## Archive for January, 2010

### What is so Geometric about the Geometric Series

January 17, 2010

Behold the picture:

Exhibit number 2, details on Wikipedia:

Exhibit number 3, calculation of the area under a power curve by Fermat, pilfered from Analysis by Its History, by E.Hairer and G.Wanner (Springer 1996), Sect. I.3, pp. 33-34:

From this picture it looks like the area we want to calculate is sandwiched between the sums of two geometric series:

$B^{k+1}(1-R)(R^k+R^{2k+1}+\dots) \le A(B) \le B^{k+1}(1-R)(1+R^{k+1}+R^{2k+2}+\dots)$

After summing the series we get

$R^k B^{k+1}(1-R)/(1-R^{k+1}) \le A(B) \le B^{k+1}(1-R)/(1-R^{k+1})$

At $R=1$ the upper and the lower bounds come together. Unfortunately, $(1-R)/(1-R^{k+1})$ is undefined for $R=1$ since it reduces to $0/0$. But fortunately Fermat knew how to make sense out of the undefined expressions like this, he knew how to differentiate, so he could figure out that $A(B)=B^{k+1}/(k+1)$.

Had he noticed this peculiar link between differentiation, i.e., making sense of $0/0$, and integration, i.e., calculating areas? I have no doubt he had, Fermat was Fermat. Had he pointed it out to his high-esteemed correspondents? I don’t know. But it was up to Isaac Barrow to expose differentiation and integration as two operations inverse to each other, and up to Newton, Leibniz and Bernoulli brothers to put his remarkable observation to work.

January 3, 2010

The standard way to derive “the quadratic formula” is by completing the square, as explained, for example, at cut-the-knot. As most standard things, it is a bit boring, maybe because everybody have seen it so many times and everyone knows it.

Once upon a time I thought about a more entertaining way to derive this well-known formula for the roots of $x^2+bx+c=0.$ Here is what I had come up with.

To find the roots $x_1$ and $x_2$ of our equation is the same as to factor its left-hand side like $x^2+bx+c=(x-x_1)(x-x_2)$. By expanding the right-hand side of this identity, we can see that $x_1+x_2=-b$ and $x_1 x_2 = c$.

Now, it would be nice to have a formula for $x_1-x_2$ in terms of $b$ and $c.$ Then we could have found the roots. But the trouble is that the equation doesn’t know which one of its roots is $x_1$ and which one is $x_2$, $b$ and $c$ don’t change when we interchange $x_1$ and $x_2$, while $x_1-x_2$ changes sign. It looks like we are out of luck.

Still, getting $x_1-x_2$ up to a sign is good enough, and $(x_1-x_2)^2$ doesn’t change when we flip $x_1$ and $x_2$; finding a formula for that would be good enough. Let us work on this idea.

We already know that $b^2=(x_1+x_2)^2=(x_1-x_2)^2+4x_1 x_2$, but $x_1 x_2=c$, whence $(x_1-x_2)^2=b^2-4c$, and now we can figure out the roots. The quadratic formula will appear, of course.

I have written a mathematica script that shows how the roots of $z^n+bz+1=0$ depend on $b$, you may find it amusing.