Constructing Newton iteration converging to non-root











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Is it possible to construct a Newton sequence $x_{n+1} := x_{n} - f(x_n)/f'(x_{n})$ such that ${x_{n}}$ is a Cauchy sequence converging to $x^*$, but $x^{*}$ is not a root of $f$? (Perhaps because $f$ has no roots?)






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    Is it possible to construct a Newton sequence $x_{n+1} := x_{n} - f(x_n)/f'(x_{n})$ such that ${x_{n}}$ is a Cauchy sequence converging to $x^*$, but $x^{*}$ is not a root of $f$? (Perhaps because $f$ has no roots?)






    newton-raphson







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      favorite











      Is it possible to construct a Newton sequence $x_{n+1} := x_{n} - f(x_n)/f'(x_{n})$ such that ${x_{n}}$ is a Cauchy sequence converging to $x^*$, but $x^{*}$ is not a root of $f$? (Perhaps because $f$ has no roots?)






      newton-raphson







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      Is it possible to construct a Newton sequence $x_{n+1} := x_{n} - f(x_n)/f'(x_{n})$ such that ${x_{n}}$ is a Cauchy sequence converging to $x^*$, but $x^{*}$ is not a root of $f$? (Perhaps because $f$ has no roots?)






      newton-raphson




      newton-raphson






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      asked Nov 28 at 21:46









      user14717

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          accepted










          It is not possible if we assume that $x^star$ lies in the domain of definition of $f$ and that $f$ and $f'$ are continuous at that point. For a counterexample where $f'$ is not continuous at $x^star$ see the nice answer by Oscar Lanzi.



          In my setting we use
          $$ f(x_n) = f'(x_n) (x_n - x_{n+1}) $$
          and take the limit (I assume $f'$ to be continuous at $x^*$).



          If we take the limit in the equation above we get (using the continuity of $f'$ and $f$ at $x^star$)
          $$ f(x^star) = f(lim_{nrightarrow infty} x_n) = lim_{nrightarrow infty} f(x_n) = lim_{nrightarrow infty} f'(x_n) (x_n - x_{n+1})
          = lim_{nrightarrow infty} f'(x_n) cdot lim_{nrightarrow infty} (x_n - x_{n+1}) = f'(x^star) cdot (x^star - x^star) = 0.$$






          share|cite|improve this answer























          • It looks like this can be weakened from "continuous derivative" to "derivative exists and is bounded in a neighborhood of $x^{*}$".
            – user14717
            Nov 28 at 22:39










          • Indeed, we could do that. But I like to have some a priori regularity (I love my $C^1$-functions) :)
            – Severin Schraven
            Nov 28 at 22:44


















          up vote
          1
          down vote













          Try this function:



          $f(x)=max(1-sqrt{|x|},|x|)$



          For most initial guesses, and for all initial guesses more than one-half in absolute value, you converge to zero. But the function has no real zeroes.






          share|cite|improve this answer























          • That sounds interesting. Is it easy to see that it converges indeed to zero?
            – Severin Schraven
            Nov 28 at 22:10






          • 1




            Truly fabulous example. Both of the answers given are required for full understanding, so I'm accepting @SeverinSchraven's answer as accepted (because it looks like Oscar is well beyond the range of caring about the point system).
            – user14717
            Nov 28 at 22:20










          • I'm now wondering if a more classical derivative discontinuity can cause the problem; say $f'$ has a pole at $x^{*}$ but $f$ still the restriction to the real line of a meromorphic function.
            – user14717
            Nov 28 at 22:25










          • In fact, I am not sure you can say this converges to zero. If you start in the range of the absolute value you are in one step at zero, but there the scheme fails to work, as you have no derivative there.
            – Severin Schraven
            Nov 28 at 22:37










          • @SeverinSchraven: The question was intentionally ambiguous regarding the function spaces--defining the function space is half the fun.
            – user14717
            Nov 28 at 22:41











          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          It is not possible if we assume that $x^star$ lies in the domain of definition of $f$ and that $f$ and $f'$ are continuous at that point. For a counterexample where $f'$ is not continuous at $x^star$ see the nice answer by Oscar Lanzi.



