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A Scheme for Finding Several Zeros of a Function
The Next CEO of Stack OverflowFinding all roots of polynomial system (numerically)Root Finding Algorithm for Discrete FunctionsWhy does Fixed Point Iteration work?Fixed Point for finding a rootHow to guess initial intervals for bisection method in order to reduce the no. of iterations?Is there an example of “unfindable” interaction function?Question on Fixed Point Iteration and the Fixed Point Theorem.Finding N roots of an oscillating function with infinite roots on interval [0,1]Can we determine the existence of real solutions of a sixth order polynomial?Using Muller's method to find ALL roots ( real and complex) with three initial guesses.
$begingroup$
Let $f(x) = x-tan(x)$.
I am trying to develope a scheme to find its zeros using a particular numerical technique.
Let:
$$
g(x) = x -mf(x)
$$
then $g(r)=r-f(r)=r-0=r$, where $r$ is any of the zeroes of $f$. So $r$ is a fixed point of $g$.
For a given $r$, let $I$ be an interval containing $r$, where $|g'(x)|<1$ in $I$. If we pick any $x_0 in I$, then it is guaranteed that the sequence $x_n = g(x_n-1)$ will converge to the (unique) fixed point of $g$ in $I$.
So, for each root, my goal is to find a suitable $m$, a suitable interval, and a $x_0$ in that interval to guarantee convergence.
Issue: Other than the root $r=0$, I am having trouble derive a general way to find an interval for each of the other roots.
It can be seen that $f$ has a root in every neighboorhood $npi, n=0,pm1,pm2,...$ But I find it hard to estimate their values unless using a graphing calculator, which is not what I want to do.
Moreover, after estimating the other roots, I still have to derive a general way to pick the corresponding $m$'s and the intervals.
Could you show me a general way to find the roots of this function, using the method above?
numerical-methods
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
$endgroup$
|
show 5 more comments
$begingroup$
Let $f(x) = x-tan(x)$.
I am trying to develope a scheme to find its zeros using a particular numerical technique.
Let:
$$
g(x) = x -mf(x)
$$
then $g(r)=r-f(r)=r-0=r$, where $r$ is any of the zeroes of $f$. So $r$ is a fixed point of $g$.
For a given $r$, let $I$ be an interval containing $r$, where $|g'(x)|<1$ in $I$. If we pick any $x_0 in I$, then it is guaranteed that the sequence $x_n = g(x_n-1)$ will converge to the (unique) fixed point of $g$ in $I$.
So, for each root, my goal is to find a suitable $m$, a suitable interval, and a $x_0$ in that interval to guarantee convergence.
Issue: Other than the root $r=0$, I am having trouble derive a general way to find an interval for each of the other roots.
It can be seen that $f$ has a root in every neighboorhood $npi, n=0,pm1,pm2,...$ But I find it hard to estimate their values unless using a graphing calculator, which is not what I want to do.
Moreover, after estimating the other roots, I still have to derive a general way to pick the corresponding $m$'s and the intervals.
Could you show me a general way to find the roots of this function, using the method above?
numerical-methods
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
$endgroup$
$begingroup$
Do you have to use this $f$? Could you alternatively use $f(x)=sin(x)-xcos(x)$ or $f_n(x)=npi+arctan(x)-x$?
$endgroup$
– LutzL
Mar 27 at 20:20
$begingroup$
@LutzL I would prefer to use the original one. However, if it is concluded that there is no way to derive a general method to solve the original one, then using the ones in your comment is ok, as long as we can show that they have the same zeroes.
$endgroup$
– A Slow Learner
Mar 27 at 20:26
$begingroup$
By the second formula used as fixed point iteration, the solutions are close to $x_n=npi+arctan(npi)approx(n+frac12)pi-frac1npi$. You might want to use a value close to $-f'(x_n)^-1$ for $m$, with the original $f$. At a first glance, this gives $m=-n^2pi^2$. This large value suggests that the interval of convergence is extremely small.
$endgroup$
– LutzL
Mar 27 at 20:36
$begingroup$
@LutzL Could you explain a bit more about the second formula? Why are its solutions close to the original formula's solutions?
$endgroup$
– A Slow Learner
Mar 27 at 20:43
$begingroup$
Because the formula $x_k+1=g(x_k)=npi+arctan(x_k)$ is a contracting fixed point iteration. Start with $x_0=0$ then $x_1=npi$, $x_2=npi+arctan(npi)=npi+fracpi2-arctan(frac1npi)approx (n+frac12)pi-frac1npi$.
$endgroup$
– LutzL
Mar 27 at 21:01
|
show 5 more comments
$begingroup$
Let $f(x) = x-tan(x)$.
I am trying to develope a scheme to find its zeros using a particular numerical technique.
Let:
$$
g(x) = x -mf(x)
$$
then $g(r)=r-f(r)=r-0=r$, where $r$ is any of the zeroes of $f$. So $r$ is a fixed point of $g$.
