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- Apr 14, 2013

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**Question 1 :**

Let $g(x)-=x-x^3$. The point $x=0$ is a fixed point for $g$. Show that if $x^{\star}$ is a fixed point of $g$, $g(x^{\star})=x^{\star}$, then $x^{\star}=0$. If $(x_k)$ the sequence $x_{k+1}=g(x_k)$, $k=0,1,2,\ldots$ show that if $0>x_0>-1$ then $(x_k)$ is increasing and converges to $0$.

__My solution :__$g(x)=x-x^3$

$g(0)=0$

$g(x^{\star})=x^{\star} \Rightarrow x^{\star} -{x^{\star}}^3=x^{\star} \Rightarrow {x^{\star}}^3=0 \Rightarrow x^{\star} =0$

$x_{k+1}=g(x_k)$

$0>x_0>-1$

We have that $g'(x)=1-3x^2$ and $g'(x)=0 \Rightarrow 3x^2=1 \Rightarrow x^2=\frac{1}{3} \Rightarrow x=\pm\frac{1}{\sqrt{3}}$.

Then $g'(x)>0$ for $-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$ and $g'(x)\leq 0$ for $x\leq -\frac{1}{\sqrt{3}}$and $x\geq \frac{1}{\sqrt{3}}$.

So $g$ is increasing at $\left (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right )$ so the sequence $(x_k)$ is increasing in that interval.

Then $x_0<0 \Rightarrow x_{k+1}g(x_k)<g(x^{\star})=x^{\star}$. Increasing and upper bounded sequence, so it converges to $x^{\star}=0$.

**Question 2 :**

We consider the linear system \begin{align*}&3x+y=1 \\ &x+2y=3\end{align*} If we know that if we apply Jacobi Method tosolve the above linear system, it would converge, then check if Jacobi mathod converges if we apply it for the linear system \begin{align*}&x+2y=3 \\ &3x+y=1\end{align*}

__My solution :__At the initial system we have the symmetric matrix $A=\begin{pmatrix}3 & 1 \\ 1 & 2\end{pmatrix}$. At the second system we have the matrix $\tilde{A}=\begin{pmatrix}1 & 2 \\ 3 & 1\end{pmatrix}$. This matrix is not symmetric neither it is diagonally dominant. So we don't have convergence.

**Question 3 :**

Let $x,y,\epsilon_1, \epsilon_2\in \mathbb{R}$ such that \begin{align*}&3x+y=7+\epsilon_1 \\ &4x+2y=10+\epsilon_2\end{align*} Show that $|x-2|+|y-1|\leq 3(|\epsilon_1|+|\epsilon_2|)$.

*My solution :*From the first equation we get $y=7+\epsilon_1-3x\ \ \ \ \ (\star)$.

Substituting this in the second equation we get \begin{align*}4x+2(7+\epsilon_1-3x)=10+\epsilon_2 &\Rightarrow 4x+14+2\epsilon_1-6x=10+\epsilon_2 \\ & \Rightarrow -2x=-4+\epsilon_2-2\epsilon_1 \\ & \Rightarrow x=2-\frac{\epsilon_2}{2}+\epsilon_1\end{align*}

Substituting this $(\star)$ we get \begin{equation*}y=7+\epsilon_1-6+\frac{3}{2}\epsilon_2-3\epsilon_1=1-2\epsilon_1+\frac{3}{2}\epsilon_2\end{equation*}

Then \begin{align*}|x-2|+|y-1|&=\left |2-\frac{\epsilon_2}{2}+\epsilon_1-1\right |+\left |1-2\epsilon_1+\frac{3}{2}\epsilon_2-1\right |\\ & = \left |\epsilon_1-\frac{\epsilon_2}{2}\right |+\left |-2\epsilon_1+\frac{3}{2}\epsilon_2\right |\\ & \leq |\epsilon_1|+\frac{|\epsilon_1|}{2}+2|\epsilon_1|+\frac{3}{2}|\epsilon_2| \\ & = 3|\epsilon_1|+2|\epsilon_2| \\ & \leq 3|\epsilon_1|+3|\epsilon_2| = 3\left (|\epsilon_1|+|\epsilon_2| \right )\end{align*}

**Question 4 :**

Let $f(x)=x-x^3$. Let $(x_k)$ be the sequence that we get if we consider Newton's method to approximate a root. If $x_0=-\frac{1}{\sqrt{5}}$, then does the sequence converge? Is yes, find the limit.

*My solution :*We have that $f'(x)=1-3x^2$.

Then \begin{equation*}x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \Rightarrow x_{n+1}=x_n-\frac{x_n-x_n^3}{1-3x_n^2}=-\frac{2x_n^3}{1-3x_n^2}\end{equation*}

Let $g(x)=-\frac{2x_n^3}{1-3x_n^2}$.

$x_1=g(x_0)=g\left (-\frac{1}{\sqrt{5}}\right )$.

\begin{equation*}\left |g'(x)\right |=\left |-\frac{6x^2}{(1-3x^2)^2}\right |=\frac{6x^2}{(1-3x^2)^2}\end{equation*}

We have that \begin{equation*}|g'(x)|<1 \Rightarrow 6x^2<1-6x^2+9x^4 \Rightarrow 9x^4-12x^2+1<0\end{equation*}

Are my solutions correct ?