From Schmid.wiki
Jump to: navigation, search

Intuition

A duck

This theorem concerns the sound of poultry. It is attempted to deduce the sound of geese and swans from the well-known sounds of ducks and ostriches via strict mathematical reasoning. A goose is larger than a duck, a swan is larger than a goose, and an ostrich is larger than the other 3 birds. If the volume of a bird determines the sound of the bird, it is possible to accurately determine the sound of geese and swans from the sounds of the other birds.

Prerequisites

It is intuitively clear that ducks are smaller than geese, geese are smaller than swans, and ostriches are larger than all of the others. This is stated in the Poultry Size Axiom.

In the following, it is assumed that d is a duck, g is a goose, s is a swan, and o is an ostrich.

Poultry Size Axiom:

<math>|d| < |g| < |s| < |o|</math>

The Even Mix Lemma states the relationship between size and the sound of a bird, defined from the sounds of other birds of different sizes.

Sound is represented as a function yielding sound pressure level from time.

Even Mix Lemma:

If <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math> are birds, and <math>|\alpha| < |\beta| < |\gamma|</math>, the sound <math>s</math> of bird <math>\beta</math> is given by

<math>s(\beta) = \frac{s(\alpha)+s(\gamma)}{2}</math>

where

<math>s : \mathrm{time} \to \mathrm{SPL}</math>

The Even Mix Lemma will be proven by avoidance, see 36 Methods of Mathematical Proof.

The Poultry Sound Theorem

The Poultry Sound Theorem intuitively states that a goose sounds like 2/3 duck and 1/3 ostrich, and a swan sounds like 2/3 ostrich and 1/3 duck.

The sound of a goose is given by

<math>s(g)=\frac{2}{3}s(d)+\frac{1}{3}s(o)</math>

and the sound of a swan is given by

<math>s(s)=\frac{2}{3}s(o)+\frac{1}{3}s(d)</math>

Proof

From the Poultry Size Axiom we have that <math>|d| < |g| < |s|</math>, which enables the use of the Even Mix Lemma. The sound of a goose is then given by

<math>s(g) = \frac{s(d) + s(s)}{2}</math>

By similar reasoning, the sound of a swan is given by

<math>s(s) = \frac{s(g) + s(o)}{2}</math>

if we substitue the value of <math>s(s)</math> into the equation for <math>s(g)</math>, we get

<math>

s(g) = \frac{s(d) + \frac{s(g) + s(o)}{2}}{2}

    = \frac{1}{2}s(d) + \frac{1}{4}s(o) + \frac{1}{4}s(g)

</math>

We subtract <math>\frac{1}{4}s(g)</math> from both sides

<math>\frac{3}{4}s(g) = \frac{1}{2}s(d) + \frac{1}{4}s(o)</math>

And get

<math>s(g) = \frac{2}{3}s(d) + \frac{1}{3}s(o)</math>

The swan sound equation is proven similarly.<math>\Box</math>

Discussion

Another way of proving the Poultry Sound Theorem is by finding the limit of the infinite recursion of the equation:

<math>s(g) = \frac{1}{2}s(d) + \frac{1}{4}s(o) + \frac{1}{4}s(g)</math>

A few lemmas are needed

Lemma 1:

<math>\sum_{i=1}^{n} a^{n-i} = \frac{a^n-1}{a-1}</math>

Proof: Observe the standard arithmetic equation

<math>a^n-b^n=(a-b)\sum_{i=1}^n a^{n-i}\;b^{i-1}</math>

If we fix <math>b=1</math>, we get:

<math>a^n -1 = (a-1) \sum_{i=1}^n a^{n-i} \quad\Leftrightarrow\quad

\sum_{i=1}^{n}a^{n-i} = \frac{a^n - 1}{a-1}</math>

Lemma 2:

<math> \sum_{i=0}^{n-1}\frac{1}{a^i} = \frac{1}{a^{n-1}}\sum_{i=1}^{n} a^{n-i} </math>

