Continued fraction for e (1 Viewer)

[tywebb]

See if you can prove this:

 

[Sy123]

Just noting so I can return to this later (and maybe it is useful to others)

You can reduce the problem into finding:

 

[dan964]

would you use Maclarin series of e^x?

 

[GoldyOrNugget]

Incomplete but I think it gets you to the solution if you actually work out the steps.

is a well-known formula for . Also,



From http://en.wikipedia.org/wiki/Engel_...sions.2C_continued_fractions.2C_and_Fibonacci (probably has a simple inductive proof but I'm lazy)

Setting and taking the limit as n goes to infinity, the left hand side is and the right-hand side is an 'upside-down' continued fraction. The continued fraction is equivalent to the one given in the question and the transformation between the two is probably easy to do.

Give me a break, I haven't done maths in a year

 

[seanieg89]

It's a little cumbersome to type out, (typesetting continued fractions is gruelling), but the basic idea can be summarised.

1. Chuck in x=-1 to the convergent Maclaurin series for e^x, this gives you an alternating series expression for e^(-1).

2. Find a finite continued fraction representation of a finite series like on Goldy's LHS but alternating. This comes from induction and the identity This lets us approximate 1-e^(-1) arbitrarily well by a finite continued fraction.

3. Notice what happens when you invert a continued fraction. The result comes from manipulating the continued fraction in 2. We invert, subtract 1, invert again, and add 1. This gives us the desired representation of e.

Note that we can't just subtract 1 from 1-e^(-1) and invert, because the continued fraction in question has no whole part outside the nested fractions, so subtraction is not easy (without getting negative numbers involved).