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Integrating polynomials on polygons

September 13, 2022 beni22sof Leave a comment

Given the coordinates ${(x_i,y_i)_{i=1}^n}$ of a polygon ${P}$ , compute the following quantities:

$\displaystyle \text{Area(P)}, \int_P x\ dxdy, \int_P y\ dxdy, \int_P x^2\ dxdy, \int_Py^2\ dxdy.$

Therefore, the area and the centroid of ${P}$ can be computed analytically in terms of the coordinates. More generally, integrals of all polynomials in the coordinates can be evaluated. To convince yourself that this is possible, recall Green’s formula. If ${L,M}$ are functions of ${x}$ and ${y}$ , ${D}$ is a two dimensional domain and ${C}$ is its boundary, then

$\displaystyle \oint_C (Ldx+Mdy) = \iint_D \left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy. \ \ \ \ \ (1)$

In the following I will omit the double integral, for simplicity and write simply ${\int_D}$ . The path integration along ${C}$ is anticlockwise. To compute an integral of the form ${\oint C L dx}$ consider a parametrization ${\gamma : [a,b] \rightarrow C, \gamma(t)=(x(t),y(t))}$ and use the formula

$\displaystyle \oint_C L(x,y) dx = \int_a^b L(x(t),y(t)) x'(t).$

The analogue formula holds for the ${dy}$ variant. Let us now turn to the case of polygons. Consider ${P=A_0...A_{n-1}}$ a polygon, the indices of the vertices being considered modulo ${n}$ whenever necessary. Observe that in order to compute the area of ${P}$ , it is enough to consider ${M(x,y)=x}$ and ${L\equiv 0}$ in (1). The coordinates of ${A_i}$ are denoted ${(x_i,y_i)}$ , ${i=0,...,n-1}$ . We parametrize the segment ${[A_iA_{i+1}]}$ using ${t \mapsto (1-t)(x_i,y_i)+t(x_{i+1},y_{i+1})}$ . Thus we are ready to compute:

$\displaystyle |P| = \oint_{\partial C} x dy = \sum_{i=0}^{n-1} \int_0^1 (x_i+t(x_{i+1}-x_i))dt (y_{i+1}-y_i).$

Computing the integral and regrouping the terms we obtain the familiar formula

$\displaystyle |P| = \frac{1}{2}\sum_{i=0}^{n-1} (x_iy_{i+1}-x_{i+1}y_i).$

We have the following analogue computations:

To compute ${\int_P x\ dxdy}$ use ${M = x^2/2, L\equiv 0}$ . We obtain
$\displaystyle \int_P x\ dx dy = \frac{1}{6} \sum_{i=0}^{n-1}(x_i+x_{i+1})(x_iy_{i+1}-x_{i+1}y_i).$
To compute ${\int_P y\ dxdy}$ use ${M = xy, L\equiv 0}$ . We obtain
$\displaystyle \int_P y\ dx dy = \frac{1}{6} \sum_{i=0}^{n-1}(y_i+y_{i+1})(x_iy_{i+1}-x_{i+1}y_i).$
To compute ${\int_P x^2\ dxdy}$ use ${M = x^3/3, L\equiv 0}$ . We obtain
$\displaystyle \int_P x^2\ dx dy = \frac{1}{6} \sum_{i=0}^{n-1}(x_i^2+x_ix_{i+1}+x_{i+1}^2)(x_iy_{i+1}-x_{i+1}y_i).$
To compute ${\int_P y^2\ dxdy}$ use ${M = xy^2, L\equiv 0}$ . We obtain
$\displaystyle \int_P y^2\ dx dy = \frac{1}{6} \sum_{i=0}^{n-1}(y_i^2+y_iy_{i+1}+y_{i+1}^2)(x_iy_{i+1}-x_{i+1}y_i).$

I guess, up to this point you can convince yourself that the general case ${\int_P x^py^q}$ is just a matter of choosing the right functions in formula (1) and performing the one dimensional integrations.

Categories: Analysis, Curves & Surfaces, Differential Geometry, Geometry, Real Analysis Tags: area, centroid, formula, integral, moment, polygon, polynomial

Function with zero average on vertices of all regular polygons

July 4, 2022 beni22sof Leave a comment

Fix a positive integer ${n \geq 3}$ . Let ${f:\Bbb{R}^2 \rightarrow \Bbb{R}}$ be a function such that for any regular ${n}$ -gon ${A_1...A_n}$ we have

$\displaystyle f(A_1)+...+f(A_n) = 0.$

Prove that ${f}$ is identically equal to zero.

