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Monthly Archives: July 2011

An Important Announcement

Sunday, July 31st, 2011 18:18 / 3 Comments

I have decided to change the style of posts and the posting frequency. Worry not! I am not leaving math, physics and chemistry. I am now an experienced campaigner and I have just arranged the things as follow.

I ‘will’ post atleast 3 posts a week.
Atleast one post will be a ‘problem’ post in which problem solving skills are tested. Another one of the posts would be about applications of mathematical sciences.
On 5th day of every month, I shall chose a blog from World Wide Web and write a review on it.
Other things that I would try in my posts are:
- Mathematical Puzzles
- Valuable Images
- Success Stories of Mathematicians / Physicists / Chemists and students.
- Learning Techniques

I have thought a lot about this blog. This is going to be a ‘wow’ in education blogging. God help me.

Gaurav Tiwari

Please let me know if you have any suggestions about the structure of posts and writing style.

How many apples did each automattician eat?

Friday, July 22nd, 2011 10:36 / 6 Comments

Česky: Jablka jsou všeamericky úspěšná potravi...

Image via Wikipedia

Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to:

Matt asked: Did you eat more apples than I did, James?

James: I don’t know. Did you, Ian, eat more apples than I did?

Ian: I don’t know.

Barry: Aha!! I figured out..

So, Barry figured out how many apples each person ate. Can you do the same?

Answer:

Matt: 1 Apple

James: 2 Apples

Ian: 3 Apples

Barry: 5 Apples

The Logic

Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian must have eaten 3 or 4 apples. Barry can only deduce the other amounts if he ate 5 apples. And the rest, in order to add up to 11 , must have eaten 1, 2 and 3.

Inspired from a childhood heard puzzle.

Fox – Rabbit Chase Problems

Wednesday, July 13th, 2011 07:22 / 2 Comments

Part I:

A fox chases a rabbit. Both run at the same speed $v$ . At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $\alpha$ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is $l$ .

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?

Part II:

Similarly think about the same situation, except now let the rabbit always move in the straight line of its initial direction in above part of the question.

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?

Solutions

Part I

The relative speed of the fox and the rabbit, along the line connecting them, is always $v_{\text{rel}}= v- v \cos \alpha$ . Therefore, the total time needed to decrease their separation from $l$ to zero is $T=\dfrac{l}{v-v \cos \alpha} =\dfrac{l}{v(1-\cos \alpha)} \ \ldots (1)$ which is valid unless $\alpha=0$ , in which case the fox never catches the rabbit.
The location of their meeting is a little trickier to obtain. We have two methods to do :

SLICK METHOD

Imagine that the rabbit chases another rabbit, which chases another rabbit, etc. Each animal runs at an angle $\alpha$ relative to the direction directly away from the animal chasing it. The initial positions of all the animals lie on a circle, which is easily seen to have radius $R=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (2)$ .
The center of the circle is the point, O, which is the vertex of the isosceles triangle with vertex angle $\alpha$ , and with the initial fox and rabbit positions as the othe two vertices. By symmetry, the positions of the animals at all times must lie on a circle with center O. Therefore, O, is the desired point where they meet. The animals simply spiral into O.

Remark

An equivalent solution is the the following:

At all times, the rabbit’s velocity vector is obtained by rotating the fox’s velocity vector by angle $\alpha$ . The meeting point O, is therefore the vertex of the above mentioned isosceles triangle,

