/sci/ - Science & Math

I accept your concession

.

what's le secret?

>>10927340
It's still a+b

>>9751854

So I was in a thread on /b/ and saw pic related. My question is if it is true. (please provide evidence to support your answer)

What is the problem in this argument?

Let <span class="math">\gamma \colon S^1 \to C[/spoiler] be the limit of the outer curves. I guess we have [eq]\gamma(S^1)=C[/eq] where C is the circle. However, <span class="math">\gamma \ne C[/spoiler] since the <span class="math">\gamme[/spoiler] isn't smooth. Hence it is not necessarily the case that [eq]l(C)=l(\gamma)[/eq]. In fact, <span class="math">l(\gamma)=4, l(C)=\pi[/spoiler].

Why isn't Pi 4?

I know this image is a comic and it's not true. But why isn't it?

Infinity and Pi are both theoretical numbers.

If you would go to infinity, you would reach 4. Not 3.14 or something.

This has been beaten to death but please help me out. Most of you have probably seen this image which is obviously wrong, but it led me to think of something.

Say we plot the graph of a semi circle with a radius of 1. y = sqrt(1-(x-1)^2) and integrate from 0 to 2, which leads us to an answer of pi/2 which is half the area of a circle of radius 1. It's what we expect; the area is in fact A = pi*d.

But here's the kicker. Integration is essentially slicing up the area below the curve into an infinite number of infinitely thin slices and adding them. As it often is with calculus and what most people have been taught, when something approaches infinity it appears smooth and not stepped. Or at least, as we increase the number of steps, the resulting error decreases to 0.

So I guess my question is why is it that in this specific case we're not allowed to approach an infinite number of steps to find the circumference of a circle? If it works for integration and other mathematical concepts, why not this? The explanation of "if you zoom in, it's still not a circle" feels like a cop out, for the reasons above. I feel like it should be something more obvious than that.