- #1

mathmari

Gold Member

MHB

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Hey!

We have the function $f(x)=x^5-\frac{5}{16}$.

I have approximated the root of that function using three steps of Newton's method with initla value $x_0=\frac{1}{2}$ :

\begin{align*}x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\approx \frac{7}{5} \\ x_2&=x_1-\frac{f(x_1)}{f'(x_1)} \approx 1.1362692628 \\ x_3&=x_2-\frac{f(x_2)}{f'(x_2)}\approx 0.9465088238\end{align*}

So, $x_3=0.9465$ is an approximation of the root of $f(x)=x^5-\frac{5}{16}$. Now, I want to write the Newton's method for the above function as a fixed point iteration $x_{k+1}=g(x_k)$ on $\Omega=\left [\frac{3}{5}, \infty\right )$ and to show that $g$ is a contraction mapping, i.e. to show that it holds that $g( \Omega )\subset\Omega$ and $|g(x)-g(y)|\leq L|x-y|$ for $x,y\in \Omega$. The Newton's method is $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^5-\frac{5}{16}}{5\cdot x_n^4}$$ Does this mean that $g(x)=x-\frac{x^5-\frac{5}{16}}{5\cdot x^4}$ ?

To show that $g( \Omega )\subset\Omega$ do we have to check if the function is monotone?

About the second property, we have the following:

\begin{align*}|g(x)-g(y)|&=\left |\left (x-\frac{x^5-\frac{5}{16}}{5\cdot x^4}\right )-\left (y-\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right )\right |\\ & = \left |x-\frac{x^5-\frac{5}{16}}{5\cdot x^4}-y+\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right | \\ & = \left |\left (x-y\right )-\left (\frac{x^5-\frac{5}{16}}{5\cdot x^4}-\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right )\right |\\ & \leq \left |x-y\right |+\left |\frac{x^5-\frac{5}{16}}{5\cdot x^4}-\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right | \end{align*}

Is the last step correct? How can we bound the second term? (Wondering)

We have the function $f(x)=x^5-\frac{5}{16}$.

I have approximated the root of that function using three steps of Newton's method with initla value $x_0=\frac{1}{2}$ :

\begin{align*}x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\approx \frac{7}{5} \\ x_2&=x_1-\frac{f(x_1)}{f'(x_1)} \approx 1.1362692628 \\ x_3&=x_2-\frac{f(x_2)}{f'(x_2)}\approx 0.9465088238\end{align*}

So, $x_3=0.9465$ is an approximation of the root of $f(x)=x^5-\frac{5}{16}$. Now, I want to write the Newton's method for the above function as a fixed point iteration $x_{k+1}=g(x_k)$ on $\Omega=\left [\frac{3}{5}, \infty\right )$ and to show that $g$ is a contraction mapping, i.e. to show that it holds that $g( \Omega )\subset\Omega$ and $|g(x)-g(y)|\leq L|x-y|$ for $x,y\in \Omega$. The Newton's method is $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^5-\frac{5}{16}}{5\cdot x_n^4}$$ Does this mean that $g(x)=x-\frac{x^5-\frac{5}{16}}{5\cdot x^4}$ ?

To show that $g( \Omega )\subset\Omega$ do we have to check if the function is monotone?

About the second property, we have the following:

\begin{align*}|g(x)-g(y)|&=\left |\left (x-\frac{x^5-\frac{5}{16}}{5\cdot x^4}\right )-\left (y-\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right )\right |\\ & = \left |x-\frac{x^5-\frac{5}{16}}{5\cdot x^4}-y+\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right | \\ & = \left |\left (x-y\right )-\left (\frac{x^5-\frac{5}{16}}{5\cdot x^4}-\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right )\right |\\ & \leq \left |x-y\right |+\left |\frac{x^5-\frac{5}{16}}{5\cdot x^4}-\frac{y^5-\frac{5}{16}}{5\cdot y^4}\right | \end{align*}

Is the last step correct? How can we bound the second term? (Wondering)

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