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MathGroup Archive 2006

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Smarter way to calculate middle-right terms of continued fraction partial sums

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70024] Smarter way to calculate middle-right terms of continued fraction partial sums
  • From: "Diana" <diana.mecum at gmail.com>
  • Date: Sat, 30 Sep 2006 05:13:05 -0400 (EDT)

Math folks

I am trying to calculate the iterative terms of a continued fraction.

It is a limit of partial sums of continued fractions defined as
follows:

x[n_]:= Sum as j goes from 1 to n of (z^q^j)/d[j], where d is a
function of j.

When n = 1, z = 1, and q = 2,

x[1]:= 1/d[1]

I then

Solve[x[1] == 1/a1, a1]

So, x[1] = {0, a1}, a two-term continued fraction.

The second term looks like:

x[2]:= 1/d[1] + 1/d[2]

I then

Solve[x[2] == 1/(a1 + (1/-a2 + (1/-a1))), a2]

And get a partial sum which looks like

x[2] = {0, a1, -a2, -a1}, a four-term continued fraction.

x[3]:= 1/d[1] + 1/d[2] + 1/d[3]

Solve[x[3] == 1/(a1 + (1/-a2 + (1/-a1 + (1/-a3 + (1/a1 + (1/a2 +
(1/-a1))))))), a3]

x[3] = {0, a1, -a2, -a1, -a3, a1, a2, -a1}, an eight-term continued
fraction.

So, each successive x[n] term is two times as long as the one before
it, and I am always solving for the middle right term. After I have
found the x[3] term, for example, I can plug the solved value of a3
into the Solve equation and solve for the x[4] term.

The terms eventually look like the following:

{0, a1, -a2, -a1, -a3, a1, a2, -a1, -a4, a1, -a2, -a1, a3, a1, a2, -a1,
-a5, a1, -a2, -a1,

-a3 a1, a2, -a1, a4, a1, -a2, -a1, a3, a1, a2, -a1, a6, ...}

Note that the signs of a1 alternate between positive and positive.
Likewise, all the variables alternate in sign.

I have been coding the Solve equation as fractions of 1 over a1 + 1
over -a2 + 1 over, etc.

By the fifth term, I have 63 fractions to code. I am sure there is a
smarter way to do this, so I am posting this question to the news
group.

Can someone help?

Diana


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