Transcendental Numbers

Date: 5/31/96 at 14:24:27
From: David Lewis
Subject: Transcendental Numbers

I am doing a school project and one of the questions is:

	Joesph Louville's candidate for a transcendental number was
	a=.110001000000000000000001..... Where is the next 1?

My partner and I have looked all over and have not been able to come 
up with the answer.  Can you help us?

Date: 5/31/96 at 15:21:19
From: Doctor Darrin
Subject: Re: Transcendental Numbers

The next 1 would be in the 120th place after the decimal. Liouville
proved that numbers of the form:

1/n+1/n^2+1/n^6+1/n^24+1/n^120+1/n^720+...

are trancendental (where the exponents are factorials (usually written 
with an exclamation point):
1!=1
2!=2*1=2
3!=3*2*1=6
4!=4*3*2*1=24
5!=5*4*3*2*1=120
...
n!=n*(n-1)*(n-2)*...*2*1

The number that you gave is the case of the formula above where n=10.
So you would have 1 in the 1st, 2nd, 6th, 24th, 120th, 720th, 5040th, 
etc, places and zeroes everywhere else.

-Doctor Darrin,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/