202110110958 Sidenote (Section 3.8.2)

Given that the binomial coefficient is so defined as

\begin{pmatrix} N \\ n \end{pmatrix} \stackrel{\textrm{def}}{=} \displaystyle{\frac{N!}{n!(N-n)!}}.

Prove the identities (3.49), (3.50), and (3.51) in Section 3.8.2 Useful Identities for the Binomial Coefficients. Namely,

Eq. (3.49):

\begin{pmatrix} N \\ 0 \end{pmatrix} = \begin{pmatrix} N \\ N \end{pmatrix} = 1

Eq. (3.50):

\begin{pmatrix} N - 1 \\ n \end{pmatrix} + \begin{pmatrix} N - 1 \\ n-1 \end{pmatrix} = \begin{pmatrix} N \\ n \end{pmatrix}

Eq. (3.51):

\begin{pmatrix} N \\ n+1 \end{pmatrix} = \displaystyle{\frac{N-n}{n+1}}\begin{pmatrix} N \\ n \end{pmatrix}

R. H. Swendsen. (2012). An Introduction to Statistical Mechanics and Thermodynamics


Proof of Eq. (3.49).

\begin{aligned} \begin{pmatrix} N \\ 0 \end{pmatrix} & = \frac{N!}{0!(N-0)!} \\ & = \frac{N!}{1\times N!} \\ & = 1 \end{aligned}
\begin{aligned} \begin{pmatrix} N \\ N \end{pmatrix} & = \frac{N!}{N!(N-N)!} \\ & = \frac{N!}{N!\times 1} \\ & = 1 \end{aligned}

Proof of Eq. (3.50).

\begin{aligned} & \quad \begin{pmatrix} N-1 \\ n \end{pmatrix} + \begin{pmatrix} N-1 \\ n-1 \end{pmatrix} \\ & = \frac{(N-1)!}{(n)!(N-1-n)!} + \frac{(N-1)!}{(n-1)!\big((N-1)-(n-1)\big)!} \\ & = \frac{(N-1)!}{(n)!(N-1-n)!} + \frac{(N-1)!}{(n-1)!(N-n)!} \\ & = \frac{(N-n)(N-1)!+(n)(N-1)!}{(n)(n-1)!(N-n)(N-n-1)!} \\ & = \frac{N(N-1)!}{n!(N-n)!} \\ & = \frac{N!}{n!(N-n)!} \\ & = \begin{pmatrix} N \\ n \end{pmatrix} \\ \end{aligned}

Proof of Eq. (3.51).

\begin{aligned} &\quad \begin{pmatrix} N \\ n+1 \end{pmatrix} \\ & = \frac{N!}{(n+1)!\big(N-(n+1)\big) !} \\ & = \frac{(N!)(N-n)}{\big( (n+1)(n!)\big) \big( (N-n)(N-n-1)! \big)} \\ & = \bigg(\frac{N-n}{n+1}\bigg)\bigg(\frac{N!}{n!(N-n)!}\bigg) \\ & = \frac{N-n}{n+1}\begin{pmatrix} N \\n \end{pmatrix} \\ \end{aligned}

QED

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202109161207 Exercises 3.5.A (Q1)

Find the volume V inside the paraboloid z=x^2+y^2 for 0\le z\le 4.


Background. (Triple integral in cylindrical coordinates)

\displaystyle{\iiint\limits_{S}f(x,y,z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=\iiint\limits_{S'}f(r\cos\theta ,r\sin\theta ,z)\, r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z} where the mapping x=r\cos\theta, y=r\sin\theta, z=z maps the solid S' in r\theta z-space onto the solid S in xyz-space in a one-to-one manner.

pg.122, Chapter 3.5, Michael Corral. (2008). Vector Calculus


Setup.

\begin{aligned} x & = r\cos\theta \\ y & = r\sin\theta \\ z & = z\\ & = x^2+y^2 \\ & = r^2\cos^2\theta + r^2\sin^2\theta \\ & = r^2 \\ \dots\enspace\textrm{if }& z=0\textrm{ then }r=0\enspace\dots \\ \dots\enspace\textrm{if }& z=4\textrm{ then }r=2\enspace\dots \\ \end{aligned}

hence,

\begin{aligned} & 0\le r\le 2 \\ & 0\le \theta \le 2\pi \\ & 0\le z\le 4\\ \end{aligned}


Solution.

Using vertical slices, we see that

\begin{aligned} V & =\iint\limits_{R}(4-z)\,\mathrm{d}A \\ & = \iint\limits_{R}\big( 4-(x^2+y^2)\big)\,\mathrm{d}A\end{aligned}

where R=\{ (x,y):x^2+y^2\le 4\} is the disc in \mathbb{R}^2. In polar coordinates (r,\theta ) we know that x^2+y^2=r^2 and that R'=\{ (r,\theta ): 0\le r\le 2, 0\le\theta\le 2\pi\}.

Thus,

\begin{aligned} V & = \int_{0}^{2\pi}\int_{0}^{2} (4-r^2)\, r\,\mathrm{d}r\,\mathrm{d}\theta \\ & = \int_{0}^{2\pi}\int_{0}^{2} (4r-r^3)\,\mathrm{d}r\,\mathrm{d}\theta \\ & = \int_{0}^{2\pi} \bigg[ 2r^2-\frac{r^4}{4}\bigg]\bigg|_{0}^{2}\,\mathrm{d}\theta \\ & = 4\int_{0}^{2\pi} \mathrm{d}\theta \\ & = 4(2\pi )\\ & = 8\pi\quad \textrm{(cubic units)} \end{aligned}