Proof:

The order of the summation can of course be reversed:

<math>\sum_{i=0}^{n-1}\frac{1}{a^i} = \sum_{i=0}^{n-1}\frac{1}{a^{(n-1)-i}}</math>

Using standard exponentiation rules, we get:

<math>\sum_{i=0}^{n-1}\frac{a^i}{a^{n-1}}</math>

We can factor out the denominator:

<math>\frac{1}{a^{n-1}}\sum_{i=0}^{n-1} a^i</math>

and reverse the order of the summation back:

<math>\frac{1}{a^{n-1}}\sum_{i=1}^{n} a^{n-i}</math>

<math>\Box</math>

Alternative Proof of the Poultry Sound Theorem:

<math>s(g) = \frac{1}{2}s(d) + \frac{1}{4}s(o) + \frac{1}{4}s(g)</math>

If we substitute <math>s(g)</math> into the equation again, we get:

<math>s(g) = \frac{1}{2}s(d) + \frac{1}{4}s(o)

+ \frac{1}{8}s(d) + \frac{1}{16}s(o) + \frac{1}{16}s(g) )</math>

This procedure can be repeated:

<math>s(g) = \frac{1}{2}s(d) + \frac{1}{4}s(o)

+ \frac{1}{8}s(d) + \frac{1}{16}s(o) + \frac{1}{32}s(d) + \frac{1}{64}s(o) + \frac{1}{128}s(d) + \frac{1}{256}s(o) + \ldots </math>

This sum can be written as

<math>s(g) = \lim_{N \to \infty} \sum_{i=0}^{N} \frac{\frac{s(d)}{2}+\frac{s(o)}{4}}{4^i}</math>

As the sounds of the duck and the ostrich can be considered constant for all <math>i</math> and <math>n</math>, they may be put outside the summation and the limit

<math>s(g) = (\frac{s(d)}{2}+\frac{s(o)}{4}) \lim_{N \to \infty} \sum_{i=0}^{N} \frac{1}{4^i}</math>

Consider the expression

<math>\sum_{i=0}^{N} \frac{1}{4^i}</math>

If we use Lemma 2 and select <math>n=N+1</math> and <math>a=4</math>, we get:

<math>\sum_{i=0}^{n-1}\frac{1}{4^i} = \frac{1}{4^{n-1}}\sum_{i=1}^{n} 4^{n-i}</math>

According to Lemma 1, <math>\sum_{i=1}^{n} 4^{n-i} = \frac{4^n-1}{4-1}</math>, which gives us:

<math>\sum_{i=0}^{n-1}\frac{1}{4^i}

= \frac{4^n-1}{4-1} \cdot \frac{1}{4^{n-1}} = \frac{4^n-1}{4^{n-1}(4-1)} = \frac{4^{n-(n-1)}-\frac{1}{4^{n-1}}}{4-1} = \frac{4-\frac{1}{4^{n-1}}}{4-1} </math> It is well-known that <math>\lim_{n \to \infty} \frac{1}{a^x} = 0</math>, which implies that <math>\lim_{n \to \infty} \frac{1}{4^{n-1}} = 0</math>, so

<math>\lim_{N \to \infty} \sum_{i=0}^{N} \frac{1}{4^i}

= \lim_{n \to \infty} \frac{4-\frac{1}{4^{n-1}}}{4-1} = \frac{4}{4-1} = \frac{4}{3} </math>

By substituting the value <math>\frac{4}{3}</math> for <math>\lim_{N \to \infty} \sum_{i=0}^{N} \frac{1}{4^i}</math>, we get

<math>

s(g) = \frac{4}{3} \cdot (\frac{s(d)}{2}+\frac{s(o)}{4})

    = \frac{2}{3}s(d) + \frac{1}{3}s(o)

</math>

Which is the same result as the original proof (only slightly more complicated ;) <math>\Box</math>

References

Here's a few other silly theorems: