Ngô Quốc Anh

Tháng Chín 26, 2009

How to calculate limit by using definition of definite integral?

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 14:23

Let f : [a, b] \to \mathbb R be a continuous function, not necessarily nonnegative. Partition [a, b] into n consecutive sub-intervals [x_{i-1}, x_i] (i = 1, 2, ..., n) each of length \Delta x = \frac{b-a}{n}, where we set a=x_0, b=x_n and x_1, x_2,...,x_{n-1} to be successive points between a and b with x_k-x_{k-1}=\Delta x. Let c_k be any intermediate point in the sub-interval [x_{k-1},x_k]. Then the sum

\displaystyle\sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x}

is called a Riemann sum for f on [a, b].

Suppose we let the number of partition in P tends to infinity.

\displaystyle\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x} = I.

We call I the Riemann integral (or definite integral) of f over [a, b] and we write

\displaystyle I = \int_a^b {f(x)dx} .

In other words,

\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)}

if the limit on the right side exists.

If we put c_k=x_{k-1} we the obtain

\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + (k - 1)\frac{{b - a}}{n}} \right)} .

Example 1. Find

\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right).

Solution. Clearly

\displaystyle\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}} = \sum\limits_{k = 1}^n {\frac{1}{{n + (k - 1)}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + \frac{{k - 1}}{n}}}} .

Then if we choose a=0, b=1 we then get

\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \int_0^1 {\frac{{dx}}{{1 + x}}} .

With this it is easy to see that

\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \ln 2

since

\displaystyle\int_0^1 {\frac{{dx}}{{1 + x}}} = \ln 2.

If we put c_k=x_k we the obtain

\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + k\frac{{b - a}}{n}} \right)} .

Example 2. Find

\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {{(n - 1)}^2}}} + \frac{n}{{{n^2} + {n^2}}}} \right).

Solution. Clearly,

\displaystyle\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}}

which yields

\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}} \right) = \int_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \frac{\pi }{4}.

Remark. It is worth mentioning that in general it is not true that

\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right).

For example, we all know that for each fixed k

\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}} = 0

but

\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{n + k}}} } \right) = \ln 2 \ne 0 = \sum\limits_{k = 1}^n {\left( {\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}}} \right)} .

The point is

\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right)

holds true only for finite summation.

Tháng Chín 13, 2009

QE in Department of Mathematics, National University of Singapore, August 2009

Chuyên mục: Giải Tích Cổ Điển, Giải Tích Hiện Đại, Linh Tinh, Nghiên Cứu Khoa Học, Đề Thi — Ngô Quốc Anh @ 14:53

I have just passed QE held in August 2009 for my first attendance, I hereby show you the analysis paper

Question 1 [10 marks]. Suppose f and g are both measurable functions on the interval (0,1) such that for all t \in \mathbb R^1

| \{ x \in(0,1):f(x)\geq t\}| = |\{ x\in (0,1):g(x)\geq t\}|

Assume that f and g both are monotone decreasing and continuous from left. Can you conclude that f(x)=g(x) for all x \in (0,1)? Give the reason to support your answer.

Question 2 [10 marks]. Compute the volume of the region bounded by

{\left( {{a_{11}}x + {a_{12}}y + {a_{13}}z} \right)^2} + {\left( {{a_{21}}x + {a_{22}}y + {a_{23}}z} \right)^2} + {\left( {{a_{31}}x + {a_{32}}y + {a_{33}}z} \right)^2} = 1

where the determinant of the 3 \times 3 matrix (a_{ij}) is NOT equal to zero.

Question 3 [10 marks]. Let D be a measureable set in \mathbb R^n with finite measure. Suppose \phi(x,t) is a real valued continuous function on D \times \mathbb R^1 such that for almost every x \in D, \phi(x,t) is a continuous function of t and for every real number t, \phi(x,t) is measurable function of x. If \{f_n\} is a sequence of measurable functions on D that converges to f in measure, show that \{\phi(x,f_n(x))\} converges to \phi(x,f(x)) in measure.

Question 4 [10 marks]. Find the function

I\left( y \right) =\displaystyle\int\limits_0^\infty {{e^{ - a{x^2}}}\cos \left( {yx} \right)dx}

if a>0 is a constant. Justify your answer.

