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Saturday, September 16, 2006
I disgraced myself in the eyes of my ancestors

Just now, a young friend of mine asked of me some wisdom for O level maths. I thought, how hard could that be? So I agreed to take a look at it. Binomial theorem question. Cool, I hadn't touched it in, oh, more than ten years, but hey, I'm God, right? I can do it all. Convinced of my own awesome ability to assimilate and apply, I wiki-ed binomial theorem, and here's what I found.

Great, I thought, given the formula, was there any related problem I could not solve? Fuck, no! So here's the question (more or less). Given (2+x)^6, find the first three coefficients of its binomial expansion in ascending degrees of x and, hence or otherwise, find the values of p and q in the equation:

(2 + px -x^2)^6 = 64 + qx + 48x^2 + ...

The first thing I did was to rearrange the original expression like so.


This would, when I applied the binomial theorem, expand nicely into ascending degrees of x, right? So, I expanded the first 3 terms, and I got

64 + 192x + 240x^2

Cool. Easy part over. So far so good. Then the shit hit the fan. I tried to factorise 2+px-x^2 into something resembling either x+2 or 2+x. I will not go into the gory details of the exotic methods I used or the wrong turns I made. Suffice it to say that trying to do math problems with only your memory, a text edit buffer and the calculator in your computer while IM-ing other people simultaneously is never a good idea.

You seldom see the whole picture.

By then, my friend was getting visibly discontented with my lack of omniscience and I bade the sprout to go to sleep, feeling that I had disgraced myself in the eyes of my ancestors with my lack of brilliance before a minor.

After that, I finally got out some paper and wrote the question down, determined to subdue the dragons of ignorance once and for all. I noticed something immediately.

(2 + px - x^2)^6 = (px - x^2 + 2)^6 = ((px - x^2) + 2)^6

Expand LHS using binomial theorem treating (px - x^2) as one variable, you get:

(px - x^2 + 2)^6 = 64 + 192(px - x^2) + 240(px - x^2)^2 + . . .
(px - x^2 + 2)^6 = 64 + 192px - 192x^2 + 240px^2 + (unimportant stuff where the power of x is greater than 2)

Equating LHS and RHS, you get the following:

64 + 192px - 192x^2 + 240px^2 + . . . = 64 + qx + 48x^2 + . . .

Comparing coefficients of x^2, you have

240p - 192 = 48 => p = 1

Substituting p = 1 and comparing coefficients of x, we get

192 = q

Forgive me, O honourable ancestors, for I have disgraced myself by failing to notice the obvious immediately. As penance, I shall stop declaring that I am God for half a day and promise to always write the fucking problem on a fucking piece of paper with a fucking pen.

p.s. If you're reading this, kid, trust me. Write it down on a piece of paper. It looks way clearer.
ew... it's been 4 years since i've taken a math course. please stop making the painful flashbacks return.
Haha, yeah I don't have any particular love for math either. I just get a masochistic pleasure from occasionally solving math problems.
Cool. Didn't think of it that way first... Ah, you're genius. (taking into account that you haven't done this for ten years or so)
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