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# Monthly Archives: March 2011

## Natural Hazards: Impacts of Catastrophic Collisions

“I have grown, looking at sky and stars.” Many people can say this, including me. We loved to watch stars in night. We enjoyed the Sunrise and Sunsets. And offcourse, millions of people loves NASA’s Astronomy Pic Of the Day and they like to add APOD as their wallpapers. There are many more beautiful phenomena in the universe, like Solar Eclipse, Eagle Nebula, Aurora etc. We feel safe, cheer and enjoy our life. However, it does not mean that a smaller strike of any tragedy cannot happen even as you are reading this blog. For a horrible example, who was knowing about the latest Tsunami in Japan, before the date it occured?

The theory of gravity is not enough developed that you can exactly state the motion of a planet or star or any other material in the space. Some bodies, like dwarf planets, sattelites, meteoroids, have random motions. The most fearful motion is of meteoroids, the rock like structures. Usually these are found in between Mars and Jupiter as a belt. But when a meteoroid leaves the belt, the gravitational attraction of the earth, nearest planet to mars, can pull that into it. When a meteoroid enters the atmosphere, friction causes it to heat up and glow. Then we call it a shooting star or a meteor. Small meteors burn up completely as they pass through the atmosphere. Larger ones end up on earth as meteoroid. Impact craters are formed when a large meteoroid or comet crashes into a planet. One such massive natural hazard happened in past, which destroyed all the Dinosaurs.

## Problem

1. Prove that ${(x+y)}^n-x^n-y^n$ is divisible by $xy(x+y) \times (x^2+xy+y^2)$ if $n$ is an odd number not divisible by $3$.
2. Prove that ${(x+y)}^n-x^n-y^n$ is divisible by $xy(x+y) \times {(x^2+xy+y^2)}^2$ if $n \equiv \pmod{6}$

## Solution

1.Considering the given expression as a polynomial in $y$, let us put $y=0$. We see that at $y=0$ the polynomial vanishes (for any $x$). Therefore our polynomial is divisible by $y$. Similarly, it is divisible by $x$ as well. Thus the polynomial is divisible by $xy$.
To prove that it is divisible by $x+y$ , put $x+y=0 \ {or} \ y=-x$. It is evident that for odd n we have : ${(x+(-x)}^n-x^n-{(-x)}^n = 0$ for $y=-x$.
Consequently, our polynomial is divisible by $x+y$. It only remains to prove the divisibility of the polynomial by $x^2 +xy+y^2$ , which also be written as $(y-x\epsilon)(y-x{\epsilon}^2 )$ where $\epsilon^2+\epsilon+1=0$.
For this purpose it only remains to replace $y$ first by $x \epsilon$ and then by $x\epsilon^2$ to make sure that with these substitutions the polynomial vanishes. Since, by hypothesis, $n$ is not divisible by 3, it follows that $n=3l+1 \ or \ 3l+2$, for every $l \in \mathbb{Z}$, in which $3l+1$ is not acceptable since $n$ is odd from the problem. At $y=x\epsilon$ the polynomial attains the following value
${(x+x\epsilon)}^n-x^n-{(x\epsilon)}^n=x^n [{(1+\epsilon)}^n-1-\epsilon^n] \\ =x^n {(-\epsilon^2)}^n -1 -\epsilon^n ....$ since ($1+\epsilon + \epsilon^2=0$) substituting $n=3l+2$ we get
$1+\epsilon+\epsilon^2 =0$
Likewise we prove that at $y=x\epsilon^2$ the polynomial vanishes as well, and consequently, its by divisibility by $xy(x+y) \times (x^2+xy+y^2)$ is proved.
2.To prove the second statement, let us proceed as follows. Let the quantities ${-x, -y, \, and \, x+y}$ be the roots of a cubic equation $X^3-rX^2-pX-q=0$. Then by virtue of the known relations between the roots of an equation and its coefficients we have $r=-x-y-(x+y)=0 \\ -p=xy-x(x+y)-y(x+y)$ or $p=x^2+xy+y^2$ and $q=xy(x+y)$.
Thus, $-x, \, -y \, x+y$ are the roots of the equation $X^3-pX-q=0$ where $p=x^2+xy+y^2$ and $q=xy(x+y)$
Put ${(-x)}^n-{(-y)}^n+{(x+y)}^n=S_n$. Among successive values of $S_n$, there exist the relationship $S_{n+3}=pS_{n+1}+qS_n$,: $S_1$ being equal to zero.
Let us prove that $S_n$ is divisible by $p^2$ if $n \equiv 1 \pmod{6}$ using the method of mathematical induction. Suppose $S_n$ is divisible by $p^2$ and prove that then $S_{n+6}$ is also divisible by $p^2$.
So, using this relation we get that
$S_{n+6}=p(pS_{n+2} + qS_{n+1}) + q(pS_{n+1}+qS_n) \\ =p^2S_{n+2}+2pqS_{n+1}+q^2S_n$.
Since, by supposition, $S_n$ is divisible by $p^2$ , it suffices to prove that $S_{n+1}$ is divisible by $p$. Thus we only have to prove than $S_n={(x+y)}^n+(-x)^n+(-y)^n$ is divisible by $p=x^2+xy+y^2$ if $n \equiv 2 \pmod{6}$, we easily prove our assertion. And so, assuming that $S_n$ is divisible by $p^2$, we have proved that (from induction) $S_{n+6}$ is also divisible by $p^2$. Consequently $S_n ={(x+y)}^n+(-x)^n+(-y)^n={(x+y)}^n-x^n-y^n$ for any $n \equiv 1 \pmod{6}$ is divisible by $p^2={(x^2+xy+y^2)}^2$.
Now it only remains to prove its divisibility by $x+y$ and by $xy$ , which is quite elementary.

