## Archive for April, 2009

### Example of the Solution of a Quartic Polynomial

Thursday, 30 April 2009

Here’s a worked example of the solution of a quartic polynomial, as described in my earlier post on solving polynomials algebraically.

### Differentiation Formulae

Wednesday, 29 April 2009

Here I summarise the recursive rules for deriving (differentiating) compound expressions.

### Further Calculator Statistics

Tuesday, 28 April 2009

This is further to my post on calculator statistics.

If the calculator also stores the six values $\Sigma x^3$, $\Sigma x^4$ and $\Sigma x^2y$, then it can calculate the linear regression to a parabola.

### MSVC Projects, Source Control and GUIDs

Tuesday, 28 April 2009

Further to my post on build events in MSVC, I’ve noticed that sometimes MSDEV sometimes re-orders the GUIDs in a solution file.

I know a mouse and he hasn’t got a house;
I don’t know why I call him Gerald.
He’s getting rather old, but he’s a good mouse.

I have a simple Python script (sort of attached) that can read these files and sort those GUID sections into order.

### Cross-section of Fuel

Monday, 27 April 2009

Here’s another, brief, post on the subject of dimensional analysis; it concerns a curious interpretation of dimensions. [Previous posts concerned statistics and kinematics.]

Consider cars. A typical fuel efficiency is
≈ 40 mpg (miles / gallon [UK])
≈ 14 km / litre = 14 m / cc = 14 mm / mm3

Expressed differently (taking the reciprocal), this is about
≈ 4 fl oz [UK] / mi
≈ 70 ml / km = 70 cc / km = 70 mm3 / m
≈ 70,000 μ2 = 70,000 μm2 = 0.07 mm2

What’s that? It’s a unit of area.

If you imagine forming a tube of fuel as the car uses it (as the car goes along), then that’s the tube’s cross-sectional area.

### Forgotten Functions

Sunday, 26 April 2009

Here are some functions you no longer see much.

 coversine covers(x) = 1 − sin(x) exsecant exsec(x) = sec(x) − 1 hacoversine, cohaversine, havercosine hacov(x) = covers(x) / 2 haversine (half versine) hav(x) = vers(x) / 2 versine (versed sine) vers(x) = 1 − cos(x)

### Colour Mixing

Sunday, 26 April 2009

There seems to be some confusion about what the primary and secondary colours are. Well, it depends.

### Dimensional Analysis of the Kinematic Equations

Sunday, 26 April 2009

To help to remember the kinematic equations, it may be useful to know a little dimensional analysis.

### Catalan and Fibonacci Formulae

Sunday, 26 April 2009

Using the difference table method described in my previous post, I discovered the following formulae.

The first is not new. I do not know whether the last two are:

$\bigtriangleup_d(n) = \dbinom{n+d-1}{d} = \sum_{k=1}^{d} \dbinom{d-1}{k-1} \dbinom{n}{k}$;

$C_n = \sum_{i=0}^{n} \dbinom{n}{i} \bigtriangleup(F_{i-1})$;

$\sum_{i=0}^{n-1} C_i = \sum_{i=0}^{n} \dbinom{n}{i} \bigtriangleup(F_{i-2})$.

Here, $C_n = \dfrac{1}{n+1}\dbinom{2n}{n}$ is the nth Catalan number, $F_n$ is the nth Fibonacci number, $\bigtriangleup_d(n)$ is the nth d-dimensional tetrahedral number, and $\bigtriangleup(n) = \bigtriangleup_2(n)$ is the nth triangular number.

### Difference Tables

Saturday, 25 April 2009

Consider a sequence C = (C0, C1, C2, …). Define:

A0,j = Cj,
Ai,j = Ai,j+1 − Ai,j, and
Ri = Ai,o.

This might be pictured:

 C0 C1 C2 C3 C4 C5 || || || || || || R0 = A0,0 A0,1 A0,2 A0,3 A0,4 A0,5 … R1 = A1,0 A1,1 A1,2 A1,3 A1,4 … R2 = A2,0 A2,1 A2,2 A2,3 … R3 = A3,0 A3,1 A3,2 …

R = (R0, R1, R2, …) is another sequence. We might call C the sequence generated by R and R the generator of C.

It turns out that

$C_n = \sum_{k=0}^{n} \dbinom{n}{k} R_k$;

$R_n = \sum_{k=0}^{n} (-1)^{n-k} \dbinom{n}{k} C_k$.

The first of these formulae is particularly handy if the the sequence R is finite (or rather, is eventually zero):

$C_n = \sum_{k} R_k \dbinom{n}{k}$.

Given some unknown sequence, this formula may be used to find empirically the power-series closed form, if it has one.

For example, consider the tetrahedral numbers C = (0, 1, 4, 10, 20, 35, …). We find that R = (0, 1, 2, 1).

Therefore $C_n = \dbinom{n}{1} + 2 \dbinom{n}{2} + \dbinom{n}{3} = \dbinom{n+2}{3}$.

This is not a proof of course, but it can be a useful experimental tool. It is also useful for manipulating sum sequences and difference sequences.