heeellpp
#im realy neeed it pls guys
restive current
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heeellpp
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topaz tiger
What do you need help with?
topaz tiger
Can you translate it?
restive current
sure
\documentclass{article}
\usepackage{amsmath, amssymb}
\begin{document}
\section*{Part B}
Let ( F ) be the numerical function defined on the interval ( \mathbb{R}_+ ) by:
[
F(x) = \int_1^{e^x} \frac{\ln(1+t)}{t} dt.
]
\subsection*{1.}
\begin{itemize}
\item[a.] Show that:
[
\forall x \in \mathbb{R}+, \quad \ln(2)x \leq F(x).
]
\item[b.] Deduce:
[
\lim{x \to +\infty} F(x).
]
\end{itemize}
\subsection*{2.}
\begin{itemize}
\item[a.] Show that ( F ) is differentiable on ( \mathbb{R}+ ), then calculate ( F'(x) ) for all ( x \in \mathbb{R}+ ).
\item[b.] Show that ( F ) is strictly increasing on ( \mathbb{R}_+ ).
\end{itemize}
\subsection*{3.}
\begin{itemize}
\item[a.] Show that, for every ( n \in \mathbb{N}^* ), the equation ( F(\ln(x)) = n ) has a unique solution in the interval ( ]1, +\infty[ ), which we will denote as ( x_n ).
\item[b.] Establish that:
[
\forall t \in \mathbb{R}_+^, \quad 0 < \frac{\ln(1+t)}{t} < 1.
]
\item[c.] Deduce that:
[
\forall n \in \mathbb{N}^, \quad x_n > n.
]
\item[d.] Determine the limit of the sequence ( (x_n) ).
\end{itemize}
\subsection*{4.}
\begin{itemize}
\item[a.] Show that:
[
\forall n \in \mathbb{N}^, \quad \frac{\ln(2) \ln(n)}{n} \leq \frac{F(\ln(n))}{n} \leq \frac{\ln(1+n)}{n}.
]
\item[b.] Deduce that:
[
\lim_{n \to +\infty} \frac{F(\ln(n))}{n} = 0.
]
\item[c.] Show that:
[
\forall n \in \mathbb{N}^, \quad 0 < \frac{n}{x_n} < \frac{\ln(1+n)}{n} + \frac{F(\ln(n))}{n}.
]
\item[d.] Deduce:
[
\lim_{n \to +\infty} \frac{n}{x_n}.
]
\end{itemize}
\end{document}
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badremouusaid
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restive current
4/c
restive current
**badremouusaid**
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@topaz tiger
topaz tiger
Hm, ok.
(1). Note that since the interval of integration starts with 1, we can notice that ln(1 + x) ≤ ln(1 + 1) = ln(2) on that interval. Then you get an easy integral.
(2). ln(1 + t)/t is smooth, as are the limits of integration. Differentiation should be easy - just use the Lebnitz's formula.
(3). F(x) has to be increasing and unbounded from two previous points, so that should help. Not sure about (c) and (d), though.
(4) seems to be harder, can't say anything about it yet.
restive current
Hm, ok.
(1). Note that since the interval of integration starts with 1, we can n...
btw you can use 4/a and sustitue n by xn
cuz F(ln(xn))=n
restive current
\documentclass{article}
\usepackage{amsmath, amssymb}
\begin{document}
Résolution par Badre Moussaid :
[
\frac{b^2}{(z - b)(z' - b)}
]
Or, par calcul :
[
b^2 = 2a
]
et
[
2b = a \bar{b}
]
Donc :
[
\frac{b^2}{(z - b)(z' - b)} = \frac{2a}{(z - b)(a \bar{z} - 2b)}
]
[
= \frac{2a}{(z - b)(a \bar{z} - a \bar{b})}
]
[
= \frac{2}{|z - b|} \quad \text{appartient à } \mathbb{R}
]
Ainsi, (\frac{b^2}{(z - b)(z' - b)}) est réel.
De plus :
[
\frac{b - 0}{(z - b)} \times \frac{b - 0}{(z' - b)} \text{ est réel}
]
Donc :
[
\arg\left(\frac{b - 0}{z - b}\right) = -\arg\left(\frac{b - 0}{z' - b}\right) \mod 2\pi
]
Ainsi :
[
\arg(OB, MB) = -\arg(OB, M'B) \mod 2\pi
]
[
\arg(OB, MB) = \arg(M'B, OB) \mod 2\pi
]
Donc, la bissectrice de ((MB, M'B)) est la droite (\delta).
Remarque : (O) et (B) appartiennent à (\delta).
\end{document}
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badremouusaid
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tardy dove
btw you can use 4/a and sustitue n by xn
cuz F(ln(xn))=n
So denote G(x) = F(ln(x))
$G(x_n) = G(n) + \int_{n}^{x_n} \frac{\ln(1+t)}{t}dt$
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Miguel
tardy dove
\begin{align}
\frac{G(x_n)}{x_n} = \frac{G(n)}{x_n} + \frac{1}{x_n} \int_{n}^{x_n} \frac{\ln(1+t)}{t}dt \\
\leq \frac{G(n)}{n} + \frac{1}{x_n} \int_{n}^{x_n} \frac{\ln(1+t)}{t}dt
\end{align}
Miguel
tardy dove
You could try upper-bounding the integral first
By considering that ln(1+t)/t is decreasing on R+
,w plot ln(1+x)/x
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tardy dove
(but honestly you don't need all that to conclude, because the result from question 4a applies to any x instead of any natural integer n, so you can plug x_n too and let n tend to infinity)
tardy dove
(but honestly you don't need all that to conclude, because the result from quest...
I am assuming you did 4a already, so to clarify what I mean:
$\forall x \geq 1, G(x) = \int_{1}^{x} \frac{\ln(1+t)}{t}dt \leq \int_{1}^{x} \frac{\ln(1+x)}{t}dt = \ln(1+x)\ln(x)$
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Miguel
tardy dove
So $\forall x \geq 1, \frac{G(x)}{x} \leq \frac{\ln(1+x)\ln(x)}{x}$, hence $\lim_{x \to \infty} \frac{G(x)}{x} = 0$ (this is what you should have found for the question 4b, but replacing $n$ by $x$)
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Miguel
tardy dove
Anyway good luck with this
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