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Continued fraction for ex
By Brad Rodgers on Monday, December 09, 2002 - 09:50 pm:

How is the continued fraction for ex

contfrace

proven?

Thanks,

Brad

By Andre Rzym on Monday, December 09, 2002 - 10:17 pm:

Brad, have a look here first. When you are comfortable with it, I'll show you how to expand it for ez

Andre

By Brad Rodgers on Saturday, December 14, 2002 - 05:06 pm:

Sorry to take so long to get back; I'm comfortable with the article.

Brad

By Andre Rzym on Monday, December 16, 2002 - 08:25 am:

Firstly, we define

F(a,b,x) = 1 + (a)/(b).x/1! + (a(a+1))/(b(b+1)).x2/2! + (a(a+1)(a+2))/(b(b+1)(b+2)).x3/3! + …

This is analogous to (a generalisation of) (8) of the article.

Next we need two recurrence relationships. I claim that

F(a,b,x) = F(a+1,b+1,x) + q(a,b).x.F(a+1,b+2,x)

Also that

F(a,b,x) = F(a,b+1,x) + r(a,b).x.F(a+1,b+2,x)

By matching up the powers of x, in the power series, can you find q(a,b) & r(a,b)?

Andre

By Brad Rodgers on Monday, December 23, 2002 - 11:48 pm:

I get

q(a,b) = (a-b)/[b(b+1)]

r(a,b) = a/[b(b+1)].

I'm still unsure what to do next. We're naturally looking for a relationship to the effect of either

1/rn = 2n -1 - x/(2 + rn+1)

or

1/rn = 2 + x/(2n - 1 - rn+1),

where rn is a combination of F's, but I've yet to find one...

Brad

By Andre Rzym on Tuesday, December 24, 2002 - 09:43 am:

Brad, using your q(a,b) we have

F(a,b,x) = F(a+1,b+1,x) – (b-a)/[b(b+1)].x.F(a+1,b+2,x)

Rearranging:

F(a+1,b+1,x) / F(a,b,x) = 1 / 1 – (b-a)/[b(b+1)].x.F(a+1,b+2,x) / F(a+1,b+1,x) (1)

However, using your r(a,b) we also have:

F(a,b,x) = F(a,b+1,x) + a/[b(b+1)].x.F(a+1,b+2,x)

Rearranging:

F(a,b+1,x) / F(a,b,x) = 1 / 1 + a/[b(b+1)].x.F(a+1,b+2,x) / F(a,b+1,x) (2)

If you look at the rightermost part of (1), i.e. F(a+1,b+2,x) / F(a+1,b+1,x), this is just the LHS of (2), with a replaced by a+1, b replaced by b+1. So we substitute (2) into (1). Of the resulting equation, we can then substitute (1) into it and so on, until we get:

F(a+1,b+1,x) / F(a,b,x) = 1 / 1 – x(b-a)/b(b+1) / 1 + x(a+1)/(b+1)(b+2) / 1 - …

Now set a=0:

F(1,b+1,x) = 1 / 1 – x / (b+1) / 1 + x/(b+1)(b+2) / 1 - x (b+1)/[(b+2)(b+3)] / 1 + …
F(1,b+1,x) = 1 / 1 – x / (b+1) + x / (b+2) - x(b+1) / (b+3) + …

Now let b ® 0 [why can’t we just set it to zero?]

F(1, 1,x) = 1 / 1 – x / 1 + x / 2 - x / 3 + …

Now the LHS is just ex. To get it in the form you originally wrote, replace x by –x and take the reciprocal of both sides.

Andre

By Andre Rzym on Sunday, December 29, 2002 - 10:25 am:

One last point, Brad. If you set z=1 in your continued fraction you get:

e = 1 + 1 / 1- 1 / 2+ 1 / 3-

However you will also find that

e = 2 + 1 / 1+ 1 / 2+ 1 / 1+

where the denominators follow the pattern [1,2,1,1,4,1,1,6,1,1,8,..]

You may wish to try to derive the second formula from the first.

Andre

By Brad Rodgers on Wednesday, January 01, 2003 - 04:10 am:

Andre - the result you give is interesting; I've seen it before, but assumed that it's proven in an entirely different manner. The proof of it still puzzles me, though. I tried using recursive formulae, but I didn't get too far with that.

Could you give me a hint?

Thanks,

Brad

By Andre Rzym on Wednesday, January 01, 2003 - 09:19 am:



Brad, one thing you might look at is successive approximants to the continued fractions. Thus look at:

1, 1+1/(1-1), 1+1/(1-1/(2+1)), 1+1/(1-1/(2+1/(3-1))), etc

and

2+1, 2+1/(1+1), 2+1/(1+1/(2+1)) etc.

I’d do it on a spreadsheet. Notice any commonality? Now try replacing the last ‘1’ in each sequence by another number. What happens to the commonality?

Andre

By Andre Rzym on Sunday, January 26, 2003 - 02:00 pm:

Brad,

You may not be looking at this thread any more but if you are, you may be interested in the generalisation of the second continued fraction above for e. The question that has been bothering me for a while is how it generalises to ex. I’ve finally come up with an answer:

ex = (1+x) + x^2/(2-x)+ x/(3-x)+ x/1+ 1/1+ x/(5-x)+ x/1+ 1/1+ x/(7-x)+

Putting x=1 gives the series we had above.

This is effectively a ‘rearrangement’ of the one that you originally posted.

Andre