Wissety52

2022-05-02

Understanding an Approximation:

$1-{(1-\frac{\mu}{n})}^{\mu -1+(\mu -1{)}^{2}}\approx \frac{{\mu}^{3}}{n}.$

$1-{(1-\frac{\mu}{n})}^{\mu -1+(\mu -1{)}^{2}}\approx \frac{{\mu}^{3}}{n}.$

Tatairfzk

Beginner2022-05-03Added 12 answers

Note that $\mu -1+(\mu -1{)}^{2}=\mu (\mu -1).$. Using the approximation $(1+x{)}^{\alpha}\approx 1+\alpha x$ for small $x$, we obtain

${(1-\frac{\mu}{n})}^{\mu -1+(\mu -1{)}^{2}}\approx 1-\frac{{\mu}^{2}(\mu -1)}{n}.$

Hence

$1-{(1-\frac{\mu}{n})}^{\mu -1+(\mu -1{)}^{2}}\approx \frac{{\mu}^{2}(\mu -1)}{n}=\frac{{\mu}^{3}-{\mu}^{2}}{n}.$

If we can assume that $\mu $ is sufficiently large (and indeed we assume that $p>3\frac{\mathrm{log}n}{n}$ so that $\mu =np>3\mathrm{log}n$), then the ${\mu}^{2}$ term pales in comparison to ${\mu}^{3}$, so the approximation

$\frac{{\mu}^{3}-{\mu}^{2}}{n}\approx \frac{{\mu}^{3}}{n}$

can also be made. Another way to view this last approximation is to say that $\mu -1\approx \mu $ for large $\mu $ so that ${\mu}^{2}(\mu -1)\approx {\mu}^{2}(\mu )={\mu}^{3}.$.

${(1-\frac{\mu}{n})}^{\mu -1+(\mu -1{)}^{2}}\approx 1-\frac{{\mu}^{2}(\mu -1)}{n}.$

Hence

$1-{(1-\frac{\mu}{n})}^{\mu -1+(\mu -1{)}^{2}}\approx \frac{{\mu}^{2}(\mu -1)}{n}=\frac{{\mu}^{3}-{\mu}^{2}}{n}.$

If we can assume that $\mu $ is sufficiently large (and indeed we assume that $p>3\frac{\mathrm{log}n}{n}$ so that $\mu =np>3\mathrm{log}n$), then the ${\mu}^{2}$ term pales in comparison to ${\mu}^{3}$, so the approximation

$\frac{{\mu}^{3}-{\mu}^{2}}{n}\approx \frac{{\mu}^{3}}{n}$

can also be made. Another way to view this last approximation is to say that $\mu -1\approx \mu $ for large $\mu $ so that ${\mu}^{2}(\mu -1)\approx {\mu}^{2}(\mu )={\mu}^{3}.$.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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