Euler’s formula e^ix = cos x + i sin x: a geometric approach

Today I mentioned the famous Euler’s formula briefly in my calculus class (when discussing hyperbolic functions, lecture notes here):

\displaystyle \begin{array}{rl} \displaystyle   \boxed{e^{ix}=\cos x+ i \sin x}  \ \ \ \ \ (1)\end{array}

where {i\in \mathbb C} is a solution to {z^2=-1} (usually denoted by “{i=\sqrt{-1}}”, but indeed there is no single-valued square root for complex numbers, or even negative real numbers).

One of the usual ways to derive this formula is by comparing the power series of the exponential function and the trigonometric functions {\sin} and {\cos}:

\displaystyle \begin{array}{rl} \displaystyle   \begin{cases} e^z=& \displaystyle 1+\frac{z^1}{1!}+\frac{z^2}{2!} +\frac{z^3}{3!} +\frac{z^4}{4!}+\cdots\\ \sin x=& \displaystyle  \frac{x^1}{1!} -\frac{x^3}{3!} +\frac{x^5}{5!}-\cdots\\ \cos x=& \displaystyle 1 -\frac{x^2}{2!} +\frac{x^4}{4!}-\cdots \end{cases} \end{array}

Putting {z=ix} in the first expansion and comparing with the remaining two, it’s easy to see that

\displaystyle \begin{array}{rl} \displaystyle   e^{ix} =& \displaystyle  \left(1-\frac{x^2}{2!} +\frac{x^4}{4!}-\cdots\right) +i\left(\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)\\ =& \displaystyle \cos x+i\sin x. \end{array}

Here, I am going to give another approach which does not require any knowledge of series (thus avoid the problem of convergence), but only basic knowledge in complex number (complex addition and multiplication). I am sure this approach must has been taken before but I couldn’t find a suitable reference, especially an online one.

Let us agree that we define the Euler’s number to be

\displaystyle \begin{array}{rl} \displaystyle   e:=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n. \end{array}

From this it is easy to see that

\displaystyle \begin{array}{rl} \displaystyle   e^x=\lim_{n\rightarrow \infty}\left(1+\frac{x}{n}\right)^n\textrm{ for }x\in \mathbb R. \end{array}

It is then natural to define the complex exponential function by

\displaystyle \begin{array}{rl} \displaystyle   e^z :=\lim_{n\rightarrow \infty}\left(1+\frac{z}{n}\right)^n\textrm{ for }z\in \mathbb C. \end{array}

Here I am cheating a little bit because I have implicitly assumed that this limit exists.

Now recall the geometry of the complex plane. We can identify a complex number {x+iy} with the point {(x, y)} on the plane. We can write a complex number in its polar form {z=r(\cos \theta+i \sin\theta)}, which is identified with {(r, \theta)} in polar coordinates. We call {r=|z|} and {\theta=\mathrm{arg}(z)} the modulus (or length) and the argument (or angle) of {z} respectively.

The complex plane

The complex multiplication of {z_1=r_1(\cos \theta_1+i\sin \theta_1)} and {z_2=r_2(\cos \theta_2+i\sin \theta_2)} is then

\displaystyle \begin{array}{rl} \displaystyle   z_1z_2 =r_1r_2\left(\cos (\theta_1+\theta_2)+i\sin (\theta_1+\theta_2)\right). \end{array}

i.e. the modulus of {z_1z_2} is the product of the two moduli and the argument of {z_1z_2} is the sum of the two arguments.