          In my setting we use
          $$ f(x_n) = f'(x_n) (x_n - x_{n+1}) $$
          and take the limit (I assume $f'$ to be continuous at $x^*$).



          If we take the limit in the equation above we get (using the continuity of $f'$ and $f$ at $x^star$)
          $$ f(x^star) = f(lim_{nrightarrow infty} x_n) = lim_{nrightarrow infty} f(x_n) = lim_{nrightarrow infty} f'(x_n) (x_n - x_{n+1})
          = lim_{nrightarrow infty} f'(x_n) cdot lim_{nrightarrow infty} (x_n - x_{n+1}) = f'(x^star) cdot (x^star - x^star) = 0.$$






          share|cite|improve this answer























          • It looks like this can be weakened from "continuous derivative" to "derivative exists and is bounded in a neighborhood of $x^{*}$".
            – user14717
            Nov 28 at 22:39










          • Indeed, we could do that. But I like to have some a priori regularity (I love my $C^1$-functions) :)
            – Severin Schraven
            Nov 28 at 22:44















          up vote
          2
          down vote



          accepted










          It is not possible if we assume that $x^star$ lies in the domain of definition of $f$ and that $f$ and $f'$ are continuous at that point. For a counterexample where $f'$ is not continuous at $x^star$ see the nice answer by Oscar Lanzi.



          In my setting we use
          $$ f(x_n) = f'(x_n) (x_n - x_{n+1}) $$
          and take the limit (I assume $f'$ to be continuous at $x^*$).



          If we take the limit in the equation above we get (using the continuity of $f'$ and $f$ at $x^star$)
          $$ f(x^star) = f(lim_{nrightarrow infty} x_n) = lim_{nrightarrow infty} f(x_n) = lim_{nrightarrow infty} f'(x_n) (x_n - x_{n+1})
          = lim_{nrightarrow infty} f'(x_n) cdot lim_{nrightarrow infty} (x_n - x_{n+1}) = f'(x^star) cdot (x^star - x^star) = 0.$$






          share|cite|improve this answer























          • It looks like this can be weakened from "continuous derivative" to "derivative exists and is bounded in a neighborhood of $x^{*}$".
            – user14717
            Nov 28 at 22:39










          • Indeed, we could do that. But I like to have some a priori regularity (I love my $C^1$-functions) :)
            – Severin Schraven
            Nov 28 at 22:44













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          It is not possible if we assume that $x^star$ lies in the domain of definition of $f$ and that $f$ and $f'$ are continuous at that point. For a counterexample where $f'$ is not continuous at $x^star$ see the nice answer by Oscar Lanzi.



          In my setting we use
          $$ f(x_n) = f'(x_n) (x_n - x_{n+1}) $$
          and take the limit (I assume $f'$ to be continuous at $x^*$).



          If we take the limit in the equation above we get (using the continuity of $f'$ and $f$ at $x^star$)
          $$ f(x^star) = f(lim_{nrightarrow infty} x_n) = lim_{nrightarrow infty} f(x_n) = lim_{nrightarrow infty} f'(x_n) (x_n - x_{n+1})
          = lim_{nrightarrow infty} f'(x_n) cdot lim_{nrightarrow infty} (x_n - x_{n+1}) = f'(x^star) cdot (x^star - x^star) = 0.$$






          share|cite|improve this answer














          It is not possible if we assume that $x^star$ lies in the domain of definition of $f$ and that $f$ and $f'$ are continuous at that point. For a counterexample where $f'$ is not continuous at $x^star$ see the nice answer by Oscar Lanzi.



          In my setting we use
          $$ f(x_n) = f'(x_n) (x_n - x_{n+1}) $$
          and take the limit (I assume $f'$ to be continuous at $x^*$).



