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On the convergence rate of Newton's method
Bairstow's method: Rate of convergenceNewton's Method - Slow ConvergenceHow to show Newton's method has quadratic convergence rate with an example?What have I done wrong? in Newton's method calculation?Numerical method with convergence greater than 2Why linear convergence in this example?Convergence of a variant of Newton's MethodNewton's method and fixed pointIf fixed-point iteration has linear convergence, how can Newton's Method have quadratic convergence?Is modified Newton's Raphson method redundant?
$begingroup$
Is that possible for some function whose convergence rate is linear by using Newton's method?
I am solving the function $$f(x) = sin^2(x) - x sin(x) + frac 14 x^2$$ by Newton's Method. I got the error with $0.11$, $0.05$, $0.024$, $0.012$ in the first few iterations, which looks like it has linear convergence rate with $1/2$. Is that possible?
numerical-methods newton-raphson
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user135671user135671
1
$endgroup$
add a comment |
$begingroup$
Is that possible for some function whose convergence rate is linear by using Newton's method?
I am solving the function $$f(x) = sin^2(x) - x sin(x) + frac 14 x^2$$ by Newton's Method. I got the error with $0.11$, $0.05$, $0.024$, $0.012$ in the first few iterations, which looks like it has linear convergence rate with $1/2$. Is that possible?
numerical-methods newton-raphson
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user135671user135671
1
$endgroup$
$begingroup$
Yes, this is possible and expected when you are approaching a "double" root with Newton iterations. Note that each of the terms in $f(x)$ has a root of multiplicity two at $x=0$, so if that is the root your iterates converge to, linear convergence is quite possible.
$endgroup$
– hardmath
2 days ago
add a comment |
$begingroup$
Is that possible for some function whose convergence rate is linear by using Newton's method?
I am solving the function $$f(x) = sin^2(x) - x sin(x) + frac 14 x^2$$ by Newton's Method. I got the error with $0.11$, $0.05$, $0.024$, $0.012$ in the first few iterations, which looks like it has linear convergence rate with $1/2$. Is that possible?
numerical-methods newton-raphson
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user135671user135671
1
$endgroup$
Is that possible for some function whose convergence rate is linear by using Newton's method?
I am solving the function $$f(x) = sin^2(x) - x sin(x) + frac 14 x^2$$ by Newton's Method. I got the error with $0.11$, $0.05$, $0.024$, $0.012$ in the first few iterations, which looks like it has linear convergence rate with $1/2$. Is that possible?
numerical-methods newton-raphson
numerical-methods newton-raphson
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user135671user135671
1
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user135671user135671
1
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
13k41960
asked 2 days ago
user135671user135671
1
asked 2 days ago
user135671user135671
1
asked 2 days ago
user135671user135671
1
1
$begingroup$
Yes, this is possible and expected when you are approaching a "double" root with Newton iterations. Note that each of the terms in $f(x)$ has a root of multiplicity two at $x=0$, so if that is the root your iterates converge to, linear convergence is quite possible.
$endgroup$
– hardmath
2 days ago
add a comment |
$begingroup$
Yes, this is possible and expected when you are approaching a "double" root with Newton iterations. Note that each of the terms in $f(x)$ has a root of multiplicity two at $x=0$, so if that is the root your iterates converge to, linear convergence is quite possible.
$endgroup$
– hardmath
2 days ago
$begingroup$
Yes, this is possible and expected when you are approaching a "double" root with Newton iterations. Note that each of the terms in $f(x)$ has a root of multiplicity two at $x=0$, so if that is the root your iterates converge to, linear convergence is quite possible.
$endgroup$
– hardmath
2 days ago
$begingroup$
Yes, this is possible and expected when you are approaching a "double" root with Newton iterations. Note that each of the terms in $f(x)$ has a root of multiplicity two at $x=0$, so if that is the root your iterates converge to, linear convergence is quite possible.
$endgroup$
– hardmath
2 days ago
add a comment |
1 Answer
1
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oldest
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$begingroup$
As your function is a square, $f(x)=(sin x-frac12x)^2$, all roots will have even multiplicity. And indeed, Newton's method has linear convergence towards multiple roots, with factor $1-frac1m$ for multiplicity $m$.