MESSIER METHOD

The speed of the rabbit in the direction orthogonal to the line connecting the two animals in $v \sin \alpha$ . Therefore, during a time $dt$ , the direction of the fox’s motion changes by an angle $d\theta =\dfrac {v \sin \alpha}{l_t} dt$ , where $l_t$ is the separation at time $t$ . Hence the change in the fox’s velocity has magnitude $|d\overrightarrow{v}|=v d\theta =v (v \sin \alpha dt/l_t)$ . The vector $d\overrightarrow{v}$ is orthogonal to $\overrightarrow{v}$ , therefore, to get the $x$ -component of $d\overrightarrow{v}$ , we need to multiply $|d\overrightarrow{v}|$ by $v_y/v$ . Similar reasoning holds for $y$ -component of $d\overrightarrow{v}$ , so we arrive at the two equations $\dot{v_x}= \frac{vv_y \sin \alpha}{l_t} \ \ldots (3)$ $\dot{v_y}=- \frac{vv_x \sin \alpha}{l_t} \ \ldots (4)$
Now, we know that $l_t =\{ l-v(1-\cos \alpha) t \}$ . Multiplying the above equations (3) and (4) by $l_t$ , and integrating from the initial to final times, yields $v_{x,0}l+v(1-\cos \alpha)X=v \sin \alpha \, Y \ \ldots (5)$ $v_{y,0}l+v(1-\cos \alpha)Y=-v \sin \alpha \, X \ \ldots (6)$
where (X,Y) is the total displacement vector and $(v_{x,0},v_{y,0})$ is the initial velocity vector. Putting all the X and Y terms on the right sides, and squaring and adding the equations, we get $l^2v^2=(X^2+Y^2)(v^2 \sin^2 \alpha +v^2{(1-\cos \alpha)}^2). \ \ldots (7)$ Therefore , the net displacement is
$R=\sqrt{X^2+Y^2}=\dfrac{l}{\sqrt{2(1-\cos \alpha)}}=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (8)$
To find the exact location, we can, without loss of generality, set $v_{x,0} =0$ , in which case we find $Y/X=(1-\cos \alpha)/\sin \alpha =\tan \alpha/2$ . This agrees with the result of the first solution. $\Box$

Part II:

SLICK METHOD

Let $A(t)$ and $B(t)$ be the positions of the fox and the rabbit respectively. Let $C(t)$ be the foot of the perpendicular dropped from $A$ to the line of the rabbit’s path. Let $\alpha_t$ be the angle, dependent to the time, at which the rabbit moves relative to the direction directly away from the fox (so at $t=0, \ \alpha_0=\alpha$ and at $t=\infty , \ \alpha_{\infty}=0$ ).

The speed at which the distance AB decreases is equal to $v-v \cos \alpha_t$ . Therefore, the sum of the distances AB and CB doesn’t change. Initially, the sum is $l+l \cos \alpha$ and in the end , it is $2d$ where $d$ is desired eventual separation. Therefore, the desired eventual separation
$d=\dfrac{l(1+\cos \alpha)}{2} \ \ldots (9)$

STRAIGHT FORWARD METHOD

Let $\alpha_t$ be defined as in the first solution, and let $l_t$ be the separation at time $t$ . The speed of the rabbit in the direction orthogonal to the line connecting the two animals is $v \sin \alpha_t$ . The separation is $l_t$ , so the angle $\alpha_t$ changes at a rate $\dot{\alpha_{t}}= - \dfrac{v \sin \alpha_t}{l_t} \ \ldots (10)$ . And $l_t$ changes at a rate $\dot{l_t}=-v(1-\cos \alpha_t) \ \ldots (11)$ .
Taking the quotient of the above two equations, separating variables, gives a differential equation $\dfrac{\dot{l_t}}{l_t} = \dfrac {\dot{\alpha_t}(1-\cos \alpha_t)}{\sin \alpha_t} \ \ldots (12)$ which on solving gives $\ln (l_t) = -\ln {(1+\cos \alpha_t)} + \ln (k) \ \ldots (13)$ . Where $k$ is the constant of integration. Which gives $k=l_t (1+\cos \alpha_t) \ \ldots (14)$ . Applying initial conditions $k_0 = l_0 (1+\cos \alpha_0)= l(1+\cos \alpha) \ \ldots (15)$ . Therefore from (14), we get $l(1+\cos \alpha) = l_t(1+\cos \alpha_t)$ , or $l_t= \dfrac{l(1+\cos \alpha)}{1+\cos \alpha_t}$ .
Setting $t= \infty$ and using $\alpha_{\infty} =0$ , gives the final result $l_{\infty} = \dfrac{l(1+\cos \alpha)}{2}$ . $\Box$ .

Remark:

The solution of Part II is valid for all $\alpha$ except $\alpha= \pi$ . If $\alpha = \pi$ , the rabbit run directly towards the fox and they will meet halfway in time $l/2v$ .

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.

