bac-s-maths 2025 Q2

bac-s-maths · France · bac-spe-maths__amerique-nord_j2 5 marks Sequences and series, recurrence and convergence Convergence proof and limit determination
One of the objectives of this exercise is to determine an approximation of the real number $\ln ( 2 )$, using one of the methods of the English mathematician Henry Briggs in the XVI${}^{\text{th}}$ century.
We denote by ( $u _ { n }$ ) the sequence defined by:
$$u _ { 0 } = 2 \quad \text { and, for every natural number } n , \quad u _ { n + 1 } = \sqrt { u _ { n } }$$
Part A
  1. a. Give the exact value of $u _ { 1 }$ and $u _ { 2 }$. b. Make a conjecture, using a calculator, about the direction of variation and the possible limit of the sequence.
  2. a. Show by induction that for every natural number $n , \quad 1 \leqslant u _ { n + 1 } \leqslant u _ { n }$. b. Deduce that the sequence ( $u _ { n }$ ) is convergent. c. Solve in the interval [ $0 ; + \infty$ [ the equation $\sqrt { x } = x$. d. Determine, by justifying, the limit of the sequence $\left( u _ { n } \right)$.

Part B
We denote by ( $\nu _ { n }$ ) the sequence defined for every natural number $n$ by $\nu _ { n } = \ln \left( u _ { n } \right)$.
  1. a. Prove that the sequence ( $v _ { n }$ ) is geometric with common ratio $\frac { 1 } { 2 }$. b. Express $v _ { n }$ as a function of $n$, for every natural number $n$. c. Deduce that, for every natural number $n , \quad \ln ( 2 ) = 2 ^ { n } \ln \left( u _ { n } \right)$.
  2. We have traced below in an orthonormal coordinate system the curve $\mathscr { C }$ of the function ln and the tangent T to the curve $\mathscr { C }$ at the point with abscissa 1. An equation of the line T is $y = x - 1$. The points $\mathrm { A } _ { 0 } , \mathrm {~A} _ { 1 } , \mathrm {~A} _ { 2 }$ have abscissas $u _ { 0 } , u _ { 1 }$ and $u _ { 2 }$ respectively and ordinate 0. We decide to take $x - 1$ as an approximation of $\ln ( x )$ when $x$ belongs to the interval $] 0,99 ; 1,01 [$. a. Using a calculator, determine the smallest natural number $k$ such that $u _ { k }$ belongs to the interval $] 0,99 ; 1,01 [$ and give an approximate value of $u _ { k }$ to $10 ^ { - 5 }$ near. b. Deduce an approximation of $\ln \left( u _ { k } \right)$. c. Deduce from questions 1.c. and 2.b. of Part B an approximation of $\ln ( 2 )$.
  3. We generalize the previous method to any real number $a$ strictly greater than 1.
    Copy and complete the algorithm below so that the call Briggs(a) returns an approximation of $\ln ( a )$.
    We recall that the instruction in Python language sqrt (a) corresponds to $\sqrt { a }$.
    \begin{verbatim} from math import* def Briggs(a): n = 0 while a >= 1.01: a = sqrt(a) n = n+1 L =... return L \end{verbatim}
One of the objectives of this exercise is to determine an approximation of the real number $\ln ( 2 )$, using one of the methods of the English mathematician Henry Briggs in the XVI${}^{\text{th}}$ century.

We denote by ( $u _ { n }$ ) the sequence defined by:

$$u _ { 0 } = 2 \quad \text { and, for every natural number } n , \quad u _ { n + 1 } = \sqrt { u _ { n } }$$

\section*{Part A}
\begin{enumerate}
  \item a. Give the exact value of $u _ { 1 }$ and $u _ { 2 }$.\\
b. Make a conjecture, using a calculator, about the direction of variation and the possible limit of the sequence.
  \item a. Show by induction that for every natural number $n , \quad 1 \leqslant u _ { n + 1 } \leqslant u _ { n }$.\\
b. Deduce that the sequence ( $u _ { n }$ ) is convergent.\\
c. Solve in the interval [ $0 ; + \infty$ [ the equation $\sqrt { x } = x$.\\
d. Determine, by justifying, the limit of the sequence $\left( u _ { n } \right)$.
\end{enumerate}

\section*{Part B}
We denote by ( $\nu _ { n }$ ) the sequence defined for every natural number $n$ by $\nu _ { n } = \ln \left( u _ { n } \right)$.

\begin{enumerate}
  \item a. Prove that the sequence ( $v _ { n }$ ) is geometric with common ratio $\frac { 1 } { 2 }$.\\
b. Express $v _ { n }$ as a function of $n$, for every natural number $n$.\\
c. Deduce that, for every natural number $n , \quad \ln ( 2 ) = 2 ^ { n } \ln \left( u _ { n } \right)$.
  \item We have traced below in an orthonormal coordinate system the curve $\mathscr { C }$ of the function ln and the tangent T to the curve $\mathscr { C }$ at the point with abscissa 1.\\
An equation of the line T is $y = x - 1$.\\
The points $\mathrm { A } _ { 0 } , \mathrm {~A} _ { 1 } , \mathrm {~A} _ { 2 }$ have abscissas $u _ { 0 } , u _ { 1 }$ and $u _ { 2 }$ respectively and ordinate 0.\\
We decide to take $x - 1$ as an approximation of $\ln ( x )$ when $x$ belongs to the interval $] 0,99 ; 1,01 [$.\\
a. Using a calculator, determine the smallest natural number $k$ such that $u _ { k }$ belongs to the interval $] 0,99 ; 1,01 [$ and give an approximate value of $u _ { k }$ to $10 ^ { - 5 }$ near.\\
b. Deduce an approximation of $\ln \left( u _ { k } \right)$.\\
c. Deduce from questions 1.c. and 2.b. of Part B an approximation of $\ln ( 2 )$.
  \item We generalize the previous method to any real number $a$ strictly greater than 1.

Copy and complete the algorithm below so that the call Briggs(a) returns an approximation of $\ln ( a )$.

We recall that the instruction in Python language sqrt (a) corresponds to $\sqrt { a }$.

\begin{verbatim}
from math import*
def Briggs(a):
    n = 0
    while a >= 1.01:
        a = sqrt(a)
        n = n+1
    L =...
return L
\end{verbatim}
\end{enumerate}