Convergence of Vector Combinations
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I have $n$ vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
, where $||v_i(t)||$ denotes the Euclidean norm of $v_i(t)$.
It turns out that this procedure converges for $t rightarrow infty $ and in the end all vectors are the same. However, I cannot predict the limit vector.
Does anyone have an idea?
linear-algebra vectors vector-analysis heat-equation
asked Nov 24 at 16:19
Jannik
587
add a comment |
up vote
0
down vote
favorite
I have $n$ vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
, where $||v_i(t)||$ denotes the Euclidean norm of $v_i(t)$.
It turns out that this procedure converges for $t rightarrow infty $ and in the end all vectors are the same. However, I cannot predict the limit vector.
Does anyone have an idea?
linear-algebra vectors vector-analysis heat-equation
asked Nov 24 at 16:19
Jannik
587
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have $n$ vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
, where $||v_i(t)||$ denotes the Euclidean norm of $v_i(t)$.
It turns out that this procedure converges for $t rightarrow infty $ and in the end all vectors are the same. However, I cannot predict the limit vector.
Does anyone have an idea?
linear-algebra vectors vector-analysis heat-equation
asked Nov 24 at 16:19
Jannik
587
I have $n$ vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
, where $||v_i(t)||$ denotes the Euclidean norm of $v_i(t)$.
It turns out that this procedure converges for $t rightarrow infty $ and in the end all vectors are the same. However, I cannot predict the limit vector.
Does anyone have an idea?
linear-algebra vectors vector-analysis heat-equation
linear-algebra vectors vector-analysis heat-equation
asked Nov 24 at 16:19
Jannik
587
asked Nov 24 at 16:19
Jannik
587
edited Nov 26 at 21:55
asked Nov 24 at 16:19
Jannik
587
asked Nov 24 at 16:19
Jannik
587
asked Nov 24 at 16:19
Jannik
587
587
add a comment |
add a comment |
1 Answer
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Here's my idea...
It reminds me a discrete version of the heat equation for a vector valued function on the interval $(1,2,...,n)$, with Von Neumann conditions imposing the norm of the function at the boundary to be $1$.
I think this since the discrete version of the laplacian operator at one point of a graph-like domain is just the mean taken on all the adiacent vertexes.
That said we can understand that the limit will have norm equal to $1$, but the exact value seems harder to determine. We may also note that all vectors will have the same limit value if the norm we use in the recursion is not the $infty$-norm, otherwhise they may take different values.
answered Nov 24 at 18:29
Dinisaur
1009
I wouldn't know how to prove it formally, I was rather using intuition. The idea is that, without considering the normalization part for the first and last vector, we have there a diffusive process (we are averaging the neighbours) like the heat equation, which conserves the total heat (which in this analogy is the sum af all vectors) and converges to a uniform solution. In fact in that case would converge to the baricenter.
– Dinisaur
Nov 26 at 15:06
In this analogy our domain is an interval which consider a discrete sequence of n points, and our quantity is vector valued. When we add the normalization at the extremal points (which would be the boundary of the domain) we are forcing our vectors to have norm 1, but since the process is diffusive all points tend to have the same value (i.e. all vectors tend to be the same) the limit will have norm 1.
– Dinisaur
Nov 26 at 15:10
We may conjecture that the limit will be the normalization of the baricenter, which is the case for two points, but I'm not really sure about it.
– Dinisaur
Nov 26 at 15:15
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here's my idea...
It reminds me a discrete version of the heat equation for a vector valued function on the interval $(1,2,...,n)$, with Von Neumann conditions imposing the norm of the function at the boundary to be $1$.
I think this since the discrete version of the laplacian operator at one point of a graph-like domain is just the mean taken on all the adiacent vertexes.
That said we can understand that the limit will have norm equal to $1$, but the exact value seems harder to determine. We may also note that all vectors will have the same limit value if the norm we use in the recursion is not the $infty$-norm, otherwhise they may take different values.
answered Nov 24 at 18:29
Dinisaur
1009
I wouldn't know how to prove it formally, I was rather using intuition. The idea is that, without considering the normalization part for the first and last vector, we have there a diffusive process (we are averaging the neighbours) like the heat equation, which conserves the total heat (which in this analogy is the sum af all vectors) and converges to a uniform solution. In fact in that case would converge to the baricenter.
– Dinisaur
Nov 26 at 15:06
In this analogy our domain is an interval which consider a discrete sequence of n points, and our quantity is vector valued. When we add the normalization at the extremal points (which would be the boundary of the domain) we are forcing our vectors to have norm 1, but since the process is diffusive all points tend to have the same value (i.e. all vectors tend to be the same) the limit will have norm 1.
– Dinisaur
Nov 26 at 15:10
We may conjecture that the limit will be the normalization of the baricenter, which is the case for two points, but I'm not really sure about it.
– Dinisaur
Nov 26 at 15:15
add a comment |
up vote
1
down vote
Here's my idea...
It reminds me a discrete version of the heat equation for a vector valued function on the interval $(1,2,...,n)$, with Von Neumann conditions imposing the norm of the function at the boundary to be $1$.
I think this since the discrete version of the laplacian operator at one point of a graph-like domain is just the mean taken on all the adiacent vertexes.
That said we can understand that the limit will have norm equal to $1$, but the exact value seems harder to determine. We may also note that all vectors will have the same limit value if the norm we use in the recursion is not the $infty$-norm, otherwhise they may take different values.
answered Nov 24 at 18:29
Dinisaur
1009
I wouldn't know how to prove it formally, I was rather using intuition. The idea is that, without considering the normalization part for the first and last vector, we have there a diffusive process (we are averaging the neighbours) like the heat equation, which conserves the total heat (which in this analogy is the sum af all vectors) and converges to a uniform solution. In fact in that case would converge to the baricenter.
