g(u) ? -1 - \frac{1}{2}(u - 1)^2 Substitute into the Integral: Substituting this expansion into the integral for n!: n! ? n^{n+1}e^{-n}\int_{0}^{?} e^{-\frac{n}{2}(u-1)^2} du Approximate the Integral as a Gaussian: The integral is now of Gaussian form. Extending the limits to infinity (since the integrand is sharply peaked near u_0 : $\int_{-?}^{?} e^{-\frac{n}{2}(u-1)^2} du = \sqrt{\frac{2\pi}{n}}$
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Step 1: The integral is of the form $\int_{-\infty}^\infty e^{-\frac{1}{2}(x-a)^2} dx = \sqrt{2\pi}$. Show more…
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The Stirling approximation is useful in a variety of different settings. The goal of the present problem is to work through a more sophisticated treatment of this approximation than the simple heuristic argument given in the chapter. Our task is to find useful representations of $n !$ since terms of the form In $n !$ arise often in reasoning about entropy. (a) Begin by showing that $$n !=\int_{0}^{\infty} x^{n} e^{-x} d x$$ To demonstrate this, use repeated integration by parts. In particular, demonstrate the recurrence relation $$\int_{0}^{\infty} x^{n} \mathrm{e}^{-x} \mathrm{d} x=n \int_{0}^{\infty} x^{n-1} \mathrm{e}^{-x} \mathrm{d} x$$ and then argue that repeated application of this relation leads to the desired result. (b) Make plots of the integrand $x^{n} \mathrm{e}^{-x}$ for various values of $n$ and observe the peak width and height of this integrand. We are interested now in finding the value of $x$ for which this function is a maximum. The idea is that we will then expand about that maximum. To carry out this step, consider $\ln \left(x^{n} \mathrm{e}^{-x}\right)$ and find its maximum-argue why it is acceptable to use the logarithm of the original function as a surrogate for the function itself, that is, show that the maxima of both the function and its logarithm are at the same $x$, Also, argue why it might be a good idea to use the logarithm of the integrand rather than the integrand itself as the basis of our analysis. Call the value of $x$ for which this function is maximized $x_{0}$. Now expand the logarithm about $x_{0}$. In particular, examine \[ \ln \left[\left(x_{0}+\delta\right)^{n} \mathrm{e}^{-\left(x_{0}+\delta\right)}\right]=n \ln \left(x_{0}+\delta\right)-\left(x_{0}+\delta\right) \] and expand to second order in $8 .$ Exponentiate your result and you should now have an approximation to the original integrand that is good in the neighborhood of $x_{0}$. Plug this back into the integral (be careful with limits of integration) and, by showing that it is acceptable to send the lower limit of integration to $-\infty,$ show that \[ n ! \approx n^{n} \mathrm{e}^{-n} \int_{-\infty}^{\infty} \mathrm{e}^{-\delta^{2} / 2 n} \mathrm{d} \delta \] Evaluate the integral and show that in this approximation \[ n !=n^{n} e^{-n}(2 \pi n)^{1 / 2} \] Also, take the logarithm of this result and make an argument as to why most of the time we can get away with dropping the $(2 \pi n)^{1 / 2}$ term.
Factorial growth rate The factorial function is defined for positive integers as $n !=n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1 .$ For example $5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120 .$ A valuable result that gives good approximations to $n !$ for large values of $n$ is Stirling's formula, $n ! \approx \sqrt{2 \pi n} n^{n} e^{-n} .$ Use this formula and a calculator to determine where the factorial function appears in the ranking of growth rates given in Theorem 4.15.
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In calculus, it can be shown that $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !} \approx 1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots+\frac{x^{n}}{n !}$$ where the larger $n$ is, the better the approximation. Refer to this series. Note that $n !$, read "n factorial," is defined by $0 !=1$ and $n !=1 \cdot 2 \cdot 3 \cdots \cdots n$ for $n \in N$. Approximate $e^{-0.5}$ using the first five terms of the series. Compare this approximation with your calculator evaluation of $e^{-0.5}$.
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