Monday, 24 November 2014

Analysis for Realsies: 8 (final)

Okay, as promised - my last words. =P

In my previous post, I promised you guys that I'd show you how to differentiate \(\sin(x)\) from first principles and that I'd also prove a cool geometric fact to do with the lunes of Alhazen.

So, you suddenly wake up in the dead of night with cold beads of sweat on your forehead aperch'd, and there is but one solitary thought plaguing your addled-mind: how the heck do you actually differentiate \(\sin(x)\)? Well, when in doubt...write it out in first principles: \begin{align}\frac{\mathrm{d}\sin(x)}{\mathrm{d}x}=&\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}\\=&\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\=&\lim_{h\to0}\sin(x)\frac{\cos(h)-1}{h}+\cos(x)\frac{\sin(h)}{h}.\end{align} So, we see that we need to calculate these two limits: \begin{align}\lim_{h\to0}\frac{\sin(h)}{h}\text{ and }\lim_{h\to0}\frac{1-\cos(h)}{h}.\end{align} To calculate the former, consider the following picture:
The height of the smaller triangle is \(\sin(x)\). This is shorter than the circular arc, which is of length \(x=\frac{x}{2\pi}\times2\pi\). Moreover, the area of the circular sector is \(\frac{1}{2}x=\frac{x}{2\pi}\times\pi\times1^2\), which is less than the area of the big triangle: \(\frac{1}{2}\tan(x)\). Putting these two inequalities together, we have: \begin{align}\forall x\in(0,\frac{\pi}{2}), \sin(x)\leq x\leq\tan(x).\end{align} Dividing by \(\sin(x)\) and taking the reciprocal, we see that: \begin{align}\cos(x)\leq\frac{\sin(x)}{x}\leq 1.\end{align} Since \(\lim_{h\to0^+}\cos(h)=0\), the sandwich theorem tells us that: \begin{align}\lim_{h\to0^+}\frac{\sin(h)}{h}=1.\end{align}Since \(\frac{\sin(h)}{h}\) is an even function, the limit from the negative side of \(h=0\) must also approach \(1\). Knowing this first limit allows us to compute the second:\begin{align}\lim_{h\to0}\frac{1-\cos(h)}{h}=&\lim_{h\to0}\frac{1-\cos^2(h)}{h(1+\cos(h))}\\=&\lim_{h\to0}\frac{\sin^2(h)}{h(1+\cos(h))}=\frac{1}{2}\times0=0.\end{align} Shoving this into our first principles computation of the derivative of \(\sin(x)\) and we see that \(\frac{\mathrm{d}\sin(x)}{\mathrm{d}x}=\cos(x)\). W00000t! Having assured yourself that you're perfectly capable of differentiating \(\sin(x)\), you drift back to sleep...

Only to be awoken by an infernal desire to prove that the area of the two dark shaded burgundy "lunes" (the crescent shaped moon things) in the following diagram add up to the area of the white triangle in the middle:

Orrite, let's prove this. First note that the white triangle in the middle is a right-angled triangle (this is true of all triangles circumscribed by a circle whose hypotenuse is the diameter), and we denote its side-lengths by \(a\leq b\leq c\), where \(c\) is the diameter of the big circle. Pythagoras' theorem tells us that \begin{align} a^2+b^2=c^2,\end{align} hence, the area of the yellow semicircle (\(\frac{1}{8}\pi c^2\)) is equal to the total area of the two (duo-toned) smaller red semicircles (\(\frac{1}{8}\pi a^2+\frac{1}{8}\pi b^2\)). Now, the area of the yellow semicircle is the same as the area of white triangle and the two lighter red sections, and therefore:

Area of the white triangle+area of the light red bits=area of the dark red bits+area of the light red bits.

Cancelling the light red bits from both sides then yields the desired result. =)

Well, it's been a good semester guys, I hope that you have an awesome Christmas break!

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