- Deriving Pi
- 17 Sep 2019 10:17:36 pm Permalink
- Last edited by Sam on 20 Oct 2019 06:24:27 pm; edited 2 times in total

Link to the Nspire CX II document containing my work

I'm sure Euler is rolling in his grave about now.

About three weeks ago, following my purchase of the new TI-Nspire CX II CAS, I embarked upon a mathematical journey with the intent of doing the impossible. Literally, I'm pretty sure this is impossible.

So what exactly am I trying to do?

Pi is an irrational number, meaning it cannot be represented as a fraction, nor do its decimal digits follow any pattern. Similarly, sqrt(2) is an irrational number, also unable to be represented as a fraction. Who's to say that pi cannot be expressed as a combination of other irrational numbers (in this case, roots)? I am attempting to express pi as a finite expression containing only operations made on rational numbers. Finite means that my expression cannot be an infinite series. One should be able to write pi, exactly, with one pencil and one sheet of paper.

Consider the following system of equations:

y=x

y=0

x=0

(x-1)^2+(y-1)^2=1

Since the radius of the circle is 1, the area must be pi. Therefore, the area bounded by the circle, y=x, and y=0 must be (4-pi)/8

Now imagine that shape combined with another identical shape tacked onto the back, but flipped.

graph y=x, y=0, x=0, (x-1)^2+(y-1)^2=1, y=x+1, (x+1)^2+y^2=1

What if we warped this shape to fall upon y=-x, thus eliminating the bottom curve?

graph y=x, x=0, (x-1)^2+(y-1)^2=1, y=x+1, y=-x, (x-(sqrt(2) - 1)/sqrt(2))^2+(y-(1/2+(sqrt(2) - 1)/sqrt(2)))^2=.25

Now, if we enlarge that shape by a factor of four, it should fit right back on to the original circle.

Since this new shape should be four times the area of the original shape, and we know the area of the original shape is equal to corner of the square, we should be able to slice the edges off the new shape (the edges of which we know the lengths of), measure that area, divide it by three, and have an expression that is exactly equal to (4-pi)/8

Except it doesn't work, of course, because deriving pi is clearly not that easy. So why didn't this work? I would post all the math, but it's a pain since I wrote it all on paper at the time. In essence, it doesn't work because curve on the new shape created when we dropped the shape onto y=-x is not a segment of a circle. It's really close, but it isn't quite there. The shape is a little bit too short to touch the circle, such that when it is pressed up to the circle, there are only two points of contact. (interestingly, the tangents of the shape and the circle at those two points are identical.)

But I haven't given up yet! Surely there's a way to create a shape that fulfills our requirements! We now have a new objective: create a shape of an area which is some known ratio to any part of our circle-square, which contains exactly one circular section of a curve, which was created using two or more circular curves. Basically, we have to meld two curves into one, while maintaining their circular-ness.

My attempt at doing so has led me down a math rabbit hole I dare not try to write out completely here, so I'll skip to where I am right now. Consider the following two systems of equations:

This is a modified version of the previous set I've been showing, but with a different origin, among a few things. Note the wedge shape made by the blue circle, the yellow graph, and the pink graph. This shape is similar to the one we "tacked on to the back, but flipped." Both systems are the same, but with a modified coefficient for three of the functions (thanks for OS 5.0, TI!) Notice that the solution of the yellow graph and circle, as well as the brown graph and the circle, is in the same point on both graphs, whereas the solution of the black graph and circle changes its point around the circle. This means that we can alter the ratio of (the length of a side of our shape)/(the length of the midpoint of the base of our shape to the circle) to get closer to an actual circular curve (after we fix the aspect ratio of the shape), fulfilling our requirements. I decided to solve this with a nice fat system of equations, so here it is:

(((abs(√(−2*(m^(2)*(2*√(2)-5)-2*m*(√(2)+1)-7))+m^(2)*(√(2)-2)-m*√(2)-4))/(4*√(m^(2)+1))))/(((abs(m*(√(2)-2)-√(2)))/(2*√(m^(2)+1))))=sqrt(2)/2-1/2

If you dissect this equation, it comes out to mean (segment of black graph between the circle and pink graph)/(segment of orange graph between circle and pink graph)=(optimal ratio) where m=the slope of the three parallel linears.

I put it in Wolfram Alpha and it had a heart attack, so I haven't gotten much farther yet. I'm trying to logically come up with a reason why this wouldn't work, but I can't beyond the fact that I stumped cOmPuTaTiOnAl InTeLlIgEnCe with a one-var equation. Somebody who knows what they're doing, please tell me why I'm a moron, quick before I pull any more all nighters trying to do what I know ful well is impossible.