Source: Romanian Team Selection Test 1996, see also the Putnam Contest Problems from 2009.

Solution: The solution comes by looking at some examples:

1. Consider an equilateral triangle ${A_1A_2A_3}$ . It is possible to produce another two equilateral triangles ${A_1B_2B_3}$ and ${A_1C_2C_3}$ such that ${A_2B_2C_2}$ , ${A_3B_3C_3}$ are equilateral. Note that we kept a common vertex and we rotated the initial triangle by ${2\pi/3}$ and ${4\pi/3}$ . Applying the result for all the small triangles and summing we obtain

$\displaystyle 3f(A_1)+... = 0$

where the missing terms are again sums of values of ${f}$ on some equilateral triangles. It follows that ${f(A_1)=0}$ .

2. For a square things are even simpler, since considering rotations of a square around one vertex one ends up with a configuration containing a square, its midpoints and its center. A similar reasoning shows that the value of the function ${f}$ at the center needs to be equal to zero, summing the values of the function ${f}$ on the vertices of all small squares.

In the general case, the idea is the same. Consider an initial polygon ${A_1...A_n}$ and rotate it around ${A_1}$ with angles ${2k\pi/n}$ , ${1\leq k \leq n-1}$ . Then sum all the values of the function on the vertices of these regular polygons. Observe that the vertex ${A_1}$ is repeated ${n}$ times while all other vertices are part of some regular polygon. In the end we get

$\displaystyle n f(A_1)+0 = 0$

where the zeroes correspond to sums over vertices of regular polygons.

The same type of reasoning should hold when the sum over vertices of regular polygons is replaced by an integral on a circle. The proof would follow the same lines. Fix a point ${A}$ , then integrate ${f}$ on all rotations of the circle ${C}$ through ${A}$ . On one side this integral should be equal to zero. On the other it contains the value of ${f}$ in ${A}$ and values on concentric circles in ${A}$ . This should imply that ${f(A)}$ is zero for any point ${A}$ .

Categories: Geometry, IMO, Olympiad, puzzles Tags: IMO, integral, olympiad, regular polygons

IMC 2016 – Day 2 – Problem 7

July 28, 2016 beni22sof Leave a comment

Problem 7. Today, Ivan the Confessor prefers continuous functions ${f:[0,1]\rightarrow \Bbb{R}}$ satisfying ${f(x)+f(y) \geq |x-y|}$ for all ${x,y \in [0,1]}$ . Fin the minimum of ${\int_0^1 f}$ over all preferred functions.

Categories: Analysis, Inequalities, Uncategorized Tags: IMC, Inequalities, integral, minimal, optimization

SEEMOUS 2016 Problem 4 – Solution

March 6, 2016 beni22sof 3 comments

Problem 4. Let ${n \geq 1}$ be an integer and set

$\displaystyle I_n = \int_0^\infty \frac{\arctan x}{(1+x^2)^n}dx.$

Prove that

a) $\sum_{i=1}^\infty \frac{I_n}{n} =\frac{\pi^2}{6}.$

b) $\int_0^\infty \arctan x \cdot \ln \left( 1+\frac{1}{x^2}\right) dx = \frac{\pi^2}{6}$ .

Categories: Analysis, Olympiad, Real Analysis, Uncategorized Tags: Analysis, integral, olympiad, SEEMOUS, uniform convergence

Variations on Fatou’s Lemma – Part 2

November 29, 2014 beni22sof Leave a comment

As we have seen in a previous post, Fatou’s lemma is a result of measure theory, which is strong for the simplicity of its hypotheses. There are cases in which we would like to study the convergence or lower semicontinuity for integrals of the type $\int_\Omega f_ng_n$ where ${f_n}$ converges pointwise to ${f}$ and ${g_n}$ converges to ${g}$ in some fashion, but not pointwise. For example, we could have that ${g_n}$ converges to ${g}$ in ${L^1}$ . In this case we could write the integral $\int_\Omega f_ng_n$ as $\int_\Omega f d\mu_n$ where ${\mu_n}$ is the measure defined by ${\mu_n(A) = \displaystyle\int_A g_n}$ . All measures considered in this post will be positive measures.

