| Author | Message | ||
| Gillian Stansfield |
I read with great interest the article on Infinite Continued Fractions by Andre Rzym (NRich September 2002) This article showed how the continued fraction for tanh(x) may be found. For some time I have been trying to discover an analytic method for finding the Infinite continued fraction for exp(x), sin(x) ln(x) etc, and also for pi. Can you help; and/or recommend any (reasonably easy to follow) book or article that could do so? | ||
| Ian Short |
I'm sure Andre will be able to answer this with his own ez continued fraction expansion. There is not a UNIQUE continued fraction expansion for any of these functions, there are many. Power series are generally better as they are unique, easy to manipulate, differentiate, convergence issues are clear. Regarding books, there are three modern expensive books on continued fractions. (1) H.S.Wall- not actually very modern. (2) Jones and Thron (3) Lorentzen and Waadeland These are in chronological order. I think they are all easy to read but I wouldn't recommend buying any unless you are really interested. I could put some of the expansions you ask for here, but not now as I don't have the books with me (I've not memorised them). Although there do exist lots of random continued fraction expansions of functions such as those you describe, or p, the whole subject (of CF representation of analytic functions) is a mess and lacks order, it needs sorting out! Ian | ||
| Ian Short |
There's a lot of information in mathworld. | ||
| Andre Rzym |
Gillian, I have been away - could you confirm you are still looking at this thread - if so I'll sort you out with ln(x), pi etc. As Ian said, those books are expensive! Andre | ||
| Andre Rzym |
Gillian, You might also have a look here (although the first equation doesn't show up on my browser - you may have to guess it from what follows). Andre | ||
| Emma McCaughan |
The picture wasn't where it should be. It should now appear next time the site is updated. In the meantime, maybe I can put it here:  | ||
| Gillian Stansfield |
Thank you very much Ian and Andre. I am still looking. I find continued fractions fascinating - and fustrating. Of course I agree with Ian that power series are better in many senses, but that's not the point. I have looked up the books on Amazon.co.uk, and it seems that the H S Wall book has been recently republished by the American Mathematical Society - in 2001 - as The Analytic Theory of Continued Fractions, and costs £21.50 - so not too bad. I have found that Brunel Univ have a copy of the Lorentzen and Waadeland so I may try and get hold of that. Thanks too for the other links Andre. I should have looked at the "Asked Enrich" before posting my question. I found that very helpful. I would be delighted if you have the time to sort me out with continued fractions for pi and ln(x) etc, Andre, and what about the Ramanujan cf you started your excellent article with?! Many thanks Gillian | ||
| Andre Rzym |
Gillian, I agree with you about CF’s being fascinating and frustrating – when you first look at them (e.g. the expansion for e) they just seem to defy attack. The plan of assault on ln(x) and p [actually we will go for ln(1+x) and tan-1(1)] will be to generalise the approach of this. The approach is that of Gauss, and allows us to capture the CF expansions for a range of functions. Firstly, we need to generalise equation (8) of the article. Define the hypergeometric function: The series never terminates, although obviously if a or b are non-positive integers then the number of terms becomes finite. Also, if a or b are zero, the function is particularly simple! Note that the ‘2’ prefix (in front of ‘F’) denotes the number of parameters in the numerator (‘a’ and ‘b’ in this case) and the ‘1’ suffix denotes the number of parameters in the denominator (‘c’ in this case). So, being consistent, we would define (for example): etc. Secondly, note some limit relationships: Thirdly, let’s look at how various functions can be represented in terms of hypergeometric functions. For example: Here’s a couple of things to think about: Can you represent ln(1+x) in terms of 2F1 Can you represent tan-1x in terms of 2F1 Once you have had a look at them, we’ll develop a recurrence relationship involving 2F1. Andre | ||
| Gillian Stansfield |
Thanks again Andre; I will have a go when time permits - hopefully very soon. Gillian | ||
| Ian Short |
There is a general scrappy way of getting a continued fraction expansion from the power series. Power series: a0+a1z+a2z2+... Then the continued fraction is: a0 + a1z(1+a2z/a1 +....) = a0 + a1z/(1+a2z/a1 +....)-1 Set x=a2z/a1 +.... then use the power series expansion of (1+x)-1 to get another power series and repeat the procedure. I think Euler is originally credited with this. Ian | ||
| Gillian Stansfield |
