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On Ramanujan’s Nested Radicals

Friday, December 30th, 2011 08:50 / 8 Comments

Ramanujan (1887-1920) discovered some formulas on algebraic nested radicals. This article is based on one of those formulas. The main aim of this article is to discuss and derive them intuitively. Nested radicals have many applications in Number Theory as well as in Numerical Methods.

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Four Math Problems

Thursday, September 1st, 2011 11:44 / 2 Comments

Solve the equation
${(\dfrac{1}{10})}^{\log_{\frac{x}{4}} {\sqrt [4] {x} -1}} -4^{\log_{10} {\sqrt [4] {x} +5}} =6, \forall x \ge 1$
The function $f: \mathbb{R} \to \mathbb{R}$ has $n+1$ derivatives. Show that if $a < b$ and $\log [f(b)+f'(b)+f"(b)+ \ldots +f^n(b)] - \log [f(a)+f'(a)+f"(a)+ \ldots +f^n(a)] =b-a$ then we can find $c \in (a,b)$ such that $f^{n+1} (c) = f (c)$ .
Let $A = \{(x,y) : 0 \le < 1 \}$ .
For $(x,y) \in A$ , let $\mathbf{S} (x,y)= \displaystyle{\sum_{\frac{1}{2} \le \frac{m}{n} \le 2}} x^m y^n$ where the sum ranges over all pairs $(m,n)$ of positive integer satisfying the indicated inequalities. Evaluate:
$\displaystyle {\lim_{{(x,y) \to (1,1)}_{(x,y) \in A}}} (1-xy^2)(1-x^2y) \mathbf{S} (x,y)$ .
This problem deals to elementary functional analysis and is taken from very old paper of Putnam Competitions.
$f: [0, \mathbf{N}] \to \mathbf{R}$ has continuous second derivative and $|f'(x)| < 1$ , $f"(x) > 0$ for all $x$ .
$0 \le m_0 < m_1 < m_2 < \ldots < m_k \le \mathbf{N}$ are integers such that $f(m_i)$ are all integers. Put $a_i=m_i-m_{i-1}$ and $b_i=f(m_i)-f(m_{i-1})$ .
•Prove that
$-1 < \frac {b_1}{a_1} < \frac {b_2}{a_2} < \ldots < < \frac {b_k}{a_k} < 1$ .
• Show that for $A > 1$ , there are atmost $\dfrac{\mathbf{N}}{A}$ such indices $i$ .
• Show that there are atmost $3 {(\mathbf{N})}^{2/3}$ lattice points on the curve $y=f(x)$ .

Free Online Algebra Books

Monday, May 2nd, 2011 18:49 / 3 Comments

Internet is full of knowledge. There are many professors who have shared their works online. I have listed a few books on Algebra and Related Mathematics in this article. I am not very serious, but I think these are very useful for Undergraduate and Graduate Students, including me.

Abstract Algebra OnLine by Prof. Beachy
Understanding Algebra by James Brennan
Abstract Algebra : Theory and Applications by Tom Judson
Elements of Abstract and Linear Algebra by E H Connell
Linear Algebra by Jim Hefferon
Elementary Linear Algebra by Keith Matthews
Linear Algebra, Infinite dimensions and Mapleby James Herod
Elementary Number Theory by William Stein
Abstract Algebra: The Basic Graduate Year, A Course in Algebraic Number Theory, and, A Course in Commutative Algebra are three ebooks by Robert Ash and are available here on his website.
A Course in Universal Algebra by Stanley Burris & H. P. Sankappanvar
An Introduction to the Theory of Numbers by Leo Moser
A Computational introduction to Number Theory and Algebra by Victor Shoup
Sets, Relations, Functions by Ivo Duentsch and Günther Gedigo
Group Theory by Pedrag Civitanovic
Linear & Multilinear Algebra by C C Wang & R M Bowen
Abelian Categories by Peter Freyd
Categories and Groupoids by P. J. Higgins
Lie Algebras by Prof. Sternberg

All books, except few are in Acrobat Portable Document Format. You may need acrobat pdf reader to read these. However, if you don’t have that- or hate pdf readers – try online pdf/ps reader at view.samurajdata.se.

This list is expandable. If you know any other book on Algebra which is available online for free, then please give a few seconds and put that into the Comment-Box below, with the link. (It supports HTML.)

Life after Differential Equations (worldscientists.wordpress.com)
The Lie Algebra of a Lie Group (unapologetic.wordpress.com)

A General Problem on functions

Saturday, April 2nd, 2011 09:18 / Leave a Comment

Problem

Let $\mathbb{Z}$ denote the set of all integers (as usually it do ).

Consider a function $f : \mathbb{Z} \rightarrow \mathbb{Z}$ with the following properties:
$f (92+x) = f (92-x)$
$f (19 \times 92+x) = f (19 \times 92 -x)$
$f (1992+x)=f (1992-x)$ for all $x \in \mathbb{Z}$ . Is it possible that all positive divisors of $92$ occur as values of $f$ ?
A happy note: $19 \times 92$ is actually 1748 and it is written to retain symmetry in problem.

