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/
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