# Chaitanya's Random Pages

## September 18, 2013

### Spacecraft missions to outer planets, minor planets or comets

Filed under: science — ckrao @ 11:50 am

After recently hearing news about Voyager 1 being the first human-made object to leave our solar system, I looked up other spacecraft that have visited the outer planets, minor planets (e.g. asteroids) and comets.

#### Outer Planets

 Spacecraft Launch Date Agency Remarks Pioneer 10 3-Mar-72 NASA / ARC flyby of Jupiter in Dec 1973, lost contact in Jan 2003 when 12b km from earth Pioneer 11 6-Apr-73 NASA /ARC flyby of Jupiter in Dec 1974, Saturn in May 1979, lost contact in 1995 Voyager 2 20-Aug-77 NASA / JPL flyby of all outer planets, the only one to fly by Uranus and Neptune Voyager 1 5-Sep-77 NASA / JPL flyby of Jupiter in Sep 1977, Saturn in Mar 1979, now 19b km from earth Galileo 18-Oct-89 NASA flyby of Venus, Earth, asteroids Gasra and Ida (both incidental) before detailed study of Jupiter and its moons, probe dropped into its atmosphere Ulysses 6-Oct-90 NASA / ESA flyby of Jupiter (Feb 1992), orbiting sun, measuring solar wind and gamma ray bursts Cassini–Huygens 15-Oct-97 NASA/ESA flyby of Jupiter & Saturn, Huygens deployed from Cassini, landed on Saturn’s moon Titan in Jan 2005 New Horizons 19-Jan-06 NASA expected to fly by Pluto in July 2015 Juno 5-Aug-11 NASA expected to reach Jupiter in Aug 2016

#### Minor Planets and comets (other than Halley)

 Spacecraft Launch Date Agency Remarks Dawn 27-Sep-07 NASA orbited Vesta July ’11-Sep ’12, expected to reach Ceres in Feb ’15 Hayabusa 9-May-03 JAXA visited asteroid Itokawa bringing back to earth tiny grains of asteroidal material in June 2010 NEAR Shoemaker 17-Feb-96 NASA studied near-earth asteroid Eros, touched down in Feb 2001 Deep Space 1 24-Oct-98 NASA/JPL flybys of asteroid Braille and Comet Borelly Stardust 7-Feb-99 NASA/JPL returned dust samples from Comet Wild 2, also intercepted comet Tempel 1 in Feb 2011 Deep Impact 12-Jan-05 NASA/JPL impactor collided with nucleus of comet Tempel in July 2005, extended mission EPOXI flew by Comet 103P/Hartley, now on its way to asteroid 2002 GT Rosetta 2-Mar-04 ESA flybys of asteroids Steins and Lutetia in 2008, 2010; expected to reach comet 67P/Churyumov-Gerasimenko in mid 2014 deploying the lander Philae Chang’e-2 1-Oct-10 CNSA flew by asteroid 4179 Toutatis in Dec 2012 after orbiting moon International Cometary Explorer (ICE) 12-Aug-78 NASA/ESA first spacecraft to pass through comet tail (Comet Giacobini-Zinner)

#### Halley’s Comet (during its most recent near-earth encounter in 1985-6)

 Spacecraft Launch date Agency Remarks Giotto 2-Jul-85 ESA passed within 600km of Halley’s comet, then flew by Comet Grigg-Skjellerup in Jul 1992 Vega 1 15-Dec-84 USSR flyby of Venus (descent craft surfaced on Venus) and Halley’s Comet (within 9,000km Mar ’86) Vega 2 21-Dec-84 USSR flyby of Venus (descent craft surfaced on Venus) and Halley’s Comet (Mar ’86) Suisei 18-Aug-85 ISAS (now part of JAXA) within 151,000km of Comet Halley in Mar ’86 Sakigake 7-Jan-85 ISAS (now part of JAXA) within 7m km of Comet Halley

## September 16, 2013

### Polynomial extrapolation of finite exponential sequences

Filed under: mathematics — ckrao @ 11:40 am

Here is a cute problem I first encountered many years ago:

A polynomial $P(x)$ of degree $n$ satisfies $P(k) = 2^k$ for $k = 0, 1, 2, \ldots, n$. Determine $P(n+1)$.

Recall that a polynomial of degree $n$ is uniquely determined by its value at $n+1$ points. Hence $P(n+1)$ could have only one possible value. Is it $2^{n+1}$? Let’s see: here is the solution I saw at the time.

Consider the polynomial $\displaystyle P(x) = \binom{x}{0} + \binom{x}{1} + \ldots + \binom{x}{n}$. Then $P(x)$ is a polynomial of degree $n$ (the $k$‘th term has degree $k+1$). Furthermore, since $\binom{x}{k} = 0$ for $x = 0, 1, \ldots, k-1$, we have for $k = 0, 1, 2, \ldots, n$ the following:

$\displaystyle P(k) = \binom{k}{0} + \binom{k}{1} + \ldots + \binom{k}{n} = \binom{k}{0} + \binom{k}{1} + \ldots + \binom{k}{k} = 2^k.$

(Here we recall that the sum of a row of Pascal’s triangle is a power of 2.)

This shows that $P(x)$ is our unique polynomial with the desired properties, and we find

$\displaystyle P(n+1) = \binom{n+1}{0} + \binom{n+1}{1} + \ldots + \binom{n+1}{n} = 2^{n+1} - 1.$

Neat isn’t it that the polynomial almost climbs up to $2^n$ but ends up one short. 🙂 Let’s see how this result generalises. Modify the above question so that $P(k) = a^k$ for $k = 0, 1, 2, \ldots, n$ (where $a$ is a constant not equal to 1). What is $P(n+1)$? This time we let

$\displaystyle P(x) := (a-1)^0 \binom{x}{0} + (a-1)^1 \binom{x}{1} + \ldots + (a-1)^n \binom{x}{n} = \sum_{i=0}^n (a-1)^i \binom{x}{i}.$

As before $P(x)$ is clearly a degree $n$ polynomial. Furthermore, for $k = 0, 1, 2, \ldots, n$ we apply the binomial theorem to obtain

\begin{aligned} P(k) &= \sum_{i=0}^n (a-1)^i \binom{k}{i}\\ &= \sum_{i=0}^k (a-1)^i \binom{k}{i}\\ &= (a-1+1)^k\\ &= a^k. \end{aligned}

This constructed polynomial has the desired properties, so by uniqueness we simply substitute $x=n+1$ to find $P(n+1)$:

\begin{aligned} P(n+1) &= \sum_{i=0}^n (a-1)^i \binom{n+1}{i}\\ &= \sum_{i=0}^{n+1} (a-1)^i \binom{n+1}{i} - (a-1)^{n+1} \binom{n+1}{n+1}\\ &= a^{n+1} - (a-1)^{n+1}.\end{aligned}

The way such polynomials can be constructed (i.e. not pulled out of thin air!) is to write down a finite difference table – refer to an earlier blog post here: if the first diagonal of the table is $a_0, a_1, \ldots, a_n$ where $a_0 = P(0)$, then

$\displaystyle p(x) = \sum_{i=0}^n a_i \binom{x}{i}.$

In the first problem it is easy to verify that $a_i = 1$, in the generalisation $a_i = (a-1)^i$.

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