k/n$ for sufficently small $k/n$ ...
$endgroup$
– tired
Jan 5 at 16:09
$begingroup$
this is cool indeed. I wish i would have done more of this stuff back at University. Another Thing that i found nice is to divide everything by $n^n$ and use Stirling. We then see that the probability to get sequences with freshness $k$ is given by $k/n$ for sufficently small $k/n$ ...
$endgroup$
– tired
Jan 5 at 16:09
1
1
$begingroup$
@tired That probability distribution is cool, too! Letting $X_n$ be the freshness, then $P(X_n/sqrt{n}>t)to e^{-t^2/2}$ as $ntoinfty$, so $X_n$ has an asymptotically Weibull distribution. Then $E[X_n]sim sqrt{npi/2}$, so in a random function, around $sqrt{npi/2}$ elements will be contained in a cycle, meaning $f^m(x)=x$ for some $m$.
$endgroup$
– Mike Earnest
Jan 5 at 16:22
$begingroup$
@tired That probability distribution is cool, too! Letting $X_n$ be the freshness, then $P(X_n/sqrt{n}>t)to e^{-t^2/2}$ as $ntoinfty$, so $X_n$ has an asymptotically Weibull distribution. Then $E[X_n]sim sqrt{npi/2}$, so in a random function, around $sqrt{npi/2}$ elements will be contained in a cycle, meaning $f^m(x)=x$ for some $m$.
$endgroup$
– Mike Earnest
Jan 5 at 16:22
add a comment |
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2
$begingroup$
You don't choose the last $n-k$ numbers freely. You have to make sure that the first of them violates the distinctness condition, i.e., equals one of the preceding $k$ numbers. That's where the $k$ factor will come from.
$endgroup$
– darij grinberg
Jan 5 at 14:15
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@darijgrinberg thanks: i was already suspecting that. See my edit!
$endgroup$
– tired
Jan 5 at 14:17
$begingroup$
@darijgrinberg do you have any idea how to see the last Formula (after edit) directly without resorting to annoying Algebra? Thx!
$endgroup$
– tired
Jan 5 at 14:22
1
$begingroup$
You don't have to subtract anything. Just fix your argument as I suggested in the first comment.
$endgroup$
– darij grinberg
Jan 5 at 14:43
$begingroup$
apart from the sign, edit 2 is correct.
$endgroup$
– G Cab
Jan 5 at 14:57