Quod Erat Demonstrandum

2008/05/07

Filed under: HKALE,Pure Mathematics — johnmayhk @ 4:29 下午

In AL pure mathematics syllabus, we come across the following formulae.

$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$
$1^3 + 2^3 + 3^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$

More?

Just post the following 10 formulae (copied from mathworld) for your reference.

We may observe that, when the power of $k$ is even, there must be a factor $n(n+1)(2n+1)$ in the expressions on R.H.S. It is a must?

For any natural numbers $m$ and $n$, consider a polynomial $P(x)$ such that

$P(n) = \sum^n_{k=1}k^m$ for $k \in \mathbb{N}$

That is, values of polynomial $P(x)$ match the values of so-called power sums (i.e. $\sum_{k=1}^nk^m$) when $x$ is a natural number.

Urm, forming a polynomial out of the power sum is just extending the domain from set of natural numbers to that real numbers. And the extension is extremely easy, just write (say)

$\frac{n(n+1)}{2}$ for $n \in \mathbb{N}$

as

$\frac{x(x+1)}{2}$ for $x \in \mathbb{R}$

So, to prove that $\sum_{k=1}^nk^m$ has a “factor" of $n(n+1)(2n+1)$ for even number $m$, we may show that $P(0) = P(-1) = P(-\frac{1}{2})$.

As Marco_Dick said in the mathdb blog, this may be the idea of setting one of questions in AL Pure Math (I) 2008.

By the definition of $P(x)$, we have

$P(1) = 1$
$P(x) = P(x - 1) + x^m$ for any natural number $n > 1$

After “extension", the second statement could be written as

$P(x) = P(x - 1) + x^m$ for any real number $x$ – – – – – – (*)

It is extremely easy to show $P(0) = P(-1) = 0$, just put $x = 1$ and then $x = 0$ into (*). For showing $P(-\frac{1}{2}) = 0$, we replace $x$ by $-x$ in (*), yield

$P(-x) = P(-x - 1) + x^m$ for any real number $x$ – – – – – – (#)

See, the “$m$" plays a very important role here since $(-x)^m \equiv x^m$ for EVEN NUMBER $m$.

Now (*) – (#), and rearrange, we have

$P(x) + P(-x - 1) = P(x - 1) + P(-x)$ for any real number $x$ – – – – – – (@)

When $x = 0$, $P(x - 1) + P(-x) = P(-1) + P(0) = 0$. Then, by M.I. and (@), we have

$P(x) + P(-x - 1) = 0$ for any natural number $x$

Now, we “extend" the above result from natural number to real number by knowing the fact that $P(x) + P(-x - 1)$ is a polynomial (i.e. finite degree), but the result just above is saying that $P(x) + P(-x - 1) = 0$ has infinitely many roots (0,1,2,3,…), then $P(x) + P(-x - 1)$ is identical to the zero polynomial, that is $P(x) + P(-x - 1) \equiv 0$

Now, take $x = -\frac{1}{2}$, immediately, $P(-\frac{1}{2}) = 0$ as expected.

P.S. The whole story is based on the “method of difference", could you see the linkage among the articles and the original question? Also, it is always my bad habbit to say something like “還有更多想說，有機會再談。" (as in the so-called article posted), in most of the cases, I actually forget about what stuff I wanna say…sorry all!

2 則迴響 »

1. Hello, I am a S.7 student from a school (very close to where you works…hmm)
I’d like to share something about the topic here.

As m is even, we know that x(x+1)(2x+1) is a factor of p(x).
Then I wonder what happens when m is odd.

From the list of power sums, we can easily observe that x^2(x+1)^2 is a factor of p(x) as m is odd.

And it is fun for some interested student to prove it~~haha
Or teachers can design a question similar to that of 2008….

迴響 由 rai — 2008/05/14 @ 12:26 上午 | 回應

2. Good idea rai!

迴響 由 johnmayhk — 2008/05/14 @ 8:07 上午 | 回應