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So I was trying to derive a way to approximate $$\pi$$ using my compass and straightedge, and then use algebra. I began by creating a $$30$$ degree angle:

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

This becomes: $\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos(30 \cdot 0)} \cdot 24 \cdot 2^{r-2} = 0$ Obviously, $$\pi \neq 0$$. So, I tested with two values of $$r$$ that my calculator could handle. For $$r =17$$, $$\pi = 3.14159265358$$ (correct to 11 decimal places.) However, at $$r = 18$$, $$\pi = 3.1415$$ (correct to only 4 decimal places.) So why does this equation get close to $$\pi$$, 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
1 year ago

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Extremely interesting! How are you saying that limit is 0? (Hint: It's an indeterminate form, ($$0 \cdot \infty$$))
The correct limit is $$\pi$$, just like you wanted. · 1 year ago

Well, it seemed like as r approached infinity, the cosine would approach 0. of course, cos(0) = 1 and so we would get $$0 * 24 * 2^{r-2}$$. Since the limit is $$\pi$$, 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 $$0 * \infty$$ case. I just assumed for any number, you would get zero. Thanks for the help! :) · 1 year ago

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

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

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

So, we've seen that the limit can be anything really. This is why we call $$0 \cdot \infty$$ an indeterminate form, it can 'evaluate' to anything.

Now, to your question, how do we evaluate $$\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{30}{2^{r}}} \cdot 24 \cdot 2^{r-2}$$?
Here, $$f(x) = 24 \cdot 2^{r - 2}$$ and $$g(x) = \sqrt{2 - 2 \cdot \cos \frac{30}{2^r}}$$ (Understand why.)
So what is L, here?
(Convert $$30$$ to $$\frac{\pi}{6}$$ radians)
L = $$\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{\pi}{6 \cdot 2^{r}}} \cdot 6 \cdot 2^{r}$$
Let $$t = \dfrac{1}{6\cdot 2^r}$$.
L = $$\lim_{t \rightarrow 0} \dfrac{\sqrt{2- 2 \cdot \cos \pi t }}{t} = \lim_{t \rightarrow 0} \dfrac{\sqrt{2}\sqrt{1 - \cos \pi t }}{t} = \lim_{t \rightarrow 0} \dfrac{\sqrt{2}\sqrt{2 \sin^2 \frac{\pi t}{2}}}{t} = \lim_{t \rightarrow 0} \dfrac{2 sin \frac{\pi t}{2}}{t} = \pi$$

I used the well-known facts that $$1 - \cos 2x = 2 \sin^2 x$$ and $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. · 1 year ago

Well explained.

It's a very common misconception to "Let $$n \rightarrow \infty$$ in a specific portion of the expression, while ignoring the rest of it". Staff · 12 months ago

Thanks a lot, I think I understand it now! Very clear explanation, also! +1 · 1 year ago

Thanks! Also, the reason your calculator drops off in precision is because of rounding off errors and approximate values for the cosines and sines. · 1 year ago