          If we take the limit in the equation above we get (using the continuity of $f'$ and $f$ at $x^star$)
          $$ f(x^star) = f(lim_{nrightarrow infty} x_n) = lim_{nrightarrow infty} f(x_n) = lim_{nrightarrow infty} f'(x_n) (x_n - x_{n+1})
          = lim_{nrightarrow infty} f'(x_n) cdot lim_{nrightarrow infty} (x_n - x_{n+1}) = f'(x^star) cdot (x^star - x^star) = 0.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 at 22:34

























          answered Nov 28 at 21:59









          Severin Schraven

          5,6131832




          5,6131832












          • It looks like this can be weakened from "continuous derivative" to "derivative exists and is bounded in a neighborhood of $x^{*}$".
            – user14717
            Nov 28 at 22:39










          • Indeed, we could do that. But I like to have some a priori regularity (I love my $C^1$-functions) :)
            – Severin Schraven
            Nov 28 at 22:44


















          • It looks like this can be weakened from "continuous derivative" to "derivative exists and is bounded in a neighborhood of $x^{*}$".
            – user14717
            Nov 28 at 22:39










          • Indeed, we could do that. But I like to have some a priori regularity (I love my $C^1$-functions) :)
            – Severin Schraven
            Nov 28 at 22:44
















          It looks like this can be weakened from "continuous derivative" to "derivative exists and is bounded in a neighborhood of $x^{*}$".
          – user14717
          Nov 28 at 22:39




          It looks like this can be weakened from "continuous derivative" to "derivative exists and is bounded in a neighborhood of $x^{*}$".
          – user14717
          Nov 28 at 22:39












          Indeed, we could do that. But I like to have some a priori regularity (I love my $C^1$-functions) :)
          – Severin Schraven
          Nov 28 at 22:44




          Indeed, we could do that. But I like to have some a priori regularity (I love my $C^1$-functions) :)
          – Severin Schraven
          Nov 28 at 22:44










          up vote
          1
          down vote













          Try this function:



          $f(x)=max(1-sqrt{|x|},|x|)$



          For most initial guesses, and for all initial guesses more than one-half in absolute value, you converge to zero. But the function has no real zeroes.






          share|cite|improve this answer























          • That sounds interesting. Is it easy to see that it converges indeed to zero?
            – Severin Schraven
            Nov 28 at 22:10






          • 1




            Truly fabulous example. Both of the answers given are required for full understanding, so I'm accepting @SeverinSchraven's answer as accepted (because it looks like Oscar is well beyond the range of caring about the point system).
            – user14717
            Nov 28 at 22:20










          • I'm now wondering if a more classical derivative discontinuity can cause the problem; say $f'$ has a pole at $x^{*}$ but $f$ still the restriction to the real line of a meromorphic function.
            – user14717
            Nov 28 at 22:25










          • In fact, I am not sure you can say this converges to zero. If you start in the range of the absolute value you are in one step at zero, but there the scheme fails to work, as you have no derivative there.
            – Severin Schraven
            Nov 28 at 22:37










          • @SeverinSchraven: The question was intentionally ambiguous regarding the function spaces--defining the function space is half the fun.
            – user14717
            Nov 28 at 22:41















          up vote
          1
          down vote













          Try this function:



          $f(x)=max(1-sqrt{|x|},|x|)$



          For most initial guesses, and for all initial guesses more than one-half in absolute value, you converge to zero. But the function has no real zeroes.






          share|cite|improve this answer























          • That sounds interesting. Is it easy to see that it converges indeed to zero?
            – Severin Schraven
            Nov 28 at 22:10






          • 1




            Truly fabulous example. Both of the answers given are required for full understanding, so I'm accepting @SeverinSchraven's answer as accepted (because it looks like Oscar is well beyond the range of caring about the point system).
            – user14717
            Nov 28 at 22:20










          • I'm now wondering if a more classical derivative discontinuity can cause the problem; say $f'$ has a pole at $x^{*}$ but $f$ still the restriction to the real line of a meromorphic function.
            – user14717
            Nov 28 at 22:25










          • In fact, I am not sure you can say this converges to zero. If you start in the range of the absolute value you are in one step at zero, but there the scheme fails to work, as you have no derivative there.
            – Severin Schraven
            Nov 28 at 22:37










          • @SeverinSchraven: The question was intentionally ambiguous regarding the function spaces--defining the function space is half the fun.
            – user14717
            Nov 28 at 22:41













          up vote
          1
          down vote










          up vote
          1
          down vote









          Try this function:



          $f(x)=max(1-sqrt{|x|},|x|)$



          For most initial guesses, and for all initial guesses more than one-half in absolute value, you converge to zero. But the function has no real zeroes.






          share|cite|improve this answer














          Try this function:



