Saturday, May 28, 2016

Generating functions of sequences

The generating function of a sequence
$\{a(n)\}_{n=0}^\infty = \{a_0, a_1, a_2, \cdots\}$ for integers $n\geq 0$ is defined as $f(x) = a_0 + a_1x + a_1x^2 + \cdots$.

For a sequence that satisfies the linear recurrence relation
$a(n) = c_0a(n-1) + c_1a(n-2) + \cdots + c_{m}a(n-m-1)$
with initial terms $a(0) = a_0, a(1) = a_1, \cdots , a(m) = a_m$,
the general form of the generating function is given by:

$$ f(x) = \frac{a_mx^m + \sum_{i=0}^{i< m}\left(a_ix^i-  c_ix\sum_{j=0}^{j < m-i}a_jx^{i+j}\right)}{
1-x\sum_{i=0}^{i \leq m}c_ix^{i}}

The sequence $a(n) = bc^n+d$ satisfies the recurrence relation $a(n) = (c+1)a(n-1)-ca(n-2)$ with
generating function:

$$ (b(1-x) + d(1- cx ))/((1-x )(1-cx)) $$.

How to lose weight without working out

Today, my son said: "Since we are spinning along with the earth, shouldn't the centrifugal force make us feel lighter?"  The answer is yes, since the centrifugal force cancels out a little bit of the gravitational pull of the Earth.  In fact, that is exactly what a satellite does.  It spins so fast around the earth that the centrifugal force cancels out the gravitational pull so it is "floating" in space. The centrifugal force is a virtual or fictitious force denoting the opposite of the force (the centripetal force) needed to achieve the circular motion and this centripetal force is supplied by the force of gravity. There is another interpretation where the satellite is forever falling towards earth, but the motion also has a velocity component that is perpendicular to the force of gravity and its direction changes (due to the gravitational acceleration), but the magnitude doesn't.

How much lighter do you weight on the equator, where you are spinning the fastest, versus on the poles where you are not spinning at all.  There are 2 reasons why you weight is different on the poles versus the equator.  The first reason is the centrifugal force described above.
This force is given by Newton's second law: $F = ma$ along with $a = \omega^2r$.  $\omega$ is the angular velocity of the earth rotation, i.e. $\frac{2\pi}{T}$, where $T$ is about 24 hours = $86400s$. The radius at the equator is about $r$ = 6378137 m. This results in a=0.03373 which is about 0.34% of the gravitational acceleration of $g =9.8 \frac{m}{s^2}$

The second reason is that the earth is not a perfect sphere but is squashed on the poles since the centrifugal force of the spinning earth pulls the matter of the earth from the center of mass and this happens more at the equator than at the poles.  Thus you are further away from the center of mass of the Earth on the surface of the equator than on the surface of the pole.  This affects the value of $g = \frac{MG}{r^2}$ where $M$ is the mass of the Earth and $G$ is the universal gravitational constant.  This adds another 0.67% to the difference resulting in you (and any mass) weighting about 1% less on the equator than at the poles.

Tuesday, May 3, 2016

Ironic song while driving this morning

While driving to work this morning, the sky was gray and overcast and there is a light drizzle. Sometime during the commute, the song "I can see clearly now" by Johnny Nash came on the music streaming service. I got a chuckle out of it as the lyrics are in direct contradiction to what I was seeing around me.