Ngô Quốc Anh

April 1, 2011

Several interesting limits from a paper by Chang-Qing-Yang

Filed under: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 2, Giải Tích Cổ Điển, Linh Tinh — Ngô Quốc Anh @ 16:42

Recently, I have learnt from my friend, ZJ, the following result

Assume that F:\mathbb R \to \mathbb R is absolutely integrable. Then

\displaystyle\begin{gathered} \mathop {\lim }\limits_{t \to \pm \infty } {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = 0, \hfill \\ \mathop {\lim }\limits_{t \to \pm \infty } {e^{ - 2t}}\int_{ - \infty }^t {F(x){e^{ - 2x}}dx} = 0. \hfill \\ \end{gathered}

The result seems reasonable by the following observation, for example, we consider the first identity when t \to +\infty. Then the factor

\displaystyle\int_t^{ + \infty } {F(x){e^{ - 2x}}dx}

decays faster then the exponent function \exp (2t). This may be true, of course we need to prove mathematically, because the integrand contains the term \exp (-2x) which turns out to be a good term since x \geqslant t. So here is the trick in order to solve such a problem.

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January 8, 2011

A funny limit involving sine function

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 2:32

Today, I have been asked to calculate the following limit

\displaystyle \mathop {\lim }\limits_{n \to + \infty } \sin (\sin \overbrace {(...(}^n\sin x)...))

for each fixed x \in [0,2\pi]. From the mathematical point of view, we can assume x \in (-\frac{\pi}{2}, \frac{\pi}{2}) as we just replace x by \sin (\sin x)) if necessary.

There are three possible cases

Case 1. x \in (0, \frac{\pi}{2}). In this case, it is well known that function \frac{\sin x}{x} is monotone decreasing since

\displaystyle {\left( {\frac{{\sin x}}{x}} \right)^\prime } = \frac{{x\cos x - \sin x}}{{{x^2}}} = \frac{{\cos x}}{{{x^2}}}\left( {x - \tan x} \right) \leqslant 0

in its domain. Consequently, it holds

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November 2, 2010

Jacobi’s formula for the differential of the determinant of matrices

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 15:34

In matrix calculus, Jacobi’s formula expresses the differential of the determinant of a matrix A in terms of the adjugate of A and the differential of A. The formula is

\displaystyle d\, \mbox{det} (A) = \mbox{tr} (\mbox{adj}(A) \, dA).

It is named after the mathematician C.G.J. Jacobi.

We first prove a preliminary lemma.

Lemma. Given a pair of square matrices A and B of the same dimension n, then

\displaystyle\sum_i \sum_j A_{ij} B_{ij} = \mbox{tr} (A^\top B).

Proof. The product AB of the pair of matrices has components

\displaystyle (AB)_{jk} = \sum_i A_{ji} B_{ik}.

Replacing the matrix A by its transpose A^\top is equivalent to permuting the indices of its components

\displaystyle (A^\top B)_{jk} = \sum_i A_{ij} B_{ik}.

The result follows by taking the trace of both sides

\displaystyle \mbox{tr} (A^\top B) = \sum_j (A^\top B)_{jj} = \sum_j \sum_i A_{ij} B_{ij} = \sum_i \sum_j A_{ij} B_{ij}.

Theorem. It holds

\displaystyle d \, \mbox{det} (A) = \mbox{tr} (\mbox{adj}(A) \, dA).

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October 18, 2010

1/infinity = 0 is equivalent to 1/0=infinity?

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 12:05

It is now the time to discuss some funny thing. I just learn from GR class this morning a proof of the following statement

\displaystyle \frac{1}{\infty}=0 \quad \Longleftrightarrow \quad \frac{1}{0}=\infty.

Okay, let us start with the left hand side. By rotating 90 degrees counter-clockwise both sides of

\displaystyle \frac{1}{\infty}=0

we get

\displaystyle -18=0.

Now adding both sides by 8 we arrive at

\displaystyle -10=8.

