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  • Brother Daniel
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Fun with pi
So does anyone want to see a proof of the irrationality of pi?

(The one proof that I've read requires some basic differential calculus in order to follow it.)

hit us up brah.
OK!  I'll try to go step by step through a proof that pi is irrational.

I'm taking this proof from An Introduction to Classical Real Analysis by Karl Stromberg (1981).  He attributes the proof to Ivan Niven, and his bibliography includes a 1956 monograph by Niven called Irrational Numbers, so I assume he got it from there.

The method of proof is "proof by contradiction".  That is, we'll start by assuming that pi is rational, then without making any other questionable assumptions, we'll derive something impossible from that premise.

So.  First step:  Assume that π is rational.  Then π 2 is rational.  And we can find natural numbers (i.e., positive integers) a and b such that π 2 = a/b.

Next, let N be a natural number that is big enough so that a N/N! < 1/π.

Why can we do this?  Because as N grows, N! ultimately grows faster than a N.  So we can make a N/N! arbitrarily small by picking a big enough value of N.

(In fact, you can make an infinite series, summing the terms a n/n! as n goes from 0 to infinity.  This series converges to e a, which is a real number for any real a.  In order for that convergence to come about, the individual terms have to get arbitrarily small.  If anyone is unhappy with this part, we can have a little(?) digression about infinite series.)
  • Last Edit: August 25, 2016, 05:35:33 PM by Brother Daniel

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Re: Fun with pi
Reply #1
Next, let's define a function f on the real numbers, as follows.

f(x) = x N(1 - x) N/N!

So f is a polynomial of degree 2N.

We can expand the expression for f to give
f(x) = (1/N!) Σ cnx n,
where the sum is over n from N to 2N, and the cn are integers (whose exact values we don't care about).

Let f (k) denote the k th derivative of f.  Then
f (k)(x) = (1/N!) Σ n(n - 1)...(n - k + 1)cnx n - k,
where the sum is over n from N to 2N in the case where kN, and over n from k to 2N in the case where Nk ≤ 2N.
(If k > 2N, then f (k)(x) = 0.)

Now f (k)(0) and f (k)(1) are both integers, for any nonnegative integer k.

[edit:  This last assertion wasn't adequately justified here.  See reply #24, below.]
  • Last Edit: August 26, 2016, 05:48:21 AM by Brother Daniel

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Re: Fun with pi
Reply #2
Now let's define another function g on the real numbers as follows.

g(x) = b N Σ (-1) jπ 2N - 2jf (2j)(x)
where the sum is over j from 0 to N.

Since we have π 2 = a/b, we can write this as
g(x) = Σ (-1) ja N - jb jf (2j)(x)
where the sum is over j from 0 to N.

From the second expression for g, we can see that g(0) and g(1) are integers.  So g(1) + g(0) is an integer.
  • Last Edit: August 25, 2016, 05:49:43 PM by Brother Daniel

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Re: Fun with pi
Reply #3
That looks like a useful set of curves.

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Re: Fun with pi
Reply #4
Sorry, saw a shiny thing.

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Re: Fun with pi
Reply #5
Now let's define another function h on the real numbers as follows.

h(x) = g'(x) sin(πx) - π g(x) cos(πx)

Taking the derivative, we get
h'(x) = (g''(x) + π 2g(x)) sin(πx)
  • Last Edit: August 25, 2016, 05:52:19 PM by Brother Daniel

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Re: Fun with pi
Reply #6
Sorry, saw a shiny thing.
No problem.  (But TBH, I don't know what you were plotting there.)

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Re: Fun with pi
Reply #7
Now if we look at the derivatives of g, we can see that
g''(x) = b N Σ (-1) jπ 2N - 2jf (2j + 2)(x)
where the sum is over j from 0 to N.

We can shift the summation variable in order to write this as
g''(x) = b N Σ (-1) j - 1π 2N - 2j + 2f (2j)(x)
where the sum is now over j from 1 to N + 1.

But we can throw away the j = N + 1 term from the sum, because we know that f (2N + 2)(x) = 0.  So the sum really just takes j from 1 to N.

And with a tiny rearrangement, we can write this as
g''(x) = -π 2b N Σ (-1) jπ 2N - 2jf (2j)(x)
where again the sum is over j from 1 to N.

