User:Karlhahn/user pi-irrational

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$\pi \ne \frac {p} {q}$

This user can prove that

π

is irrational.

Usage: {{{User:Karlhahn/user pi-irrational}}

PROOF:

If $\scriptstyle \pi$ were rational, then

$\pi = \frac {p} {q}$

where $\scriptstyle p$ and $\scriptstyle q$ are positive integers.

The function, $\scriptstyle x (\pi - x)$ has zeros at $\scriptstyle x = 0$ and at $\scriptstyle x = \pi$ . So does $\scriptstyle x^n (\pi - x)^n$ , where $\scriptstyle n$ is an arbitrary positive integer -- which is to say that $\scriptstyle n$ can be chosen to be arbitrarily large. Now we scale the function by $\scriptstyle q^n$ to get $\scriptstyle q^n x^n (\pi - x) = x^n (p - qx)^n$ , which also has zeros at $\scriptstyle x = 0$ and at $\scriptstyle x = \pi$ . Observe that this function is a $\scriptstyle 2n$ th degree polynomial with integer coefficients.

Finally we scale this function by $\scriptstyle \frac {1} {n!}$ to form

$f(x) = \frac {x^n (p - qx)^n} {n!}$

Looking at the derivatives of $\scriptstyle f(x)$ , we find that the first $\scriptstyle n-1$ derivatives also have zeros at $\scriptstyle x=0$ and $\scriptstyle x= \pi$ . At the $\scriptstyle n$ th derivative, Leibniz' rule yields the following:

$f^{(n)}(x) = \frac {1} {n!} \sum_{k=0}^n \binom {n} {k} \frac {n!} {k!} x^{n-k} \frac {n!} {(n-k)!} (-q)^{n-k} (p - qx)^k =$

$\frac {1} {n!} \sum_{k=0}^n {\binom {n} {k}}^2 n! (-q)^{n-k} x^{n-k} (p - qx)^k =$

$\sum_{k=0}^n {\binom {n} {k}}^2 (-q)^{n-k} x^{n-k} (p - qx)^k$

Observe that all of the coefficients in the last expression are integers. Further observe that

f (n) (0) = p n

which is an integer, and that

f (n) (π) = ( - q n)π n = ( - p) n

which is also an integer. The same kind of analysis on higher derivatives up to the $\scriptstyle 2n$ th derivative shows that they too have all integer coefficients, and more importantly, at $\scriptstyle x=0$ and at $\scriptstyle x=\pi$ , they too have integer values. Beyond the $\scriptstyle 2n$ th derivative, all derivatives are identically zero. Why? Because $\scriptstyle f(x)$ is a polynomial of degree $\scriptstyle 2n$ .

Now define a new polynomial function, $\scriptstyle F(x)$ , which is an alternating sum of even derivatives of $\scriptstyle f(x)$

F (x) = f (x) - f''(x) + f (4) (x) - f (6) (x) + ... f (2 n) (x)

It's quite easy to see that $\scriptstyle F(x) + F''(x) = f(x)$ . With only a little more effort you can see that

(F'(x)sin(x) - F (x)cos(x))' = f (x)sin(x)

This means that

$\int_0^\pi f(x) \sin(x) dx = {\lbrack F'(x) \sin(x) - F(x) \cos(x)} \rbrack_0^\pi$

Since $\scriptstyle \sin(x)$ is zero at $\scriptstyle x=0$ and at $\scriptstyle x=\pi$ , the right hand side of the above is simply equal to $\scriptstyle F(\pi) + F(0)$ . The previous analysis of derivatives of $\scriptstyle f$ requires that $\scriptstyle F(\pi) + F(0)$ be an integer. That means that the area under the curve,

$g(x) = \frac {x^n (p - qx)^n} {n!} \sin(x) = \frac {q^n x^n (\pi - x)^n} {n!} \sin(x)$

between $\scriptstyle x=0$ and $\scriptstyle x=\pi$ is an integer. We know that $\scriptstyle g(x)$ is positive throughout that open interval. We also know that $\scriptstyle g(x)$ can be bounded in the interval to as small a positive value as you like simply by choosing $\scriptstyle n$ large enough. The area under the curve can be no greater than $\scriptstyle \pi$ times that bound. Hence the area can also be bounded to as small a positive value as you like by choosing $\scriptstyle n$ large enough. That means the area can be bounded to less than unity. We are left with an area whose value is an integer that is strictly between zero and unity, which is clearly impossible. Hence $\scriptstyle \pi$ cannot be rational.

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User:Karlhahn/user pi-irrational

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