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Take a circle with diameter $1$. Obviously its circumference is $\pi$.

Draw a square inside this circle (biggest such). Since the circle's diameter is 1, we can work out that each side of the square would be $\frac{1}{\sqrt{2}}$. Its perimeter, therefore, is $2\sqrt{2}$.

Imagine a triangle being drawn over each side of the square, so as to create a regular octagon. Using the law of sines, we can work out the measure of each side of the octagon. The base of each triangle would be $\frac{1}{\sqrt{2}}$, and the angle opposite to it $135^o$. Each of the other angles would be $\frac{180-135}{2} = 22.5^o$. So,

$$ \frac{1}{\sin135(\sqrt2)} = \frac{x}{\sin22.5} $$ $$ 1 = \frac{x}{\sin22.5} $$ $$ x=\sin22.5 $$

Therefore the perimeter of this octagon is $8\sin22.5$.

We can continue drawing triangles on this octagon, and more triangles on that, and so on. Once the number of sides reaches infinity (which, of course, is not possible in the physical world), the perimeter of the said figure would be $\pi$.

I am trying to write a similar function $f(x)$ such that $$ \lim_{x \to 0}f(x) = \pi $$

Note that $f(x)$ is a function of levels. Level 1 is for 4 sides, level 2 is for 8 sides, then 16 sides, 32 sides and so on.

Insofar: $$f(1) = 2\sqrt2 = 4 \sin 45$$ $$f(2) = 8 \sin22.5$$ $$...$$

I work out that

$$f(x) = 2^x \sin\frac{180}{2^x}$$

When graphing that function: Fig 1

Given this, it doesn't seem to limit on the graph. Say, if it does limit to $\pi$, would it approximate $\pi$ for a relatively large $x$?

Please note. There might be a lot of mistakes here. I'm not studying math at the college-level and is a careless person.

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    $\begingroup$ the problem is degree vs. radians as argument is sin $\endgroup$ – gimusi Mar 22 '18 at 14:24
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You're quite right, there doesn't seem to be a limit on the graph. But that's partly because the graph doesn't really look at what's happening for large values of $x$. It stops at around $x=8$, which turns out to be much too small.

The main problem here, though, is that you're conflating radians and degrees. Some graphing calculators can be instructed to compute $\sin\theta$ with $\theta$ entered in degrees, but most use radians as their default setting. So when you compute something like $2^6\sin(180/2^6)=64\sin(2.8125)$, the calculator returns $\sin(2.8125)\approx0.323$ instead of $\sin(2.8125^\circ)\approx0.049$, so you get a point at height around $20.7$ instead of around $3.14$.

As others have remarked, if you did continue your graph for truly large values of $x$, you would find it converges not to $\pi$ but to $180$. You might find it instructive to do so, to see exactly when the limiting behavior becomes clear. (Ah, I see Ross Millikan and kingW3 have done this, and given essentially the same answer, while I was composing this. I was referring to other "others" at the top of this paragraph.)

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  • $\begingroup$ I set the sin(x) function to receive input in degrees and now it does indeed limit to pi. Thanks a lot! $\endgroup$ – Aravind Suresh Mar 23 '18 at 11:15
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Try with

$$f(x) = 2^x \sin\frac{\pi}{2^x} =\pi\frac{\sin\frac{\pi}{2^x} }{\frac{\pi}{2^x} }\to \pi$$

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  • $\begingroup$ The same argument works for OP's problem showing the limit should be $180$. $\endgroup$ – Ross Millikan Mar 22 '18 at 18:14
  • $\begingroup$ Yes of course it works with any number by the standard limit, the fact is that the OP was try to find out $\pi$. $\endgroup$ – gimusi Mar 22 '18 at 18:19
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enter image description here This is the same graph enlarged, you can see that the graph has a limit point somewhere below 200(at 180) now why isn't it $\pi$? It's because some calculators use radians if you had used $2^x\sin(\pi/2^x)$,you would have gotten a similar graph which would converge to $\pi$ instead to $180$

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The limit should be $180$ if you are taking the sine in radians. Yes, it should be $\pi$ if you are taking the sine in degrees. You should be graphing for larger values of $x$ as even $\frac {180}{2^8} \gt \frac 12$ is not so small. Below is a plot from Alpha that shows nice convergence to $180$ using radians

enter image description here

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