O wow! I like it! And I had wondered how Euler found his continued fraction for e. I actually once tried something like it myself - but gave up too quickly because it seemed so messy. I'll try again. Thanks for that, Ian. Thanks too for the link to mathworld. There is some useful stuff there about the use of hypergeometric series to find the cfs of analytic functions. Gillian | ||
| Ian Short |
Incidentally, if you do buy a CF book then I'd recommend Lisa Lorentzen's book above the other two. Wall's is dated (the new version has no significant changes). Ian | ||
| Andre Rzym |
I also prefer Lorentzen's book to Wall - it is far less dry. It also has stuff that Wall does not - e.g. the relationship to Riccati equations. I would caution you that (unless you have a symbolic maths package) the route Ian describes is a lot of work. Computing reciprocals of series is tedious. What is worse, I doubt that (in gereral) you would recognise the power series for the "x's" in the formula above, so computing the first few terms of the series and guessing the rest is not likely to work. Andre | ||
| Ian Short |
Yeah, it's a right mess. What it shows is that every function with a power series has a continued fraction associated. Ignoring convergence issues. Ian | ||
| Andre Rzym |
Gillian, If you are still reading this thread, you may be interested in the following rather bizarre continued fraction that I computed today: It takes a few seconds to spot the pattern, but you should notice the two sequences of 0,0,1,1,2,2 etc. Everything else repeats itself. I would reiterate Ian’s comment that CFs are not unique – they multiply as a result of different methods of generation (hypergeometric [or other] recurrence relationships, inversion of power series, differential equations etc), manipulation of a given series (expansion/contraction formulae) and specific manipulations arising from the function itself (e-x=1/ex etc). In that context, the above CF is not interesting, but it does converge rather quickly – I’m pretty sure that you gain about 0.4 decimal digits for each additional ‘nested fraction’ you add. Andre | ||
| Gillian Stansfield |
Hallo Andre. I am still reading this thread. I've finally carved out some time, and I've finally found hypergeometric series for ln(1+x) and tan-1(x) as you suggested on 3 September. I get ln(1+z)= z2F1(1,1,2,-z) and tan-1(z)= z2F1(1,1/2,3/2,-z2) It took me ages. I had hoped I would just spot the answer when I equated coefficients, but no. I also wondered if I could use the hypergeometric series you gave for the binomial expansion, and put r to convenient values and integrate, etc - but of course - no joy, as for both ln(1+z) and tan-1(z), r is negative, and so the hypergeometric series terminates. How do you know what sort of hypergeometric series to equate your function to. ie how do you choose m and n in mFn? | ||
| Andre Rzym |
Gillian, I agree with your two expressions. tan-1(x) is particularly tricky. Next we need a recurrence relationship for 2F1. This will generalise equation (11) of the article. I claim that (note that I am omitting the x argument to save space): The question is, what is the function q(a,b,c)? Note that it is a function of a,b,c only. Specifically, it has no dependence on x. The way to find it is to compare powers of xn on both sides [it’s an exercise worth doing for yourself], yielding: Dividing by F(a,b+1,c+1) and taking the reciprocal of both sides we get: (1) But note that F(a,b,c,x)=F(b,a,c,x). Therefore we can take (1), swap the variable names a & b, then reverse the order in which they are ‘supplied’ to F, to give: (2) Now we have everything that we need: Firstly, start with (1). Now take (2), but with b replaced by b+1 & c replaced by c+1. This can then be substituted into (1) to give: (3) We repeatedly substitute (1) and (2) until we get: (4) This is the continued fraction of gauss. Note that, at the time of writing, there is a minus sign that should be a plus for this formula on the wolfram page here. Note also that the (more basic) recurrence relation in the article can be derived from this one via the limit approach mentioned a few posts ago. Now it remains to use this relationship to churn out some results. As an example, The right hand side is in the format of (4) [with a=1,b=0,c=1], so we have I’ll leave you to produce similar expansions for the other functions we listed above. Let me know how you get on. You should be able to get (1+x)k, tan-1x, ln[(1+x)/(1-x)], ex, tan(x), tanh(x) etc. Andre | ||
| Gillian Stansfield |
Thanks Andre. The way looks much clearer now, and I'll go through what you wrote carefully when I next have a chance. I'm not sure about the limit relations though. Should the first one, for example, say limit as n tends to infinity of 2F1(n,b,c,x/n) = 1F1(b,c,x)? (you had written 2F1(an,b,c,x/n) on the LHS? Gillian | ||
| Andre Rzym |
Gillian, Thanks for pointing out my carelessness. Let me try again: Andre |