The other day, I found myself with an interesting problem of approximating a circle with the enclosing square which seems to prove pi = 4.
The paradox was forwarded by a most interesting puzzle collector, Surajit Basu, a friend and life long inspiration. See Sonata for Unaccompanied Tortoise for why!
Here is the offending paradox:
This is an example of how counterintuitive questions can be answered with a little calculus.
The key is to realize that no matter how closely we approximate the circle, the orthogonal lines of the approximation formed by inverting the square corners will never actually be tangential to the circle.
Note carefully that as you get closer to 90 degrees, the horizontal line is much longer than the vertical. Same goes with the approximation at 0 and 180 - the vertical line is much larger than the horizontal component.
If we take a quadrant of the circle - let's say the top left quadrant, moving counter clockwise from top to left - we can imagine that each infinitesimal arc (at an angle theta) is approximated by a horizontal line that is the approximate length of arc times the cosine of the angle, and the vertical line is the same arc times the sine of the angle.
Here's the rough visual:
Thus, each arc is being approximated by two lines, and we merely add all the approximations. This is where calculus and limits come in. For the one quadrant from 0 to 90 degrees, here is the result:
Multiply by four and you get 8*r (or 4*Diameter).
Voila!
PI is not 4, because the approximate figure is never really the same as the circle, even in the limit of infinite number of approximations.
An interesting result arises from this: Most circles in digital representation (B&W) should have a brightness (or color density) of 4/pi - or about 27% brighter than a real circle of the same dimension in the real world.
Also, can anti-aliasing can be done in a more clever way to not only do edge smoothening, but also reducing the brightness so that the circle's relative brightness is same as physical reality - when the resolution of the picture is less than human eye's resolution?
Is this one of the reasons why Apple's move to Retina display - where the pixel resolution is better than retinal resolution - makes the iPhone (and now iPad3) different?
More questions than answers.
The paradox was forwarded by a most interesting puzzle collector, Surajit Basu, a friend and life long inspiration. See Sonata for Unaccompanied Tortoise for why!
Here is the offending paradox:
This is an example of how counterintuitive questions can be answered with a little calculus.
The key is to realize that no matter how closely we approximate the circle, the orthogonal lines of the approximation formed by inverting the square corners will never actually be tangential to the circle.
Note carefully that as you get closer to 90 degrees, the horizontal line is much longer than the vertical. Same goes with the approximation at 0 and 180 - the vertical line is much larger than the horizontal component.
If we take a quadrant of the circle - let's say the top left quadrant, moving counter clockwise from top to left - we can imagine that each infinitesimal arc (at an angle theta) is approximated by a horizontal line that is the approximate length of arc times the cosine of the angle, and the vertical line is the same arc times the sine of the angle.
Here's the rough visual:
Thus, each arc is being approximated by two lines, and we merely add all the approximations. This is where calculus and limits come in. For the one quadrant from 0 to 90 degrees, here is the result:
Voila!
PI is not 4, because the approximate figure is never really the same as the circle, even in the limit of infinite number of approximations.
An interesting result arises from this: Most circles in digital representation (B&W) should have a brightness (or color density) of 4/pi - or about 27% brighter than a real circle of the same dimension in the real world.
Also, can anti-aliasing can be done in a more clever way to not only do edge smoothening, but also reducing the brightness so that the circle's relative brightness is same as physical reality - when the resolution of the picture is less than human eye's resolution?
Is this one of the reasons why Apple's move to Retina display - where the pixel resolution is better than retinal resolution - makes the iPhone (and now iPad3) different?
More questions than answers.
Circles on digital displays have diagonal steps as well as horizontal and vertical steps so the length will be closer to pi than 4 is and any intensity error will be less.
ReplyDeletePure (abstract) math concerns itself only with a single dimension line of a circle though it is defined by two dimensions (on the cartesian plane) in the calculation of the ratio of pi.
ReplyDeleteThe dimensional quality and difference between a drawn line and an abstract line, a straight line and a curved line are lost in the abstract. The shape of the point on the drawn line is different depending on its curvature which is why it can't be comprised of a single dimension. The inside of the line and the outside are different (e.g. the surface of an egg's exterior is greater than the surface area of its interior).
The abstract, purely theoretical circle ignores this entirely, that the inside of the line and the outside are different - there's no similarity between the one dimensional straight line and the two dimensional curve, which limits the application of π (3.14…) for calculation of the circumference of an infinitesimally narrow line to abstract expressions only.
Regarding the shape of the points along the curve, I predict lining up the arc/wedge-shaped (the shape of the "points" along the curve), will add up to 4.00000 when lined up point-to-point at the furthest part of the wedge diameter as the width of individual segments approaches zero.
BTW, the diagram of a stepped (bit-mapped) circle of increasing granularity works for both area and circumference and is a proper application of limits, and is not a fractal. It requires a regression of limits for area, but not for circumference.
Just another note, the rebuttal by Calculus (at least one which includes SIN and COS, doesn't actually remove the paradox, because both SIN and COS assume π = 3.14, making it circular reasoning and no proof at all. If SIN and COS operated under the assumption that π = 4.00, then the same derivative would prove that π = 4.
ReplyDeleteI think you can, however, prove it using Pythagorean's theorem to calculate the length of the tangent (versus the right angle sides). Then I think the limit properly approaches 3.14 and you can now go back and prove that COS and SIN can assume π=3.14.
However, the devil is in the details, and assuming a one dimensional line moving through, and described by two dimensional space (the cartesian plane) creates the paradox because in real space, the line inscribing the circle has a width, "stretching" the outside, and "compressing" the inside. If you lined up the arc segments of infinitely small widths (an infinite number of them), point to point (the widest portion of the pseudo-trapezoid), then you'd get the π=4, but you'd have an infinite number of triangular gaps in the resulting line. Subtracting that, using Pythagorean's theorem to calculate the gap width, and you're back to π=3.14.
thanks for the explanation. I was thinking about the circular reasoning. The way the square is folded to touch the circle already assumes that pi=3.1415. There are infinite ways to fold a corner. For example if dy=dx always, you get the paradox that 4 = 2*sqrt(2) - the square becomes 'arbitrarily close' to the smaller square formed from the midpoints of each side... I assumed the original puzzle was "assume pi=3.1415... (circle definition), and prove that pi=4 using a tricky construction" :)
ReplyDeleteAwesome content ,clears all the concept and nice pictures too.You can also check our work on visiting architecture firms in dubai .
ReplyDelete