Certain hypotheses on the measures ${\mu_n,\mu}$ allow us to find a result similar to Fatou’s Lemma for varying measures. In the following, we define a type of convergence for the measures ${\mu_n,\mu}$ , named setwise convergence, which will allow us to prove the lower semicontinuity result. We say that ${\mu_n}$ converges setwise to ${\mu}$ if ${\mu_n(A) \rightarrow \mu(A)}$ for every measurable set ${A}$ . The following proof is taken from Royden, H.L., Real Analysis, Chapter 11, Section 4. It is very similar to the proof of Fatou’s lemma given here.

Theorem A. Let ${\mu_n}$ be a sequence of measures defined on ${\Omega}$ which converges setwise to a measure ${\mu}$ and ${(f_n)}$ a sequence of nonnegative measurable functions which converge pointwise (or almost everywhere in ${\Omega}$ ) to the function ${f}$ . Then

$\displaystyle \int_\Omega f d\mu \leq \liminf_{n \rightarrow \infty} \int_\Omega f_n d\mu_n.$

Categories: Calculus of Variations, Measure Theory, Real Analysis Tags: convergence, integral, measurable, measure

Variations on Fatou’s Lemma – Part 1

November 13, 2014 beni22sof 1 comment

One of the most basic and powerful results of measure theory is Fatou’s Lemma. It provides information on the convergence of the integrals of functions which converge pointwise. The power of this lemma is the simplicity of its hypothesis. One of its interpretations is that the functional which to a positive function associates its integral is lower semicontinuous with respect to the pointwise convergence. Among other things, it allows us to prove the monotone convergence theorem, or Lebesgue’s dominated convergence theorem in just one line.

Fatou’s Lemma Suppose ${(X,\mathcal{M},\mu)}$ is a measure space and ${f_n : X \rightarrow [0,\infty]}$ is a sequence of integrable functions converging pointwise ${\mu}$ almost everywhere to ${f}$ . Then ${f}$ is measurable and

$\displaystyle \int_X f d\mu \leq \liminf_{n \rightarrow \infty} \int_X f_n d\mu.$

Proof: One of the classical proofs of this result goes as follows. Using the definition

$\displaystyle \int_X fd\mu = \sup\{ \int_X \varphi : \varphi \leq f,\ \varphi \text{ is a simple function} \}$

we note that it is enough to prove that

$\displaystyle \int_X \varphi d\mu \leq \liminf_{n \rightarrow \infty} \int_X f_n d\mu,$

for every simple function ${\varphi \leq f}$ .

Pick ${\varphi\leq f}$ and ${\varepsilon>0}$ . Suppose ${\int_X \varphi d\mu<\infty}$ . Denote ${E_n = \{x : (1-\varepsilon)\varphi \leq f_k(x),\ \text{ for every }k\geq n\}}$ . Note that ${E_n \subset E_{n+1}}$ and ${X = \bigcup_{n \geq 1} E_n}$ . If we denote ${E}$ the support of ${\varphi}$ , we may choose ${n_0}$ such that for every ${n \geq n_0}$ we have ${\mu(E\setminus E_n)<\varepsilon}$ . Thus

$\displaystyle \int_X f_k d\mu_k \geq \int_{E_k} f_k d\mu \geq (1-\varepsilon)\int_{E_k}\varphi d \mu \geq (1-\varepsilon) \int_X \varphi d\mu - (1-\varepsilon) \int_{E\setminus E_k} \varphi d\mu.$

Since ${\varphi}$ is bounded above by a constant ${M<\infty}$ and ${\mu(E\setminus E_k)<\varepsilon}$ , we have

$\displaystyle \int_X f_k d\mu \geq (1-\varepsilon) \int_E \varphi d\mu -(1-\varepsilon)\varepsilon M.$

Taking ${k \rightarrow \infty}$ we have

$\displaystyle \liminf_{k \rightarrow \infty} \int_X f_k d\mu \geq (1-\varepsilon) \int_E \varphi d\mu -(1-\varepsilon)\varepsilon M.$

Take ${\varepsilon \rightarrow 0}$ and then supremum over ${\varphi\leq f}$ to finish the proof. \hfill ${\square}$

Monotone convergence Suppose that ${f_k}$ is a sequence of function defined on ${X}$ such that

$\displaystyle 0 \leq f_k(x) \leq f_{k+1}(x),$

for every ${k \geq 1}$ and ${x \in X}$ . Denote ${f(x) = \lim_{k \rightarrow \infty} f_k(x)}$ . Then ${f}$ is measurable and

$\displaystyle \int_X fd\mu = \lim_{n \rightarrow \infty} \int_X f_n d\mu.$

Proof: The ${\limsup}$ comes from the monotonicity, while the ${\liminf}$ is just Fatou’s Lemma.

