Upload
others
View
20
Download
0
Embed Size (px)
Citation preview
Egyptian Fraction Expansions
Elvia Nidia Gonzalez & Julia Bergner PhDDepartment of Mathematics
University of California, Riverside
• The Egyptians expressed their rational numbers as sums of unique unit fractions.
• A unit fraction has the number one in the numerator.
What is an Egyptian Fraction Expansion?
• The Egyptians expressed their rational numbers as sums of unique unit fractions.
• A unit fraction has the number one in the numerator.
What is an Egyptian Fraction Expansion?
= 110= 1
3
Were the Egyptians limited?
⇡ = 18 + 1
61 + 15020 + 1
128541455 + . . .
e = 12 + 1
5 + 155 + 1
9999 + 13620211523 . . .
dre
d 12e = 1
d5.32e = 6
Ceiling Function
The ceiling function is the least integer no less than than r
dre
d 12e = 1
d5.32e = 6
Ceiling Function
The ceiling function is the least integer no less than than r
d� 12e = 0
dre
d 12e = 1
d5.32e = 6
Ceiling Function
The ceiling function is the least integer no less than than r
d� 12e = 0
d3e = 3
Engel Series expansions with lengths equal to their numerators
x
x
n
{a1, a2, . . . , ax}
2 23 {2, 3}
3 37 {3, 4, 7}
4 413 {4, 5, 7, 13}
5 561 {13, 16, 21, 31, 61}
6 661 {11, 13, 16, 21, 31, 61}
The Pattern
We conjectured and proved that
n = lcm(x, x� 1, . . . , 3, 2) + 1 will
produce a length x Engel expansion.
Is this the only “n” that works?
37 = 1
3 + 13·4 + 1
3·4·7313 = 1
5 + 15·7 + 1
5·7·13319 = 1
7 + 17·10 + 1
7·10·19
The Second Pattern
x 2 N where N = {1, 2, 3, 4, . . . }
n = k · lcm(x, x� 1, . . . , 3, 2) + 1for
k 2 N
The Second Pattern
We conjectured and proved that
n = k · lcm(x, x� 1, . . . , 3, 2) + 1 will
produce an entire family of length x
Engel expansions for
x
n
2 (0, 1).
Families of expansions whose lengths equal their numerators
3n n Engel Expansion
37 7 = 1 · lcm(3, 2) + 1 1
3 + 13·4 + 1
3·4·7
313 13 = 2 · lcm(3, 2) + 1 1
5 + 15·7 + 1
5·7·13
319 19 = 3 · lcm(3, 2) + 1 1
7 + 17·10 + 1
7·10·19
325 25 = 4 · lcm(3, 2) + 1 1
9 + 19·13 + 1
9·13·25
331 31 = 5 · lcm(3, 2) + 1 1
11 + 111·16 + 1
11·16·31
337 37 = 6 · lcm(3, 2) + 1 1
13 + 113·19 + 1
13·19·37
Conjectures to Prove
• We would like to prove
n = lcm(x, x� 1, . . . , 3, 2) + 1 is the
least n that will produce a length x
Engel expansion for
x
n
2 (0, 1).
Related Questions
• What conditions produce
“best case” scenarios. That is,
expansions of length 2?
• Does n = k · lcm(x, x� 1, . . . , 3, 2) + 1produce a length x expansion for
x
n
< 0? Or x
n
> 1?