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# Monthly Archives: March 2011

## Essential Steps of Problem Solving in Mathematical Sciences

### Learning how to solve problems in mathematics is simply to know what to look for.

Math problems often require established procedures and one must know What & When to apply them. To identify procedures, you have to be familiar with the different problem situations, and be able to collect the appropriate information, identify a strategy or strategies and use the strategy/strategies appropriately. But exercise is must for problem solving. It needs practice!! The more you practice, the better you get. The great mathematical wizard G Polya wrote a book titled How to Solve It in 1957. Many of the ideas that worked then, do still continue to work for us. Given below are the four essential steps of problem solving based on the central ideas of Polya.

## A Problem on Ordinary Nested Radicals

### Problem

Prove or disprove that
$\sqrt {7+\sqrt{33}} + \sqrt {6+\sqrt{35} } = \sqrt {5+\sqrt{21}} + \sqrt {8+\sqrt{55}}$

### Solution

In order to simplify the radicals, the radicands should be forced to equal square numbers (e.g., $7+\sqrt{33}$ should be a square of some number). Numbers whose squares have a rational and radical part are usually in the form $x+y$.
So let $\sqrt{7 +\sqrt{33}} = x+y = \sqrt{(x+y)^2} = \sqrt {x^2+y^2+2xy}$
and set
$x^2+y^2=7$ and $2xy=\sqrt{33} \, , i.e., y=\sqrt{33}/2x$
Thus $x^2 + \left ( \frac {\sqrt{33}}{2x} \right )^2 =7$
which on simplification yields $x=\sqrt{22}/2$
And also $y=\sqrt{6}/2$

Thus,
$\mathbf {\sqrt{7+\sqrt{33}} = \frac {\sqrt{22}+\sqrt{6}}{2} }$
Using the same process for other radicals:
$\mathbf {\sqrt{6+\sqrt{35}} = \frac{\sqrt{10}+\sqrt{14}}{2} }$
$\mathbf {\sqrt{8+\sqrt{55}} = \frac{\sqrt{10}+\sqrt{22}}{2} }$
$\mathbf {\sqrt{5+\sqrt{21}} = \frac{\sqrt{6}+\sqrt{14}}{2} }$
Thus, now we can easily prove (by addition) that

$\sqrt {7+\sqrt{33}} + \sqrt {6+\sqrt{35}} =\sqrt {5+\sqrt{21}} + \sqrt {8+\sqrt{55}}$

## Classical Theory of Raman Scattering

The classical theory of Raman effect, also called the polarizability theory, was developed by G. Placzek in 1934. I shall discuss it briefly here. It is known from electrostatics that the electric field $E$ associated with the electromagnetic radiation induces a dipole moment $\mu$ in the molecule, given by
$\mu = \alpha E$ …….(1)
where $\alpha$ is the polarizability of the molecule. The electric field vector $E$ itself is given by
$E = E_0 \sin \omega t = E_0 \sin 2\pi \nu t$ ……(2)
where $E_0$ is the amplitude of the vibrating electric field vector and $\nu$ is the frequency of the incident light radiation.

Thus, from Eqs. (1) & (2),
$\mu= \alpha E_0 \sin 2\pi \nu t$ …..(3)
Such an oscillating dipole emits radiation of its own oscillation with a frequency $\nu$, giving the Rayleigh scattered beam. If, however, the polarizability varies slightly with molecular vibration, we can write
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q$ …..(4)
where the coordinate q describes the molecular vibration. We can also write q as:
$q=q_0 \sin 2\pi \nu_m t$ …..(5)
Where $q_0$ is the amplitude of the molecular vibration and $\nu_m$ is its (molecular) frequency. From Eqs. 4 & 5, we have
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q_0 \sin 2\pi \nu_m t$ …..(6)
Substituting for $\alpha$  in (3), we have
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {d\alpha}{dq} q_0 E_0 \sin 2\pi \nu t \sin 2\pi \nu_m t$ …….(7)
Making use of the trigonometric relation $\sin x \sin y = \frac{1}{2} [\cos (x-y) -\cos (x+y) ]$ this equation reduces to:
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {1}{2} \frac {d\alpha}{dq} q_0 E_0 [\cos 2\pi (\nu - \nu_m) t - \cos 2\pi (\nu+\nu_m) t]$ ……(8)
Thus, we find that the oscillating dipole has three distinct frequency components:

1• The exciting frequency $\nu$ with amplitude $\alpha_0 E_0$
2• $\nu - \nu_m$
3• $\nu + \nu_m$ (2 & 3 with very small amplitudes of $\frac {1}{2} \frac {d\alpha}{dq} q_0 E_0$. Hence, the Raman spectrum of a vibrating molecule consists of a relatively intense band at the incident frequency and two very weak bands at frequencies slightly above and below that of the intense band.

If, however, the molecular vibration does not change the polarizability of the molecule then $(d\alpha / dq )=0$ so that the dipole oscillates only at the frequency of the incident (exciting) radiation. The same is true for the molecular rotation. We conclude that for a molecular vibration or rotation to be active in the Raman Spectrum, it must cause a change in the molecular polarizability, i.e., $d\alpha/dq \ne 0$ …….(9)

Homonuclear diatomic molecules such as $\mathbf {H_2 \, N_2 \, O_2}$ which do not show IR Spectra since they don’t possess a permanent dipole moment, do show Raman spectra since their vibration is accompanied by a change in polarizability of the molecule. As a consequence of the change in polarizability, there occurs a change in the induced dipole moment at the vibrational frequency.

###### Related Articles

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Google is looking for the brightest, best young scientists from around the world to submit interesting, creative projects that are relevant to the world today.

To help make today’s young scientists the rock stars of tomorrow, in partnership with CERN, The LEGO Group, National Geographic and Scientific American, Google is introducing the first global online science competition: the Google Science Fair. It’s open to students around the world who are between the ages of 13-18. All you need is access to a computer, the Internet and a web browser.

To enter, register online and create your project as a Google Site. Registration is open through April 4, 2011.

## 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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## Principle of Equivalence Led to the Formulation of Einstein’s General Theory of Relativity

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The mass of a body, when subjected to a gravitational attraction but no acceleration (i.e., its gravitational mass) is the same when it is subjected to an acceleration but no gravitational attraction (i.e., its inertial mass).

This gave Einstein the idea that a gravitational field can be imitated by a field of acceleration and this, ultimately, led to the formulation of his general theory of relativity, wherein if showed that a non-accelerating or inertial frame of reference in which there is a gravitational field is physically equivalent to a reference frame accelerating uniformly with reference to the inertial frame but in which there is no gravitational field. This means, in other words, that experiments carried out in the two frames, under the same conditions, will yield identical results. This is called the Principle of Equivalence.

## Ultimate Speed of A Material Particle – Denying the Concept of Infinite mass – Photons and More

In classical mechanics, there being no upper limit to velocity it is possible that as a particle is given more and more acceleration, its speed may go on increasing progressively and may well become greater than $c$, –in fact, it may have any velocity whatever.

This is firmly denied by the theory of relativity. It may legitimately be asked, therefore, as to what will happen if the particle is continually accelerated. Certainly, its velocity v goes on increasing and hence also its mass in accordance with the mass-velocity relation $m= \frac {m_0} { \sqrt {1- \frac {v^2} {c^2} } } = \gamma m_0$ . But as $v$ approaches $c$, $\frac {v^2} {c^2} \longrightarrow 1$ and therefore $\sqrt {1- \frac {v^2} {c^2} } \longrightarrow 0$ and hence the mass of the particle $m \longrightarrow \infty$, as shown graphically, from which it is clear that for velocities right up to 50% of $c$, the increase in mass from the value of the rest mass or inertial mass $m_0$ is quite inappreciable. (more…)