Question 5 [10 marks]. Compute the intergal

\displaystyle\int\limits_0^\pi {\frac{{x\sin x}}{{1 + {a^2} - 2a\cos x}}dx}

where a>0 is a constant.

Question 6 [10 marks]. Supposet f(z) is a holomorphic function on the complex plane \mathbb C. If f locally keeps the area invariant, what will the function f be?

Question 7 [10 marks]. Is there an analytic function f on \Delta (unit disk in the complex plane with center 0) such that |f(z)|<1 for |z|<1 with f(0)=\frac{1}{2} and f'(0)=\frac{3}{4}? If so, find such an f. Is it unique?

Question 8 [10 marks]. Let m<n be two positive integers and \Omega and G be open subsets in \mathbb R^n and \mathbb R^m respectively. Does there exist a map f :\Omega \to G such that f and the inverse of f are both C^1? Justify your answer.

Question 9 [10 marks]. Is there a square integrable function f on [0,\pi] such that both inequalities

\displaystyle\int\limits_0^\pi{{{\left( {f\left( x \right)-\sin x} \right)}^2}dx}\leq\frac{4}{9}

and

\displaystyle\int\limits_0^\pi{{{\left( {f\left( x \right)-\cos x} \right)}^2}dx}\leq\frac{1}{9}

hold? Justify your answer.

Question 10 [10 marks]. Let \alpha_k for k=1,2,...,n be n real numbers such that 0<\alpha_k<\pi for any k. Define

\alpha=\displaystyle\frac{1}{n}\sum\limits_{k=1}^{n}{\alpha_k}.

Show that

\displaystyle{\left( {\prod\limits_{k = 1}^n {\frac{{\sin {\alpha _k}}}{{{\alpha _k}}}} } \right)^{\frac{1}{n}}} \leq \frac{{\sin \alpha }}{\alpha }.

Tháng tám 18, 2009

An other limit supremum of sin function

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 12:35

In the topic we showed that for any irrational \alpha the limit \mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right) does not exist. In this topic, we consider the following limit

\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) .

To be precise, we prove that

\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1

for almost every x \in [0,2\pi].

Solution. Let

A = \left\{ {x \in \left( {0,2\pi } \right): \frac{x} {\pi } \notin \mathbb{Q}} \right\}.

Then A is a measurable set of measure 2\pi. Moreover, for any x \in A,

\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1.

Indeed for any x \in A, since

\left\{ {k\frac{x} {\pi } - 2l: l \in \mathbb{Z}} \right\}

is dense subgroup of \mathbb R there are sequences \{k_n\} and \{l_n\} of \mathbb Z such that

\mathop {\lim }\limits_{n \to \infty } \left( {{k_n}\frac{x} {\pi } - {l_n}} \right) = \frac{1} {2}.

Since

\frac{1} {2} \notin \left\{ {k\frac{x} {\pi } - 2l: k,l \in \mathbb{Z}} \right\}

\{k_n\} admits a subsequence \{k'_n\} either increasing to +\infty or decreasing to -\infty. If \mathop {\lim }\limits_{n \to \infty } {{k'}_n} = + \infty then

\mathop {\lim }\limits_{n \to \infty } \sin \left( {{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \left( {...

Otherwise \mathop {\lim }\limits_{n \to \infty } \left( { - 3{{k'}_n}} \right) = + \infty and

\mathop {\lim }\limits_{n \to \infty } \sin \left( { - 3{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \lef...

Tháng Một 31, 2009

Problems in Real Analysis: Advanced Calculus on the Real Axis

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích Cổ Điển, Sách Hay — Ngô Quốc Anh @ 15:24

Problems in Real Analysis: Advanced Calculus on the Real Axis features a comprehensive collection of challenging problems in mathematical analysis that aim to promote creative, nonstandard techniques for solving problems. This self-contained text offers a host of new mathematical tools and strategies which develop a connection between analysis and other mathematical disciplines, such as physics and engineering. A broad view of mathematics is presented throughout; the text is excellent for the classroom or self-study. It is intended for undergraduate and graduate students in mathematics, as well as for researchers engaged in the interplay between applied analysis, mathematical physics, and numerical analysis.

problems-in-real-analysis-advanced-calculus-on-the-real-axis-cover

Key features:

  • Uses competition-inspired problems as a platform for training typical inventive skills;
  • Develops basic valuable techniques for solving problems in mathematical analysis on the real axis and provides solid preparation for deeper study of real analysis;
  • Includes numerous examples and interesting, valuable historical accounts of ideas and methods in analysis;
  • Offers a systematic path to organizing a natural transition that bridges elementary problem-solving activity to independent exploration of new results and properties.