          $f(x)=max(1-sqrt{|x|},|x|)$



          For most initial guesses, and for all initial guesses more than one-half in absolute value, you converge to zero. But the function has no real zeroes.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 at 0:22

























          answered Nov 28 at 22:00









          Oscar Lanzi

          12k12036




          12k12036












          • That sounds interesting. Is it easy to see that it converges indeed to zero?
            – Severin Schraven
            Nov 28 at 22:10






          • 1




            Truly fabulous example. Both of the answers given are required for full understanding, so I'm accepting @SeverinSchraven's answer as accepted (because it looks like Oscar is well beyond the range of caring about the point system).
            – user14717
            Nov 28 at 22:20










          • I'm now wondering if a more classical derivative discontinuity can cause the problem; say $f'$ has a pole at $x^{*}$ but $f$ still the restriction to the real line of a meromorphic function.
            – user14717
            Nov 28 at 22:25










          • In fact, I am not sure you can say this converges to zero. If you start in the range of the absolute value you are in one step at zero, but there the scheme fails to work, as you have no derivative there.
            – Severin Schraven
            Nov 28 at 22:37










          • @SeverinSchraven: The question was intentionally ambiguous regarding the function spaces--defining the function space is half the fun.
            – user14717
            Nov 28 at 22:41


















          • That sounds interesting. Is it easy to see that it converges indeed to zero?
            – Severin Schraven
            Nov 28 at 22:10






          • 1




            Truly fabulous example. Both of the answers given are required for full understanding, so I'm accepting @SeverinSchraven's answer as accepted (because it looks like Oscar is well beyond the range of caring about the point system).
            – user14717
            Nov 28 at 22:20










          • I'm now wondering if a more classical derivative discontinuity can cause the problem; say $f'$ has a pole at $x^{*}$ but $f$ still the restriction to the real line of a meromorphic function.
            – user14717
            Nov 28 at 22:25










          • In fact, I am not sure you can say this converges to zero. If you start in the range of the absolute value you are in one step at zero, but there the scheme fails to work, as you have no derivative there.
            – Severin Schraven
            Nov 28 at 22:37










          • @SeverinSchraven: The question was intentionally ambiguous regarding the function spaces--defining the function space is half the fun.
            – user14717
            Nov 28 at 22:41
















          That sounds interesting. Is it easy to see that it converges indeed to zero?
          – Severin Schraven
          Nov 28 at 22:10




          That sounds interesting. Is it easy to see that it converges indeed to zero?
          – Severin Schraven
          Nov 28 at 22:10




          1




          1




          Truly fabulous example. Both of the answers given are required for full understanding, so I'm accepting @SeverinSchraven's answer as accepted (because it looks like Oscar is well beyond the range of caring about the point system).
          – user14717
          Nov 28 at 22:20




          Truly fabulous example. Both of the answers given are required for full understanding, so I'm accepting @SeverinSchraven's answer as accepted (because it looks like Oscar is well beyond the range of caring about the point system).
          – user14717
          Nov 28 at 22:20












          I'm now wondering if a more classical derivative discontinuity can cause the problem; say $f'$ has a pole at $x^{*}$ but $f$ still the restriction to the real line of a meromorphic function.
          – user14717
          Nov 28 at 22:25




          I'm now wondering if a more classical derivative discontinuity can cause the problem; say $f'$ has a pole at $x^{*}$ but $f$ still the restriction to the real line of a meromorphic function.
          – user14717
          Nov 28 at 22:25












          In fact, I am not sure you can say this converges to zero. If you start in the range of the absolute value you are in one step at zero, but there the scheme fails to work, as you have no derivative there.
          – Severin Schraven
          Nov 28 at 22:37




          In fact, I am not sure you can say this converges to zero. If you start in the range of the absolute value you are in one step at zero, but there the scheme fails to work, as you have no derivative there.
          – Severin Schraven
          Nov 28 at 22:37












          @SeverinSchraven: The question was intentionally ambiguous regarding the function spaces--defining the function space is half the fun.
          – user14717
          Nov 28 at 22:41




          @SeverinSchraven: The question was intentionally ambiguous regarding the function spaces--defining the function space is half the fun.
          – user14717
          Nov 28 at 22:41


















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