Dominated convergence Suppose ${f_k}$ is a sequence of measurable functions defined on ${X}$ which converges almost everywhere to ${f}$ . Moreover, suppose there exists an integrable function ${g}$ such that ${|f_n|\leq g}$ for every ${n}$ . Then

$\displaystyle \int_X fd\mu = \lim_{n \rightarrow \infty} \int_X f_n d\mu.$

Proof: Apply Fatou’s lemma to ${g_n = 2g-|f_n-f|\geq 0}$ .

It is straightforward to see that the positivity condition can be replaced with a domination from below. If ${f_n \geq -g}$ and ${g}$ is integrable, then apply Fatou’s lemma to ${f_n+g}$ to deduce that the conclusion still holds for ${(f_n)}$ . One may ask what happens with the ${\limsup}$ part. If ${ f_n \leq g}$ with ${g}$ integrable, then the reverse Fatou Lemma works. To see this, apply Fatou’s lemma to ${g-f_n \geq 0}$ .

Of course, every one of the few hypotheses made for the Fatou’s lemma are essential. If we allow negative functions, then ${f_n = -n \chi_{[0,1/n]}}$ is a counterexample. ${f_n}$ converges to zero almost everywhere, while the integrals of ${f_n}$ are equal to ${-1}$ . The inequality can be strict. Consider the similar functions ${f_n = n \chi_{[0,1/n]}}$ .

Seeing that the hypotheses needed for Fatou lemma to hold are simple enough, we may try and see which generalizations of Fatou lemma hold? One interesting direction is to consider not only variable functions, but also variable measures. This will be treated in a future post.

Categories: Measure Theory, Real Analysis Tags: Analysis, convergence, fatou, integral, semicontinuity

Generalized version of Lebesgue dominated convergence theorem

October 10, 2014 beni22sof 5 comments

The following variant of the Lebesgue dominated convergence theorem may be useful in the case we can not dominate a sequence of functions by only one integrable function, but by a convergent sequence of integrable functions.

Let ${(f_n)}$ be a sequence of measurable functions defined on a measurable set ${\Omega}$ with real values, which converges pointwise almost everywhere to ${f}$ . Suppose that there exists a sequence of positive, integrable functions ${g_n}$ such that the two following conditions hold:

${|f_n(x)|\leq g_n(x)}$ for almost every ${x \in \Omega}$ ;
There exists an integrable function ${g}$ such that ${(g_n) \rightarrow g}$ in ${L^1(\Omega)}$ .

Then

$\displaystyle \lim_{n \rightarrow \infty} \int_\Omega f_n = \int_\Omega f$

Categories: Analysis, Measure Theory, Real Analysis Tags: convergence, integral, lebesgue, measurable

SEEMOUS 2014

March 18, 2014 beni22sof Leave a comment

Problem 1. Let ${n}$ be a nonzero natural number and ${f:\Bbb{R} \rightarrow \Bbb{R}\setminus \{0\}}$ be a function such that ${f(2014)=1-f(2013)}$ . Let ${x_1,..,x_n}$ be distinct real numbers. If

$\displaystyle \left| \begin{matrix} 1+f(x_1)& f(x_2)&f(x_3) & \cdots & f(x_n) \\ f(x_1) & 1+f(x_2) & f(x_3) & \cdots & f(x_n)\\ f(x_1) & f(x_2) &1+f(x_3) & \cdots & f(x_n) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ f(x_1)& f(x_2) & f(x_3) & \cdots & 1+f(x_n) \end{matrix} \right|=0$

prove that ${f}$ is not continuous.

Problem 2. Consider the sequence ${(x_n)}$ given by

$\displaystyle x_1=2,\ \ x_{n+1}= \frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2},\ n \geq 2.$

Prove that the sequence ${y_n = \displaystyle \sum_{k=1}^n \frac{1}{x_k^2-1} ,\ n \geq 1}$ is convergent and find its limit.

Problem 3. Let ${A \in \mathcal{M}_n (\Bbb{C})}$ and ${a \in \Bbb{C},\ a \neq 0}$ such that ${A-A^* =2aI_n}$ , where ${A^* = (\overline A)^t}$ and ${\overline A}$ is the conjugate matrix of ${A}$ .