I'm sure Euler is rolling in his grave about now.

About three weeks ago, following my purchase of the new TI-Nspire CX II CAS, I embarked upon a mathematical journey with the intent of doing the impossible. Literally, I'm pretty sure this is impossible.

So what exactly am I trying to do?

Pi is an irrational number, meaning it cannot be represented as a fraction, nor do its decimal digits follow any pattern. Similarly, sqrt(2) is an irrational number, also unable to be represented as a fraction. Who's to say that pi cannot be expressed as a combination of other irrational numbers (in this case, roots)? I am attempting to express pi as a finite expression containing only operations made on rational numbers. Finite means that my expression cannot be an infinite series. One should be able to write pi, exactly, with one pencil and one sheet of paper.

Consider the following system of equations:

y=x

y=0

x=0

(x-1)^2+(y-1)^2=1

Since the radius of the circle is 1, the area must be pi. Therefore, the area bounded by the circle, y=x, and y=0 must be (4-pi)/8

Now imagine that shape combined with another identical shape tacked onto the back, but flipped.

graph y=x, y=0, x=0, (x-1)^2+(y-1)^2=1, y=x+1, (x+1)^2+y^2=1

What if we warped this shape to fall upon y=-x, thus eliminating the bottom curve?

graph y=x, x=0, (x-1)^2+(y-1)^2=1, y=x+1, y=-x, (x-(sqrt(2) - 1)/sqrt(2))^2+(y-(1/2+(sqrt(2) - 1)/sqrt(2)))^2=.25

Now, if we enlarge that shape by a factor of four, it should fit right back on to the original circle.

Since this new shape should be four times the area of the original shape, and we know the area of the original shape is equal to corner of the square, we should be able to slice the edges off the new shape (the edges of which we know the lengths of), measure that area, divide it by three, and have an expression that is exactly equal to (4-pi)/8

Except it doesn't work, of course, because deriving pi is clearly not that easy. So why didn't this work? I would post all the math, but it's a pain since I wrote it all on paper at the time. In essence, it doesn't work because curve on the new shape created when we dropped the shape onto y=-x is not a segment of a circle. It's really close, but it isn't quite there. The shape is a little bit too short to touch the circle, such that when it is pressed up to the circle, there are only two points of contact. (interestingly, the tangents of the shape and the circle at those two points are identical.)

But I haven't given up yet! Surely there's a way to create a shape that fulfills our requirements! We now have a new objective: create a shape of an area which is some known ratio to any part of our circle-square, which contains exactly one circular section of a curve, which was created using two or more circular curves. Basically, we have to meld two curves into one, while maintaining their circular-ness.

My attempt at doing so has led me down a math rabbit hole I dare not try to write out completely here, so I'll skip to where I am right now. Consider the following two systems of equations:

This is a modified version of the previous set I've been showing, but with a different origin, among a few things. Note the wedge shape made by the blue circle, the yellow graph, and the pink graph. This shape is similar to the one we "tacked on to the back, but flipped." Both systems are the same, but with a modified coefficient for three of the functions (thanks for OS 5.0, TI!) Notice that the solution of the yellow graph and circle, as well as the brown graph and the circle, is in the same point on both graphs, whereas the solution of the black graph and circle changes its point around the circle. This means that we can alter the ratio of (the length of a side of our shape)/(the length of the midpoint of the base of our shape to the circle) to get closer to an actual circular curve (after we fix the aspect ratio of the shape), fulfilling our requirements. I decided to solve this with a nice fat system of equations, so here it is:

(((abs(√(−2*(m^(2)*(2*√(2)-5)-2*m*(√(2)+1)-7))+m^(2)*(√(2)-2)-m*√(2)-4))/(4*√(m^(2)+1))))/(((abs(m*(√(2)-2)-√(2)))/(2*√(m^(2)+1))))=sqrt(2)/2-1/2

If you dissect this equation, it comes out to mean (segment of black graph between the circle and pink graph)/(segment of orange graph between circle and pink graph)=(optimal ratio) where m=the slope of the three parallel linears.

I put it in Wolfram Alpha and it had a heart attack, so I haven't gotten much farther yet. I'm trying to logically come up with a reason why this wouldn't work, but I can't beyond the fact that I stumped cOmPuTaTiOnAl InTeLlIgEnCe with a one-var equation. Somebody who knows what they're doing, please tell me why I'm a moron, quick before I pull any more all nighters trying to do what I know ful well is impossible.