MHB How Does the Order of μ Affect the Height Function h(μ)?

[ra_forever8]

Consider non-dimensional differential equation for the height at the highest point is given by
\begin{equation} h(\mu)= \frac{1}{\mu}- \frac{1}{\mu^2} log_e(1+\mu) \end{equation}
$0<\mu<<1.$
Deduce an estimate to $O(\mu)$ for $h(\mu)$ and compare with $t_h(\mu)=1-\frac{\mu}{2}+...$
=> I really don't how to start this question. please help me.

 

[I like Serena]

Consider non-dimensional differential equation for the height at the highest point is given by
\begin{equation} h(\mu)= \frac{1}{\mu}- \frac{1}{\mu^2} log_e(1+\mu) \end{equation}
$0<\mu<<1.$
Deduce an estimate to $O(\mu)$ for $h(\mu)$ and compare with $t_h(\mu)=1-\frac{\mu}{2}+...$
=> I really don't how to start this question. please help me.

Start with a Taylor expansion...

 

[ra_forever8]

Start with a Taylor expansion...

$\log_e(1+\mu) = \mu - \dfrac{\mu^2}{2} + \dfrac{\mu^3}{3} - \dfrac{\mu^4}{4}+\cdots$ and plug that in,
to get $h(\mu) =\dfrac12- \dfrac\mu3+\dfrac{\mu^2}{4}+\cdots.$
now, comparing $\log_e(1+\mu)$ with $t_h(\mu)=1-\frac{\mu}{2}+\cdots$
what can I say?

 

[I like Serena]

$\log_e(1+\mu) = \mu - \dfrac{\mu^2}{2} + \dfrac{\mu^3}{3} - \dfrac{\mu^4}{4}+\cdots$ and plug that in,
to get $h(\mu) =\dfrac12- \dfrac\mu3+\dfrac{\mu^2}{4}+\cdots.$

The problem asked for $h(\mu)$ up to $O(\mu)$, so that would be:
$$h(\mu) = \frac 1 2 + O(\mu)$$

now, comparing $\log_e(1+\mu)$ with $t_h(\mu)=1-\frac{\mu}{2}+\cdots$
what can I say?

The time is $t_h(\mu) = 1 + O(\mu)$.
So with small enough $\mu$ the maximum height is approximately $h \approx \dfrac 1 2$ which is reached at a time of approximately $t \approx 1$.