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## Tuesday, December 1, 2009

### Edition3: JEE help

In this article we will discuss few problems of Physics and Mathematics. The topics relates to Mechanics in Physics and concepts of infinite series of Mathematics. You may find that these are out of syllabus of IIT JEE, but as you will see that they don't need any additional knowledge other than what you already know. These problems will help you strengthen your understanding even more.

Problem 1
Let's start with the first problem of Mathematics. You have studies (or will study) infinite series and natural logarithm in Algebra. Consider the following series of Natural Logarithm of +2.
Ln 2 = 1 1/2 + 1/3 1/4 + 1/5 1/6 + 1/7 … to infinity.
If you rearrange the terms you can write
Ln 2 = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 …. to infinity) - 2 * ( 1/2 + 1/4 + 1/6 + 1/8 …. to infinity)
Ln 2 = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 …. to infinity) - (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 …. to infinity)

At this juncture, you may tend to say that the right hand side of the equation evaluates to Zero. This would be wrong for two reasons, which has a profound implication in Set theory of numbers (called Axiomatic Set Theory proposed by Cantor) and concepts of different sizes of infinity. These are explained below

1. (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 …. to infinity) = the harmonic series, which is divergent. It diverges to infinity and addition subtraction using infinity is not always a valid mathematical operation. Mathematical physicists often stumble across such infinities in Quantum Mechanics and they have a nice tool called 'Normalization' to work around these thorny issues. But don't equate this situation with what you learn in calculus. Example, what is the value of this expression?
(X ^ 2 9)/(x+3) (X-3) where X -> infinity
The expression simplifies to:-
(X 3) (X-3) = X X = 0.
Even in this expression we are subtracting 'INFINITY' from 'INFINITY', but this operation is valid because these two infinities are of 'same size'. In the next Para, you will learn why the expression of Ln 2 doesn't fall in this category.
2. Let's observe carefully the formation of the confusing expression of Ln2
Ln 2 = 1 1/2 + 1/3 1/4 + 1/5 1/6 + 1/7 … to infinity.
Ln 2 = 1 1/2 + 1/3 1/4 + 1/5 1/6 + 1/7 … + 1/ (odd) 1/(even) …. To infinity
Ln 2 = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 …. + 1/n + 1/(n+1) + …. + .. 1/2n)
- 2 ( 1/2 + 1/4 + 1/6 + 1/8 + … + 1/2n) … where n -> infinity

The first part of the expression has 2n terms, while the second part has only n terms. When evaluated separately, both diverge to infinity but the second part is a 'smaller infinity than the first', hence you can't subtract one from the other. According to Cantor's theory, both parts have equal number of terms (called Aleph 0 = the smallest infinity), but their sum is not of same size.

But then mathematicians don't leave problems half-solved and confusing. We will not do it either. Check this out
Ln 2 = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 …. + 1/n + 1/(n+1) + …. + .. 1/2n) - (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 …. + 1/n) (after multiplying all the terms by 2 in the second part)
Ln 2 = [(1 + 1/2 + 1/3 …. + 1/n) + {1/(n+1) + 1/(n+2) + … + 1/2n)}] - (1 + 1/2 + 1/3 + …. + 1/n) (after splitting the first part into two)
Ln 2 = {1/(n+1) + 1/(n+2) + … + 1/2n)} (this is what remains after subtraction) where n-> infinity.
Ln 2 = Definite integral of 1/(1+x) with respect to x, within limits of 0 and 1 (this comes from Newton's Summation formula of convergent infinite series) which is same as Ln 2 .
So we are back to square one
Ln 2 = Ln 2

Problem 2
Now let's look at some interesting problem of elementary mechanics. You have studied the equations on a freely falling body in uniform gravitational field 'g'. Now we want to study a raindrop falling under gravity but getting a resistance due to friction with air. It is often observed that the resistance is typically proportional to the kinetic energy of the rain drop = k.m.v^2. So the equation of the raindrop looks like

m. D^2 (x) = m.g kmv^2 (where D^2 stands for second order derivative with respect to time, v is the velocity after falling a distance x from the cloud)
First you observe that

1. Initially the velocity is 0, so there is no resistance at all.

2. Once the raindrop starts accelerating due to gravity, the
resistance increases due to increase
in velocity
3. Assuming that the cloud is formed
very high in the sky, there is enough
height available for the rain drop to
attain a steady state where, the
resistance cancels out gravitation
force and the net force on the
raindrop will be zero. In that case,
the raindrop will not accelerate any
further but will continue to move
with a constant velocity, called the
'terminal speed (S). The expression
for this is
m.g kmS^2 = net force on
raindrop = 0
=> S = sqrt ( g/k ), which is a
constant.
=> k = g / ( S^2 )
So now, our original equation becomes (after cancelling 'm' from both sides)
D^2 (x) = g ( 1 v^2 / S^2 ) . Please make the following observations
1. The equation is now independent
of the mass of the rain drop. In
general, all the equations of motion
under gravity always end up being
independent of mass of the moving
body. This is what Galileo
demonstrated, and Newton made a
theory of and later (year 1915)
Einstein based his General Theory
of Relativity on, to correct the
shortcomings of Newton's theory
of Gravitation.
2. To solve this equation you need to
integrate both sides and when you
do that (correctly !) you get
v^2 = S^2 {1 exp ( - 2gx / S^2 ) },
which gives velocity 'v' after falling
distance 'x' from the cloud.
3. From the above you can say that
a. The raindrop is always in catch-up
mode with the terminal speed S.
b. Theoretically, it will take infinite
distance and time to attain S.
c. The maximum velocity that a
raindrop can ever achieve is S,
which is sqrt (g/k).
Given that 'g' is more or less constant, and 'k' is almost constant unless there is major change in the composition of the atmosphere, no raindrop can ever hit the ground with a velocity > S, not even during 'Alia' or 'Phyan' !