Prove that for large $n$ the following equation:
$$\ln n ! \approx n \ln n - n $$
holds.
Solution
We note that
$$\begin{aligned}
\ln n! &= \ln \left ( 1\cdot 2 \cdot 3 \cdots n \right ) \\
&= \ln \left [ \prod_{i=1}^{n}i \right ]\\
&= \sum_{i=1}^{n}\ln i\\
&\approx \int_{1}^{n}\ln x \, {\rm d}x \\
&=\left [ x \ln x -x \right ]_1^n \\
&= n \ln n -n +1 \\
&\approx n \ln n -n
\end{aligned}$$
which is what we wanted.
The above result is a powerful one when needed to estimate series or integrals involving logs.
$$\ln n ! \approx n \ln n - n $$
holds.
Solution
We note that
$$\begin{aligned}
\ln n! &= \ln \left ( 1\cdot 2 \cdot 3 \cdots n \right ) \\
&= \ln \left [ \prod_{i=1}^{n}i \right ]\\
&= \sum_{i=1}^{n}\ln i\\
&\approx \int_{1}^{n}\ln x \, {\rm d}x \\
&=\left [ x \ln x -x \right ]_1^n \\
&= n \ln n -n +1 \\
&\approx n \ln n -n
\end{aligned}$$
which is what we wanted.
The above result is a powerful one when needed to estimate series or integrals involving logs.
No comments:
Post a Comment