(a) Show that ${|\det(A)| \geq |a|^n}$ .

(b) Show that if ${|\det(A)|=|a|^n}$ then ${A=aI_n}$ .

Problem 4. a) Prove that $\lim_{n \rightarrow \infty} n \int_0^n \frac{\arctan(x/n)}{x(x^2+1)}dx=\frac{\pi}{2}$ .

b) Find the limit $\lim_{n \rightarrow \infty} n\left(n \int_0^n \frac{\arctan(x/n)}{x(x^2+1)}dx-\frac{\pi}{2} \right)$

Categories: Algebra, Analysis, Linear Algebra, Olympiad Tags: Algebra, continuity, integral, limit, matrix, SEEMOUS, sequence

SEEMOUS 2012 Problem 4

March 8, 2012 beni22sof Leave a comment

a) Compute $\displaystyle \lim_{n \to \infty} n \int_0^1 \left(\frac{1-x}{1+x} \right)^n dx$ .

b) Let $k \geq 1$ be an integer. Compute

$\displaystyle\lim_{n \to \infty}n^{k+1}\int_0^1 \left( \frac{1-x}{1+x}\right)^n x^k dx$ .

Categories: Analysis, Olympiad, Problem Solving Tags: integral, limit

Integrals

May 5, 2011 beni22sof Leave a comment

a) Let $A=(a_{ij})_{i,j=1}^n$ be a real matrix. Show that

$\displaystyle \int_{|x|<1} (x,Ax) dx = \frac{\omega_n}{n(n+2)} tr(A),$ where $(\cdot ,\cdot)$ is the usual dot product and $\omega_n$ is the area of the unit sphere.

b) Show that for all $u \in C_0^2 (\Bbb{R}^n)$ we have

$\displaystyle \int_{\Bbb{R}^n} (\Delta u)^2 dx =\sum_{i,j=1}^n \int_{\Bbb{R}^n} |D_{ij}u|^2 dx.$

PHD Iowa (6101)

Categories: Analysis Tags: Analysis, integral

Useful continuity

February 18, 2011 beni22sof Leave a comment

Suppose $f: \Bbb{R}^p \to \Bbb{R}$ is integrable and bounded. Prove that the mapping
$(\Bbb{R}^p)^n \ni (a_1,a_2,...,a_n) \mapsto \displaystyle \int_{\Bbb{R}^p}f(x+a_1)f(x+a_2)...f(x+a_n)f(x)dx$ is continuous.

Then use this result to solve the following problem

Categories: Analysis, Measure Theory Tags: continuity, continuous, integral

p-integrable product

December 13, 2010 beni22sof 1 comment

Let $(X,\mathcal{M},\mu)$ be a fixed measure space, and $p_i,\ 1\leq i\leq n$ be positive nmbers such that $p_i >1,\ 1\leq i\leq n$ and $\displaystyle \sum_{i=1}^n\frac{1}{p_i}=\frac{1}{p}$ , with $p>1$ (works for $p=1$ also).
For each $i \in \{1,2,...,n\}$ consider $f_i \in \mathcal{L}^{p_i}(\mu)$ . Is it true that $f_1f_2...f_n \in \mathcal{L}^p(\mu)$ ?
PHD 4302
Read more…

Categories: Analysis, Measure Theory, Problem Solving Tags: integral, measure

Hard & Easy Inequality

November 18, 2009 beni22sof 1 comment

Let $a_i,\ i=1..n$ be positive numbers with their sum equal to 1. Prove that $\displaystyle \sum_{i=1}^n \frac{a_i}{1+a_1+...+a_i} < \frac{1}{\sqrt{2}}$ .
Read more…

Categories: Analysis, Inequalities, Problem Solving Tags: Inequalities, integral, olympiad

Beni Bogoşel's blog

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Integrating polynomials on polygons

Function with zero average on vertices of all regular polygons

IMC 2016 – Day 2 – Problem 7

SEEMOUS 2016 Problem 4 – Solution

Variations on Fatou’s Lemma – Part 2

Variations on Fatou’s Lemma – Part 1

Generalized version of Lebesgue dominated convergence theorem

SEEMOUS 2014

SEEMOUS 2012 Problem 4

Integrals

Useful continuity

p-integrable product

Hard & Easy Inequality

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