Again, rotating 90 degrees clockwise both sides we reach to

\displaystyle \frac{1}{0}=\infty.

The reverse case can be treated similarly.

September 22, 2010

An identity of differentiation involving the Kelvin transform

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Tags: — Ngô Quốc Anh @ 15:47

This short note is to prove the following

\displaystyle {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = - {\nabla _y}\left( {u\left( y \right)} \right) \cdot y

where x and y are connected by

\displaystyle y = \frac{x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^2}.

The proof is straightforward as follows.

  • Calculation of \frac{\partial}{\partial x_1}.

We see that

\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_1}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_1} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_1} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_1^2}}{{{{\left| x \right|}^4}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_1}. \hfill \\ \end{gathered}

  • Calculation of \frac{\partial}{\partial x_2}.

Similarly, we get

\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_2}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_2} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_2} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_2^2}}{{{{\left| x \right|}^4}}}} \right){x_2}. \hfill \\ \end{gathered}

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July 19, 2010

On the determinant of a matrix

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 20:10

Several days ago, I placed a question on MathLinks asking the relation between \det A and \det(A-\lambda I). The point is how to evaluate

\displaystyle\det\begin{bmatrix}1+|x|^{2}-2x_{1}^{2}&-x_{1}x_{2}&\cdots&-x_{1}x_{n}\\ -x_{1}x_{2}&1+|x|^{2}-2x_{2}^{2}&\cdots&-x_{2}x_{n}\\ \vdots &\vdots &\ddots &\vdots\\ -x_{n}x_{1}&-x_{n}x_{2}&\cdots&1+|x|^{2}-2x_{n}^{2}\end{bmatrix}.

Interestingly, K.M. showed me a new way to attack such a problem but slightly different from the original one. He proved

\displaystyle\det\begin{bmatrix}1+|x|^{2}-x_{1}^{2}&-x_{1}x_{2}&\cdots&-x_{1}x_{n}\\ -x_{1}x_{2}&1+|x|^{2}-x_{2}^{2}&\cdots&-x_{2}x_{n}\\ \vdots &\vdots &\ddots &\vdots\\ -x_{n}x_{1}&-x_{n}x_{2}&\cdots&1+|x|^{2}-x_{n}^{2}\end{bmatrix}=\left(1+|x|^{2}\right)^{n-1}.

Let us discuss the proof of this modified problem.

Let

x=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}

and let

A=xx^T.

The determinant we are trying to compute is

\displaystyle \det\left((1+|x|^2)I-A\right),

which is the characteristic polynomial of A evaluated at 1+|x|^2.

Now, A is certainly diagonalizable (which doesn’t even matter, but it makes it easier to think about), and we know its eigenvalues. Why do we know its eigenvalues? Because A is a matrix of rank 1, hence nullity n-1, hence n-1 of its n eigenvalues are zero. What is the other eigenvalue? It’s the same as the sum of the eigenvalues, which is the trace of A, which is |x|^2. Put that information together, and we have that the characteristic polynomial of A is

\det(\lambda I-A)=\left(\lambda-|x|^2\right)\lambda^{n-1}=\lambda^n-|x|^2\lambda^{n-1}.

Substitute 1+|x|^2 for \lambda to get the result quoted.

May 1, 2010

A useful identity in a book due to L. Ahlfors

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 3:12

Let \mathbf{x},\mathbf{y} be points in \mathbb R^n. If we denote by \mathbf{x}^\sharp the reflection point of \mathbf{x} with respect to the unit ball, i.e.

\displaystyle \mathbf{x}^\sharp = \frac{\mathbf{x}}{|\mathbf{x}|^2}

we then have the following well-known identity

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } - \mathbf{x}} \right|.

The proof of the above identity comes from the fact that

\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt {1 + |\mathbf{x}{|^2}|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}} = |\mathbf{y}|\left| {\frac{\mathbf{y}}{{|\mathbf{y}|^2}} - \mathbf{x}} \right|.

