I came across this so called proof that pi = 4
[IMG]http://2.bp.blogspot.com/_0X1ggm5ZqsA/TOvYP5KOhgI/AAAAAAAAJl4/PgZZ8ANWcKY/s1600/troll_mathemathics_pi.jpg[/IMG]
consider every horizontal line to be the x component of the shape and every y to be the vertical component.
And before you tell me wat I already know, Im gonna tell you wat I already know. And I already know pi is not 4. pi is the ratio of the circumference of a circle to its diameter.
I started this thread to offer a theory as to why its not 4.
My theory is basically as the amount of those right angle cuts approach infinity, I was thinking that the perimeter does change, and at first that sounds illogical, but hold on. I think that If I made infinite cuts around that square, my circumference would become pi. As you can see the pixel 'circle' is made up of only right angle corners, the more cuts that I make, the smaller and smaller the length of the x and y components of each corner become. At infinity, the length of the x and y components of the corner is 0, so you don't have a corner anymore, you have an infinitesimally small dot. Every infinitesimally small dot connects tangentially to the boundary of the real circle perfectly, and because the x and y component lengths are now 0 they cannot add together to give you a perimeter/circumference of 4, because they don't exist anymore, just the infinitesimally small dot. The circle of dots now have a circumference of pi instead of 4, only because the circumference is the length of all the infinite amount of dots that make up the circle.
Why am I thinking this way? well I have herd of another theory, and its that no matter how many cuts you make, you will never ever approach a perfect circle, you will always get a more pixely version. If that was the case, then in calculus I should never be able to find the exact area under a curve using an infinite amount of rectangles, my area would always be almost there because it just becomes a more pixely version of the actual area. However that is false, with an infinite amount of rectangles, I can calculate the exact area under the curve if I set the amount of rectangles to infinity. So why can I not use this for the circle in a square? Its essentially the same thing.
Am I correct in my approach? What do you think?
nah man pi's that dessert with the crust and has fruit or cream in it
Wouldn't you end up with a diamond shape than circle?
[QUOTE=CaptainSnake;43868105]Wouldn't you end up with a diamond shape than circle?[/QUOTE]
Then you'd end up with sqrt(2) = 2, which is also complete nonsense. The thing is, you can never reach infinity so you can never have "infinitesimally small dots", no matter how small each segments are, they will still be segments and they will still join at 90° corners, and that's why you're never really approaching a circle.
i see no evidence that the perimeter of the squares ever becomes the circumference of the circle
edit: if you're interested in this sort of thing OP take a look at mathematical sequences & series
the inconsistancy here is that the surface of the square approaching the circle is infinitely rough, therefore adding extra perimeter, while the circle is infinitely smooth.
wish i had some pie right now
[QUOTE=AK'z;43869101]wish i had some pie right now[/QUOTE]
Apple pie is pretty fucking good. Specially with cinnamon.
haven't had apple crumble in years either.. that stuff was the greatest thing to exist
I don't think there is a need to try and disprove pi being 4 because you can prove that with a measuring tape any anything round in a much shorter time with much more understandable vocabulary.
[QUOTE=Disseminate;43869043]i see no evidence that the perimeter of the squares ever becomes the circumference of the circle
edit: if you're interested in this sort of thing OP take a look at mathematical sequences & series[/QUOTE]
What I am confused about is why in calculus we are able to do things like this:
[IMG]http://tutorial.math.lamar.edu/Classes/CalcI/AreaProblem_files/image003.gif[/IMG]
but we cant apply the same logic to turning a square into a circle. Everyone says you cant get to infinity which is true. But there is still a concept of approaching it, and in calculus when we take the limit of something at infinity, we see what something could be if it actually was at infinity.
In the picture above to find the area under that curve the amount of rectangles are increased to infinity and their widths become 0. We add all the areas of the infinite rectangles together and we get the area under the curve. There is a point where the approximated area of the curve equals the area of the curve, and the only way for that to happen is if that shape under the curve becomes smooth. Why can the same logic not be applied to the circle?