Note that as $fracpi2>0=sin(pi)$ and $fracpi4<1=sin(fracpi2)$, there are additional (double) roots inside $[fracpi2,pi]$ and its mirrored interval.
answered yesterday
LutzLLutzL
59.5k42057
$endgroup$
$begingroup$
Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values?
$endgroup$
– user135671
9 hours ago
$begingroup$
It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)ne 0$, and that $m$ is an even number.
$endgroup$
– LutzL
9 hours ago
add a comment |
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1 Answer
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$begingroup$
As your function is a square, $f(x)=(sin x-frac12x)^2$, all roots will have even multiplicity. And indeed, Newton's method has linear convergence towards multiple roots, with factor $1-frac1m$ for multiplicity $m$.
Note that as $fracpi2>0=sin(pi)$ and $fracpi4<1=sin(fracpi2)$, there are additional (double) roots inside $[fracpi2,pi]$ and its mirrored interval.
answered yesterday
LutzLLutzL
59.5k42057
$endgroup$
$begingroup$
Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values?
$endgroup$
– user135671
9 hours ago
$begingroup$
It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)ne 0$, and that $m$ is an even number.
$endgroup$
– LutzL
9 hours ago
add a comment |
$begingroup$
As your function is a square, $f(x)=(sin x-frac12x)^2$, all roots will have even multiplicity. And indeed, Newton's method has linear convergence towards multiple roots, with factor $1-frac1m$ for multiplicity $m$.
Note that as $fracpi2>0=sin(pi)$ and $fracpi4<1=sin(fracpi2)$, there are additional (double) roots inside $[fracpi2,pi]$ and its mirrored interval.
answered yesterday
LutzLLutzL
59.5k42057
$endgroup$
$begingroup$
Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values?
$endgroup$
– user135671
9 hours ago
$begingroup$
It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)ne 0$, and that $m$ is an even number.
$endgroup$
– LutzL
9 hours ago
add a comment |
$begingroup$
As your function is a square, $f(x)=(sin x-frac12x)^2$, all roots will have even multiplicity. And indeed, Newton's method has linear convergence towards multiple roots, with factor $1-frac1m$ for multiplicity $m$.
Note that as $fracpi2>0=sin(pi)$ and $fracpi4<1=sin(fracpi2)$, there are additional (double) roots inside $[fracpi2,pi]$ and its mirrored interval.
answered yesterday
LutzLLutzL
59.5k42057
$endgroup$
As your function is a square, $f(x)=(sin x-frac12x)^2$, all roots will have even multiplicity. And indeed, Newton's method has linear convergence towards multiple roots, with factor $1-frac1m$ for multiplicity $m$.
Note that as $fracpi2>0=sin(pi)$ and $fracpi4<1=sin(fracpi2)$, there are additional (double) roots inside $[fracpi2,pi]$ and its mirrored interval.
answered yesterday
LutzLLutzL
59.5k42057
answered yesterday
LutzLLutzL
59.5k42057
answered yesterday
LutzLLutzL
59.5k42057
answered yesterday
LutzLLutzL
59.5k42057
59.5k42057
$begingroup$
Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values?
$endgroup$
– user135671
9 hours ago
$begingroup$
It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)ne 0$, and that $m$ is an even number.
$endgroup$
– LutzL
9 hours ago
add a comment |
$begingroup$
Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values?
$endgroup$
– user135671
9 hours ago
$begingroup$
It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)ne 0$, and that $m$ is an even number.
$endgroup$
– LutzL
9 hours ago
$begingroup$
Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values?
$endgroup$
– user135671
9 hours ago
$begingroup$
Thanks for the help, Can I ask one more question? What is even multiplicity? Does that mean for each $f(x) = 0$, there are two x with the same values?
$endgroup$
– user135671
9 hours ago
$begingroup$
It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)ne 0$, and that $m$ is an even number.
$endgroup$
– LutzL
9 hours ago
$begingroup$
It means that first there is an integer multiplicity at any root $a$, meaning that $f(x)=(x-a)^mg_a(x)$ with $g_a$ (at least) continuous and $g_a(a)ne 0$, and that $m$ is an even number.
$endgroup$
– LutzL
9 hours ago
add a comment |
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$begingroup$
Yes, this is possible and expected when you are approaching a "double" root with Newton iterations. Note that each of the terms in $f(x)$ has a root of multiplicity two at $x=0$, so if that is the root your iterates converge to, linear convergence is quite possible.
$endgroup$
– hardmath
2 days ago