What is Going on Here? Would like an answer, please!

So I was trying to derive a way to approximate π\pi using my compass and straightedge, and then use algebra. I began by creating a 3030 degree angle:

So CADCAD is 3030 degrees. Then, I imagined bisecting this angle into infinity. Notice if we draw a segment from CC to DD, we get an isosceles. We can calculate for this length using the law of cosines: x=22cos(302r)x = \sqrt{2-2\cos(\frac{30}{2r}}), where rr is a reiteration (another bisection). So, if I am correct in my assumption, π\pi should be about xnx \cdot n, if nn are the number of divisions in the circle. We can find nn easily enough: n=242r2n = 24 \cdot 2^{r-2}. we know this becuase, as we bisect, we get a table of values: {24,48,96,}\left \{ 24, 48, 96, \ldots \right \}. Likewise, we find θ\theta thusly: θ=302r\theta = \frac{30}{2^{r}}. This table of values is found by dividing 360 by nn: {30,15,7.5,...}\left \{ 30, 15, 7.5, ... \right \}. So to find xx, we use the law of cosine, take the square, and multiply by n. This gives me the equation: π=22cos302r242r2\pi = \sqrt{2- 2 \cdot \cos \dfrac{30}{2^{r}}} \cdot 24 \cdot 2^{r-2} This makes sense from the way I constructed it, but not here: limr22cos302r242r2\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{30}{2^{r}}} \cdot 24 \cdot 2^{r-2}

This becomes: limr22cos(300)242r2=0\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos(30 \cdot 0)} \cdot 24 \cdot 2^{r-2} = 0 Obviously, π0\pi \neq 0. So, I tested with two values of rr that my calculator could handle. For r=17r =17, π=3.14159265358\pi = 3.14159265358 (correct to 11 decimal places.) However, at r=18r = 18, π=3.1415\pi = 3.1415 (correct to only 4 decimal places.) So why does this equation get close to π\pi, as it is supposed to, and then stop, and then appear to approach zero? Thanks for the help, I am not all that great at math, and would really appreciate it!

Note by Drex Beckman
5 years, 1 month ago

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Extremely interesting! How are you saying that limit is 0? (Hint: It's an indeterminate form, (0 0 \cdot \infty ))
The correct limit is π \pi , just like you wanted.

Ameya Daigavane - 5 years, 1 month ago

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Well, it seemed like as r approached infinity, the cosine would approach 0. of course, cos(0) = 1 and so we would get 0242r20 * 24 * 2^{r-2}. Since the limit is π\pi, is there something wrong with my calculations? Since the precision seemed to degrade for higher r's. Thanks, I don't have the experience of ever learning limits in school, so I do not realize there is such a thing as indeterminate forms, but I was unsure what to do with the 00 * \infty case. I just assumed for any number, you would get zero. Thanks for the help! :)

Drex Beckman - 5 years, 1 month ago

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Oh okay, I'll try to explain then. Suppose you had two functions, f(x) f(x) and g(x) g(x) ,
limxf(x)= \lim_{x \to \infty} f(x) = \infty and limxg(x)=0 \lim_{x \to \infty} g(x) = 0 . Now what is limxf(x)g(x) \lim_{x \to \infty} f(x)\cdot g(x)?

If you think about it, you really can't say, because it could be 0 0 or \infty or some value in between.
Why? It depends on the functions f(x) f(x) and g(x) g(x) themselves.
Let me give you some examples. Let L=limxf(x)g(x) L = \lim_{x \to \infty} f(x)\cdot g(x).

  1. f(x)=xf(x) = x and g(x)=1xL=1g(x) = \frac{1}{x} \Rightarrow L = 1 as f(x)g(x)=1 f(x)g(x) = 1 always.
  2. f(x)=x2f(x) = x^2 and g(x)=1xL=g(x) = \frac{1}{x} \Rightarrow L = \infty as f(x)g(x)=x f(x)g(x) = x always.
  3. f(x)=xf(x) = x and g(x)=1x2L=0g(x) = \frac{1}{x^2} \Rightarrow L = 0 as f(x)g(x)=1x f(x)g(x) = \frac{1}{x} always.
  4. f(x)=5xf(x) = 5x and g(x)=1xL=5g(x) = \frac{1}{x} \Rightarrow L = 5 as f(x)g(x)=5 f(x)g(x) = 5 always.  

So, we've seen that the limit can be anything really. This is why we call 00 \cdot \infty an indeterminate form, it can 'evaluate' to anything.  

Now, to your question, how do we evaluate limr22cos302r242r2\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{30}{2^{r}}} \cdot 24 \cdot 2^{r-2}?
Here, f(x)=242r2 f(x) = 24 \cdot 2^{r - 2} and g(x)=22cos302r g(x) = \sqrt{2 - 2 \cdot \cos \frac{30}{2^r}} (Understand why.)
So what is L, here?
(Convert 30 30 to π6 \frac{\pi}{6} radians)
L = limr22cosπ62r62r\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{\pi}{6 \cdot 2^{r}}} \cdot 6 \cdot 2^{r}
Let t=162r t = \dfrac{1}{6\cdot 2^r} .
L = limt022cosπtt=limt021cosπtt=limt022sin2πt2t=limt02sinπt2t=π \lim_{t \rightarrow 0} \dfrac{\sqrt{2- 2 \cdot \cos \pi t }}{t} = \lim_{t \rightarrow 0} \dfrac{\sqrt{2}\sqrt{1 - \cos \pi t }}{t} = \lim_{t \rightarrow 0} \dfrac{\sqrt{2}\sqrt{2 \sin^2 \frac{\pi t}{2}}}{t} = \lim_{t \rightarrow 0} \dfrac{2 sin \frac{\pi t}{2}}{t} = \pi  

I used the well-known facts that 1cos2x=2sin2x 1 - \cos 2x = 2 \sin^2 x and limx0sinxx=1 \lim_{x \to 0} \frac{\sin x}{x} = 1.

Ameya Daigavane - 5 years, 1 month ago

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@Ameya Daigavane Well explained.

It's a very common misconception to "Let n n \rightarrow \infty in a specific portion of the expression, while ignoring the rest of it".

Calvin Lin Staff - 5 years ago

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@Ameya Daigavane Thanks a lot, I think I understand it now! Very clear explanation, also! +1

Drex Beckman - 5 years, 1 month ago

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@Drex Beckman Thanks! Also, the reason your calculator drops off in precision is because of rounding off errors and approximate values for the cosines and sines.

Ameya Daigavane - 5 years, 1 month ago

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@Ameya Daigavane Right, that was what I suspected, but I put a bit too much faith in the arbitrary precision calculations, I guess. xD

Drex Beckman - 5 years, 1 month ago

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