Comparing this expression for g'' to the expression for g, we can see that most of the terms cancel in the expression g''(x) + π 2g(x).  This latter quantity is really just equal to π 2 times the first (j = 0) term of g(x).  That is:
g''(x) + π 2g(x) = π 2N + 2b Nf(x) = π 2a Nf(x).
  • Last Edit: August 25, 2016, 05:55:50 PM by Brother Daniel

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Re: Fun with pi
Reply #8
So we have
h'(x) = π 2a Nf(x) sin(πx).

Also, from the original expression for h, we can see that h(1) - h(0) = π (g(1) + g(0)).
  • Last Edit: August 25, 2016, 05:58:41 PM by Brother Daniel

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Re: Fun with pi
Reply #9
Now in basic calculus, there's a thing called the Mean Value Theorem.  When you have a function h that is continuous on some closed interval [p, q] and differentiable at least on the corresponding open interval (p, q), then there is some number r in the open interval such that h'(r) = (h(q) - h(p))/(q - p).  Applying that theorem to our present function h (which is certainly continuous and differentiable all over the place), it follows that there is some number z in the interval (0, 1) such that h'(z) = h(1) - h(0).

In other words, we have z such that 0 < z < 1 and
π 2a N f(z) sin(πz) = π (g(1) + g(0)).

Cancelling out a factor of π, we can write this last equation as π a N f(z) sin(πz) = g(1) + g(0).
  • Last Edit: August 25, 2016, 06:02:47 PM by Brother Daniel

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Re: Fun with pi
Reply #10
From the definition of f, we can see that 0 < f(z) < 1/N! (remembering that 0 < z < 1), so 0 < aNf(z) < aN/N!.
But recall that we picked N to have the property that aN/N! < 1/π.  So 0 < π aNf(z) < 1.

Meanwhile, 0 < sin(πz) ≤ 1, so 0 < π a Nf(z) sin(πz) < 1.

So 0 < g(1) + g(0) < 1.
  • Last Edit: August 25, 2016, 06:05:09 PM by Brother Daniel

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Re: Fun with pi
Reply #11
But remember that g(1) + g(0) is an integer.  So we have found an integer between 0 and 1.

And that's obviously impossible.

The only questionable assumption we made is that π is rational, and we got an absurdity from it.  It follows that π is irrational.
  • Last Edit: August 25, 2016, 06:07:15 PM by Brother Daniel

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Re: Fun with pi
Reply #12
So yay for Ivan Niven.  There's no way I'd ever be clever enough to think of such a proof myself.

If anyone is unhappy with any of those steps, let me know.  I may have hurried in a few places.

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Re: Fun with pi
Reply #13
I'm with Knox here ....

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Re: Fun with pi
Reply #14
Mr. Knox.  Now come now.  Come now.
You don't have to be so dumb now....

Try to say this, Mr. Knox, please....

Through three cheese trees three free fleas flew.
While these fleas flew, freezy breeze blew.
Freezy breeze made these three trees freeze.
Freezy trees made these trees' cheese freeze.
That's what made these three free fleas sneeze.

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Re: Fun with pi
Reply #15
So yay for Ivan Niven.  There's no way I'd ever be clever enough to think of such a proof myself.

If anyone is unhappy with any of those steps, let me know.  I may have hurried in a few places.

Before I begin to follow all that...
(and I'm sure this is a really dumb question, but...)
can you pinpoint the step at which it becomes clear that "p" is π ?
"I understand Donald Trump better than many people because I really am a lot like him." - Dave Hawkins

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Re: Fun with pi
Reply #16
Mr. Knox.  Now come now.  Come now.
You don't have to be so dumb now....

Try to say this, Mr. Knox, please....

Now if we look at the derivatives of g, we can see that g''(x) = b N Σ (-1) jπ 2N - 2jf (2j + 2)(x) where the sum is over j from 0 to N.

We can shift the summation variable in order to write this as g''(x) = b N Σ (-1) j - 1π 2N - 2j + 2f (2j)(x) where the sum is now over j from 1 to N + 1.

But we can throw away the j = N + 1 term from the sum, because we know that f (2N + 2)(x) = 0.  So the sum really just takes j from 1 to N.

