Math stuff that hasn't been proven

[disregardthat]

No, it doesn't. The only way to "calculate" pi from a circle is to actually measure the circumference and divide by the diameter. Doing that for any number of circles and finding that you always get the same thing does NOT show that is true for all circles.

What are you talking about? What do you mean by "actually measure"?

Surely we can calculate (mathematically) the circumference by means of mathematical methods for a circle with radius r, must I show it to you?

A circle is, by the way, in this regard, a mathematical object.

 

[Fredrik]

I also find HallsofIvy's complaint a bit strange, but I have one of my own. Edit: Not anymore. See the next few posts after this one.

I would define the length of a differentiable curve C:[a,b]\rightarrow\mathbb R^2 as L(C)=\int_a^b\sqrt{C_1'(t)^2+C_2'(t)}\, dt. A circle around (a,b) with radius r is the range of the curve C defined by \begin{align}
x(t) &=a+r\cos t\\
y(t) &=b+r\sin t\\
C(t) &=(x(t),y(t))
\end{align} for all t\in[0,2\pi). So the circumference of that circle is L(C)=\int_0^{2\pi}\sqrt{x'(t)^2+y'(t)^2}\, dt=r\int_0^{2\pi}dt=2\pi r Yes, we can see that L(C)/2r is independent of a,b and r, but the argument looks circular. You might be able to fix it, but I don't immediately see how.

 

[I like Serena]

Your argument depends on cosine, sine, and pi, whose definitions should not be readily used.

A simple fix is using y(x) = \sqrt{r^2 - x^2}
L(C) = \int_{-r}^{+r} \sqrt{1 + y'(x)^2} dx

This integral measures half the circumference of a circle.
Calculate it and you will find \pi r.

 

[Dr. Seafood]

but the argument looks circular

I see what you did there

 

[Fredrik]

A simple fix is using y(x) = \sqrt{r^2 - x^2}

Cool. That approach looks correct and non-circular. \begin{align}
x(t) &=a+t\\
y(t) &=b+\sqrt{r^2-(x(t)-a)^2}=b+\sqrt{r^2-t^2}
\end{align} We want to show that L(C)/2r is independent of a,b and r.
\begin{align}
\frac{L(C)}{2r} &= \frac{1}{r}\int_{-r}^{r}\sqrt{x'(t)^2+y'(t)^2}\,dt =\frac{1}{r}\int_{-r}^{r}\sqrt{1+\frac{t^2}{r^2-t^2}}\,dt =\frac{1}{r}\int_{-r}^{r}\sqrt{\frac{r^2}{r^2-t^2}}\,dt \\
&=\frac{1}{r}\int_{-r}^{r}\frac{1}{\sqrt{1-\frac{t^2}{r^2}}}\,dt =\left[\text{Variable change: }s=\frac{t}{r}\right] = \int_{-1}^1\frac{1}{\sqrt{1-s^2}} ds
\end{align}That last integral is clearly independent of a,b and r, and Wolfram alpha says that it's equal to \pi.

 

[micromass]

Yes, but to calculate that last integral, we need cosines, sines and pi again. But we need to be careful to use these concepts here.

 

[Fredrik]

We don't need to do the calculation. We just need to see that it's independent of a,b and r.

 

[I like Serena]

Hmm, I think you need to add "a" in the limits of the integrals.
Otherwise that looks correct.

 

[Kindayr]

So would you just define \pi = \int_{-1}^1 \frac{1}{\sqrt{1-s^2}} ds with s=\frac{t}{r} and leave the approximation to someone else?

 

[I like Serena]

Yes, but to calculate that last integral, we need cosines, sines and pi again. But we need to be careful to use these concepts here.

Noooo.
If you want you can define a special function called micromass(x) that happens to be the result of that integral (and that happens to work out to pi).

Oh, and not entirely by coincidence micromass(x) = arcsin(x).

So we found in a mathematically sound argument a definition for arcsin, as well as for pi! Yay!

 

[micromass]

Noooo.
If you want you can define a special function called micromass(x) that happens to be the result of that integral (and that happens to work out to pi).

Oh, and not entirely by coincidence micromass(x) = arcsin(x).

So we found in a mathematically sound argument a definition for arcsin, as well as for pi! Yay!

Yes, but how do you know that that integral even exists??

Math people are annoying, I know.

 

[Kindayr]

If you show \frac{1}{\sqrt{1-x^2}} to be continuous between ]-1,1[, then its Riemann integrable. Would that work? Then just do the improper integral as x\to 1.

And it is continuous, so the improper integral should exist. I think at least, I could very well be wrong.

 

[HallsofIvy]

We don't need to do the calculation. We just need to see that it's independent of a,b and r.

No, you first have to show that integral exists- and it is an improper integral so that is non-trivial.

 

[I like Serena]

Yes, but how do you know that that integral even exists??

Math people are annoying, I know.

Ah, are you really forcing me to brush off my rusty integral theorems?
Reminds me of a glass bead necklace!

