In Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
asked Nov 27 '18 at 6:46
Jacob S.
19010
add a comment |
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
asked Nov 27 '18 at 6:46
Jacob S.
19010
add a comment |
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
asked Nov 27 '18 at 6:46
Jacob S.
19010
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
numerical-methods
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
asked Nov 27 '18 at 6:46
Jacob S.
19010
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
asked Nov 27 '18 at 6:46
Jacob S.
19010
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
edited Nov 27 '18 at 9:37
LutzL
56.1k42054
56.1k42054
asked Nov 27 '18 at 6:46
Jacob S.
19010
asked Nov 27 '18 at 6:46
Jacob S.
19010
asked Nov 27 '18 at 6:46
Jacob S.
19010
19010
add a comment |
add a comment |
1 Answer
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Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 '18 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 '18 at 14:39
add a comment |
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1 Answer
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Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 '18 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 '18 at 14:39
add a comment |
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 '18 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 '18 at 14:39
add a comment |
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
edited Nov 27 '18 at 9:39
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
answered Nov 27 '18 at 9:15
LutzL
56.1k42054
56.1k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 '18 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 '18 at 14:39
add a comment |
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 '18 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 '18 at 14:39
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 '18 at 14:33
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 '18 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 '18 at 14:39
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 '18 at 14:39
add a comment |
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