Indeed, by squaring both sides of

\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt {1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}}

we arrive at

\displaystyle |\mathbf{x}|^2\left( {\frac{{|\mathbf{x}|^2}}{{|\mathbf{x}|^4}} - 2\frac{{\mathbf{x} \cdot \mathbf{y}}}{{|\mathbf{x}|^2}} + |\mathbf{y}|^2} \right) = 1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}

which is obviously true. Similarly, the last identity also holds. If we replace \mathbf{y} by -\mathbf{y} we also have

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp }+ \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } + \mathbf{x}} \right|.

Generally, if we consider the reflection point of \mathbf{x} over a ball B_r(0), i.e.

\displaystyle \mathbf{x}^\sharp = \frac{r^2\mathbf{x}}{|\mathbf{x}|^2}

we still have the fact

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } - \mathbf{x}} \right|.

Indeed, one gets

\displaystyle |\mathbf{x}|\left| {\frac{{{r^2}\mathbf{x}}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = {r^2}|\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \frac{\mathbf{y}}{{{r^2}}}} \right| = {r^2}\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|\left| {\frac{{\frac{\mathbf{y}}{{{r^2}}}}}{{{{\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|}^2}}} - \mathbf{x}} \right| = \left| \mathbf{y} \right|\left| {\frac{{{r^2}\mathbf{y}}}{{|\mathbf{y}{|^2}}} - \mathbf{x}} \right|.

Similarly,

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } + y} \right| = |y|\left| {{y^\sharp } + \mathbf{x}} \right|.

Such identity is very useful. For example, in \mathbb R^n (n\geqslant 3) the following holds

\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \min \left\{ {\frac{1}{{|{\mathbf{y}}|^{n - 2}}},\frac{1}{r^{n - 2}}} \right\}.

This type of formula has been considered before when n=3 here. For a general case, Lieb and Loss introduced another method in their book published by AMS in 2001. Here we introduce a completely new proof. At first, if |\mathbf{y}|>r by the potential theory, one easily gets

\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \frac{1}{{|{\mathbf{y}}|^{n - 2}}}.

If |\mathbf{y}|<r, one needs to make use of the reflection point of \mathbf{y} and the above identity to go back to the first case. The point here is |\mathbf{y}^\sharp|>r. The integral is obviously continuous as a function of \mathbf{y}. The above argument is due to professor X.X.W.

April 29, 2010

Surface integrals over a sphere when its radius varies

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Giải Tích Cổ Điển — Tags: — Ngô Quốc Anh @ 1:53

Let us consider the following integral in \mathbb R^3

\displaystyle\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}}

when t varies.

Obviously, the sphere |{\mathbf{x}}| = t can be parametrized as the following

\displaystyle {\mathbf{x}} = t\left( {\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta } \right)

so

\displaystyle\begin{gathered} \iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi )){t^2}\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi ))\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{|{\mathbf{x}}| = 1} {f \left(t\mathbf{x} \right)d{\sigma _{\mathbf{x}}}}. \hfill \\ \end{gathered}

If we wish to work on the average, the formula is much simpler than that, precisely

\displaystyle\frac{1}{{4\pi {t^2}}}\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \frac{1}{{4\pi }}\iint\limits_{|{\mathbf{x}}| = 1} {f(t{\mathbf{x}})d{\sigma _{\mathbf{x}}}}

that means

\displaystyle \overline {\iint\limits_{|{\mathbf{x}}| = t} } f({\mathbf{x}})d{\sigma _{\mathbf{x}}}=\overline {\iint\limits_{|{\mathbf{x}}| = 1} } f({t\mathbf{x}})d{\sigma _{\mathbf{x}}}

where the bar means the average.

More general, we get

\displaystyle\iint\limits_{|{\mathbf{x}}| = {t_1}} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = {\left( {\frac{{{t_1}}}{{{t_2}}}} \right)^2}\iint\limits_{|{\mathbf{x}}| = {t_2}} {f\left( {\frac{{{t_1}}}{{{t_2}}}{\mathbf{x}}} \right)d{\sigma _{\mathbf{x}}}}.

