Mark Jacobs' Mathematical Excursions

Transcendental Algebraic Equations
Trigonometrical Addition Theorems and Repercussions
Perl Regular Expression Tester

Mathematical Theorems about Men and Women

  • A man will pay $2 for a $1 item he needs.
  • A woman will pay $1 for a $2 item that she doesn't need.
  • A woman worries about the future until she gets a husband.
  • A man never worries about the future until he gets a wife.
  • A successful man is one who makes more money than his wife can spend.
  • A successful woman is one who can find such a man.
  • To be happy with a man, you must understand him a lot and love him a little.
  • To be happy with a woman, you must love her a lot and not try to understand her at all.
  • Married men live longer than single men, but married men are a lot more willing to die.
  • Any married man should forget his mistakes, there's no use in two people remembering the same thing.
  • Men wake up as good-looking as they went to bed. Women somehow deteriorate during the night.
  • A woman marries a man expecting he will change, but he doesn't.
  • A man marries a woman expecting that she won't change, and she does.
  • A woman has the last word in any argument.
  • Anything a man says after that is the beginning of a new argument.

    Transcendental Algebraic Equations

    45/52-e/pi = 0.00012863595235031
    A long-time obsession of mine has been transcendental algebraic equations, which despite sounding a long way away from the real world, threw up a real gem of a function the other night. I have a function plotter that will map out phenomenally complex equations of the single variant variety. It looks like this (click on image to download the zip file) :-

    Mark Jacobs Graph Plotter

    Trying out a few randomisations, I began to recognise patterns and groups of curves and asymptotes. Then a level 9 function produced this :-

    Quantum Jump Graph

    The amazing thing about this curve is that it seems to go along smoothly, then flips out completely, but finitely, and then continues along smoothly in the same direction and from the same area as where the flip-out started. I call this the first portrayal of a mathematically rigorous representation of a quantum jump. For example, consider an electron orbiting a nucleus. Prior to the point of flip-out (or supposed asymptote), the electron is in a steadily-changing energy field which is drawing the electron out of its current orbit, then a sudden plummeting followed by an even more sudden rise, as the electron joins its new orbit, which continues on smoothly. Many other representations may be inferred from this graph, but it's just one of many that can be generated. This is my "pet" application and it was ported from a programmable Casio calculator. Here is another one portraying Electron Shell Mappings :-

    Electron Shell Mapping Graph

    Trigonometrical Addition Theorems and Repercussions

    First, let's do a simple proof of Pythagoras' theorem :-

    It can be seen from the diagram that the areas of triangles 1 and 2 add up to the area of triangle 3. It can be also seen that the angles in all 3 triangles are equal to each other (look at the angles θ and σ in the diagram) and therefore, the triangles are mathematically similar to each other. This means that each triangle can be scaled to exactly fit either of the other 2 triangles by multiplying each of the sides by a fixed factor. This also applies to the triangles' heights. Let's call the height of triangle 1's right-angle above its base, h. Then, using scaling, triangle 2's height would be :-
    and triangle 3's height would be :-
    In terms of areas of triangles (half base times height), we would have :-
    1) Triangle 1 Area = ½ah
    2) Triangle 2 Area = ½bhb/a
    3) Triangle 3 Area = ½chc/a
    But we already know that (1)+(2)=(3) (triangle areas 1 and 2 add up to 3), so adding equations and doubling, we get :-
    ah + b²h/a = c²h/a
    Dividing by h throughout and multiplying both sides by a, we get :-
    a² + b² = c²

    Once you have accepted that a² + b² = c² in the case of a right-angled triangle where c is the hypotenuse, it becomes quite interesting how things turn out for the relationships between the angles and sides of a triangle. The following diagram is representative of a couple of simple, but sweet theorems, which lead nicely on to the next section on sine and cosine laws.

    Addition Theorems Diagram

    Using Pythagoras we see that
    c² + d² = 1
    Using definitions of sine and cosine we get
    1) c = cos B
    2) d = sin B

    sin² B + cos² B = 1
    Continuing, we get
    3) a + b = c cos A = cos B cos A from (1) and
    4) b = d sin A = sin B sin A from (2).
    We also know
    cos (A + B) = a = (a + b) - b
    Hence from (3) and (4), we get
    cos (A + B) = cos A cos B - sin A sin B
    sin (A + B) = e + f = d cos A + c sin A = sin A cos B + sin B cos A
    Sine and Cosine Laws for any Triangle
    The next diagram shows how to derive the sine and cosine laws for triangles.

    Sine and Cosine Laws Diagram

    1) A + B = 180° - C
    2) cos (180° - C) = - cos C
    3) f = b sin A
    4) f = a sin B
    5) d = b cos A
    6) e = a cos B

    (3) and (4) yield the sine law in
    7) a / sin A = b / sin B
    Again, using Pythagoras, we can see from the diagram that
    8) d² + f² = b²
    9) e² + f² = a²

    Adding (8) and (9), we get
    a² + b² = d² + e² + 2 f²
    a² + b² = (d + e)² - 2 d e + 2 f²

    Using (4), (5), and (6), we get
    a² + b² = c² - 2 a b cos A cos B + 2 a² sin² B
    Using a rearranged (7), so that a sin B = b sin A, we get
    a² + b² = c² - 2 a b (cos A cos B - sin A sin B)
    Now, we use the Addition Theorem above for cosine to simply the bracket contents and rearrange to get
    c² = a² + b² + 2 a b cos (A + B)
    Putting (1) and (2) together, we get cos (A + B) = - cos C, giving the final equation :-
    c² = a² + b² - 2 a b cos C

    This looks like a crooked variety of Pythagoras, bent by the fact that there may not be a right angle. When C is 90°, cos 90° is zero, so it is Pythagoras.

    Circle Geometry and Angles Subtended by Chords

    The angle A in the above and below diagrams, is exactly twice the size of angle B, wherever you draw the chord or the point on the circumference (as long as it's above the chord). B is the angle subtended by the chord at the centre of the circle, and A is the angle subtended at the circumference. Lines with a fleck through them are equal in length and are radii of the circle. To prove this, we use a property of isosceles triangles : the angles opposite the equal sides are also equal. And that the angles of a triangle add up to 180°.

    1) A = x + y
    x + x + y + y + z + z = 180°

    Therefore, x + y + z = 90°
    Rearranging, 2) z = 90° - x - y
    Now, B = 180° - z - z
    So from (2), B = 180° - 90° + x + y - 90° + x + y
    Simplifying, B = 2(x + y)
    Using (1), we get B = 2A

    In this case, angle A has been subtended at a point swung round the circumference to the left. Despite this, we can still prove the same, no matter how small angle w becomes.

    A + x = y
    Hence, 3) y - x = A
    Also, B + 2z = 180°
    4) 2z = 180° - B

    Now, 5) A + y + z + w = 180°
    But, z = x + w
    6) w = z - x

    So, using (6) in (5), we get A + y + z + z - x = 180°
    7) A + y - x + 2z = 180°

    Using (4) in (7), we get A + y - x + 180° - B = 180°
    A + y - x = B

    Using (3), we obtain B = 2A

    This is true even as the angle w tends to zero, which is surprising because one would expect angle A to tend to 0 rather than remain fixed. The reason is because the point on the circumference gets ever closer to the end of the subtending chord, and so the line rendering the angle gets ever shorter, hence w can become really small, but angle A remains the same.
    Back to Main Page