For a given $r$, let $I$ be an interval containing $r$, where $|g'(x)|<1$ in $I$. If we pick any $x_0 in I$, then it is guaranteed that the sequence $x_n = g(x_n-1)$ will converge to the (unique) fixed point of $g$ in $I$.
So, for each root, my goal is to find a suitable $m$, a suitable interval, and a $x_0$ in that interval to guarantee convergence.
Issue: Other than the root $r=0$, I am having trouble derive a general way to find an interval for each of the other roots.
It can be seen that $f$ has a root in every neighboorhood $npi, n=0,pm1,pm2,...$ But I find it hard to estimate their values unless using a graphing calculator, which is not what I want to do.
Moreover, after estimating the other roots, I still have to derive a general way to pick the corresponding $m$'s and the intervals.
Could you show me a general way to find the roots of this function, using the method above?
numerical-methods
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
$endgroup$
Let $f(x) = x-tan(x)$.
I am trying to develope a scheme to find its zeros using a particular numerical technique.
Let:
$$
g(x) = x -mf(x)
$$
then $g(r)=r-f(r)=r-0=r$, where $r$ is any of the zeroes of $f$. So $r$ is a fixed point of $g$.
For a given $r$, let $I$ be an interval containing $r$, where $|g'(x)|<1$ in $I$. If we pick any $x_0 in I$, then it is guaranteed that the sequence $x_n = g(x_n-1)$ will converge to the (unique) fixed point of $g$ in $I$.
So, for each root, my goal is to find a suitable $m$, a suitable interval, and a $x_0$ in that interval to guarantee convergence.
Issue: Other than the root $r=0$, I am having trouble derive a general way to find an interval for each of the other roots.
It can be seen that $f$ has a root in every neighboorhood $npi, n=0,pm1,pm2,...$ But I find it hard to estimate their values unless using a graphing calculator, which is not what I want to do.
Moreover, after estimating the other roots, I still have to derive a general way to pick the corresponding $m$'s and the intervals.
Could you show me a general way to find the roots of this function, using the method above?
numerical-methods
numerical-methods
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
asked Mar 27 at 19:15
A Slow LearnerA Slow Learner
463213
463213
$begingroup$
Do you have to use this $f$? Could you alternatively use $f(x)=sin(x)-xcos(x)$ or $f_n(x)=npi+arctan(x)-x$?
$endgroup$
– LutzL
Mar 27 at 20:20
$begingroup$
@LutzL I would prefer to use the original one. However, if it is concluded that there is no way to derive a general method to solve the original one, then using the ones in your comment is ok, as long as we can show that they have the same zeroes.
$endgroup$
– A Slow Learner
Mar 27 at 20:26
$begingroup$
By the second formula used as fixed point iteration, the solutions are close to $x_n=npi+arctan(npi)approx(n+frac12)pi-frac1npi$. You might want to use a value close to $-f'(x_n)^-1$ for $m$, with the original $f$. At a first glance, this gives $m=-n^2pi^2$. This large value suggests that the interval of convergence is extremely small.
$endgroup$
– LutzL
Mar 27 at 20:36
$begingroup$
@LutzL Could you explain a bit more about the second formula? Why are its solutions close to the original formula's solutions?
$endgroup$
– A Slow Learner
Mar 27 at 20:43
$begingroup$
Because the formula $x_k+1=g(x_k)=npi+arctan(x_k)$ is a contracting fixed point iteration. Start with $x_0=0$ then $x_1=npi$, $x_2=npi+arctan(npi)=npi+fracpi2-arctan(frac1npi)approx (n+frac12)pi-frac1npi$.
$endgroup$
– LutzL
Mar 27 at 21:01
|
show 5 more comments
$begingroup$
Do you have to use this $f$? Could you alternatively use $f(x)=sin(x)-xcos(x)$ or $f_n(x)=npi+arctan(x)-x$?
$endgroup$
– LutzL
Mar 27 at 20:20
$begingroup$
@LutzL I would prefer to use the original one. However, if it is concluded that there is no way to derive a general method to solve the original one, then using the ones in your comment is ok, as long as we can show that they have the same zeroes.
$endgroup$
– A Slow Learner
Mar 27 at 20:26
$begingroup$
By the second formula used as fixed point iteration, the solutions are close to $x_n=npi+arctan(npi)approx(n+frac12)pi-frac1npi$. You might want to use a value close to $-f'(x_n)^-1$ for $m$, with the original $f$. At a first glance, this gives $m=-n^2pi^2$. This large value suggests that the interval of convergence is extremely small.
$endgroup$
– LutzL
Mar 27 at 20:36
$begingroup$
@LutzL Could you explain a bit more about the second formula? Why are its solutions close to the original formula's solutions?