Click Here for the Solution

A Problem on Infinite Sum and Recurrence Relations

Friday, March 18th, 2011 20:08 / Leave a Comment

Problem

Consider the infinite sum
$\mathbb{S} = \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} + \dfrac {a_2} {10^4} + .......$ where the sequence $\{a_n\}$ is defined by $a_0=a_1=1$ , and the recurrence relation $a_n=20a_{n-1} + 12 a_{n-2}$ for all positive integers $n \ge 2$ . If $\sqrt {\mathbb {S} }$ can be expressed in the form $\dfrac {a} {\sqrt{b}}$ where $a$ & $b$ are relatively prime positive integers. Determine the ordered pair $(a, \, b)$ .

Solution

As the recurrence relation states $a_n-20a_{n-1} - 12 a_{n-2} =0$ , we shall reform the infinite sum into the same pattern (have a deep look);
$\mathbb{S} - \dfrac {20 \mathbb{S} } {10^2} - \dfrac {12 \mathbb{S} } {10^4}$
$= ( \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} + \dfrac {a_2} {10^4} + \dfrac {a_3} {10^6} + .... )$ $- ( \dfrac {20a_0} {10^2} + \dfrac {20a_1} {10^4} + \dfrac {20a_2} {10^6} + \dfrac {20a_3} {10^8} + .... )$ $- ( \dfrac {12a_0} {10^4} + \dfrac {12a_1} {10^6} + \dfrac {12a_2} {10^8} + \dfrac {12a_3} {10^10} + .... )$

After Simplifying and arranging

$= \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} - \dfrac {20a_0} {10^2} + \dfrac {a_2 -20a_1-12a_0} {10^4} + \dfrac {a_3-20a_2-12a_1} {10^6} + \dfrac {a_4-20a_3-12a_2} {10^8} + . . . . . . \infty$
Now, as {the recurrence relation is}
$a_n-20a_{n-1}-12a_{n-2} =0$ for all $n \ge 2$ , all terms except first three are zero in R.H.S.
Hence we have,
$\mathbb{S} - \dfrac {20 \mathbb{S} } {10^2} - \dfrac {12 \mathbb{S} } {10^4} = \dfrac {a_0} {10^0} + \dfrac {a_1} {10^2} - \dfrac {20a_0} {10^2}$
and substituting $a_0 =a_1=1$ , we have
$\mathbb{S} - \dfrac {20 \mathbb{S} } {100} - \dfrac {12 \mathbb{S} } {10000}$
$= \dfrac {1} {1} + \dfrac {1} {100} - \dfrac {20} {100}$
or
$\dfrac {7988 \mathbb{S}} {10000} = \dfrac {81} {100}$
so,
$\mathbb{S} =2025/1997$
From the Problem,
$\sqrt {\mathbb{S}} = \sqrt{2025/1997} = 45/\sqrt{1997} =a/\sqrt{b}$
So, the desired ordered pair is $(a, b) = (45, 1997)$ .

A College Level Problem on Jensen’s Inequality

Saturday, March 12th, 2011 08:47 / 1 Comment

Problem

Given, $-1 \le a_1 \le a_2 \le ... \le a_n \le 1$ .

Prove that

$\mathbf {\sum_{i=1}^{n-1}}$ $\sqrt {1-a_i a_{i+1} - \sqrt {(1-a_i^2) (1-a_{i+1}^2)}}$ $< \frac {\pi \sqrt {2}} {2}$ .

Solution

It is natural to make trigonometric substitution $a_i= \cos{x_i}$ for some $x_i \in [0, \pi]$ , $i=1,2,. . . . . ,n.$

Note that the monotonicity of the cosine function combined with the given inequalities shows that the $x_i's$ form a decreasing sequence. The expression on the left

$\mathbf {\sum_{i=1}^{n-1}}$ $\sqrt {1-a_ia_{i+1} - \sqrt {(1-a_i^2) (1-a_{i+1}^2)}}$

$= \mathbf {\sum_{i=1}^{n-1}} \sqrt {1- \cos {x_i} \cos {x_{i+1}} - \sin {x_i} \sin {x_{i+1} } }$

$= \mathbf {\sum_{i=1}^{n-1}} \sqrt {1- \cos {(x_{i+1}-x_i)}}$

$= \sqrt{2} \mathbf {\sum_{i=1}^{n-1}} \sin {\frac {x_{i+1}-x_i} {2}}$

Here we used a subtraction and a double-angle formula. The sine function is concave down on $[0, \pi]$ ; hence we can use Jensen’s Inequality to obtain

$\frac {1}{n-1} \mathbf {\sum_{i=1}^{n-1}}$ $\sin {\frac {x_{i+1}-x_i} {2} }$ $\le \sin {(\frac {1}{n-1} \mathbf {\sum_{i=1}^{n-1}} \frac {x_{i+1}-x_i} {2} )}$