## Number Game: The Word Addition

 Consider the summation of letters : $T \ A \ M \ I \ N \ G \\ + \\ H \ E \ L \ M \ E \ T \\ = \\ 9\ 2\ 5\ 7\ 6 \ 4$ In the addition shown above, the sum $925764$ represents a word. We have to find this word — on the conditions given below: Each letter represents a different digit. No letter represents zero.

So, let we go to find the word represented by 925764.

## Solution:

T+H cannot be 9. Otherwise, A & E both are 1 [which contradicts (1)] or one of A & E is zero [which contradicts (2)]. So, T+H is 8 and 1 is carried from A+E.

Then T cannot be:

0 –[[2]]

2 or 7 – then G & T would be the same digit, contradicting [1]

4 — since $T+H=8$, H would then be 4, contradicting [1].

So the possible digits for T & H are as follows:

${T : \ 1 \ 3 \ 5 \ 6}$

${H : \ 7 \ 5 \ 3 \ 2}$

Then, continuing the table with G, so that G+T is 4 or 14:

${G : \ 3 \ 1 \ 9 \ 8}$

${T : \ 1\ 3 \ 5 \ 6}$

${H : \ 7 \ 5 \ 3 \ 2}$

Because 1 is carried from A+E, A+E is either 11 or 12.

Suppose, A+E is 11. Then using [[1]] to continue the table with A and E: [${a, b, c, d, e, . . . }$ represent cases.]

$\rightarrow \ \, a \ b \ c \ d \ e \ f \ g \ h \ i \ j \ k \ l \\ {G : \ 3\ 3\ 3\ 3\ 1\ 1\ 1\ 1\ 9\ 9\ 8\ 8} \\ {T : \ 1 \ 1 \ 1\ 1\ 3\ 3\ 3\ 3\ 5\ 5 \ 6\ 6} \\ {H : \ 7\ 7\ 7\ 7\ 5\ 5\ 5\ 5\ 3\ 3\ 2\ 2} \\ {E : \ 2\ 9\ 5\ 6\ 2\ 9\ 4\ 7 \ 4 \ 7\ 4\ 7} \\ {A : \ 9\ 2\ 6\ 5\ 9\ 2 \ 7 \ 4 \ 7 \ 4 \ 7 \ 4}$

Suppose A+E is 12. Then by using [1] to continue the table with A and E

$\rightarrow m \ n \ o \ p \ q \ r \ s \ t \ u \ v \\ {G : \ 3\ 3\ 1\ 1\ 9\ 9\ 8\ 8\ 8\ 8} \\ {T : \ 1\ 1\ 3\ 3\ 5\ 5\ 6\ 6\ 6 \ 6} \\ {H : \ 7 \ 7\ 5\ 5\ 3\ 3\ 2\ 2\ 2\ 2} \\ {E : \ 4\ 8\ 4\ 8\ 4\ 8\ 3\ 9\ 5\ 7} \\ {A : \ 8\ 4\ 8\ 4\ 8\ 4\ 9\ 3\ 7 \ 5}$