$endgroup$
– A Slow Learner
Mar 27 at 20:43
$begingroup$
Because the formula $x_k+1=g(x_k)=npi+arctan(x_k)$ is a contracting fixed point iteration. Start with $x_0=0$ then $x_1=npi$, $x_2=npi+arctan(npi)=npi+fracpi2-arctan(frac1npi)approx (n+frac12)pi-frac1npi$.
$endgroup$
– LutzL
Mar 27 at 21:01
$begingroup$
Do you have to use this $f$? Could you alternatively use $f(x)=sin(x)-xcos(x)$ or $f_n(x)=npi+arctan(x)-x$?
$endgroup$
– LutzL
Mar 27 at 20:20
$begingroup$
Do you have to use this $f$? Could you alternatively use $f(x)=sin(x)-xcos(x)$ or $f_n(x)=npi+arctan(x)-x$?
$endgroup$
– LutzL
Mar 27 at 20:20
$begingroup$
@LutzL I would prefer to use the original one. However, if it is concluded that there is no way to derive a general method to solve the original one, then using the ones in your comment is ok, as long as we can show that they have the same zeroes.
$endgroup$
– A Slow Learner
Mar 27 at 20:26
$begingroup$
@LutzL I would prefer to use the original one. However, if it is concluded that there is no way to derive a general method to solve the original one, then using the ones in your comment is ok, as long as we can show that they have the same zeroes.
$endgroup$
– A Slow Learner
Mar 27 at 20:26
$begingroup$
By the second formula used as fixed point iteration, the solutions are close to $x_n=npi+arctan(npi)approx(n+frac12)pi-frac1npi$. You might want to use a value close to $-f'(x_n)^-1$ for $m$, with the original $f$. At a first glance, this gives $m=-n^2pi^2$. This large value suggests that the interval of convergence is extremely small.
$endgroup$
– LutzL
Mar 27 at 20:36
$begingroup$
By the second formula used as fixed point iteration, the solutions are close to $x_n=npi+arctan(npi)approx(n+frac12)pi-frac1npi$. You might want to use a value close to $-f'(x_n)^-1$ for $m$, with the original $f$. At a first glance, this gives $m=-n^2pi^2$. This large value suggests that the interval of convergence is extremely small.
$endgroup$
– LutzL
Mar 27 at 20:36
$begingroup$
@LutzL Could you explain a bit more about the second formula? Why are its solutions close to the original formula's solutions?
$endgroup$
– A Slow Learner
Mar 27 at 20:43
$begingroup$
@LutzL Could you explain a bit more about the second formula? Why are its solutions close to the original formula's solutions?
$endgroup$
– A Slow Learner
Mar 27 at 20:43
$begingroup$
Because the formula $x_k+1=g(x_k)=npi+arctan(x_k)$ is a contracting fixed point iteration. Start with $x_0=0$ then $x_1=npi$, $x_2=npi+arctan(npi)=npi+fracpi2-arctan(frac1npi)approx (n+frac12)pi-frac1npi$.
$endgroup$
– LutzL
Mar 27 at 21:01
$begingroup$
Because the formula $x_k+1=g(x_k)=npi+arctan(x_k)$ is a contracting fixed point iteration. Start with $x_0=0$ then $x_1=npi$, $x_2=npi+arctan(npi)=npi+fracpi2-arctan(frac1npi)approx (n+frac12)pi-frac1npi$.
$endgroup$
– LutzL
Mar 27 at 21:01
|
show 5 more comments
0
active
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$begingroup$
Do you have to use this $f$? Could you alternatively use $f(x)=sin(x)-xcos(x)$ or $f_n(x)=npi+arctan(x)-x$?
$endgroup$
– LutzL
Mar 27 at 20:20
$begingroup$
@LutzL I would prefer to use the original one. However, if it is concluded that there is no way to derive a general method to solve the original one, then using the ones in your comment is ok, as long as we can show that they have the same zeroes.
$endgroup$
– A Slow Learner
Mar 27 at 20:26
$begingroup$
By the second formula used as fixed point iteration, the solutions are close to $x_n=npi+arctan(npi)approx(n+frac12)pi-frac1npi$. You might want to use a value close to $-f'(x_n)^-1$ for $m$, with the original $f$. At a first glance, this gives $m=-n^2pi^2$. This large value suggests that the interval of convergence is extremely small.
$endgroup$
– LutzL
Mar 27 at 20:36
$begingroup$
@LutzL Could you explain a bit more about the second formula? Why are its solutions close to the original formula's solutions?
$endgroup$
– A Slow Learner
Mar 27 at 20:43
$begingroup$
Because the formula $x_k+1=g(x_k)=npi+arctan(x_k)$ is a contracting fixed point iteration. Start with $x_0=0$ then $x_1=npi$, $x_2=npi+arctan(npi)=npi+fracpi2-arctan(frac1npi)approx (n+frac12)pi-frac1npi$.
$endgroup$
– LutzL
Mar 27 at 21:01