Feeling very chuffed for completing this online math-based, course on Special Relativity covering: time dilation, length contraction, synchronicity of clocks and the relativity of simultaneity.
Certificate of Completion
Imagine a veeerrryyyy long rocket travelling through space at “ludicrous” speed. From our stationary perspective the length of the rocket would appear shorter (length contraction), clocks on the rocket would be ticking more slowly (time dilation) and clocks spread throughout the rocket would show different times (asynchronous time).
Twelve tone music. Wow. Found this extraordinary creative, explanatory composition by ViHartViHart. I’m hooked! http://www.youtube.com/watch?v=4niz8TfY794
Is this a digraph of Reachability or dominance? And the matrix?
Answer: both or either. It depends on the context. And what you with the matrix. For total Reachability add each column. For total Dominance add each row.
Complex roots can be used to calculate the exact trigonometric value for unusual angles; for example cos(2π/5) and sin(2π/5). Choosing z5=1 will produce five complex roots aligned as shown in the first diagram. All five roots must be equally spaced over 2π, so their radial spacing will be 2π/5.
A right triangle can be superimposed on z2 as shown in diagram 2 with the angle at the origin being 2π/5.
For z5=1 the modulus for all the roots =1, so z1 … z5 are all on a unit circle (diagram 3). Thus from trigonometry, the x-coordinate of z2= cos(2π/5) and the y-coordinate of z2 = sin(2π/5).
Calculating their value requires some clever maths.
In Cartesian form let z5=(x+yi)5. If z5=1+0i, then (x+yi)5=1+0i. So the first step is to expand (x+yi)5 collect Re() and Im() parts and equate the Re() part to 1. (Diagram 4 lines 1 and 2)
This real part contains both x and y. In order to solve for x we first eliminate y by returning to the fact that z2 is on a unit circle so x2+y2=1. Rearranging this gives y2=1-x2 which can be used as a substitute for y2 and y4 since y4=(y2)2. (Diagram 4 lines 3, 4 and 5).
Now we have an equation in x only and resorting to a CAS calculator to solve it yields three exact values for x (Diagram 5 lines 1 and 2).
As z2 is in the first quadrant we choose the positive value of x as cos(2π/5) and use this value to calculate y=sin(2π/5) (diagram 5 lines 3 and 4).
Found this fascinating link to the historically famous (1748) “Witch of Agnesi” curve which I read about in Jennifer Ouolette’s very readable book The Calculus Diaries, courtesy of Mildura library. You will need java to activate the applet. Persevere; my browser initially blocked the javascript.
http://dynamicgeometry.com/JavaSketchpad/Gallery/Trigonometry_and_Analytic_Geometry/The_Witch_of_Agnesi.html
See explanation in frame below.
Referred to by Emma Ayres on ABC Classic Fm this morning. Pythagoras would be pleased http://www.bbc.co.uk/news/science-environment-26151062
aussiemathstutor 