[QUOTE=noh_mercy;43870297]In the picture above to find the area under that curve the amount of rectangles are increased to infinity and their widths become 0. We add all the areas of the infinite rectangles together and we get the area under the curve. There is a point where the approximated area of the curve equals the area of the curve, and the only way for that to happen is if that shape under the curve becomes smooth. Why can the same logic not be applied to the circle?[/QUOTE]
Because area and perimeter are completely different things.
[QUOTE=_Kilburn;43870395]Because area and perimeter are completely different things.[/QUOTE]
So by my understanding, the top part of the approximated curve is still infinitly rough(when the rectangles go to infinity), but somehow the areas under the real curve and the aproximated curve are identical. Right?
[url]http://britton.disted.camosun.bc.ca/Pi_unrolled.gif[/url]
Isn't this the right explanation of pi?
(gifs dont work for me)
We are talking about pie not circles GOD
[QUOTE=noh_mercy;43870435]So by my understanding, the top part of the approximated curve is still infinitly rough(when the rectangles go to infinity), but somehow the areas under the real curve and the aproximated curve are identical. Right?[/QUOTE]
Pretty much. The same applies for that proof you posted above, by removing corners the perimeter doesn't change but the area gradually decreases until it becomes equal to the area of the circle.
[QUOTE=_Kilburn;43870825]Pretty much. The same applies for that proof you posted above, by removing corners the perimeter doesn't change but the area gradually decreases until it becomes equal to the area of the circle.[/QUOTE]
If that is true, then there is only 1 problem. The only logical way for the approx area to truly become equal to the area under the curve and still be a different shape is if part of the approximation crosses over the boundary of the curve its approximating. But thats impossible.
For example I can draw a square and a zig zaggy square and their areas will be equal as long as one part of the zig zaggy square extends beyond the boundaries of the real square.
I feel like both sides of the counter to this proof have problems. And I found it more convenient to actually think that the perimeter goes from 4 -> pi, because ultimately it has to do with the way I look at 0. and there are 2 ways to think of 0, either 0 is truly nothing, or 0 is an infinitely small something. If you think of 0 as an infinitely small something, then the zig zaggy shape counter proof would work for you. And If you take 0 = nothing, than its easy to see how the x and y components of the approximated shape become 0 distance in length, changing the actual shape of the approximated shape. Calc just teaches you when to say "its close enough to be the same as _____"
edit
and fundamentally, all this tells me is that you literally run into logic problems if 0 and infinity are in the same sentence.
[QUOTE=noh_mercy;43870435]So by my understanding, the top part of the approximated curve is still infinitly rough(when the rectangles go to infinity), but somehow the areas under the real curve and the aproximated curve are identical. Right?[/QUOTE]
The difference is that we are measuring areas instead of an arc length. With the arc length, we should be using the formula
[IMG]http://latex.codecogs.com/gif.latex?S%20%3D%20%5Cint_%7Ba%7D%5E%7Bb%7D%20%5Csqrt%7B1+%5Cleft%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%20%29%5E2%7D[/IMG]
[editline]11th February 2014[/editline]
where dy/dx is the derivative of the function that we're trying to find out
[QUOTE=OogalaBoogal;43873411]The difference is that we are measuring areas instead of an arc length. With the arc length, we should be using the formula
[IMG]http://latex.codecogs.com/gif.latex?S%20%3D%20%5Cint_%7Ba%7D%5E%7Bb%7D%20%5Csqrt%7B1+%5Cleft%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%20%29%5E2%7D[/IMG]
[editline]11th February 2014[/editline]
where dy/dx is the derivative of the function that we're trying to find out[/QUOTE]
use this formula for calculating what the circumference of the circle?
