Fixed point iteration: how to finish proof of its convergence
$begingroup$
For this assignment we want to determine the location $r = (x, y)$ with measure distances $d_i$ from two reference points $(x_i, y_i)$. The system of equations would then be
$f(r) = sqrt{(x-x_i)^2 + (y-y_i)^2} = d$.
We are asked to evaluate this fixed point iteration:
$r_{k+1} = r_k + alpha(f(r_k) - d)$.
and determine the contraints on $alpha$ for the iteration to converge. So far I came up with the following.
$r_{k+1} = G(r_k)$ converges if there is a $q in [0,1)$ such that $||G(x) - G(y)|| leq q ||x-y||$.
begin{align}
||G(x) - G(y) || &= ||x + alpha(f(x) - d) - y - alpha(f(y) - d)||\
&= || x - y + alpha(f(x) - f(y))||\
&= || x - y + alpha(f(y) + triangledown f(z)(x-y) - f(y)) ||\
&= || x - y + alpha(triangledown f(z)(x-y))||
end{align}
with $z in [x,y]$ using the mean value theorem. This is where I'm stuck. Since we're working with the Euclidian norm and the gradient vector of $f$ I'm not sure if the following is allowed.
begin{align}
||G(x) - G(y) || &= || (x-y)(1+alphatriangledown f(z)||\
&leq ||x-y|| ||1+alphatriangledown f(z)||\
end{align}
What constraint then needs to be placed on $alpha$ such that
$||1+alphatriangledown f(z)|| < 1$, if this is allowed at all?
fixed-point-theorems
asked Jan 21 at 16:42
FleetLFleetL
114
$endgroup$
add a comment |
$begingroup$
For this assignment we want to determine the location $r = (x, y)$ with measure distances $d_i$ from two reference points $(x_i, y_i)$. The system of equations would then be
$f(r) = sqrt{(x-x_i)^2 + (y-y_i)^2} = d$.
We are asked to evaluate this fixed point iteration:
$r_{k+1} = r_k + alpha(f(r_k) - d)$.
and determine the contraints on $alpha$ for the iteration to converge. So far I came up with the following.
$r_{k+1} = G(r_k)$ converges if there is a $q in [0,1)$ such that $||G(x) - G(y)|| leq q ||x-y||$.
begin{align}
||G(x) - G(y) || &= ||x + alpha(f(x) - d) - y - alpha(f(y) - d)||\
&= || x - y + alpha(f(x) - f(y))||\
&= || x - y + alpha(f(y) + triangledown f(z)(x-y) - f(y)) ||\
&= || x - y + alpha(triangledown f(z)(x-y))||
end{align}
with $z in [x,y]$ using the mean value theorem. This is where I'm stuck. Since we're working with the Euclidian norm and the gradient vector of $f$ I'm not sure if the following is allowed.
begin{align}
||G(x) - G(y) || &= || (x-y)(1+alphatriangledown f(z)||\
&leq ||x-y|| ||1+alphatriangledown f(z)||\
end{align}
What constraint then needs to be placed on $alpha$ such that
$||1+alphatriangledown f(z)|| < 1$, if this is allowed at all?
fixed-point-theorems
asked Jan 21 at 16:42
FleetLFleetL
114
$endgroup$
add a comment |
$begingroup$
For this assignment we want to determine the location $r = (x, y)$ with measure distances $d_i$ from two reference points $(x_i, y_i)$. The system of equations would then be
$f(r) = sqrt{(x-x_i)^2 + (y-y_i)^2} = d$.
We are asked to evaluate this fixed point iteration:
$r_{k+1} = r_k + alpha(f(r_k) - d)$.
and determine the contraints on $alpha$ for the iteration to converge. So far I came up with the following.
$r_{k+1} = G(r_k)$ converges if there is a $q in [0,1)$ such that $||G(x) - G(y)|| leq q ||x-y||$.
begin{align}
||G(x) - G(y) || &= ||x + alpha(f(x) - d) - y - alpha(f(y) - d)||\
&= || x - y + alpha(f(x) - f(y))||\
&= || x - y + alpha(f(y) + triangledown f(z)(x-y) - f(y)) ||\
&= || x - y + alpha(triangledown f(z)(x-y))||
end{align}
with $z in [x,y]$ using the mean value theorem. This is where I'm stuck. Since we're working with the Euclidian norm and the gradient vector of $f$ I'm not sure if the following is allowed.
begin{align}
||G(x) - G(y) || &= || (x-y)(1+alphatriangledown f(z)||\
&leq ||x-y|| ||1+alphatriangledown f(z)||\
end{align}
What constraint then needs to be placed on $alpha$ such that
$||1+alphatriangledown f(z)|| < 1$, if this is allowed at all?
fixed-point-theorems
asked Jan 21 at 16:42
FleetLFleetL
114
$endgroup$
For this assignment we want to determine the location $r = (x, y)$ with measure distances $d_i$ from two reference points $(x_i, y_i)$. The system of equations would then be
$f(r) = sqrt{(x-x_i)^2 + (y-y_i)^2} = d$.
We are asked to evaluate this fixed point iteration:
$r_{k+1} = r_k + alpha(f(r_k) - d)$.
and determine the contraints on $alpha$ for the iteration to converge. So far I came up with the following.
$r_{k+1} = G(r_k)$ converges if there is a $q in [0,1)$ such that $||G(x) - G(y)|| leq q ||x-y||$.
begin{align}
||G(x) - G(y) || &= ||x + alpha(f(x) - d) - y - alpha(f(y) - d)||\
&= || x - y + alpha(f(x) - f(y))||\
&= || x - y + alpha(f(y) + triangledown f(z)(x-y) - f(y)) ||\
&= || x - y + alpha(triangledown f(z)(x-y))||
end{align}
with $z in [x,y]$ using the mean value theorem. This is where I'm stuck. Since we're working with the Euclidian norm and the gradient vector of $f$ I'm not sure if the following is allowed.
begin{align}
||G(x) - G(y) || &= || (x-y)(1+alphatriangledown f(z)||\
&leq ||x-y|| ||1+alphatriangledown f(z)||\
end{align}
What constraint then needs to be placed on $alpha$ such that
$||1+alphatriangledown f(z)|| < 1$, if this is allowed at all?
fixed-point-theorems
fixed-point-theorems
asked Jan 21 at 16:42
FleetLFleetL
114
asked Jan 21 at 16:42
FleetLFleetL
114
edited Jan 21 at 19:25
FleetL
asked Jan 21 at 16:42
FleetLFleetL
114
asked Jan 21 at 16:42
FleetLFleetL
114
asked Jan 21 at 16:42
FleetLFleetL
114
114
add a comment |
add a comment |
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