Then, keeping [1] & [2] in mind, to continue the table with N, so that N+E is 6 or 16 ( if G+T 10; those case that are not listed are eliminated because no digit is possible for N):

$\rightarrow \ a \ e \ f \ g \ h \ i \ j \ k \ m \ o \ q \ r \\ {G : \ 3\ 1\ 1\ 1\ 1\ 9\ 9 \ 8 \ 3\ 1\ 9\ 9} \\ {T : \ 1\ 3\ 3\ 3\ 3\ 5\ 5\ 6\ 1\ 3\ 5\ 5} \\ {H : \ 7\ 5\ 5\ 5\ 5\ 3\ 3\ 2\ 7\ 5\ 3\ 3} \\ {N : \ 4\ 4\ 7\ 2\ 9\ 1\ 8\ 1\ 2\ 2\ 1\ 7} \\ {E : \ 2\ 2\ 9\ 4\ 7\ 4\ 7\ 4\ 4\ 4\ 4\ 8} \\ {A : \ 9\ 9\ 2\ 7\ 4\ 7\ 4\ 7\ 8\ 8\ 8\ 4}$

Again, using (1) and (2) to continue the table with M and L so that M+L is 4 or 5 (if A+E=12) , or 14 or 15 (if A+E =11; those cases are not listed are eliminated because no digits are possible for M and L):

$\rightarrow G \ T \ H \ N \ E \ A \ M \ L$

$a_1 \ \, 3\ 1\ 7\ 4\ 2\ 9\ 6\ 8 \\ a_2 \ \, 3\ 1\ 7\ 4\ 2\ 9\ 8\ 6 \\ e_1 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 6\ 8 \\ e_2 \ 1\ 3\ 5\ 4\ 2\ 9\ 8\ 6 \\ e_3 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 7\ 8 \\ e_4 \ \, 1\ 3\ 5\ 4\ 2\ 9\ 8\ 7 \\ f_1 \ \, 1\ 3\ 5\ 7\ 9\ 2\ 6\ 8 \\ f_2 \ \, 1\ 3\ 5\ 7\ 9\ 2\ 8\ 6 \\ g_1\ \, 1\ 3\ 5\ 2\ 4\ 7\ 6\ 9 \\ g_2 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 9 \ 6 \\ g_3 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 6\ 8 \\ g_4 \ \, 1\ 3\ 5\ 2\ 4\ 7\ 8\ 6 \\ h_1 \ \, 1\ 3\ 5\ 9\ 7\ 4\ 6\ 8 \\ h_2 \ \, 1\ 3\ 5\ 9\ 7\ 4\ 8\ 6 \\ i_1 \ \, 9\ 5\ 3\ 1\ 4\ 7\ 6\ 8 \\ i_2 \ \, 9\ 5\ 3\ 1\ 4\ 7\ 8\ 6 \\ k_1 \ \, 8\ 6\ 2\ 1\ 4\ 7\ 5\ 9 \\ k_2 \ \, 8\ 6\ 2\ 1\ 4\ 7\ 9\ 5$

Again continuing the table with I , so that I+M is 7 or 17 (if N+E 10; those cases not listed are eliminated because no digits are possible for I), there remains only one case $g$ with two sub-cases ${(g_2, g_4)}$:

$- G \ T \ H \ N \ E \ A \ I \ M \ L$

$g_2 \ 1 \ \, 3 \ \, 5 \ \, 2 \ \, 4 \ \, 7 \ \, 8 \ \, 9 \ \, 6 \\ g_4 \ 1 \ \, 3 \ \, 5 \ \, 2 \ \, 4 \ \, 7 \ \, 9 \ \, 8 \ \ 6$

Out of which only $g_4$ is correct since M+L = 14 .

 Substituting the letters for the digits $9 \, 2 \, 5 \, 7 \, 6 \, 4$ is ${I N H A L E}$.

## Note

• This problem requires handy exercise, and this should be done manually with great concentration.