[QUOTE=noh_mercy;43874577]use this formula for calculating what the circumference of the circle?[/QUOTE]
equation of a circle centered at (0,0):
x^2+y^2=r^2
y = +- sqrt(r^2-x^2)
we'll just split it into two since a circle is symmetrical, find the arclength of half a circle and multiply that by 2.
plug that into the formula:
arclength = integral from a to b of sqrt ( 1 + (dy/dx)^2)
the deriviative of the circle equation is:
dy/dx = -x/sqrt(r^2-x^2)
squared:
x^2/(r^2-x^2)
plug it into the formula:
arclength = integral from a to b of sqrt ( 1 + x^2/(r^2-x^2))
should integrate from a=-r to b=r, but since its symmetrical we integrate from a=0 to b=r and then multiply by 2 again.
circumference = 4 arclength
circumference = 4 * integral from a=0 to b=r of sqrt(1+x^2/(r^2-x^2)) = 2*pi*r
[IMG]http://i.imgur.com/06VNnYL.png[/IMG]
Why is everyone trying to use advance formulas to prove this wrong. There will always be enough space between the circle and the squares in order for the squares to add up to 4, regardless of how many squares you cut it down to. The length of each square will decrease as the # of squares increase, keeping the permitter at 4. For the circle, if D=1 then C=3.14159...
What.
trolled softly.
[t]http://puu.sh/6SPsz.png[/t]
I should probably add that that's the limit as the corners cut approaches infinity. You can't cut those corners any more. It never becomes a circle.
This is true with your curve estimation too-- it just looks like a circle at low resolutions because of how our eyes see things traced onto other things.
[QUOTE=Itszutak;43878128][t]http://puu.sh/6SPsz.png[/t][/QUOTE]
Why does that even matter? Circle, octogon or diamond, you're starting with a shape which has a perimeter of 4 and end with a shape with a perimeter that is "seemingly" less than 4.
Now of course that circle proof is wrong, but your counterargument is wrong as well, the shape never really approaches a circle because the corners will always remain there, not because the shape becomes an octogon instead of a circle (because it doesn't).
[QUOTE=noh_mercy;43866380]I chink?[/QUOTE]
no
as it gets smaller it becomes a fractal edge with this pattern
/\/\/\/\/\/\/\/\/\/\/\/\/\
whereas a circle's edge is this non-pattern
____________________
they look like the same thing from a macroscopic observation point but they aren't
move along people
You're just gonna end up with a crinkled line with the length 4.
people. I already addressed the crinkly explanation of this troll proof. Telling me that youre gonna end up with a very tiny crinkled 'circle' is something I already know.
All I was doing was asking if it is possible to look at this another way, I was still proving it wrong but thinking of another method.
This is how I was thinking of it:
Imagine having a piece of 2d material that can be folded infinitely. Now I have a square shape of this 2d material, draw a circle of diameter 1 on to the square material.
now fold each corner of the square onto the boundary of the circle you drew.
All I'm doing is taking a point off of the square and making that point tangent to the boundary of the circle. The square and the circle are both made of an infinite amount of points, because thats what lines and curves are made of (correct me if im wrong).
So If I fold an infinite amount of times, I would be able to displace an infinite amount of points from the square to the circle. Thats what was making me think that it would be logical to assume that the perimeter of the square changes to the circumference of the circle. 4 changes to 3.14, not steadily, but instantly. But i can tell that sounds crazy just by looking at the comments you have shared with me and the dumb ratings lawl.
btw the fractal explanation makes the most sense of all that I read.
I am moving this to fast threads because it's pretty much just a question and it fits better there. You were already banned for putting it in debate. Next time, ask a mod where it ought to go before you remake it if you're unsure.
Okay, the problems here are basically:
1) A picture is not a proof. "It looks like the circle" doesn't mean the shape becomes a circle as you add more lengths.
2) You are using the wrong metric.
The first point is self-explanatory. For the second point, by restricting to horizontal and vertical lines like that, you are essentially [I]no longer in standard Euclidean space[/I]. You are in the plane with the "taxicab metric" or "Manhattan metric," where distances are measured by the sum of the differences in both coordinates of the point. Notice that in normal polygonal approximations of the circle, you've still got diagonal lines going on, because they're allowed, and you're simply taking a limit as the number of sides increases. That's why these approximations are valid, they're still in the Euclidean plane.
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