Example 1.5.4, the decimal 0.101001000100001000001..., is related to an 1844 construction by Joseph Liouville, who proved that the number is transcendental. The decimal expansion of Liouville's number can be thought of as a sequence of blocks, each block consisting of a digit 1 followed by a string of zeros. Liouville's example differs from Example 1.5.4 because now the strings of zeros increase in length extremely fast. Let j be the number of times the digit 1 appears in the first 50 decimal places of the number in Example 1.5.4, and let k be the number of times the digit 1 appears in the first 50 decimal places of Liouville's number. Then j - k = 1 1 10n! 3 5

Example 1.5.4, the decimal 0.101001000100001000001..., is related to an 1844 construction by Joseph Liouville, who proved that the number is transcendental. The decimal expansion of Liouville's number can be thought of as a sequence of blocks, each block consisting of a digit 1 followed by a string of zeros. Liouville's example differs from Example 1.5.4 because now the strings of zeros increase in length extremely fast. Let j be the number of times the digit 1 appears in the first 50 decimal places of the number in Example 1.5.4, and let k be the number of times the digit 1 appears in the first 50 decimal places of Liouville's number. Then j - k = 1 1 10n! 3 5

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Example 1.5.4, the decimal
0.101001000100001000001...,
is related to an 1844 construction by Joseph Liouville, who proved that the number
is transcendental. The decimal expansion of Liouville's number can be thought of as a sequence of
blocks, each block consisting of a digit 1 followed by a string of zeros. Liouville's example differs
from Example 1.5.4 because now the strings of zeros increase in length extremely fast.
Let j be the number of times the digit 1 appears in the first 50 decimal places of the number in
Example 1.5.4, and let k be the number of times the digit 1 appears in the first 50 decimal places of
Liouville's number. Then j - k =
1
1
10n!
3
5

Expert Solution

Step 1

0.10100100010000100000100000010000000100000000100000...

This is the number with first 50 decimal places in example 1.5.4

In which 1 appeared 9 times

So j=9

While

In Liouville's number

1/10^1! + 1/10^2! + 1/10^3! + 1/10^4! + 1/10^5! + ...

=0.11000100000000000000000100000000000000000000000000...

This is the number with first 50 decimal places in Liouville's number

In which 1 appeared 4 time's

So k=4

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