• I am not good in latex, hence was unable to make perfect matrices. Sorry! Hope you’ll compromise with it.

## Milnor wins 2011 Abel Prize

John Milnor

The Norwegian Academy of Science and Letters has decided to award the Abel Prize for 2011 to John Milnor, Institute for Mathematical Sciences, Stony Brook University, New York “for pioneering discoveries in topology, geometry and algebra”. The President of the Norwegian Academy of Science and Letters, Øyvind Østerud, announced the winner of this year’s Abel Prize at the Academy in Oslo today, 23 March.

John Milnor will receive the Abel Prize from His Majesty King Harald at an award ceremony in Oslo on 24 May. The Abel Prize recognizes contributions of extraordinary depth and influence to the mathematical sciences and has been awarded annually since 2003. It carries a cash award of NOK 6,000,000 (close to EUR 750,000 or USD 1 mill.)

(more…)

## A Brief Discussion on Participation of \mu-mesons (muons) based upon Relativistic Dynamics

 The life time of muons { $\mu$-mesons } is $2.2 \times 10^{-6}$ seconds and their speed $0.998c$, so that they can cover only a distance of $0.998c \times 2.2 \times 10^{-6}$ or 658.6 meters in their entire lifetime, and yet they are found in profusion at sea level, i.e., at a depth of 10 kilometers from the upper atmosphere where they are produced. How may this be explained on the basis of (i) Lorentz – Fitzgerald Contraction; (ii) Time Dilation.

## Consequences of Light Absorption – The Jablonski Diagram

Image via Wikipedia

According to the Grotthus – Draper Law of photo-chemistry, also called the principal of photo chemical activation, Only that light which is absorbed by a system can bring about a photochemical change. However it is not essential that the light which is absorbed must bring about a photochemical change. The absorption of light may result in a number of other phenomena as well. For instance, the light absorbed may cause only a decrease in the intensity of the incident radiation. This event is governed by the Beer-Lambert Law. Secondly, the light absorbed may be re-emitted almost instantaneously (within $10^{-8}$ second) in one or more steps. This phenomenon is known as fluorescence. The emission in fluorescence bearer with the removal of the source of light. Sometimes the light absorbed is given out slowly and even long after the removal of the source of light. This phenomenon is known as phosphorescence.
The phenomena of fluorescence and phosphorescence are best explained with the help of the Jablonski Diagram.

In order to understand this diagram, we need to define some terminology. Most molecules have an even number of electrons and thus in the ground state, all the electrons are spin paired. The quantity $\mathbf {2S+1}$, where $S$ is the total electron spin, is known as the spin multiplicity of a state. When the spins are paired $\uparrow \downarrow$ as shown in the figure, the upward orientation of the electron spin is cancelled by the downward orientation so that $\mathbf {S=0}$.

$s_1= + \frac {1}{2}$ ; $s_2= - \frac {1}{2}$ so that $\mathbf{S}=s_1+s_2 =0$.
Hence, $\mathbf {2S+1}=1$

Thus, the spin multiplicity of the molecule is 1. We express it by saying that the molecule is in the singlet ground state.

When by the absorption of a photon of a suitable energy $h \nu$, one of the paired electrons goes to a higher energy level (excited state), the spin orientations of the single electrons may be either parallel or antiparallel. [see image]

•If spins are parallel, $\mathbf {S=1}$ or $\mathbf {2S+1=3}$ i.e., the spin multiplicity is 3. This is expressed by saying that the molecule is in the triplet excited state.
• If the spins are anti-parallel, then
$\mathbf{S=0}$ so that $\mathbf {2S+1=1}$ i.e., the singlet excited state, as already mentioned.

Since the electron can jump to any of the higher electronic states depending upon the energy of the photon absorbed, we get a series of singlet excited states, $\{S_n\}$ where $n \ge 1$ and a series of triplet excited state $\{T_n\}$ where $n \ge 1$. Thus $S_1, \, S_2, \, S_3, ....$ etc are respectively known as first singlet excited state, second singlet excited state and so on. Similarly, in $T_1, \, T_2,\, .....$ are respectively known as first triplet excited state, second triplet excited state and so on.

Make sure, you are not confused in $\mathbf{S}$ & $S_n$

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