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Proofs of Irrationality

Tuesday, February 14th, 2012 19:18 / 3 Comments

“Irrational numbers are those real numbers which are not rational numbers!”

A rational number is a real number which can be expressed in the form of $\frac{a}{b}$ where $a$ and $b$ are both integers relatively prime to each other and $b$ being non-zero.
Following two statements are equivalent to the definition 1.
1. $x=\frac{a}{b}$ is rational if and only if $a$ and $b$ are integers relatively prime to each other and $b$ does not equal to zero.
2. $x=\frac{a}{b} \in \mathbb{Q} \iff \mathrm{g.c.d.} (a,b) =1, \ a \in \mathbb{Z}, \ b \in \mathbb{Z} \setminus \{0\}$ .

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The Area of a Disk

Friday, January 27th, 2012 16:22 / 1 Comment

~~[This post is under review.]~~

If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius $R$ is $\pi R^2$ .

The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘measure of surface‘ surrounded by the round edge (circumference) of the disk.

The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings -sum them till whole disk is obtained. (more…)

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 $y=f(x)$ .

Everywhere Continuous Non-derivable Function

Thursday, July 7th, 2011 13:12 / 2 Comments

Weierstrass had drawn attention to the fact that there exist functions which are continuous for every value of $x$ but do not possess a derivative for any value. We now consider the celebrated function given by Weierstrass to show this fact. It will be shown that if

$f(x)= \displaystyle{\sum_{n=0}^{\infty} } b^n \cos (a^n \pi x) \ \ldots (1) \\ = \cos \pi x +b \cos a \pi x + b^2 \cos a^2 \pi x+ \ldots$ where $a$ is an odd positive integer, $0 < b <1$ and $ab > 1+\frac{3}{2} \pi$ , then the function $f$ is continuous $\forall x$ but not finitely derivable for any value of $x$ .

G.H. Hardy improved this result to allow $ab \ge 1$ .

We have $|b^n \cos (a^n \pi x)| \le b^n$ and $\sum b^n$ is convergent. Thus, by Wierstrass’s $M$ -Test for uniform Convergence the series (1), is uniformly convergent in every interval. Hence $f$ is continuous $\forall x$ .
Again, we have $\dfrac{f(x+h)-f(x)}{h} = \displaystyle{\sum_{n=0}^{\infty}} b^n \dfrac{\cos [a^n \pi (x+h)]-\cos a^n \pi x}{h} \ \ \ldots (2)$
Let, now, $m$ be any positive integer. Also let $S_m$ denote the sum of the $m$ terms and $R_m$ , the remainder after $m$ terms, of the series (2), so that
$\displaystyle{\sum_{n=0}^{\infty}} b^n \dfrac{\cos [a^n \pi (x+h)]-\cos a^n \pi x}{h} = S_m+R_m$ . By Lagrange’s mean value theorem, we have
$\dfrac{|\cos {[a^n \pi (x+h)]} -\cos {a^n \pi x|}}{|h|}=|a^n \pi h \sin {a^n \pi(x+\theta h)}| \le a^n \pi |h|$ ,
$|S_m| \le \displaystyle{\sum_{n=0}^{m-1}} b^n a^n \pi = \pi \dfrac {a^m b^m -1}{ab-1} < \pi \dfrac {a^m b^m}{ab-1}$ . We shall now consider $R_m$ .
So far we have taken $h$ as an arbitrary but we shall now choose it as follows:

We write $a^m x=\alpha_m+\xi_m$ , where $\alpha_m$ is the integer nearest to $a^m x$ and $-1/2 \le \xi_m < 1/2$ .
Therefore $a^m(x+h) = \alpha_m+\xi_m+ha^m$ . We choose, $h$ , so that $\xi_m+ha^m=1$
i.e., $h=\dfrac{1-\xi_m}{a^m}$ which $\to 0 \ \text{as} \ m \to \infty$ for $0< h \le \dfrac{3}{2a^m} \ \ldots (3)$
Now, $a^n \pi (x+h) = a^{n-m} a^m (x+h.) \\ \ =a^{n-m} \pi [(\alpha_m +\xi_m)+(1-\xi_m)] \\ \ =a^{n-m} \pi(\alpha_m+1)$

Thus $\cos[a^n \pi (x+h)] =cos [a^{n-m} (\alpha_m-1) \pi] =(-1)^{\alpha_{m+1}}$ .
$\cos (a^n \pi x) = \cos [a^{n-m} (a^m \pi x)] \\ \ =\cos [a^{n-m} (\alpha_m+\xi_m) \pi] \\ \ =\cos a^{n-m} \alpha_m \pi \cos a^{n-m} \xi_m \pi - \sin a^{n-m} \alpha_m \pi \sin a^{n-m} \xi_m \pi \\ \ = (-1)^{\alpha_m} \cos a^{n-m} \xi_m \pi$ for $a$ , is an odd integer and $\alpha_m$ is an integer.

Therefore, $R_m =\dfrac{(-1)^{\alpha_m}+1}{h} \displaystyle{\sum_{n=m}^{\infty}} b^n [2+\cos (a^{n-m} \xi_m \pi] \ \ldots (4)$
Now each term of series in (4) is greater than or equal to 0 and, in particular, the first term is positive, $|R_m| > \dfrac{b^m}{|h|} > \dfrac{2a^m b^m}{3} \ \ldots (3)$
Thus $\left| {\dfrac{f(x+h) -f(x)}{h}} \right| = |R_m +S_m| \\ \ \ge |R_m|-|S_m| > \left({\frac{2}{3} -\dfrac{\pi}{ab-1}} \right) a^mb^m$
As $ab > 1+\frac{3}{2}\pi$ , therefore $\left({\frac{3}{2} -\dfrac{\pi}{ab-1}} \right)$ is positive.
Thus we see that when $m \to \infty$ so that $h \to 0$ , the expression $\dfrac{f(x+h)-f(x)}{h}$ takes arbitrary large values. Hence, $f'(x)$ does not exist or is at least not finite.