Illustrate the construction of the secant method by means of a figure. When $f ^ { \prime } > 0$ on $I$, express $x _ { n + 1 }$ as a function of $x _ { n - 1 } , x _ { n }$ by means of the function $H _ { f }$ defined in question 3.3 of the third part.
Recall: The secant method is defined as follows. Given $x_0, x_1 \in I$, for $n \geq 1$ we consider the line $L_n$ passing through the points $(x_{n-1}, f(x_{n-1}))$ and $(x_n, f(x_n))$ (with $L_n$ being the tangent line at $(x_n, f(x_n))$ when $x_n = x_{n-1}$). If $L_n$ intersects $\{(x,0) \mid x \in I\}$ at a unique point $(x,0)$, we define $x_{n+1} = x$.

In this question, we examine the special case of a polynomial function of degree two $f$ defined by the formula $f ( x ) = ( x - \alpha ) ( x - \beta )$ where $\alpha$ and $\beta$ are real and $\alpha > \beta$. We take $I = ] ( \alpha + \beta ) / 2 , + \infty [$. For $x \in \mathbb { R }$ we define $h ( x ) = \frac { x - \alpha } { x - \beta }$, with the convention $h ( \beta ) = \infty$.
(a) For $x \in \mathbb { R }$ show that we have $| h ( x ) | < 1$ if and only if $x \in I$.
(b) Explicitly state the recurrence relation satisfied by the sequence $u _ { n } : = h \left( x _ { n } \right)$ and deduce that the sequence $\left( x _ { n } \right) _ { n \geqslant 0 }$ is well defined for any $x _ { 0 }$ and $x _ { 1 }$ in $I$.
(c) Show that the sequence $\left( u _ { n } \right) _ { n \geqslant 0 }$ tends to 0 and deduce that $\left( x _ { n } \right) _ { n \geqslant 0 }$ tends to $\alpha$.
(d) Let $\phi = \frac { 1 + \sqrt { 5 } } { 2 }$. Show that there exists a strictly negative real number $s$ such that $$x _ { n } - \alpha = \mathrm { O } \left( e ^ { s \phi ^ { n } } \right)$$

We return to the general case, $f$ being any function of class $\mathcal { C } ^ { 3 }$. We assume that $f$ vanishes at a point $x ^ { * } \in I$, for which $f ^ { \prime } \left( x ^ { * } \right) > 0$.
(a) Show that there exists $\epsilon > 0$ such that $\left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right] \subset I$ and $f ^ { \prime } > 0$ on the interval $\left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right]$. We fix such an $\epsilon$ for the rest and we define $$M = \sup _ { ( x , y ) \in \left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right] ^ { 2 } } \left| \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( x , y ) \right|$$ (b) We assume that $x _ { n - 1 } , x _ { n } \in \left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right]$. Show that $$\left| x _ { n + 1 } - x ^ { * } \right| \leqslant M \left| x _ { n - 1 } - x ^ { * } \right| \cdot \left| x _ { n } - x ^ { * } \right| .$$ (c) We fix $\left. \epsilon ^ { \prime } \in \right] 0 , \epsilon$ ] such that $M \epsilon ^ { \prime } < 1$. Show that if $x _ { 0 } , x _ { 1 }$ belong to $\left[ x ^ { * } - \epsilon ^ { \prime } , x ^ { * } + \epsilon ^ { \prime } \right]$ then the sequence $\left( x _ { n } \right) _ { n \geqslant 0 }$ is well defined and converges to $x ^ { * }$.

Illustrate the construction of the secant method by means of a figure. When $f ^ { \prime } > 0$ on $I$, express $x _ { n + 1 }$ as a function of $x _ { n - 1 } , x _ { n }$ by means of the function $H _ { f }$ defined in question 3 of the third part.
(Recall: the secant method initializes with $x_0, x_1 \in I$, and at each step considers the line $L_n$ passing through $(x_{n-1}, f(x_{n-1}))$ and $(x_n, f(x_n))$, defining $x_{n+1}$ as the $x$-intercept of $L_n$.)

In this question, we examine the special case of a polynomial function of degree two $f$ defined by the formula $f ( x ) = ( x - \alpha ) ( x - \beta )$ where $\alpha$ and $\beta$ are real and $\alpha > \beta$. We take $I = ] ( \alpha + \beta ) / 2 , + \infty [$.
For $x \in \mathbb { R }$ we define $h ( x ) = \frac { x - \alpha } { x - \beta }$, with the convention $h ( \beta ) = \infty$.
(a) For $x \in \mathbb { R }$ show that we have $| h ( x ) | < 1$ if and only if $x \in I$.
(b) Explicitly state the recurrence relation satisfied by the sequence $u _ { n } : = h \left( x _ { n } \right)$ and deduce that the sequence $\left( x _ { n } \right) _ { n \geqslant 0 }$ is well defined for any $x _ { 0 }$ and $x _ { 1 }$ in $I$.
(c) Show that the sequence $\left( u _ { n } \right) _ { n \geqslant 0 }$ tends to 0 and deduce that $\left( x _ { n } \right) _ { n \geqslant 0 }$ tends to $\alpha$.
(d) Let $\phi = \frac { 1 + \sqrt { 5 } } { 2 }$. Show that there exists a strictly negative real number $s$ such that $$x _ { n } - \alpha = O \left( e ^ { s \phi ^ { n } } \right) .$$

We return to the general case, $f$ being any function of class $\mathcal { C } ^ { 3 }$. We assume that $f$ vanishes at a point $x ^ { * } \in I$, for which $f ^ { \prime } \left( x ^ { * } \right) > 0$.
(a) Show that there exists $\epsilon > 0$ such that $\left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right] \subset I$ and $f ^ { \prime } > 0$ on the interval $\left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right]$. We fix such an $\epsilon$ for the rest and we define $$M = \sup _ { ( x , y ) \in \left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right] ^ { 2 } } \left| \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( x , y ) \right| .$$
(b) We assume that $x _ { n - 1 } , x _ { n } \in \left[ x ^ { * } - \epsilon , x ^ { * } + \epsilon \right]$. Show that $$\left| x _ { n + 1 } - x ^ { * } \right| \leqslant M \left| x _ { n - 1 } - x ^ { * } \right| \cdot \left| x _ { n } - x ^ { * } \right| .$$
(c) We fix $\left. \epsilon ^ { \prime } \in \right] 0 , \epsilon ]$ such that $M \epsilon ^ { \prime } < 1$. Show that if $x _ { 0 } , x _ { 1 }$ belong to $\left[ x ^ { * } - \epsilon ^ { \prime } , x ^ { * } + \epsilon ^ { \prime } \right]$ then the sequence $\left( x _ { n } \right) _ { n \geqslant 0 }$ is well defined and converges to $x ^ { * }$.