Egyptian fractions

[Mr Stoofer]

This was a colleague's daughter's homework and we wasted a day trying to figure it out.

Egyptian fractions, so I am told, had to be in the form of a sum of fractions where the numerator in each case was 1 and the denominator was a unique integer.

Thus they were in the form

X=1/a + 1/b + 1/c ... etc etc

a, b, c etc are integers but all different.

For example:

3/4 = 1/2 + 1/4

2/3= 1/3 + 1/4 + 1/12

We managed to work out that any fraction in the form 2/x can be expressed as the Egyption fraction 1/x + 1/2x + 1/4x + 1/6x + 1/12x.

Becuase 2/x = 24/12x = 12/12x + 6/12x +3/12x + 2/12x +1/12x.

Questions

1. Can you work out a general formula for 3/x, 4/x, 5/x etc?
2. Can you work out a general formula for y/x?

[mith]

A better result for 2/x is:

2/x = 1/x + 1/(x+1) + 1/(x ² +x)

[Jack_Ian]

since 1/x = 1/(x+1) + 1/(x^2 + x)

for 3/x
3/x = 1/x + 1/x + 1/x
= 1/x + (1/(x+1) + 1/(x^2 + x)) + (1/(x+1) + 1/(x^2 + x))

Now for terms that are the same just repeat the process

= 1/x + 1/(x+1) + 1/(x^2 + x) + ( 1/((x+1)+1) + 1/((x+1)^2 + (x+1))) + (1/((x^2 + x)+1) + 1/((x^2 + x)^2 + (x^2 + x)))
= 1/x + 1/(x+1) + 1/(x^2 + x) + 1/(x + 2) + 1/(x^2 + 3x + 2) + 1/(x^2 + x + 1) + 1/(x^4 + 2x^3 + 2x^2 + x)

Unfortunately it is possible that some of the terms might have the same divisor when actual numbers are used so it might be necessary to repeat the process to split any of the fractions that are in conflict with an already used term. This step might even need to be performed several times. At least it can't become a cyclical process, since all terms are added and an infinite cycle would result in the sum going to infinity.

Interetingly,
1 = 3/3 = 1/3 + 1/4 + 1/12 + 1/5 + 1/20 + 1/13 + 1/156
= 1/3 + 1/4 + 1/5 + 1/12 + 1/13 + 1/20 + 1/156

I wonder if that kind of information is useful.