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# Tag Archives: Algebra

## Four Math Problems

1. Solve the equation
${(\dfrac{1}{10})}^{\log_{\frac{x}{4}} {\sqrt [4] {x} -1}} -4^{\log_{10} {\sqrt [4] {x} +5}} =6, \forall x \ge 1$
2. 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 $f^{n+1} (c) = f (c)$.
3. 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:
$\displaystyle {\lim_{{(x,y) \to (1,1)}_{(x,y) \in A}}} (1-xy^2)(1-x^2y) \mathbf{S} (x,y)$ .
4. 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

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.

1. Abstract Algebra OnLine by Prof. Beachy
2. Understanding Algebra by James Brennan
3. Abstract Algebra : Theory and Applications by Tom Judson
4. Elements of Abstract and Linear Algebra by E H Connell
5. Linear Algebra by Jim Hefferon
6. Elementary Linear Algebra by Keith Matthews
7. Linear Algebra, Infinite dimensions and Mapleby James Herod
8. Elementary Number Theory by William Stein
9. 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.
10. A Course in Universal Algebra by Stanley Burris & H. P. Sankappanvar
11. An Introduction to the Theory of Numbers by Leo Moser
12. A Computational introduction to Number Theory and Algebra by Victor Shoup
13. Sets, Relations, Functions by Ivo Duentsch and Günther Gedigo
14. Group Theory by Pedrag Civitanovic
15. Linear & Multilinear Algebra by C C Wang & R M Bowen
16. Abelian Categories by Peter Freyd
17. Categories and Groupoids by P. J. Higgins
18. 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.)

## 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.

## A Problem on Infinite Sum and Recurrence Relations

### 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 $(a, b) = (45, 1997)$.

## 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

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

Hence,

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

or,

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

Since,

$x_n-x_i \in (0,\pi)$

Using the fact that $\sin x < x$ for all $x > 0$ yields

$\frac {\sqrt{2}(n-1) \sin \pi} {2(n-1)} \le \frac {\sqrt{2} \pi}{2}$

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