Friday, January 10, 2014

Childhood Theme Songs

Goofed around pretty much all day; the post below is pretty much all I have to show for it.

Today I looked up all the theme songs of childhood TV shows that I used to like. I lived in France from 6-8 and so the theme songs from that time are of course in French.


Bioman show was really popular when I was young. By my recollection, it was basically just Mighty Morphing Power Rangers done on a lower budget (is such a thing possible?). I thought the show itself was pretty stupid, but this is basically what we would spend our time playing as when we were kids. It was dubbed into French, all the actors were Asian (I assume Japanese). It had something to do with fighting a mad scientist and gengineered humans who could each summon a portion of a fighting robot. The video above isn't actually the opening theme, but it is the one that every kid associated with Bioman. The actual opener wasn't all that great.

Le Chevalier du Zodiac

This show was a phenomena in France. Every boy I knew owned at least a few of the figures from this series. If you accidently broke a piece on the intricate and fragile toy figure, there was going to be waterworks that day. Young boys just flat out loved this series. It was an obsession among my age group. I have vague recollections of the characters fighting older men a lot of the time. By my memory, they just permuted on fight after fight after fight. Again, I don't think this was the actual opener (it is too long) but it is the one I remember from my memory. Also, interestingly enough, it is the same guy who sings the Bioman song. I don't know what is up with that. All I know is that every boy could sing this song.

Ken Le Survivant de la Terre l'enfer

In my memory, he was "The survivor of Earth", but it turns out he survived Hell. Funny how memory works. This was probably the most violent show I would watch as a child. Ken was pretty hardcore, he would get jacked up and stick his fingers into people somehow killing them with (Tai Chi?) pressure points. He also was very fond of making "ka-ka-ka-ka-ka-ka" noises whenever he rapid fire punched/kicked/whatever. Naturally, whenever any group of boys play fought, we always had to make the noises Ken made when he fought. It must have sounded like a group of birds dying.

Galaxy Express 999

I have no real memory of Galaxy Express 999 being very popular. I am pretty sure that it was an older series. I don't think that any of my friends were into it that much. I liked it a good deal, as it involved lots of robots and a woman that a young boy is basically in love with. I am not entirely sure, but I think the woman may have looked like a physical education teacher whom I fancied. Anyway, the theme was straight up metal. Whenever I try to remember how numbers work in French, I always humm "Galaxy Express neuf san quatre vingt dix neuve" (9*100 "neuf-san" + 4*20 "quatre-vingt" + "dix" 10 "neuve" 9). Not that it comes up that often, but there it is.

Saturday, January 4, 2014

Basic Properties of Numbers (Ch 1)

If \(a\), \(b\), and \(c\) are any numbers, then $$a + (b + c)
= (a + b) + c. $$

Using P1, you can derive the same rules for all higher cardinality
sums of numbers. Here are all the expansions of cardinality 4:

$$((a + b) + c) + d =\\
(a + (b + c)) + d =\\
a + ((b + c) + d) =\\
a + (b + (c + d)) =\\
(a + b) + (c + d)$$

The most important thing to take from the above is that we can
reapply this procedure to show that any cardinality of sums can be
substituted down to property \(a + (b + c) = (a + b) + c\). For
instance the 4 cardinality sum of \((d + (e + f) + g)\) is really
just \(d = a\), \((e + f) = b\) and \(g = c\).

For any cardinality-N of sums, it would seem to rely on
having the cardinality-(N-1) of sums defined in order to have a
pattern that you can substitute into?

What is this grouping thing (the parentheses)? How do I
know that I can freely add and remove parentheses around numbers?
Where was grouping defined?

Why is it that grouping only makes sense at cardinality 3
and greater sums?

If \(a\) is any number, then $$ a + 0 = 0 + a = a. $$

For every number \(a\), there is a number \(-a\) such that
$$a + (-a) = (-a) + a = 0. $$

It is then stated that if \(a + x = a\), then \(x = 0\). The proof,

a + x = a\\
(-a) + (a + x) = (-a) + a;\\
(-a) + (a + x) = 0;\\
((-a) + a) + x = 0;\\
0 + x = 0;\\
x = 0.