So now, let’s fix {x\in \mathbb R} and compute

\displaystyle \begin{array}{rl} \displaystyle   \lim_{n\rightarrow \infty}\left(1+\frac{ix}{n}\right)^n, \end{array}

which by definition would be {e^{ix}}. We will argue that its length is {1} and its argument is {x}, i.e. (1) holds:

\displaystyle \begin{array}{rl} \displaystyle   e^{ix}=\cos x+i \sin x.  \ \ \ \ \ (2)\end{array}

The geometry of the powers of 1+yi

To see this, let {z_n=1+\frac{ix}{n}}. Then {|z_n|=\left(1+\frac{x^2}{n^2}\right)^{\frac{1}{2}}} and so

\displaystyle \begin{array}{rl} \displaystyle   |z_n|^n=\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}  \end{array}

From this we have

\displaystyle \begin{array}{rl} \displaystyle   |({z_n})^n| =|z_n|^n =\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}} =& \displaystyle \left[\left(1+\frac{x^2}{n^2}\right)^{n^2}\right]^{\frac{1}{2n}}\\ \rightarrow& \displaystyle \left(e^{x^2}\right)^{0}=1  \ \ \ \ \ (3)\end{array}

as {n\rightarrow \infty}. On the other hand, the argument of {z_n} (which is well-defined up to a multiple of {2\pi}) can be chosen to be

\displaystyle \begin{array}{rl} \displaystyle   \arg(z_n)=\theta_n=\tan^{-1}\left(\frac{x}{n}\right). \end{array}

Then by the L’Hôpital’s rule,

\displaystyle \begin{array}{rl} \displaystyle   \lim_{n\rightarrow \infty} n\theta_n= \lim_{n\rightarrow \infty} n\tan^{-1}\left(\frac{x}{n}\right) =\lim_{t\rightarrow 0^+} \frac{\tan^{-1}(tx)}{t} =\lim_{t\rightarrow 0^+} \frac{x}{1+t^2x^2}=x. \end{array}

So we have

\displaystyle \begin{array}{rl} \displaystyle  \lim_{n\rightarrow \infty} \arg [(z_n)^n ]= \lim_{n\rightarrow \infty} n\arg(z_n)= \lim_{n\rightarrow \infty} n\theta_n=x. \ \ \ \ \ (4)\end{array}

Combining (3) and (4), we have

\displaystyle \begin{array}{rl} \displaystyle   e^{ix}:=\lim_{n\rightarrow \infty}\left(1+\frac{ix}{n}\right)^n=\lim_{n\rightarrow \infty}(z_n)^n= \cos x+i\sin x. \end{array}

Added Nov 12, 2017:
I found a video explaining e^{i\pi}=-1 (but without giving the full mathematical details) in the above approach:

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An inequality for functions on the plane

I accidentally came across a curious inequality for functions of two variables. I would like to know if this inequality is a special case of a more general result but I was unable to find a reference. It would also be interesting if applications can be found for this inequality.

Theorem 1 For any nontrivial {C^2} function {f=f(x,y)} on {\mathbb R^2} such that {[c_1,c_2]\subset f(\mathbb R^2)} and {f^{-1}([c_1,c_2])} is compact, we have

\displaystyle \begin{array}{rl} \displaystyle  \int_{\{c_1\le f\le c_2\}} \frac{|f_{xx}{f_y}^2-2f_{xy}f_xf_y+f_{yy}{f_x}^2|}{|\nabla f|^2} dxdy \ge 2 \pi (c_2-c_1). \ \ \ \ \ (1)\end{array}

Here we define the integrand to be zero if {\nabla f=0}.

In particular, if {f} is compactly supported, then

\displaystyle \begin{array}{rl} \displaystyle   \int_{\mathbb R^2} \frac{|f_{xx}{f_y}^2-2f_{xy}f_xf_y+f_{yy}{f_x}^2|}{|\nabla f|^2} dxdy \ge 2 \pi \left(\max f-\min f\right). \end{array}

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Weighted isoperimetric inequalities in warped product manifolds