Tháng Mười Một 6, 2008

Differentiability via weird argument

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 2 — Ngô Quốc Anh @ 16:37

Question. Suppose that is such that is continuous and is differentiable. Should also be differentiable?

Proof. Let . Since satisfies the differentiable equation , is differentiable at all points with by the implicit function theorem: , and . Where , we note that has a minimum and This gives \lim_{t\to 0}\left|\frac{f(x+t)}{t}\right|=0 and is differentiable by the definition.

Comment. The only points of interests are the zeros of since has the same sign in some neighborhood of points which are not roots. So there goes half the work. Geometrically it doesn’t make sense for to have anything but derivative 0 at roots of and this can be verified by taking left hand and right hand limits of the quotient.

Tháng Chín 17, 2008

L^2 differentiable?

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 3 — Ngô Quốc Anh @ 13:04

Let and define .

a) Must be differentiable at 0?
b) Must have any differentiable points?
c) Let , show that exists and determine what it is.

Solutions.

a) No. For example, let for so that near zero. This but does not exist.

b) Yes. In fact, must be differentiable almost everywhere, by the Lebesgue theorem on the differentiation of the integral. This theorem requires only that which is true.

c) By the Schwarz inequality,

f^2(x) = \left(\int_0^x}g(t)\,dt\right)^2\le \left(\int_0^x 1\,dt\right) \left(\int_0^xg^2(t)\,dt\right).

At least that’s it for Being careful about the other side, we determine that

0\le\phi(x) = f^2(x)\le|x|\left|\int_0^xg^2(t)\,dt\right|.

But since is an integrable function we have (by an argument that uses the Dominated Convergence Theorem) that

\lim_{x\to 0}\int_0^xg^2(t)\,dt = 0.

Hence \lim_{x\to0}\frac {\phi(x)}{x} = 0, so

Lời giải của Euler cho bài toán Basel về tính tổng của một chuỗi số

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 4 — Ngô Quốc Anh @ 12:50

The Basel problem is a famous problem in number theory, first posed by Pietro Mengoli in 1644, and solved by Leonhard Euler in 1735. Since the problem had withstood the attacks of the leading mathematicians of the day, Euler’s solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper On the Number of Primes Less Than a Given Magnitude, in which he defined his zeta function and proved its basic properties. The problem is named after Basel, hometown of Euler as well as of the Bernoulli family, who unsuccessfully attacked the problem.

The Basel problem asks for the precise summation of the reciprocals of the squares of the natural numbers, i.e. the precise sum of the infinite series:

\displaystyle\sum_{n=1}^\infty \frac{1}{n^2} = \lim_{n \to +\infty}\left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}\right).

The series is approximately equal to 1.644934 (sequence A013661 in OEIS). The Basel problem asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be \frac{\pi^2}{6} and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, and it was not until 1741 that he was able to produce a truly rigorous proof.

Proof and comments. Euler’s original “derivation” of the value \frac{\pi^2}{6} is clever and original. He essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series. Of course, Euler’s original reasoning requires justification, but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler’s argument, recall the Taylor series expansion of the sine function

\displaystyle\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.

Dividing through by x, we have

\displaystyle\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots.

Now, the roots (zeros) of \frac{\sin x}{x} occur precisely at x = n\cdot\pi where n = \pm1, \pm2, \pm3, \dots. Let us assume we can express this infinite series as a product of linear factors given by its roots, just as we do for finite polynomials

\displaystyle\begin{gathered}\frac{{\sin (x)}}{x} = \left( {1 - \frac{x}{\pi }} \right)\left( {1 + \frac{x}{\pi }} \right)\left( {1 - \frac{x}{{2\pi }}} \right)\left( {1 + \frac{x}{{2\pi }}} \right)\left( {1 - \frac{x}{{3\pi }}} \right)\left( {1 + \frac{x}{{3\pi }}} \right) \cdots\hfill \\ \qquad \quad= \left( {1 - \frac{{{x^2}}}{{{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{4{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{9{\pi ^2}}}} \right) \cdots\hfill \\ \end{gathered}

If we formally multiply out this product and collect all the x^2 terms, we see that the x^2 coefficient of \frac{\sin x}{x} is

\displaystyle -\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.