And with a tiny rearrangement, we can write this as g''(x) = -π 2b N Σ (-1) jπ 2N - 2jf (2j)(x) where again the sum is over j from 1 to N.

Comparing this expression for g'' to the expression for g, we can see that most of the terms cancel in the expression g''(x) + π 2g(x).  This latter quantity is really just equal to π 2 times the first (j = 0) term of g(x).  That is:
g''(x) + π 2g(x) = π 2N + 2b Nf(x) = π 2a Nf(x)

fyp
  • Last Edit: August 25, 2016, 06:10:11 PM by Brother Daniel

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Re: Fun with pi
Reply #17
(@ Vox)

Uh oh.

We may have a display problem.

I put "p" in font tags, for the "symbol" font.  I thought it would show up as pi for everyone.

I don't know what you did to make a pi character just now.  But whatever you did, maybe I should have done that instead.

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Re: Fun with pi
Reply #18
(@ Vox)

Uh oh.

We may have a display problem.

I put "p" in font tags, for the "symbol" font.  I thought it would show up as pi for everyone.

I don't know what you did to make a pi character just now.  But whatever you did, maybe I should have done that instead.
I used [alt]+p on my mac. 

But my question is more fundamental than that: in what step is it clear that p IS π?
I.e. what properties does it have that make it identical to the π we all know and love, and not just "pick a number, any number"?
"I understand Donald Trump better than many people because I really am a lot like him." - Dave Hawkins

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Re: Fun with pi
Reply #19
Re the display problem:  I've now edited most of the posts in the thread.  I hope it looks correct on most screens now.

Re your question:  In the last line of reply #8 (which refers back to reply #5), I make use of the property sin(π) = 0 and cos(π) = -1.  That's where the properties of π come in.

We can define π as the smallest positive number for which sin(π) = 0.  So sin(x) > 0 when 0 < x < π.

The sine and cosine functions can be defined as

sin(x) = Σ (-1) j x 2j + 1/(2j + 1)!
and
cos(x) = Σ (-1) j x 2j/(2j)!
where in both cases the sum is over j from 0 to infinity.

If you want to take a geometrical interpretation of the sine and cosine functions as defined here, then the "angle" is in radians.

From integral calculus you can get 2πr for the circumference of a circle of radius r, and all those other wonderful things that make π familiar.

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Re: Fun with pi
Reply #20
fyp
Was it ok before that point?  :)

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Re: Fun with pi
Reply #21
Was it ok before that point?  :)
Drat.  No, it wasn't ok.  In particular, I stupidly left out a few steps, with the upshot that I didn't give adequate justification for this assertion, up near the beginning:
Now f (k)(0) and f (k)(1) are both integers, for any nonnegative integer k.

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Re: Fun with pi
Reply #22
fyp
Was it ok before that point?  :)
Yes, quite.  One of my favorite books as a kid.
.....

OH, you mean the proof.  Hell if I know.

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Re: Fun with pi
Reply #23
Yes, quite.  One of my favorite books as a kid.
Mine too!
Quote from: MikeS
OH, you mean the proof.  Hell if I know.
:sadcheer:

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Re: Fun with pi
Reply #24
So back in reply #1, I didn't adequately justify the claim that f (k)(0) and f (k)(1) are integers.

f(x) is a polynomial of degree 2N.  The smallest power of x in f(x) is the N th power (see the expanded version of f(x)).

So f (k)(0) = 0 for k < N or k > 2N.

For Nk ≤ 2N, the expanded formula for f (k) gives us
f (k)(0) = (k!/N!)ck, which is clearly an integer.

So we have the result that f (k)(0) is an integer for any nonnegative integer k.

Now if we go back to the original formula for f, you can see that f(x) = f(1 - x).

Taking the derivative of both sides of this last equation, we get f '(x) = -f '(1 - x).

Taking the derivative again, we get f ''(x) = f ''(1 - x).

Continuing this process, we get the pattern f (k)(x) = (-1) kf (k)(1 - x).

So for any k, we have either f (k)(1) = f (k)(0) or f (k)(1) = -f (k)(0).

So f (k)(1) is also an integer for any nonnegative integer k.