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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?













0












$begingroup$


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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    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















0












$begingroup$


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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    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













0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Rodrigo de Azevedo

13k41960




13k41960










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
















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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











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1 Answer
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active

oldest

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1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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

















  • $begingroup$
    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


















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