I would like to be able to have some sort of mouseover for
each line of the proof. So that I could mouseover a line and it
would give me a little message like "P2 was applied" or "(-a) was
added to both side". How to do that?

I would like to have all the equals signs aligned vertically
in the above proof. How do I say that with Latex?

If \(a\) and \(b\) are any numbers, then
$$ a + b = b + a. $$

If \(a\), \(b\), and \(c\) are any numbers, then
$$ a \cdot (b \cdot c) = (a \cdot b) \cdot c.$$

If \(a\) is any number, then
$$ a \cdot 1 = 1 \cdot a = a\\
Moreover; 1 \neq 0$$

I do not understand why I do need to specify \(Moreover; 1 \neq 0\)
for multiplication, but I do not specify that \(Moreover; 0 \neq 1\)
in P2? What if P2 was a system that only had one number, namely 1?

For every number \(a \neq 0\), there is \(a\) number \(a^{-1}\)
such that
$$a \cdot a^{-1} = a^{-1} \cdot a = 1.$$

If \(a\) and \(b\) are any numbers, then
$$ a \cdot b = b \cdot a $$

There are several interesting things to note here:

  1. \(0 \cdot 0^{-1} = 1\) is undefined!

    because of the \(a \neq 0\) in P7.

    It is interesting that people say that it is undefined. I
    am not sure if being undefined is the same thing as not

  2. \(a \over b\) is just a different way of writing \(a \cdot

  3. If \(a \cdot b = a \cdot c\), it does not necessarily follow
    that \(b = c\).

    a \cdot b = a \cdot c and a \neq 0,\\
    a^{-1} \cdot (a \cdot b) = a^{-1} \cdot (a \cdot c);\\
    (a^{-1} \cdot a) \cdot b) = (a^{-1} \cdot a) \cdot c;\\
    1 \dot b = 1 \cdot c);\\
    b = c;

  4. If \(a \cdot b = 0 \) then either \(a = 0\) or \(b = 0\).

    The above point is commonly used to find the solutions of
    equations of the form \((x - 1)(x - 2) = 0.\)

    a \cdot b = 0 and a \neq 0,\\
    a^{-1} \cdot (a \cdot b) = 0;\\
    (a^{-1} \cdot a) \cdot b) = 0;\\
    1 \dot b = 0;\\
    b = 0;

It would be really neat if I could have a sort of inline link
where I could click on the statement of 3 or 4 above, which states
both the given and the conclusion of a proof, and that would cause a
row based proof of the statement to expand out. Can I write

If \(a\), \(b\), \(c\) are any numbers, then
$$ a \cdot (b + c) = a \cdot b + a \cdot c. $$

When does \(a - b = b - a\)?
a - b = b - a,\\
(a - b) + b = (b - a) + b = b + (b - a);\\
a = b + b - a;\\
a + a = (b + b - a) + a = b + b.\\
a \cdot (1 + 1) = b * (1 + 1),
a = b.

Prove that \(a \cdot 0 = 0\)?
a \cdot 0 + a \cdot 0 =\\
a \cdot (0 + 0) =\\
a \cdot 0;\\
(a \cdot 0) + (a \cdot 0) - (a \cdot 0) = (a \cdot 0) - (a \cdot 0),\\
(a \cdot 0) = 0

This proof really amazes me. You start by proving that the
summation of Z is equal to Z. And then you use this fact to take
your original premise and zero its left equality. That is crazy.

First we assert that \((-a) \cdot b = - (a \cdot b)\)

(-a) \cdot b + (a \cdot b) = \\
[(-a) + a] \cdot b = \\
0 \cdot b = \\

By adding \(-(a \cdot b)\) to both sides,

(-a) \cdot (-b) + [-(a \cdot b)] = (-a) \cdot (-b) + (-a) \cdot b \\
(-a) \cdot (-b) + [-(a \cdot b)] = (-a) \cdot [(-b) + b] \\
(-a) \cdot (-b) + [-(a \cdot b)] = (-a) \cdot 0 \\
(-a) \cdot (-b) + [-(a \cdot b)] = 0

Finally, adding \((a \cdot b)\) to both sides,

(-a) \cdot (-b) = a \cdot b.