Reference

A course of mathematical analysis
SHANTI NARAYAN
PK MITTAL
S. Chand Co.

Memory Methods

Monday, July 4th, 2011 20:51 / Leave a Comment

Image via Wikipedia

Memory, in human reference, is the ability to store retain and recall informations when needed. Without hammering the mind in the definitions, let we look into the ten methods of boosting our memory:

1. Simple Repitition Method

The classical method, very popular as in committing poems to memory by reading them over and over.

2. Full Concentration Method

Concentrate on the topic content while learning. Do not allow your mind to wander. Focus on names and numbers. Try deliberately to remember. Your approach should not be casual. Review soon after you learn, lest memory should fade away.

3. Visual Encoding Method

Translate information into visual formats like pictures, charts, diagrams, tables and graph.

4. Logical Organisation

Matter that is logically organised is retained better than the disjointed floating bits of information. Infuse meaning into what ever you learn. No “Nonsense Syllables”.

5. Mnemonics Method

Few good people try this method to learn some complicated series. There is no harm in using memory crutches like it, after grasping the spirit of the lesson. VIBGYOR is the most famous mnemonic that helps up to list the seven rainbow colors in their appropriate order. Once, I had remembered the name of the planets in the order of their distances from the sum by the acrostics “My very enlightened mother just served us noodles-in plates”, as it tells me the names of nine planets ( off course now they are eight, but I learned it earlier than the removal of pluto), in the order: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto.

6. Mock Teaching Method

If you find a particular portion of a chapter difficult to digest, imagine that you are teaching to a student sitting before you. Explain virtually the content of the lessen. The idea will get hammered into your mind.

7. Rhyming Method

Scholars say that verses are better than prose. Change any definition into verse and try to rhym them. Remember, I discuss just verses, and not poetry.

8. Chunking Method

Very popular method. You can divide a big list into a number of bits or chunks. For example, to remember a mobile number of ten digits, people usually split it to three or more chunks.

9. Pegging

This involves the principal of association. First make a list of ten or twenty convenient pegs or key words that you can easily recall in the right sequence. For example, in alphabetical order, ant, butterfly, cat, dog, elephant, fox, and so on (Living beings are arranged). Cartoons also help in pegging.

10. Blogging

This is the best method for that student, who want to learn very complicated topics. When you blog about any topic, you use all the nine methods listed above in your post. I got blogging, superior method to other nine.

What is an easy way to remember the planets (wiki.answers.com)

Bicycle Thieves – A puzzle

Friday, July 1st, 2011 09:02 / Leave a Comment

One day a man, who looked like a tourist, came to a bicycle shop and bought a bicycle from a shop for $70. The cost price of the bicycle was $60. So the shopkeeper was happy that he had made a profit of $10 on the sale. However, at the time of setting the bill, the tourist offered to pay in travellers cheques as he had no cash money with him. The shopkeeper hesitated. He had no arrangement with the banks to encash travellers cheques. But he remembered that his friend, the shopkeeper next door, has such a provision, and so he took the cheques to his friend next door and got cash from him.
The travellers cheques were all of $20 each and so he had taken four cheques from the tourist totalling to $80. On encashing them the shopkeeper paid back the tourist the balance of $10.
The tourist happily climbed the bicycle and pedaled away whistling a tune.
However, the next morning shopkeeper’s friend who had taken the travellers cheques to the bank called on him and returned the cheques which had proved valueless and demanded the refund of his money. The shopkeeper quietly refunded the money to his friend and tried to trace the tourist who had given him the worthless cheques and taken away his bicycle. But the tourist could not be found.

How much did the shopkeeper lose altogether in this unfortunate transaction?

Answer

One can think of different answers for this question, but yet the correct answer is very simple. All we have to consider is that the shop owner could not have possibly lost more than the tourist actually stole.
The tourist got away with the bicycle which cost the shop owner $60 and the $10 ‘change’ , and therefore, he made off with $70. And this is the exact amount of the shopkeeper’s loss.

Source of The Puzzle:

Puzzle 7, Puzzles to Puzzle You
© Shakuntala Devi, 1976

This puzzle is l’ll modified than the original.

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Monthly Archives: July 2011

An Important Announcement