Ngô Quốc Anh

November 5, 2011

MuPad: Heart in 3D

Filed under: Giải Tích 2, Giải Tích 5, Liên Kết — Tags: — Ngô Quốc Anh @ 0:26

This is not mathematics. I just found an equation so that we can draw a heart in 3D. Indeed, the following equation

\displaystyle {\left( {{x^2} + \frac{9}{4}{y^2} + {z^2} - 1} \right)^3} - {x^2}{z^3} - \frac{9}{{80}}{y^2}{z^3} = 0

will generate a heart. I have tried and the following pictures show that fact.


June 9, 2010

Invariance under fractional linear transformations

Filed under: Giải Tích 5, Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 5:02

I just read the following result due to Loss-Sloane published in J. Funct. Anal. this year [here]. This is just a lemma in their paper that I found very interesting.

Let f be any function in C_0^\infty(\mathbb R \setminus \{0\}). Consider the inversion x \mapsto \frac{1}{x} and set

\displaystyle g(x) = {\left| x \right|^{\alpha - 1}}f\left( {\frac{1}{x}} \right).

Then g \in C_0^\infty(\mathbb R) and

\displaystyle\iint\limits_{{\mathbb{R}^2}} {\frac{{{{\left| {g(x) - g(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy} = \iint\limits_{{\mathbb{R}^2}} {\frac{{{{\left| {f(x) - f(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy}.

Proof. For fixed \varepsilon consider the regions

\displaystyle {R_1}: = \left\{ {(x,y) \in {\mathbb{R}^2}:\left| {\frac{x}{y}} \right| > 1 + \varepsilon } \right\}

and likewise,

\displaystyle {R_2}: = \left\{ {(x,y) \in {\mathbb{R}^2}:\left|  {\frac{y}{x}} \right| > 1 + \varepsilon } \right\}.


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|.


\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)


\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



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

This system of coordinates can be seen via the picture below


March 30, 2008

Một số bài tập Giải tích của Khoa Toán đại học Florida

Filed under: Giải Tích 2, Giải Tích 3, Giải Tích 4, Giải Tích 5 — Ngô Quốc Anh @ 12:04


January 19, 2008

Steinmetz Solid

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 4:52

The solid common to two (or three) right circular cylinders of equal radii intersecting at right angles is called the Steinmetz solid. Two cylinders intersecting at right angles are called a bicylinder, and three intersecting cylinders a tricylinder. Half of a bicylinder is called a vault.

Steinmetz solid for two cylinders

For two cylinders of radius r oriented long the z– and x-axes gives the equations


which can be solved for x and y gives the parametric equations of the edges of the solid,

x = +/-z
y = +/-sqrt(r^2-z^2).

The surface area can be found as intxds, where

ds = sqrt(1+((dy)/(dz))^2)dz
= r/(sqrt(r^2-z^2))dz.

Taking the range of integration as a quarter or one face and then multiplying by 16 gives


The volume common to two cylinders was known to Archimedes (Heath 1953, Gardner 1962) and the Chinese mathematician Tsu Ch’ung-Chih (Kiang 1972), and does not require calculus to derive. Using calculus provides a simple derivation, however. Noting that the solid has a square cross section of side-half-length sqrt(r^2-z^2), the volume is given by


(Moore 1974). The volume can also be found using cylindrical algebraic decomposition, which reduces the inequalities

{x^2+y^2<1; -L<z<L; y^2+z^2<1; -L<x<L


{-1<x<1; -sqrt(1-x^2)<y<sqrt(1-x^2); -sqrt(1-y^2)<z<sqrt(1-y^2),

giving the integral


If the two right cylinders are of different radii a and b with a>b, then the volume common to them is


where K(k) is the complete elliptic integral of the first kind, E(k) is the complete elliptic integral of the second kind, and k=b/a is the elliptic modulus.

Steinmetz curve

The curves of intersection of two cylinders of radii a and b, shown above, are given by the parametric equations

x(t) = bcost
y(t) = bsint
z(t) = +/-sqrt(a^2-b^2sin^2t)

(Gray 1997, p. 204).

The volume common to two elliptic cylinders

(x^2)/(a^2)+(z^2)/(c^2)==1    (y^2)/(b^2)+(z^2)/(c^('2))==1

with c<c^' is


where k==c/c^' (Bowman 1961, p. 34).


For three cylinders of radii r intersecting at right angles, The resulting solid has 12 curved faces. If tangent planes are drawn where the faces meet, the result is a rhombic dodecahedron (Wells 1991). The volume of intersection can be computed in a number of different ways,

V_3(r,r,r) = 16r^3int_0^(pi/4)int_0^1ssqrt(1-s^2cos^2t)dsdt
= (sqrt(2)r)^3+6int_(r/sqrt(2))^r(2sqrt(r^2-z^2))^2dz
= 8(2-sqrt(2))r^3

(Moore 1974). According to the protagonist Christopher in the novel The Curious Incident of the Dog in the Night-Time, “…People go on holidays to see new things and relax, but it wouldn’t make me relaxed and you can see new things by looking at earth under a microscope or drawing the shape of the solid made when 3 circular rods of equal thickness intersect at right angles” (Haddon 2003, p. 178), which is of course precisely the Steinmetz solid formed by three symmetrically placed cylinders.

Steinmetz tetrahedra

Four cylinders can also be placed with axes along the lines joining the vertices of a tetrahedron with the centers on the opposite sides. The resulting solid of intersection has volume


and 24 curved faces analogous to a cube-octahedron compound (Moore 1974, Wells 1991).


Six cylinders can be placed with axes parallel to the face diagonals of a cube. The resulting solid of intersection has volume


and 36 curved faces, 24 of which are kite-shaped and 12 of which are rhombic (Moore 1974).


December 6, 2007

Surface integral: The symmetric property

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 4:59

Why is \iint_S(x^2+y^2-2z^2)dA=0 for S being the unit sphere.

\displaystyle \int_S x^2\,dA=\int_S y^2\,dA=\int_S z^2\,dA=C,

by rotational symmetry of the sphere. The integral is

\displaystyle C+C-2C=0.

And the easiest way to find C (should you want to) is again by symmetry

\displaystyle 3C=\iint_S(x^2+y^2+z^2)\,dS=\iint_S 1\,dS=4\pi,

hence C=\frac{4\pi}{3}.

December 2, 2007

Đề thi giữa kỳ GT5 của K51 Toán Tin A2

Filed under: Giải Tích 5, Đề Thi — Ngô Quốc Anh @ 23:04



November 30, 2007

Đề thi giữa kỳ GT5 của K51 Toán Tin A3

Filed under: Giải Tích 5 — Ngô Quốc Anh @ 20:52

Mới thi sáng nay. 

Đề thi: ktgk_k51-a3_gt5_2.pdf

Đáp án:


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