You set up an infinite series

1*x + 1*(x^2) + 2*(x^3) + 3*(x^4) + 5*(x^5) + 8*(x^6) + 13*(x^7) + ...

Call that expression A. Multiply A by x to get A*x =

1*(x^2) + 1*(x^3) + 2*(x^4) + 3*(x^5) + 5*(x^6) + 8*(x^7) + 13*(x^8) + ...

Then subtract these two expressions by subtracting each term of the second expression from the term of the first expression with the same power of x -- the resulting expression

A - A*x = 1*x + 1*(x^3) + 1*(x^4) + 2*(x^5) + 3*(x^6) + 5*(x^7) + ...

which except for the first term is exactly A*(x^2)

In other words

A - A*x = x + A*(x^2)

If you solve this for A by bringing all terms involving A to one side and then factoring out A, you end up with

A = x / (1 - x - x^2)

By now you're wondering what is the point of all of this. Well if you go back to the original series

1*x + 1*(x^2) + 2*(x^3) + 3*(x^4) + 5*(x^5) + 8*(x^6) + 13*(x^7) + ...

and substitute .001 for x, the terms become

.001 + .000001 + .000000002 + .000000000003 + .000000000000005 + .000000000000000008 + .000000000000000000013 + ...

which equals

.001001002003005008013 ...

so the same should be true if you substitute .001 for x in

A = x / (1 - x - x^2)

and if you do you get the fraction

1000 / 998,999

The decimal equal to this fraction begins with all the Fibonacci numbers with 1, 2 or 3 digits. If you want a fraction that has all the Fibonacci numbers with 1, 2, 3, or 4 digits, then you substitute .0001 for x into A (instead of .001).Joseph G. Rosenstein

Professor of Mathematics

Rutgers University

## Monday, January 23, 2012

### More interesting than 22/7

This post is based on this posting from Futility Closet.

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