Pi R squared

No, you don't.

Start from the definition of pi, which is the ratio of the circumference of a circle to its diameter. That's all you need, as well as the area of a rectangle:

en.wikipedia.org/wiki/Area_of_a_circle#/media/File:CircleArea.svg

Do that with a square, you'll see it's bullshit

First, you can easily find an integral representation of the area of the circle via basic geometry. Something like
[eqn]A_\circ &= 4\int_0^R \sqrt{R^2-x^2} \mathrm{d} x = 4 R \int_0^R \sqrt{1-\left(\frac{x}{R}\right)^2} \mathrm{d} x[/eqn]
[eqn]= R^2 \underbrace{4\int_0^1 \sqrt{1-t^2} \mathrm{d} t}_{\mathrm{const.}}[/eqn]

Now we know that [math]A_\circ = cR^2[/math]. What is left is the calculation of the constant c, which will turn out to be π. There's of course an analytical way to do it, but it's not really interesting, as it kind of requires extensive knowledge about π to do, so it's in a way a snake-eats-tail-thing.
[eqn]4\int_0^1 \sqrt{1-t^2}\mathrm{d} t = 4\int_0^{\pi/2} \cos u\sqrt{1-\sin^2 u}\mathrm{d} u = 4\int_0^{\pi/2} \cos^2(x)\mathrm{d} x = 4\int_0^{\pi/2} \sin^2(x)\mathrm{d} x [/eqn]
[eqn]\Longrightarrow 4\int_0^1 \sqrt{1-t^2}\mathrm{d} t = 2\int_0^{\pi/2} \cos^2(x)\mathrm{d} x + 2\int_0^{\pi/2} \sin^2(x)\mathrm{d} x = 2\int_0^{\pi/2} \mathrm{d} x =\pi[/eqn]

Could you please detail how to calculate that integral at the end ?

Trig sub, x=cos(theta), dx = -sin(theta) dtheta
You end up with integral of -sin^2(x) which you can solve with the identity sin^2(x)=(1-cos(2x))/2

But you're gonna need the fact that the derivative of cos is -sin ! How do you prove that ?

By the way, if you happen to use the limit of (sin h)/h at 0, i'm also gonna need a proof of that please

Fucking hell, fine.

[math]\displaystyle \lim_{h\rightarrow 0} { \frac { \cos{(x+h)} - \cos{(x)} } { h } }[/math]
Which can be expanded with the half angle formula (cos(a+b) = cos(a)cos(b)-sin(a)(sin(b))
[math]\displaystyle \lim_{h\rightarrow 0} { \frac { \cos{(x)}\cos{(h)} - \sin{(x)}\sin{(h)} -\cos{(x)} } { h } }[/math]
[math]\displaystyle = \lim_{h\rightarrow 0} {( \frac { \cos{(x)}\cos{(h)} - \cos{(x)} } { h } } + \frac {-\sin{(x)}\sin{(h)}} {h} )[/math]
[math]\displaystyle = \lim_{h\rightarrow 0} {( \cos{(x)} \frac { \cos{(h)} - 1 } { h } } + -\sin{(x)} \frac {\sin{(h)}} {h} )[/math]
So now we have to prove (cos(h)-1)/h approaches 0 and sin(h)/h approaches 1.
Let's start with the first.

[math]\displaystyle \lim_{x \to 0} \ \frac{\cos x - 1} x = \displaystyle \lim_{x \to 0} \ \frac{(\cos x - 1)(\cos x + 1)}{x(\cos x + 1)}[/math]
[math]=\displaystyle \lim_{x \to 0} \ \frac{\cos^2 x - 1}{x(\cos x + 1)} = \displaystyle \lim_{x \to 0} \ \frac{-\sin^2 x}{x(\cos x + 1)} = \displaystyle \left(\lim_{x \to 0} \ \frac{\sin x} x\right) \left(\lim_{x \to 0} \ \frac{-\sin x}{\cos x + 1}\right) = \displaystyle ? \cdot \left({\lim_{x \to 0} \ \frac{-\sin x}{\cos x + 1} }\right)[/math]
Which finally equals:
[math]\displaystyle ? \cdot \frac{\lim_{x \to 0} \ (-\sin x)}{\lim_{x \to 0} \ (\cos x + 1)} = ? \cdot \frac {0} {2} = 0[/math]
So even though it includes sin(h)/h, it's irrelevant what the solution is here. What we have left for the derivative is this:
[math]\displaystyle -\sin{(x}) \limit_{h \to 0} { \frac {\sin{h}} {h} }[/math]

I'm running out of space, so I'll derive the sin(h)/h limit in the next post. I really hope this formatted well and I don't have to redo it.

Right, the only way I know of to solve the sinh/h problem is geometrically.
If we set up a circle of radius 1, and set up three points (0,0), (cos(x),0), (cos(x),sin(x)) to get a triangle, then the area of the triangle will be cos(x)sin(x)/2.
The area of the sector from (0,0), (1,0), and (cos(x),sin(x)) would, obviously, be x/(2pi) by knowing how to take the area of a circle, which was proved here: . Since the area of a circle of radius 1 is pi, we have area = x/2 of the second triangle.
The area of the triangle from (0,0), (1,0), and (1,tan(x)) would be simply 1*tanx/2.
So now we get:
[math]\displaystyle \frac {\cos{x} \sin{x}} {2} \leq \frac {x} {2} \leq \frac {\tan{x}} {2}[/math]
[math]\Longleftrightarrow \displaystyle \frac {\cos{x} \sin{x}} {2} \leq \frac {x} {2} \leq \frac {\sin{x}} {2\cos{x}}[/math]
Multiply through by 2/sinx
[math]\Longleftrightarrow \displaystyle \cos{x} \leq \frac {x} {\sin{x}} \leq \frac {1} {\cos{x}}[/math]
Reciprocate.
[math]\Longleftrightarrow \displaystyle \frac {1} {cos{x}} \geq \frac {\sin{x}} {x} \geq \cos{x}[/math]
And when we take the limit of the left, we get 1, and the limit of the right gets 1. Therefore sin(x)/x must approach 1.

Nice; solution without calculus.

Why? The circumference of a concentric square is just 8r, where r is the minimal radius. The integral from 0 to R of 8r dr is 4R^2=(2R)^2, where R is half of the side length of the square. Easier yet, the area is just the integral from 0 to S of Sx/2 dx, which is just S^2 (where S=2R is the side length).

you do realize that's a coincidence and not a valid derivation right?

ok you do, cool

you guys haven't even proved circles exist

Well there's no other way to take the limit of sin(h)/h as h->0 with calculus if you don't know the derivative of sine, which depends on that limit. You also don't know the Taylor series form of sine because that involves knowing the derivative of sine.

There was a series already known before calculus for sine, though. It also happens to be the Taylor series but it obviously wasn't derived the same way. It's known as Madhava's sine series. I would've just used that, but I decided it'd be better if I went by the book.

It works with squares, but it (very ironically) doesn't work with rectangles. If you had a rectangle with sidelengths 2r and r, centered at the origin, then the perimeter is 6r, integrate that from 0 to r and you get 3r^2, when the area is actually supposed to be 2r^2.