So evidently, this is enough to "force upon us" that the product
of product of two negative numbers is positive. Hmm, I am not
quite seeing it. I can see that the steps all make logical
sense, but I am failing to actually grasp P1-P9 force the
product of two negative numbers to be positive.

P9 is kind of a big deal. It is basically the justification for
almost all algebraic manipulations

Most factorization is really just a repeated use of P9

Pretty neat example of how P9 is actually used whenever you multiply
Arabic numerals.

The numbers a satisfying \(a > 0\) are called positive, while
the numbers a satisfying \(a < 0\) are called negative.

Positivity can be defined in terms of \(<\), it is possible to reverse the procedure to mean that \(a < b\) means \(b - a\) is positive.

\(P\) is defined as being the set of all positive numbers. Note in
particular that \(a\) is in \(P\) iff \(a > 0\).

For every number \(a\), one and only one of the following holds:

(i) \(a = 0\)
(ii) \(a\) is in the collection \(P\),
(iii) \(-a\) is in the collection \(P\)

If \(a\) and \(b\) are in \(P\), then \(a + b\) is in \(P\).

If \(a\) and \(b\) are in \(P\), then \(a \cdot b\) is in \(P\).

There is a great deal here about inequalities, followed by
a rather elaborate proof showing that \(|a + b| \leq |a| + |b|.\)


I actually took Calculus in University; I passed it with minimal effort and a C to show for it. Post university, my Calculus education was firmly in the "Plug and Pray" camp of development. Whilst taking an exam, I would see a pattern, think I recognize it from something I have seen before, and attempt to work the formula based on some other formula.

At no point did I really understand anything that was going on. Most people use addition, subtraction, multiplication, and division without understanding number theory. You CAN do Calculus without understanding it at all (I did).

However, there is a big disadvantage to learning in this fashion; retention is awful. In order to build up a higher understanding, you need "footholds" of comprehension. Often, as you gain higher understanding of a subject matter, the footholds that you built upon to get at your current elevation are forgotten (internalized might be a better word). I think this is the reason that it is often so difficult for experts to explain their expertise to a laymen.

Anyway, getting distracted. The point is that without these footholds I had a basically flat knowledge of Calculus. I had a number of formulas memorized that were each disparate flat structures of knowledge. Nothing was associated. Nothing reinforced anything else. Naturally, I forgot these formulas. Because there is no association to spark memory, once I had forgotten the formulas, I had effectively lost them.

But, this time is going to be different. I am doing it on my own. I am doing it at my own pace. If desired, I can spend days on a single concept. My simple goal for this year is to understand Calculus at a fundamental level. Such that I can naturally express problems I see in the world in terms of Calculus. I wish to have a layman's mastery of The Calculus.


  • Calculus 4th Ed. , Michael Spivak
    I picked this book as I liked the fact that it seemed to be an attempt to teach Calculus from first principles. I am not going for "can do/use calc" I am going for "understands calc". It has been so long and my knowledge was so shallow that Calculus is effectively new to me. I want a book that is going to assume I don't know anything about Calculus before picking it up.
  • This Blog:
    On this blog, I am basically just going to outline interesting problems I worked through from the book, using this blog as a sort of notebook of my progress. I succeeded in installing a really neat library called mathjax that allows you to display Tex formulas online. Bit excited by that actually.
  • Time:
    I am going at my own pace, I am not skipping when I don't understand. This is for me. I have no class I need to pass. I have no test I need to prepare for. This is for me (although you are welcome to follow along). Until I get it, the procession does not move forward!
  • You:
    I don't know if anyone will follow along, but sometimes my pace may be crawling. If you see that, and you have the time, maybe you want to leave a comment about the current problems I am having? Maybe just a word of encouragement, that helps as well. Thanks.