1. Introduction

The classical isoperimetric inequality on the plane states that for a simple closed curve on {\mathbb R^2}, we have {L^2\ge 4\pi A}, where {L} is the length of the curve and {A} is the area of the region enclosed by it. The equality holds if and only if the curve is a circle. The classical isoperimetric inequality has been generalized to hypersurfaces in higher dimensional Euclidean space, and to various ambient spaces. For these generalizations, we refer to the beautiful article by Osserman [O] and the references therein. For a more modern account see [Ros]. Apart from two-dimensional manifolds and the standard space forms {\mathbb R^n}, {\mathbb H^n} and {\mathbb S^n}, there are few manifolds for which the isoperimetric surfaces are known. According to [BM], known examples include {\mathbb R\times \mathbb H^n}, {\mathbb RP^3}, {\mathbb S^1\times \mathbb R^2}, {T^2\times \mathbb R}, {\mathbb R\times \mathbb S^n}, {\mathbb S^1\times \mathbb R^n}, {\mathbb S^1\times \mathbb S^2}, {\mathbb S^1\times \mathbb H^2} and the Schwarzschild manifold, most of which are warped product manifolds over an interval or a circle. There are also many applications of the isoperimetric inequalities. For example, isoperimetric surfaces were used to prove the Penrose inequality [B], an inequality concerning the mass of black holes in general relativity, in some important cases.

In this post, we prove both classical and weighted isoperimetric results in warped product manifolds, or more generally, multiply warped product manifolds. We also relate them to inequalities involving the higher order mean-curvature integrals. Some applications to geometric inequalities and eigenvalues are also given.

For the sake of simplicity, let us describe our main results on a warped product manifold. The multiply warped product case is only notationally more complicated and presents no additional conceptual difficulty.

Let {\displaystyle M=[0, l)\times N} ({l\le \infty}) be a product manifold. Equip {M} with the warped product Riemannian metric {\displaystyle g=dr^2+ s(r)^2 g_{N}} for some continuous {s(r)\ge 0}, where {g_{N}} is a Riemannian metric on the {m}-dimensional manifold {N}, which we assume to be compact and oriented. Define {B_R:=\{(r, \theta)\in M: r< R\}}. We define the functions {A(r)} and {v(r)} by

\displaystyle \begin{array}{rl} \displaystyle   A(r):= s(r)^m \; \textrm{ and }\; v(r):= \int_{0}^{r} A(t) dt. \end{array}

Up to multiplicative constants, they are just the area of {\partial B_r} and the volume of {B_r} respectively. For a bounded domain {\Omega} in {M}, we define {\Omega^\#} to be the region {B_R} which has the same volume as {\Omega}, i.e. {\mathrm{Vol} (B_R) =\mathrm{Vol} (\Omega) }. We denote the area of {\partial \Omega} and {\partial \Omega^\#} by {|\partial \Omega|} and {|\partial \Omega^\#|} respectively.

One of our main results is the following isoperimetric theorem, which is a special case of Theorem 11.

Theorem 1 Let {\Omega} be a bounded open set in {(M,g)} with Lipschitz boundary. Assume that

  1. The projection map {\pi: \partial \Omega\subset \mathbb [0, l)\times N\rightarrow N} defined by {(r, \theta)\mapsto \theta} is surjective.
  2. {s(r)} is non-decreasing.
  3. {A\circ v^{-1}} is convex, or equivalently, {s(r)s''(r)-s'(r)^2\ge 0} for {r>0} if {s} is twice differentiable.

Then the isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle |\partial \Omega|\ge|\partial \Omega^\#|.\end{array}

The equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=\mathrm{constant}\}}.

We remark that our notion of convexity does not require the function to be differentiable: {f} is convex on {I} if and only if {f((1-t)x+ty)\le (1-t)f(x)+tf(y)} for any {t\in(0,1)} and {x, y\in I}.

One feature of our result is that except compactness, we do not impose any condition on the fiber manifold {N}. We will see in Section 7 that without further restriction on {N} or {s}, our conditions are optimal in a certain sense.

The expression {ss''-s'^2} comes from the observation that if {s} is twice differentiable, then as {v'=s^m}, the convexity of {A(v^{-1} (u))=s\left( v^{-1}\left( u \right)\right)^m} is equivalent to

\displaystyle \begin{array}{rl} \displaystyle  \frac{d^2}{du^2} A \left(v^{-1} (u)\right) =\frac{m}{ s(r)^{m+2}}\left(s(r)s''(r)-s'(r)^2\right)\ge 0, \ \ \ \ \ (1)\end{array}

where {r=v^{-1} (u)}.