What about:
"A function on a compact interval [a,b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). "

And if you really want, we can adjust the boundaries of the integral to \pm \frac 1 2 \sqrt 2 to eliminate the singularities.

 

[KingNothing]

- Every number can be uniquely (up to order) decomposed in prime factors.

It's my impression that this is a matter of definition rather than proof: we've co-defined "prime numbers" along with this axiom in the same sense that we co-define the speed of light and the length of a meter.

That is, they are understood together.

 

[Kindayr]

It's my impression that this is a matter of definition rather than proof: we've co-defined "prime numbers" along with this axiom in the same sense that we co-define the speed of light and the length of a meter.

That is, they are understood together.

To my knowledge, the fundamental theorem of arithmetic isn't taken as an axiom and infact has a proof found http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic" .

 

[micromass]

Ah, are you really forcing me to brush off my rusty integral theorems?
Reminds me of a glass bead necklace!

What about:
"A function on a compact interval [a,b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). "

And if you really want, we can adjust the boundaries of the integral to \pm \frac 1 2 \sqrt 2 to eliminate the singularities.

But the function is not bounded. And the function can never be Riemann integrable since it has poles in 1 and -1. So you're working with improper Riemann integrals...

 

[Dr. Seafood]

Cool. That approach looks correct and non-circular.

Only semi-circular! HAHA LOLlkasfksdlkvcslamc

 

[I like Serena]

But the function is not bounded. And the function can never be Riemann integrable since it has poles in 1 and -1. So you're working with improper Riemann integrals...

All right, so let's adjust the boundaries of the integrals.
Let's only integrate the circumference of a quarter of a circle instead of half of a circle.
This is achieved by integration from (a - \frac 1 2 r \sqrt 2) to (a + \frac 1 2 r \sqrt 2).
The resulting integral becomes:
\int_{-\frac 1 2 \sqrt 2}^{+\frac 1 2 \sqrt 2} \frac 1 {\sqrt{1-s^2}} ds

This integral has no singular points and it is bounded.
Furthermore, it is still the same micromass(x) although we have clipped its extremities.

It will evaluate to \frac \pi 2.

 

[micromass]

All right, so let's adjust the boundaries of the integrals.
Let's only integrate the circumference of a quarter of a circle instead of half of a circle.
This is achieved by integration from (a - \frac 1 2 r \sqrt 2) to (a + \frac 1 2 r \sqrt 2).
The resulting integral becomes:
\int_{-\frac 1 2 \sqrt 2}^{+\frac 1 2 \sqrt 2} \frac 1 {\sqrt{1-s^2}} ds

This integral has no singular points and it is bounded.
Furthermore, it is still the same micromass(x) although we have clipped its extremities.

It will evaluate to \frac \pi 2.

Seems ok

\begin{Annoying-mathematician}
Of course, this only shows that the length of a quarter of a circle is independent of pi. Nothing is said about the entire circle.
\end{Annoying-mathematician}

 

[I like Serena]

Only semi-circular! HAHA LOLlkasfksdlkvcslamc

LOL! And I've just made it only curvy!

 

[I like Serena]

Seems ok

\begin{Annoying-mathematician}
Of course, this only shows that the length of a quarter of a circle is independent of pi. Nothing is said about the entire circle.
\end{Annoying-mathematician}

LMAO.
Well, what about the argument that 4 times the length of a quarter of a circle is equal to the length of an entire circle?
To make it Annoying-mathematician'ally clad, I guess we should define 4 integrals for each quarter of the circle.
But as my prof liked to say in his proofs: this is left as an exercise to the reader.

 

[micromass]

LMAO.
Well, what about the argument that 4 times the length of a quarter of a circle is equal to the length of an entire circle?
To make it Annoying-mathematician'ally clad, I guess we should define 4 integrals for each quarter of the circle.
But as my prof liked to say in his proofs: this is left as an exercise to the reader.

Now you see why my profs were always annoyed to have me as a student

 

[disregardthat]

The archimedes way (incribing and circumscribing by polygons) can be made logically sound (no circularity) without assuming any analytical properties of sin(x).

 

[Kindayr]

But the function is not bounded. And the function can never be Riemann integrable since it has poles in 1 and -1. So you're working with improper Riemann integrals...

Would it be be possible to integrate it using Generalized Riemann Integrability? I just noticed that in one of my texts "Understanding Analysis" by Stephen Abbott says:

Now, improper Riemann integrals have been created to extend Riemann integration to some unbounded functions, but it is another interesting fact about the generalized Riemann integral that any function having an improper integral must already be integrable in the sense described in Definition 8.1.6.

A function f on [a,b] has generalized Riemann integral A if, for every \epsilon >0, there exists a gauge \delta (x) on [a,b] such that for each tagged partition (P,\{c_{k} \}^n_{k=1}) that is \delta (x)-fine, it is true that |R(f,P)-A|<\epsilon. In this case, we write A=\int_{a}^{b} f.

Due note that this is in a chapter I have not study (its apart of the additional topics), I'm more just wondering if its possible using this definition than actually trying to say something of merit.