But from the original infinite series expansion of \frac{\sin x}{x}, the coefficient of x^2 is -\frac{1}{3!} = -\frac{1}{6}. These two coefficients must be equal; thus,

\displaystyle -\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.

Multiplying through both sides of this equation by -\pi^2 gives the sum of the reciprocals of the positive square integers

\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.

SOURCE: http://en.wikipedia.org/wiki/Basel_problem#Euler_attacks_the_problem

Tháng Chín 11, 2008

Một bài giới hạn khó về hàm lượng giác

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 22:02

Show that for any irrational \alpha the limit

\displaystyle\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)

does not exist.

Solution. If the limit existed then we would get

\displaystyle 0 = \mathop {\lim }\limits_{n \to \infty } \left( {\sin \left( {\left( {n + 2} \right)\alpha \pi } \right) - \sin \left( {n\alpha \pi } \right)} \right) = 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)

and consequently, \mathop {\lim }\limits_{n \to \infty } \cos \left( {n\alpha \pi } \right) = 0. Similarly,

\displaystyle0 = \mathop {\lim }\limits_{n \to \infty } \left( {\cos \left( {\left( {n + 2} \right)\alpha \pi } \right) - \cos \left( {n\alpha \pi } \right)} \right) = - 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)

which is impossible because \sin^2x+\cos^2x=1 for all x \in \mathbb R. Therefore the limit does not exist.

Tháng tám 9, 2008

Tính tích phân suy rộng loại I

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 4 — Ngô Quốc Anh @ 13:16

Tính

\displaystyle \int_{0}^{+\infty}{\frac{e^{-x^2}}{\left(x^2+\frac{1}{2}\right)^2}\; dx}.

Lời giải. Đặt

\displaystyle I(\alpha) = - \int_{0}^{\infty} \frac {e^{ - x^2}}{x^2 + \alpha} \, dx.

Khi đó ta có

\displaystyle\begin{gathered} \int_0^\infty {\frac{{{e^{ - {x^2}}}}}{{{x^2} + \alpha }}} \,dx = \int_0^\infty {{e^{ - {x^2}}}} \int_0^\infty {{e^{ - ({x^2} + \alpha )t}}} \,dtdx \hfill \\\qquad \qquad \qquad \; \,= \int_0^\infty {{e^{ - \alpha t}}} \int_0^\infty {{e^{ - (1 + t){x^2}}}} \,dxdt \hfill \\ \qquad \qquad \qquad \; \,= \int_0^\infty {\frac{{\sqrt \pi }}{2}} \frac{{{e^{ - \alpha t}}}}{{\sqrt {1 + t} }}\,dt, \hfill \\ \end{gathered}

ta thấy

\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = I'(\tfrac{1}{2}) = \frac {\sqrt {\pi}}{2} \int_{0}^{\infty} \frac {t}{\sqrt {1 + t}} \, e^{ - \frac {t}{2}} \, dt.

Nhưng

\displaystyle \begin{gathered} \int_0^\infty {\frac{t}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt = \int_0^\infty {\sqrt {1 + t} } \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= \left[ { - 2\sqrt {1 + t} \,{e^{ - \frac{t}{2}}}} \right]_0^\infty + \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= 2. \hfill \\\end{gathered}

Vậy

\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = \sqrt {\pi}.

Tháng Sáu 5, 2008

An application of Chebyshev integral inequality

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 5:06

From http://mathworld.wolfram.com/ChebyshevIntegralInequality.html by setting , , , f_1 (x) = f_2(x) = x^\frac {x}{2} we get

\sqrt {\int^2_1x^x\,dx} \geq \int^2_1\sqrt {x^x}\,dx

It’s easy to prove that is nonnegative and monotonic increasing.

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