The expression {ss''-s'^2} also has a number of geometric and physical meanings. It is related to the stability of the slice {\Sigma=\{r=r_0\}} as a constant-mean-curvature (CMC) hypersurface (i.e. whether it is a minimizer of the area among nearby hypersurfaces enclosing the same volume). Indeed, it can be shown ([GLW] Proposition 6.2) that {\lambda_1(g_N)\ge m\left(s(r_0)'^2-s(r_0)s''(r_0)\right)} if and only if {\Sigma} is a stable CMC hypersurface, where {\lambda_1(g_N)} is the first Laplacian eigenvalue of {(N, g_N)}. It is also related to the so called “photon spheres” in relativity (see [GLW] Proposition 6.1).

From (1), {A\circ v^{-1}} is convex if and only if {s} is {\log}-convex. So if {f(r)} is non-decreasing and convex, then {s(r)=\exp(f(r))} satisfies the convexity condition. One example of such a function is {s(r)=e^r}.

If {\partial \Omega} is star-shaped in the sense that it is a graph over {N}, i.e. of the form {\partial \Omega=\{(r, \theta): r=\rho(\theta), \theta\in N\}}, we can remove the assumption on the monotonicity of {s(r)}.

Theorem 2 Suppose {\Omega} is a bounded open set in {(M,g)} with Lipschitz boundary. Assume that

  1. {\partial \Omega} is star-shaped.
  2. {A\circ v^{-1}} is convex, or equivalently, {ss''-s'^2\ge 0} if {s} is twice differentiable.

Then the isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle |\partial \Omega|\ge|\partial \Omega^\#|.\end{array}

The equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=\mathrm{constant}\}}.

If the classical isoperimetric inequality already holds on {(M,g)}, we can extend it by the following result, which is a special case of Theorem 8.

Theorem 3 (Weighted isoperimetric inequality) Let {\Omega} be a bounded open set in {(M,g)} with Lipschitz boundary. Assume that

  1. The classical isoperimetric inequality holds on {(M,g)}, i.e. {|\partial \Omega|\ge|\partial \Omega^\#|}.
  2. {a(r)} is a non-negative function such that {\psi(r):=b(r)A(r)} is non-decreasing and non-negative, where {b(r):=a(r)-a(0)}.
  3. The function {\psi\circ v^{-1} } is convex.


\displaystyle \begin{array}{rl} \displaystyle   \int_{\partial \Omega} a(r)dS \ge\int_{\partial \Omega^{\#}} a(r)dS. \end{array}

If {b(r)A(r)>0} for {r>0}, then the equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=\mathrm{constant}\}}.

Using a volume preserving flow, recently Guan, Li and Wang [GLW] (see also [GL2]) proved the following related result, assuming {s} is smooth ({\partial \Omega} is called “graphical” in [GLW]):

Theorem 4 ([GLW][Theorem 1.2]) Suppose {\Omega} is a domain in {(M,g)} with smooth star-shaped boundary. Assume that

  1. The Ricci curvature {\mathrm{Ric}_N} of {g_N} satisfies {\mathrm{Ric}_N\ge (m-1)K g_N}, where {K>0} is constant.
  2. {0\le s'^2-ss''\le K}.

Then the isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle |\partial \Omega|\ge|\partial \Omega^\#|.\end{array}

If { s'^2-ss''<K}, then the equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=r_0\}}.

We note that our assumption {ss''-s'^2\ge 0} in Theorem 2 complements that of [GLW]. This does not contradict the result in [GLW]. In fact, we will show the necessity of this and other conditions in Section 7 (cf. Proposition 28). We also notice that except the obvious case that {M} has constant curvature, the equality holds only when {\partial \Omega} is a coordinate slice. Indeed, combining Theorem 4 with Theorem 3, we can generalize Theorem 4 as follows.

Theorem 5 (Theorem 10) Suppose {\Omega} is a domain in {(M,g)} with smooth star-shaped boundary. Assume that the Ricci curvature {\mathrm{Ric}_N} of {g_N} satisfies {\mathrm{Ric}_N\ge (m-1)K g_N} and {0\le s'^2-ss''\le K}, where {K>0} is constant. Suppose {a(r)} is a positive function such that {b( v^{-1} (u))s( v^{-1} (u))^m} is convex, where {b(r):=a(r)-a(0)}. Then the weighted isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle \int_{\partial \Omega}a(r)dS\ge \int_{\partial \Omega^\#}a(r)dS.\end{array}

The equality holds if and only if either

  1. {(M,g)} has constant curvature, {a(r)} is constant on {\partial \Omega}, and {\partial \Omega} is a geodesic hypersphere, or
  2. {\partial \Omega} is a slice {\{r=r_0\}}.

Combining Theorem 4, Theorem 2 and the proof of Proposition 28, we get the following picture for the isoperimetric problem in warped product manifolds:

Theorem 6 Let {M} be the product manifold {[0, l)\times N} equipped with the warped product metric {g=dr^2+s(r)^2 g_N}.

  1. Suppose {s'^2-ss''\le 0}. Then the star-shaped isoperimetric hypersurfaces are precisely the coordinate slices {\{r=r_0\}}.
  2. Suppose {0\le s'^2-ss''\le K} and {\mathrm{Ric}_N\ge (m-1)Kg_N} where {K>0} is constant. Then the star-shaped isoperimetric hypersurfaces are either geodesic hyperspheres if {(M,g)} has constant curvature, or the coordinate slices {\{r=r_0\}}.
  3. Suppose {s'^2-ss''>K} and {\lambda_1(g_N)\le mK} where {K>0} is constant. Then the coordinate slices {\{r=r_0\}} cannot be isoperimetric hypersurfaces.

We also prove isoperimetric type theorems involving the integrals of higher order mean curvatures {(H_k)} in warped product manifolds. For simplicity, let us state the result when the ambient space is {\mathbb R^n} (Corollary 18), which follows from a more general theorem (Theorem 17)

Theorem 7 (Corollary 18) Let {\Sigma} be a closed embedded hypersurface in {\mathbb R^{m+1}} which is star-shaped with respect to {0} and {\Omega} is the region enclosed by it. Assume that {H_k>0} on {\Sigma}. Then for any integer {l\ge 0},

\displaystyle \begin{array}{rl} \displaystyle   n \beta_n ^{-\frac{l-1}{n}}\mathrm{Vol} (\Omega)^{\frac{n-1+l}{n}} \le\int_{\Sigma} H_k r^{l+k}dS, \end{array}

where {\beta_n} is the volume of the unit ball in {\mathbb R^n}. If {l\ge 1}, the equality holds if and only if {\Sigma} is a hypersphere centered at {0}.

Note that when {k=l=0}, this reduces to {n {\beta_n}^{\frac{1}{n}}\mathrm{Vol} (\Omega)^{\frac{n-1}{n}} \le \mathrm{Area} (\Sigma) }, which is the classical isoperimetric inequality. In fact, we prove a stronger result (14) :

\displaystyle \begin{array}{rl} \displaystyle   n \beta_n^{-\frac{l-1}{n}}\mathrm{Vol} (\Omega)^{\frac{n-1+l}{n}} \le \int_{\partial \Omega} H_0 r ^{l} dS \le \int_{\partial \Omega} H_1 r ^{l+1} dS \le \cdots \le \int_{\partial \Omega} H_k r ^{l+k} dS. \end{array}

This can be compared to the following result of Guan-Li [GL][Theorem 2]:

\displaystyle \begin{array}{rl} \displaystyle   \left(\frac{1}{ \beta_n }\mathrm{Vol} (\Omega)\right)^{\frac{1}{n}}\le \left(\frac{1}{n\beta_n}{\int_\Sigma H_0dS}\right)^{\frac{1}{n-1}}\le \cdots \le \left(\frac{1}{n\beta_n} \int_\Sigma H_k dS \right) ^{\frac{1}{n-k-1}} \end{array}

under the same assumption.

Some applications of the weighted isoperimetric inequalities will also be given in Section 4 and Section 6.

The rest of this post is organized as follows. In Section 2, we first prove Theorem 3. In Section 3, we prove the isoperimetric inequality involving a weighted volume (Theorem 11), which implies Theorem 1 and Theorem 2. Although Theorem 3 can also be stated using the weighted volume, we prefer to prove the version involving only the ordinary volume for the sake of clarity, and indicates the changes needed to prove Theorem 11. In Section 4, we illustrate how we can obtain interesting geometric inequalities in space forms by using Theorem 8. In Section 5, we introduce the weighted Hsiung-Minkowski formulas in warped product manifolds, and combine them with the isoperimetric theorem to obtain new isoperimetric results involving the integrals of the higher order mean curvatures. In Section 6, we give further applications of our results to obtain some sharp eigenvalue estimates for some second order differential operators related to the extrinsic geometry of hypersurfaces and an eigenvalue estimate for the Steklov differential operator (also known as Dirichlet-to-Neumann map). A Pólya-Szegö inequality and a Faber-Krahn type theorem are also derived. Finally in Section 7, we show that the conditions of Theorem 1 are necessary by giving counterexamples where the isoperimetric inequality fails if any one of the conditions is violated.

Acknowledgements. We would like to thank Professor Frank Morgan for pointing out the reference [H] to us, and Professor Mu-Tao Wang for useful comments and discussion.
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Faber-Krahn inequality

I record a proof of the Faber-Krahn inequality here, mainly for my own benefit.

Let {M} be one of the standard space forms: the Euclidean space {\mathbb R^n}, the unit sphere {\mathbb S^n}, or the hyperbolic space {\mathbb H^n}. Suppose {\Omega } is a bounded domain in {M} with smooth boundary {\partial \Omega} which is a closed hypersurface. Then we can define {\Omega^\#} to be the geodesic ball in {M} which has the same volume as {\Omega}.

Let {\lambda_1(\Omega)} be the first Laplacian eigenvalue of {\Omega} under the Dirichlet boundary condition. i.e. {\lambda_1(\Omega)>0} is the smallest number {\lambda} such that there exists a non-zero smooth function {f} with

\displaystyle \begin{array}{rl} \displaystyle   \Delta f =-\lambda f\;  \displaystyle \textrm{in }\Omega, \quad f =0 \;\textrm{on }\partial \Omega. \end{array}

Theorem 1 (Faber-Krahn inequality) Suppose {\Omega } is a bounded domain in {M} with smooth boundary {\partial \Omega}. Then

\displaystyle \begin{array}{rl} \displaystyle   \lambda_1(\Omega)\ge \lambda_1(\Omega^\#).  \end{array}

The equality holds if and only if {\Omega} is a geodesic ball.

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Why is a² + b² ≥ 2ab ?

This post can be regarded as a sequel to my previous (and very ancient) post on 1+2+3+…. Though these two posts are not quite logically related, they share the same spirit (I’m asking a dumb question again).

How can one prove the following?

Theorem 1

\displaystyle \begin{array}{rl} \displaystyle a^2+b^2\ge 2ab.\end{array}

This extremely simple inequality (just next to the simplest but arguably the most important inequality {a^2\ge 0}) turns out to be the testing ground for a lot of more advanced inequalities (e.g. Cauchy-Schwarz, or Newton’s inequality).

In this post, I record some proofs of this inequality that I can think of. Some of them are quite similar, or may be considered almost the same depending how you measure the level of similarity. I try to group proofs of roughly the same idea together. In some of these proofs, we have actually used more advanced inequalities (which perhaps are even proved from this simple inequality). For such a simple result, I guess there may be over a hundred proofs.

