B splines recursion
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Given that by definition the i-th B-spline of order k is:
$$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$
where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$
We can define the spline space as
$$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$
The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
$$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.
Any hint will be greatly appreciated.
functional-analysis recursion spline
asked Nov 19 at 17:59
Ramiro Scorolli
62813
add a comment |
up vote
0
down vote
favorite
Given that by definition the i-th B-spline of order k is:
$$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$
where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$
We can define the spline space as
$$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$
The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
$$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.
Any hint will be greatly appreciated.
functional-analysis recursion spline
asked Nov 19 at 17:59
Ramiro Scorolli
62813
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given that by definition the i-th B-spline of order k is:
$$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$
where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$
We can define the spline space as
$$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$
The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
$$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.
Any hint will be greatly appreciated.
functional-analysis recursion spline
asked Nov 19 at 17:59
Ramiro Scorolli
62813
Given that by definition the i-th B-spline of order k is:
$$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$
where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$
We can define the spline space as
$$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$
The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
$$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.
Any hint will be greatly appreciated.
functional-analysis recursion spline
functional-analysis recursion spline
asked Nov 19 at 17:59
Ramiro Scorolli
62813
asked Nov 19 at 17:59
Ramiro Scorolli
62813
asked Nov 19 at 17:59
Ramiro Scorolli
62813
asked Nov 19 at 17:59
Ramiro Scorolli
62813
asked Nov 19 at 17:59
Ramiro Scorolli
62813
62813
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The two expressions coincide:
$$
begin{split}
&phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
&= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
&= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
end{split}
$$
answered Nov 19 at 18:09
Federico
2,639510
but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
– Ramiro Scorolli
Nov 19 at 18:13
1
It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
– Federico
Nov 19 at 18:18
1
In both series the same terms are appearing, just shifted.
– Federico
Nov 19 at 18:18
You are right, don't know why I didn't see that. Thanks
– Ramiro Scorolli
Nov 19 at 18:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The two expressions coincide:
$$
begin{split}
&phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
&= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
&= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
end{split}
$$
answered Nov 19 at 18:09
Federico
2,639510
but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
– Ramiro Scorolli
Nov 19 at 18:13
1
It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
– Federico
Nov 19 at 18:18
1
In both series the same terms are appearing, just shifted.
– Federico
Nov 19 at 18:18
You are right, don't know why I didn't see that. Thanks
– Ramiro Scorolli
Nov 19 at 18:19
add a comment |
up vote
1
down vote
accepted
The two expressions coincide:
$$
begin{split}
&phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
&= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
&= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
end{split}
$$
answered Nov 19 at 18:09
Federico
2,639510
but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
– Ramiro Scorolli
Nov 19 at 18:13
1
It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
– Federico
Nov 19 at 18:18
1
In both series the same terms are appearing, just shifted.
– Federico
Nov 19 at 18:18
You are right, don't know why I didn't see that. Thanks
– Ramiro Scorolli
Nov 19 at 18:19
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The two expressions coincide:
$$
begin{split}
&phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
&= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
&= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
end{split}
$$
answered Nov 19 at 18:09
Federico
2,639510
The two expressions coincide:
$$
begin{split}
&phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
&= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
&= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
end{split}
$$
answered Nov 19 at 18:09
Federico
2,639510
answered Nov 19 at 18:09
Federico
2,639510
answered Nov 19 at 18:09
Federico
2,639510
answered Nov 19 at 18:09
Federico
2,639510
2,639510
but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
– Ramiro Scorolli
Nov 19 at 18:13
1
It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
– Federico
Nov 19 at 18:18
1
In both series the same terms are appearing, just shifted.
– Federico
Nov 19 at 18:18
You are right, don't know why I didn't see that. Thanks
– Ramiro Scorolli
Nov 19 at 18:19
add a comment |
but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
– Ramiro Scorolli
Nov 19 at 18:13
1
It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
– Federico
Nov 19 at 18:18
1
In both series the same terms are appearing, just shifted.
– Federico
Nov 19 at 18:18
You are right, don't know why I didn't see that. Thanks
– Ramiro Scorolli
Nov 19 at 18:19
but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
– Ramiro Scorolli
Nov 19 at 18:13
but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
– Ramiro Scorolli
Nov 19 at 18:13
1
1
It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
– Federico
Nov 19 at 18:18
It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
– Federico
Nov 19 at 18:18
1
1
In both series the same terms are appearing, just shifted.
– Federico
Nov 19 at 18:18
In both series the same terms are appearing, just shifted.
– Federico
Nov 19 at 18:18
You are right, don't know why I didn't see that. Thanks
– Ramiro Scorolli
Nov 19 at 18:19
You are right, don't know why I didn't see that. Thanks
– Ramiro Scorolli
Nov 19 at 18:19
add a comment |
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