– Dinisaur
Nov 26 at 15:06
In this analogy our domain is an interval which consider a discrete sequence of n points, and our quantity is vector valued. When we add the normalization at the extremal points (which would be the boundary of the domain) we are forcing our vectors to have norm 1, but since the process is diffusive all points tend to have the same value (i.e. all vectors tend to be the same) the limit will have norm 1.
– Dinisaur
Nov 26 at 15:10
We may conjecture that the limit will be the normalization of the baricenter, which is the case for two points, but I'm not really sure about it.
– Dinisaur
Nov 26 at 15:15
add a comment |
up vote
1
down vote
up vote
1
down vote
Here's my idea...
It reminds me a discrete version of the heat equation for a vector valued function on the interval $(1,2,...,n)$, with Von Neumann conditions imposing the norm of the function at the boundary to be $1$.
I think this since the discrete version of the laplacian operator at one point of a graph-like domain is just the mean taken on all the adiacent vertexes.
That said we can understand that the limit will have norm equal to $1$, but the exact value seems harder to determine. We may also note that all vectors will have the same limit value if the norm we use in the recursion is not the $infty$-norm, otherwhise they may take different values.
answered Nov 24 at 18:29
Dinisaur
1009
Here's my idea...
It reminds me a discrete version of the heat equation for a vector valued function on the interval $(1,2,...,n)$, with Von Neumann conditions imposing the norm of the function at the boundary to be $1$.
I think this since the discrete version of the laplacian operator at one point of a graph-like domain is just the mean taken on all the adiacent vertexes.
That said we can understand that the limit will have norm equal to $1$, but the exact value seems harder to determine. We may also note that all vectors will have the same limit value if the norm we use in the recursion is not the $infty$-norm, otherwhise they may take different values.
answered Nov 24 at 18:29
Dinisaur
1009
answered Nov 24 at 18:29
Dinisaur
1009
answered Nov 24 at 18:29
Dinisaur
1009
answered Nov 24 at 18:29
Dinisaur
1009
1009
I wouldn't know how to prove it formally, I was rather using intuition. The idea is that, without considering the normalization part for the first and last vector, we have there a diffusive process (we are averaging the neighbours) like the heat equation, which conserves the total heat (which in this analogy is the sum af all vectors) and converges to a uniform solution. In fact in that case would converge to the baricenter.
– Dinisaur
Nov 26 at 15:06
In this analogy our domain is an interval which consider a discrete sequence of n points, and our quantity is vector valued. When we add the normalization at the extremal points (which would be the boundary of the domain) we are forcing our vectors to have norm 1, but since the process is diffusive all points tend to have the same value (i.e. all vectors tend to be the same) the limit will have norm 1.
– Dinisaur
Nov 26 at 15:10
We may conjecture that the limit will be the normalization of the baricenter, which is the case for two points, but I'm not really sure about it.
– Dinisaur
Nov 26 at 15:15
add a comment |
I wouldn't know how to prove it formally, I was rather using intuition. The idea is that, without considering the normalization part for the first and last vector, we have there a diffusive process (we are averaging the neighbours) like the heat equation, which conserves the total heat (which in this analogy is the sum af all vectors) and converges to a uniform solution. In fact in that case would converge to the baricenter.
– Dinisaur
Nov 26 at 15:06
In this analogy our domain is an interval which consider a discrete sequence of n points, and our quantity is vector valued. When we add the normalization at the extremal points (which would be the boundary of the domain) we are forcing our vectors to have norm 1, but since the process is diffusive all points tend to have the same value (i.e. all vectors tend to be the same) the limit will have norm 1.
– Dinisaur
Nov 26 at 15:10
We may conjecture that the limit will be the normalization of the baricenter, which is the case for two points, but I'm not really sure about it.
– Dinisaur
Nov 26 at 15:15
I wouldn't know how to prove it formally, I was rather using intuition. The idea is that, without considering the normalization part for the first and last vector, we have there a diffusive process (we are averaging the neighbours) like the heat equation, which conserves the total heat (which in this analogy is the sum af all vectors) and converges to a uniform solution. In fact in that case would converge to the baricenter.
– Dinisaur
Nov 26 at 15:06
I wouldn't know how to prove it formally, I was rather using intuition. The idea is that, without considering the normalization part for the first and last vector, we have there a diffusive process (we are averaging the neighbours) like the heat equation, which conserves the total heat (which in this analogy is the sum af all vectors) and converges to a uniform solution. In fact in that case would converge to the baricenter.
– Dinisaur
Nov 26 at 15:06
In this analogy our domain is an interval which consider a discrete sequence of n points, and our quantity is vector valued. When we add the normalization at the extremal points (which would be the boundary of the domain) we are forcing our vectors to have norm 1, but since the process is diffusive all points tend to have the same value (i.e. all vectors tend to be the same) the limit will have norm 1.
– Dinisaur
Nov 26 at 15:10
In this analogy our domain is an interval which consider a discrete sequence of n points, and our quantity is vector valued. When we add the normalization at the extremal points (which would be the boundary of the domain) we are forcing our vectors to have norm 1, but since the process is diffusive all points tend to have the same value (i.e. all vectors tend to be the same) the limit will have norm 1.
– Dinisaur
Nov 26 at 15:10
We may conjecture that the limit will be the normalization of the baricenter, which is the case for two points, but I'm not really sure about it.
– Dinisaur
Nov 26 at 15:15
We may conjecture that the limit will be the normalization of the baricenter, which is the case for two points, but I'm not really sure about it.
– Dinisaur
Nov 26 at 15:15
add a comment |
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