Pi Formulae

[Kriminal99]

lim x -> inf

 

x / (2*sin((180-(180/(x/2))/2)) / sin(180/(x/2)))

[Qfwfq]

lim x -> inf

 

x / (2*sin((180-(180/(x/2))/2)) / sin(180/(x/2)))

How about something much simpler:

 

[math]\pi=\lim_{x\rightarrow\infty}x\sin\frac{\pi}{x}[/math]

[C1ay]

It seems a bit circular to use π in a formula for π... :shrug:

[CraigD]

lim x -> inf

 

x / (2*sin((180-(180/(x/2))/2)) / sin(180/(x/2)))

How about something much simpler:

 

[math]\pi=\lim_{x\rightarrow\infty}x\sin\frac{\pi}{x}[/math]

 

:thumbs_up Very cool, despite, as C1ay notes, both being formulae for [math]\pi[/math] in terms of [math]\pi[/math], so not being of much practical computational value

 

Krim, using 180 degrees instead of [math]\pi[/math] radians doesn’t make the angle measure of half a circle any more rational, as, to the best of my knowledge, it’s gotta be converted back before it’s sine can be evaluated. If you know of a computational formula for sin in degrees without any irrational constants, please broaden my knowledge, but I’ve a hunch there can be no such thing.

 

Krim, I think your post is cooler in an easier-to-read way written in LaTeX as:

 

[math] \lim_{x \to \infty} \frac{x}{\left( \frac{2 \sin\left( \frac{\pi-\frac{\pi}{\frac{x}{2}}}{2} \right)}{\sin\left( \frac{\pi}{\frac{x}{2}} \right)} \right)} [/math]

 

Or simplified a little to

 

[math]\lim_{x \to \infty} \frac{ x \sin\left( \frac{2\pi}{x} \right) }{ 2 \sin\left( \frac{(x-2)\pi }{ 2x } \right)} [/math]

 

Now, would you all be so kind as to explain why they work and/or how you found them :QuestionM

Edited October 29, 2010 by CraigD
Fixed missing term in formulae

[Qfwfq]

Craig, I believe you haven't properly translated Krim's notation into LaTeX. For one thing your expressions have a zero limit.

 

It seems a bit circular to use π in a formula for π... :shrug:

I knew someone would say that! :lol:

 

And...

Krim, using 180 degrees instead of [math]\pi[/math] radians doesn’t make the angle measure of half a circle any more rational, as, to the best of my knowledge,

...that is exactly where I was leading to, except I would leave out the "to the best of my knowledge" because there's no way to knock out the implicit circularity with sine or cosine; perhaps with arctan there's more of a hope.

[CraigD]

Craig, I believe you haven't properly translated Krim's notation into LaTeX. For one thing your expressions have a zero limit.

 

Oops!

 

I did leave out a "2", wrote something that converged on [math]2\pi[/math] rather than [math]\pi[/math], but otherwise had it right. I've fixed my post, and simplified the second expression a little more.

 

Evaluating it, it really does appear to converge to [math]2\pi[/math]. I'm curious why (and too lazy/talentless to work on explaining it), so hope someone who knows can explain.

 

Please?

[Qfwfq]

I'm curious why (and too lazy/talentless to work on explaining it), so hope someone who knows can explain.

 

Please?

Are you familiar with the de l'Hôpital rule for computing limits?

 

For my expression it is simpler and, in applying the de l'Hôpital rule, the [imath]\pi[/imath] hops out of the sine. Further, it is equivalent to:

 

[math]\lim_{x\rightarrow0}\frac{\sin\pi x}{x}[/math] analogous to the more familiar [math]\lim_{x\rightarrow0}\frac{\sin x}{x}=1[/math]

 

In Krim's case not the one of the sine terms has the trivial limit: [imath]\sin\frac{\pi}{2}=1[/imath] so you can knock it out and what remains is again like above, with the 2's cancelling out once the de l'Hôpital rule is applied.