April 27, 2010

Surface integral: The symmetric property, 2

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Tags: — Ngô Quốc Anh @ 16:07

Today, we try to evaluate the following surface integral in \mathbb R^3. I found this result in a paper published in Math. Z. 198, 277-289 (1988). This topic can be considered as a continued part to the following topic.

Proposition. Let f(x) be a continuous function. Then we have

\displaystyle\frac{1}{t}\iint\limits_{|{\mathbf{y}} - {\mathbf{x}}| = t} {f(|{\mathbf{y}}|)d{\sigma _{\mathbf{y}}}} = \frac{{2\pi }}{r}\int\limits_{|r - t|}^{r + t} {sf(s)ds}

here we denote r=|\mathbf{x}|.

Proof. Remark that the integral only depends on |\mathbf{x}|, so that we may assume \mathbf{x} = (0, 0, r) and introduce the following spherical coordinate

\mathbf{y}=\mathbf{x}+t\omega

where

\displaystyle \omega = (\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta ).

This system of coordinates can be seen via the picture below

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April 23, 2010

Comparison of some subspaces associated to A and A^TA

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:10

Let us start with a given n\times m matrix A

\displaystyle {A_{n \times m}} = \left( {\begin{array}{*{20}{c}} {{a_{11}}} & \cdots & \cdots & {{a_{1m}}} \\ \vdots & \ddots & \ddots & \vdots \\ {{a_{n1}}} & \cdots & \cdots & {{a_{nm}}} \\ \end{array} } \right).

The aim of this entry is to compare nullspace, column space and row space between A, A^TA and AA^T. Obviously, A^TA and AA^T are m\times m and n \times n matrices respectively.

Nullspaces. We start with the following result

Proposition 1. The following

\displaystyle N(A)=N(A^TA)

holds.

Proof. Pick an arbitrary element x \in N(A), i.e. Ax=0, we can see that

\displaystyle A^TAx=A^T(Ax)=A^T0=0

so

\displaystyle N(A) \subset N(A^TA).

Conversely, assume x is such that A^TAx=0, as a consequence, x^TA^TAx=0. This gives us the fact

\displaystyle (Ax) \cdot (Ax)= (Ax)^T(Ax)=x^TA^TAx=0.

Consequently, Ax=0 which proves

\displaystyle N(A^TA)\subset N(A).

In other words, \displaystyle N(A)=N(A^TA).

Remark.  \displaystyle N(A)=N(AA^T) is no longer true since these two matrices have different dimension.

Regarding to matrix A, one has

\displaystyle {\rm rank}(A)+\dim N(A)=m.

Similarly, involving matrix A^TA, one gets

\displaystyle {\rm rank}(A^TA)+\dim N(A^TA)=m.

It now follows from Proposition 1 that

\displaystyle {\rm rank}(A)={\rm rank}(A^TA).

In other words, column and row spaces associated to A and A^TA have the same dimension respectively.

Row spaces. We prove the following

Proposition 2. The following

\displaystyle RS(A)=RS(A^TA)

holds.

Proof. The way to compare column spaces is to use the following facts

\displaystyle RS(A)=CS(A^T)

and

\displaystyle CS(A^T)=A^T(\mathbb R^n).

Equivalently, from the first fact we need to show that

\displaystyle CS(A^T)=CS(A^TA).

In term of the second fact, once you have a suitable matrix Q the column space of AQ is indeed contained in the column space of A. Therefore

\displaystyle CS(A^TA) \subset CS(A^T)

which turns out to be

\displaystyle RS(A^TA) \subset RS(A)

since they have the same dimension, equality occurs.

Remark. This comes from the proof above. If you have a good matrix Q, the following is true

\displaystyle CS(AQ) \subset CS(A).

Column spaces. We prove the following

Proposition 3. The following

\displaystyle CS(A)=CS(AA^T)

holds.

Proof. This is trivial by using Proposition 2.

Remark. If you have a good matrix P, the following is true

\displaystyle RS(PA) \subset RS(A).

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