Induction proof for stirling of first kind.
$begingroup$
I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.
Now we assume $forall n in mathbb{N} $ with $ n geq 2$
that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.
It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.
Now let $n rightarrow n+1$.
$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.
If we insert our assumption, that leaves us with
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.
If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.
The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?
induction
asked 3 hours ago
D idsea JD idsea J
325
$endgroup$
add a comment |
$begingroup$
I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.
Now we assume $forall n in mathbb{N} $ with $ n geq 2$
that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.
It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.
Now let $n rightarrow n+1$.
$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.
If we insert our assumption, that leaves us with
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.
If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.
The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?
induction
asked 3 hours ago
D idsea JD idsea J
325
$endgroup$
add a comment |
$begingroup$
I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.
Now we assume $forall n in mathbb{N} $ with $ n geq 2$
that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.
It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.
Now let $n rightarrow n+1$.
$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.
If we insert our assumption, that leaves us with
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.
If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.
The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?
induction
asked 3 hours ago
D idsea JD idsea J
325
$endgroup$
I have to show that for every stirling number of the first kind $forall n geq 2 : s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $frac{1}{24}2(2-1)(2-2)(3*2-1) = frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.
Now we assume $forall n in mathbb{N} $ with $ n geq 2$
that $s_{n,n-2} = frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.
It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.
Now let $n rightarrow n+1$.
$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.
If we insert our assumption, that leaves us with
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.
If we insert $n=n+1$ into $frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.
The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?
induction
induction
asked 3 hours ago
D idsea JD idsea J
325
asked 3 hours ago
D idsea JD idsea J
325
asked 3 hours ago
D idsea JD idsea J
325
asked 3 hours ago
D idsea JD idsea J
325
asked 3 hours ago
D idsea JD idsea J
325
325
add a comment |
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2 Answers
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$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.
Substitute this in the prior equation we get
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.
which simplifies to
$s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.
Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
$endgroup$
$begingroup$
Could you clarify what you mean by identity(2)?
$endgroup$
– D idsea J
2 hours ago
1
$begingroup$
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
$endgroup$
– Satish Ramanathan
2 hours ago
$begingroup$
Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
$endgroup$
– D idsea J
2 hours ago
add a comment |
$begingroup$
Induction step:
$begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
& =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
& =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
& =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
& =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
end{aligned}
$
In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.
answered 2 hours ago
drhabdrhab
99k544130
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2 Answers
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$begingroup$
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.
Substitute this in the prior equation we get
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.
which simplifies to
$s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.
Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
$endgroup$
$begingroup$
Could you clarify what you mean by identity(2)?
$endgroup$
– D idsea J
2 hours ago
1
$begingroup$
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
$endgroup$
– Satish Ramanathan
2 hours ago
$begingroup$
Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
$endgroup$
– D idsea J
2 hours ago
add a comment |
$begingroup$
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.
Substitute this in the prior equation we get
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.
which simplifies to
$s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.
Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
$endgroup$
$begingroup$
Could you clarify what you mean by identity(2)?
$endgroup$
– D idsea J
2 hours ago
1
$begingroup$
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
$endgroup$
– Satish Ramanathan
2 hours ago
$begingroup$
Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
$endgroup$
– D idsea J
2 hours ago
add a comment |
$begingroup$
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.
Substitute this in the prior equation we get
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.
which simplifies to
$s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.
Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
$endgroup$
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} tag1$.
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.
Substitute this in the prior equation we get
$s_{n+1,n-1} = frac{1}{24}n(n-1)(n-2)(3n-1)+frac{n^2(n-1)}{2} $.
which simplifies to
$s_{n+1,n-1} = frac{1}{24}(n+1)n(n-1)(3n+2)tag 3 $.
Thus proved by induction. You will have to use the identity(2) to get the final proof. I don't think there is any way out.
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
answered 2 hours ago
Satish RamanathanSatish Ramanathan
9,51231323
9,51231323
$begingroup$
Could you clarify what you mean by identity(2)?
$endgroup$
– D idsea J
2 hours ago
1
$begingroup$
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
$endgroup$
– Satish Ramanathan
2 hours ago
$begingroup$
Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
$endgroup$
– D idsea J
2 hours ago
add a comment |
$begingroup$
Could you clarify what you mean by identity(2)?
$endgroup$
– D idsea J
2 hours ago
1
$begingroup$
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
$endgroup$
– Satish Ramanathan
2 hours ago
$begingroup$
Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
Could you clarify what you mean by identity(2)?
$endgroup$
– D idsea J
2 hours ago
$begingroup$
Could you clarify what you mean by identity(2)?
$endgroup$
– D idsea J
2 hours ago
1
1
$begingroup$
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
$endgroup$
– Satish Ramanathan
2 hours ago
$begingroup$
$s_{n,n-1} = frac{n(n-1)}{2} tag 2$.. This is what I mean.
$endgroup$
– Satish Ramanathan
2 hours ago
$begingroup$
Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
Alright, even though I thought the final form after induction should be $frac{1}{24}(n+1)(n)(n-1)(3n)$ cause that's what we get if we put n=n+1 into the formula that we need to proof.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
$endgroup$
– D idsea J
2 hours ago
$begingroup$
It might be a stupid question. However, when doing induction last time it was always the case that you could just put n+1 into the formula you have to proof to see what the end should look like. I don't see why this is a proof if it's not the same in the end to be honest.
$endgroup$
– D idsea J
2 hours ago
add a comment |
$begingroup$
Induction step:
$begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
& =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
& =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
& =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
& =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
end{aligned}
$
In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.
answered 2 hours ago
drhabdrhab
99k544130
$endgroup$
add a comment |
$begingroup$
Induction step:
$begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
& =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
& =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
& =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
& =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
end{aligned}
$
In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.
answered 2 hours ago
drhabdrhab
99k544130
$endgroup$
add a comment |
$begingroup$
Induction step:
$begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
& =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
& =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
& =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
& =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
end{aligned}
$
In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.
answered 2 hours ago
drhabdrhab
99k544130
$endgroup$
Induction step:
$begin{aligned}sleft(n+1,n-1right) & =nsleft(n,n-1right)+sleft(n,n-2right)\
& =frac{1}{2}n^{2}left(n-1right)+frac{1}{24}nleft(n-1right)left(n-2right)left(3n-1right)\
& =frac{1}{24}nleft(n-1right)left[12n+left(n-2right)left(3n-1right)right]\
& =frac{1}{24}nleft(n-1right)left(n+1right)left(3n+2right)\
& =frac{1}{24}left(n+1right)nleft(n-1right)left(3n+2right)
end{aligned}
$
In the second equality we use the equality $s(n,n-1)=frac12n(n-1)$ and also the induction hypothesis.
answered 2 hours ago
drhabdrhab
99k544130
answered 2 hours ago
drhabdrhab
99k544130
answered 2 hours ago
drhabdrhab
99k544130
answered 2 hours ago
drhabdrhab
99k544130
99k544130
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