Is there any simple explanation behind the expression shown in the attachment below?
I also have a proof for it using circles and trigonometry.
Use radians or explicit degrees
In[162]:= Limit[2^(n - 1)*Sin[Pi/2*1/2^(n - 2)], n -> Infinity]
Out[162]= \[Pi]
In[163]:= Limit[2^(n - 1)*Sin[90 Degree*1/2^(n - 2)], n -> Infinity]
Out[163]= 180 °
One simple “explanation” is the limit is the derivative of 2\sin(90t) at t=0 (or 2\sin(\pi t/2) if we change to radians to get an answer of \pi instead of 180).
Aside from pointing out the substitution h=1/2^{x-2} so that the limit is as h\rightarrow0,
I would say the foregoing explanation needs an explanation why the derivative of sine is cosine.
And the explanation of the derivative of sine needs an explanation of essentially why the limit of (\sin h)/h as h\rightarrow0 is 1 (if h is in radians), which is equivalent to the original limit. And we’re back where we started.
One way to break this chain of circular reasoning, is to use a circle and some trigonometry, which I think is the simplest way.
Another way is to use the series expansion of sine. That is built on a lot of analysis, and while it’s simple and I sometimes use it to think through things in my head, it’s not as elementary as the geometrical proof.
This is not a proof but a plausible explanation: 2 \sin(\theta/2) is the length of the chord
joining the ends of an arc of a unit circle cut by an angle of (radian) measure \theta. If the angle \theta is small, the lengths of the arc and chord are approximately equal. If we divide a unit semicircle into n=2^{x-2} equal arcs, each cut out by an angle of measure \theta = \pi/2^{x-2}, then their chords have a total length of n \times 2 \sin(\theta/(2n)) = 2^{x-1} \sin({1\over2}\pi/2^{x-2}). And the sum of the chords is approximately the perimeter of the semicircle, which is \pi.
[Update notice: Fixed some typos.]