## 2008年7月29日 星期二

### 循環論證 (II) : 圓周公式

Continue from 循環論證 (I) : 圓周公式

There should be no controversy before the integral arises,

let's verify each steps carefully starting from the arc length.

$\int_{0}^{r}\sqrt{1+\frac{dy}{dx}^2}\,dx$

The arc length formula itself is correct.

Its proof is based on Mean Value Theorem and Riemann Sum.

Here, what we need to verify is the differentiability of $\inline y=\sqrt{r^2-x^2}$ on the interval $\inline \left(0,r\right)$ and the continuity on $\inline \left[0,r\right]$.

Both of them are clear except the continuity at the end-point 0.

But, since we are considering the first quadrant only, the existence of one hand limit is enough.

For the symmetry part, there should be no problem.

{If you really think about it, just consider the properties of the odd function and even function}

Afterwards, the simplification of integral is obviously correct.

Then, we investigate the computation of integral.

$L=4r\left[\sin^{-1} \frac{x}{r} \right]_{0}^{r} \cdots (1)$

$L=4r\left[\frac{\pi}{2}-0\right]=2\pi r \cdots (2)$

As I mentioned in the first part, there must be a flaw in the proof, so the flaw should arise in the computation part.

Before we proceed, let's recall the following high-school definitions:

$\inline \pi$ is defined as the common ratio of circumference to diameter of any circle.

Radian is defined as the ratio of arc length to the radius of a circle, in particular, it is the arc length of the unit circle.

$\inline \left(\sin x,\cos x\right)$ is defined as the point lying on the unit circle making an angle $x$ (in radian) with the x-axis.

$\inline \sin^{-1} x$ is then defined as the inverse mapping on the right half of the unit circle

Let's consider the second equation first, as this is easier to explain.

Here, we use the common sense that $\inline \sin^{-1}1=\tfrac{\pi}{2}$ and $\inline\sin^{-1}0=0$ , but why!?

somehow we know $\inline \sin\tfrac{\pi}{2}=1$

(we must rely on the fact that $\inline \pi$ is the ratio)

So, we found that the second one contains flaw.

However, this does not mean that the first one contains no circular reasoning.
For those who knows a little bit of analysis,

For the first equation, we used Fundamental theorem of Calculus, which itself require the function $\inline \sin^{-1}x$ to be integrable, being the inverse function of $\inline \sin x$, it is continuous, and hence integrable.

To show that $\inline \sin x$ is a continuous function, from epsilon-delta definition, the inequality $\inline \left|\sin x - \sin y\right| \leq \left|x - y\right|$ must be involved, which is often proved using geometric argument