Estimate order O(h)

[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 $t \approx 1$.

 

[CellCoree]

As a scientist, it is important to be able to analyze and estimate the order of a given equation in order to understand its behavior and make accurate predictions. In this case, we are given a non-dimensional differential equation for the height at the highest point, which is given by h(μ) = 1/μ - 1/μ^2 log_e(1+μ), where 0<μ<<1. Our goal is to estimate the order of this equation in terms of μ and compare it with the given equation t_h(μ) = 1-μ/2+...

To begin, let us first recall the definition of order in terms of variables. The order of a function or equation is the highest power of the independent variable that appears in the equation. In this case, we have μ appearing in both the numerator and denominator of h(μ), which suggests that the order of this equation is O(μ). This means that the behavior of h(μ) is mainly dependent on the value of μ.

Next, we can deduce an estimate of h(μ) in terms of μ by using the Taylor series expansion of log_e(1+μ) around μ=0. This gives us log_e(1+μ) = μ - μ^2/2 + O(μ^3), where O(μ^3) represents higher order terms that are negligible for small values of μ. Substituting this into the equation for h(μ), we get h(μ) = 1/μ - 1/μ^2 (μ - μ^2/2 + O(μ^3)). Simplifying this further, we get h(μ) = 1/μ - 1 + μ/2 + O(μ^2).

Comparing this with the given equation t_h(μ) = 1-μ/2+..., we can see that they are similar up to the term μ/2. This suggests that the leading behavior of h(μ) is of the same order as t_h(μ), which is O(μ). However, the higher order terms in h(μ) are different from t_h(μ), which means that they may have different behaviors for larger values of μ.

In conclusion, we can estimate the order of h(μ) to be O(μ) and compare it with t_h(μ). While they have similar leading behaviors, the higher