 

[I like Serena]

Now you see why my profs were always annoyed to have me as a student

So when is it going to be professor micromass?
I believe there is already a thread with how you are going to behave.

 

[Fredrik]

So would you just define \pi = \int_{-1}^1 \frac{1}{\sqrt{1-s^2}} ds with s=\frac{t}{r} and leave the approximation to someone else?

My approach was as follows. Step 1: Prove that L(C)/r is independent of a,b and r. Step 2: Define \pi=L(C)/(2r). (Without step 1, step 2 doesn't make sense). However, as HallsofIvy and micromass has already mentioned, it's not obvious that my way of doing step 1 makes sense. To be more specific, it's not obvious that the function x\mapsto\frac{1}{\sqrt{1-x^2}} is integrable on [-1,1]. It is, but that must be proved separately. If we can prove that, my calculation would then show that \int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,dx=\pi. This means that one of many possible ways to find approximations of the value of \pi would be to do that integral numerically.

I should also mention that the words "with s=t/r" in your statement don't make sense. (The relationship between the new variable and the old doesn't tell us anything about the value of the integral).

What about:
"A function on a compact interval [a,b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). "

The function x\mapsto\frac{1}{\sqrt{1-x^2}} defined on [-1,1] doesn't satisfy that requirement. I'm guessing that the null set you had in mind was {-1,1}, but the restriction of the function to the open interval (-1,1) isn't bounded.

 

[I like Serena]

The function x\mapsto\frac{1}{\sqrt{1-x^2}} defined on [-1,1] doesn't satisfy that requirement. I'm guessing that the null set you had in mind was {-1,1}, but the restriction of the function to the open interval (-1,1) isn't bounded.

See the reply I posted after the one you quoted.

 

[disregardthat]

Actually, it's not that hard to prove that the integral of 1/sqrt(1-x^2) exists on [-1,1].

First, it's integrable on [-1/2,1/2] of course, so we prove that it's integrable on [1/2,1] and [-1,-1/2]:

[1/2,1]: We want a constant C such that 1/sqrt(1-x^2) <= C/sqrt(1-x)

It is equivalent to 1/C <= sqrt(1+x), or 1/C^2-1 <= x, so if we let 1/C^2-1 = 1/2, or C = sqrt(2/3), this will be satisfied.

Now, the integral of 1/sqrt(1-x) on [1/2,1] exists, since it can be easily computed (it's sqrt(2)).

Likewise, for [-1,-1/2], find a constant C such that 1/sqrt(1-x^2) <= C/sqrt(1+x), and show that the integral of the expression to the right exists.

 

[Hurkyl]

The archimedes way (incribing and circumscribing by polygons) can be made logically sound (no circularity) without assuming any analytical properties of sin(x).

I do believe that's true. However, Archimedes did make two significant assumptions (in addition to the assumption the perimeter exists):

• The length of the circular arc truly is somewhere between the perimeters of the inscribed and circumscribed polygons
• The perimeters of the circumscribed polygons converges to the same number as the perimeters of the inscribed polygons

Which may or may not be difficult; I'm not sure.

 

[disregardthat]

I do believe that's true. However, Archimedes did make two significant assumptions (in addition to the assumption the perimeter exists):

• The length of the circular arc truly is somewhere between the perimeters of the inscribed and circumscribed polygons
• The perimeters of the circumscribed polygons converges to the same number as the perimeters of the inscribed polygons

Which may or may not be difficult; I'm not sure.

Yes, I believe Archimedes himself made that first assumption explicitly, (which boils down to that a curve (in this case: the circle arc) has length and that a straight line is the shortest path from one point to another). The second one can be proved, I think. It is basically to show that the difference between the circumscribed and inscribed polygon converges to 0 (after showing that they converge), and that simply relies on that cos(x) --> 1 as x --> 0, if I'm not mistaken.

If I remember correctly, I read that his assumption was that the length of a convex curve from point A to B is larger than the length of a straight line from A to B.

 

[DarkDrag0nite]

Is there anyone prove that circle and square have 360 degree ?

 

[Hurkyl]

a straight line is the shortest path from one point to another

This part, IIRC, isn't hard to prove, but it only works for the inscribed polygon. Some other method is required to show the circle's perimeter is less than that of the circumscribed polygon.

There's a heuristic explanation that, between two consecutive points where they meet, the circular arc is "straighter" than the circumscribed polygon; I'm not sure how easily that can be converted into a rigorous proof.

 

[HallsofIvy]

What in the world does "circle and square has 360 degrees" mean? What does it mean for a figure to have a certain number of degrees?

 

[Char. Limit]

I believe that in the case of the square, he's referring to the sum of the interior angles being 360. For the circle, I have no idea.

 

[Fredrik]

That a rotation by 360 degrees is equivalent to no rotation at all? (Of course, that's just the definition of "degree").

 

[DarkDrag0nite]

That a rotation by 360 degrees is equivalent to no rotation at all? (Of course, that's just the definition of "degree").

If that's the definition, that's the answer. Thank you :)