So what’s the point of doing all these, since only one proof is all that suffices? Because not many people would have the patience to read till the end (or even this paragraph), let me put the summary here:

Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis? (Paul Halmos, “I want to be a mathematician: an automathography”)

So, let’s begin.

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A remark on the divergence theorem

The divergence theorem states that for a compact domain {D} in {\mathbb R^3} with piecewise smooth boundary {\partial D}, then for a smooth vector field {\textbf{F}} on {D}, we have

\displaystyle \begin{array}{rl} \displaystyle \iint_{\partial D}\textbf{F}\cdot \textbf{n}\,dS=\iiint_D \nabla \cdot\textbf{F}\,dV,\end{array}

where {\textbf{n}} is the unit outward normal and {\nabla \cdot\textbf{F}} is the divergence of {\textbf{F}}.

In most textbooks, the divergence theorem is proved by the following strategy:

  1. Prove that

    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(0, 0, R)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial R}{\partial z}\;dV \ \ \ \ \ (1)\end{array}

    where {D} is a so called type {1} domain, i.e. {D=\{(x, y, z): (x, y)\in \Omega, \underline f(x, y)\le z\le \bar f(x, y)\}} for some region {\Omega\subset \mathbb R^2} and {\bar f, \underline f\in C^1(R)}. i.e. {D} is bounded by the graphs of two functions with variables {x, y}. This can be quite easily done using the fundamental theorem of calculus.

  2. Similarly prove

    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(P, 0, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial P}{\partial x}\;dV \ \ \ \ \ (2)\end{array}


    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(0, Q, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial Q}{\partial y}\;dV \ \ \ \ \ (3)\end{array}

    for the so called type {2} domain and type {3} domain respectively.

  3. Argue that for sufficiently nice {D}, it can be cut into finitely many pieces {D_i}, each of which are of both type {1}, {2}, and {3} (the so called type {4} domain). Thus we can compute the divergence integral {\displaystyle \iiint_D \nabla \cdot \textbf{F}\,dV} by adding the integrals of the pieces and apply the previous result (using linearity). Argue that the orientation of the common face of these {D_i} (if any) is opposite to each other and thus a cancellation argument will give the surface integral {\displaystyle \iint _{\partial D}\textbf{F}\cdot \textbf{n}\;dS}.

I am slightly disappointed by this approach because even for the so called type {1} domain, we still cannot prove the divergence theorem very directly (say, by reducing to a double integral on some planar domain {\Omega} and apply Green’s theorem), but have to further cut it into even smaller type {4} domains.

In this note we modify the strategy above by proving (1), (2), (3) all hold on a type {1} domain. In particular, I will show that (2) and (3) can be proved for a type {1} domain by using Green’s theorem.
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The Cauchy-Schwarz inequality and the Lagrange identity

The classical Lagrange identity is the following:

\displaystyle \begin{array}{rl} \displaystyle \left(\sum_{i=1}^n a_ib_i\right)^2+\sum _{1\le i<j\le n} \left(a_ib_j-a_jb_i\right)^2 =\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{j=1}^{n}b_j^2\right).\end{array}

This can be proven by expanding {\displaystyle \sum _{1\le i<j\le n}\left(a_ib_j-a_jb_i\right)^2 } and separating the terms into the cross-terms part and the non cross-terms part.

The Lagrange identity implies the Cauchy-Schwarz inequality in {\mathbb R^n}. And when {n=3}, this can be rephrased as

\displaystyle \begin{array}{rl} \displaystyle |{a}|^2|{b}|^2= \left({a}\cdot {b}\right) ^2+|{a}\times {b}|^2\end{array}

for {{a}, {b}\in \mathbb R^3}. In general, the term {\displaystyle \sum _{1\le i<j\le n} \left(a_ib_j-a_jb_i\right)^2} can be identified as the norm squared of the wedge product {{a}\wedge {b}}.

In this note, we give the less well-known extension of this identity and the corresponding Cauchy-Schwarz type inequality.
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