1. I've been looking for this anywhere I can but I'm not sure how I would even search for it since it is more of a question. If you had 10 numbers and the first number was say 1.978 and the last number was 19.335 is there a function to produce the correct 100th number(if of cousre you had a way to in fact check or knew the answer ahead of time)? I've tried using log but it obviously just goes from 1.978 to 19.78. Am I missing something to logs that can account for a run of numbers that increases or decreases at a non uniform rate? I see why it doesnt work but I honestly just dont know enough yet to see how to correct for this.

Its hard to ask this question on wikipedia or math world so what little the search returns is mostly about logs. It seems that this would be quite more difficult than just using logarithms since at each step the difference isnt the same for the last number. I'm getting frustrated but I'm still learning. It seems like the problems that hold you down for a while teach you the most so I guess its ok.

Thanks for any help,
jason  2.

3. What exactly would the 100th correct number be, as this question does not have an unique answer. The starting numbers of a sequence to not fix that sequence in any way - so there is no "correct" way to do what you are doing.

That said, what you are asking for is an extrapolation method - and the "easiest way" to do that for your type of problem is to construct a langrangian polynomial for the data points you gave (i.e. {1, a<sub>1</sub>}, {2, a<sub>2</sub>}...) and then use that to find the other terms.  4. I'm sorry I really didnt explain it correctly. I should have said if you already knew what the 100th number should be but was wanting to find a function that gave you that number. I've found that I seem to learn better by taking known equations or answers and see why they work. I should have given an example such as how much stirlings factorial equation is off from the correct amount and how that amount isnt uniform for each successive n!.
Oddly enough and while not even searching for it I believe I found what I was looking for. I just dont know its name or how you arrive at it since it was just there without any info. It reminds me of some of the series or expansions, if that is the proper use of the word, that I've only recently been reading about.

I happened to see some factorial equations by various people and noticed that some contained the sqrt2pi*n that I'm used to seeing but also contained a small series, I guess you would call it. One example had fractions such as (1/12*1/n)+(1/1440*1/n^3)+((239/362880*1/n^5). I see how this works to build a number that wont increase uniformly and I recognize the 362880 but are the other numbers significant beside this equation? I know sometimes to find a number you need you can start plugging away with trial and error but there seems to be more logic than that here.

Obviously thats not the whole equation but does it look like he is using langrangian polynomials river rat? Am I correct in calling the above an expansion?
Thanks again  5. So are you working on the higher order expanions for Stirlings approximation?

If you have a recurence relation for the sequence in question then there are ways to get formula's for the series (if you are lucky), i could walk you through the higher order approximations for stirling if you think you are up to it.  6. Sure I'd love it. The stirling equation really never had any significance with me but came about by chance reading through a book on probability, lol. Since then for fun I've tried to get a better aproximation(mean while not knowing how close others had already came). With all the reading and studying I've done it doesnt even compare to what I've learned while doing this. It caused me to learn how to find the root of any number even if the root may be 1.0844..., made me study intensely and now showed me the difference and approach others take and why certain equations are built as they are.

I'm pretty bummed cause I really thought I was on to something and spent a lot of time and energy on it, lol. I cant even imagine what someone would feel like if they worked on something for years, became friends with it and then find out someone has already been there or figured it out. If anything though I cant complain since I learned more valuable information by trying to work on this then any amount of reading on wikipedia or mathworld I've done in the past.  7. The more I mess with the above equation the more it makes sense. I've seen numerous times in different books or online types of series and it looks like the bottom number n or whatever it may be is raised to the next odd power like this.
(1/n)+(1/n^3)+(1/n^5).

To just look at this didnt mean anything to me and for the most part I would go right past it knowing that once I learned more I would come back. Well I started messing with it by doing this:
(1/2*1/n)+(1/4*1/n^3)+(1/8*1/n^5)

n=1 gives .875
n=10 gives.05025125

I can directly see how the 1/n^1,3,5,n moves the answer back along and how each answer adds directly to the overall answer.

.05(1) 025(3) 125(5)

Obviously this doesnt show me how to find what function makes n=1=.875 or n=10=.05025125 but I now have a much better understanding of why these work.

I looked up recurrence equations and it pointed me to the taylor series. My question is say you started with .875 for 1 and wanted 10(or I guess the 10th step?) to be 389, 551,45 or whatever number. It seems you would have to give the equation a sort of 'answer at this point' for lack of better terms so it could work its way there. Am I making sense? I know what I'm trying to say I'm just having trouble saying it, lol.  8. Okay, im trying to work out what the easiest way to show you how to derive this series is. Have you heard of the Euler summation formula?  9. I can imagine it is hard knowing something and trying to bring it down to a lower level for explanation for someone, lol.

I have read over quite often from other studies but only recently been homing in on some of these area. I really had to crawl before I could walk. I just started a book yesterday entitled 'Gamma' and it seems to center around Euler and his work so I'll try looking it up and reading it over before I repost.

I did take the nemes' factorial equation and break each part down and started with n=1 to see what the given numbers were doing. Makes sense but I know that it doesnt help me to know why if I cant find the mechanism that allows you to get numbers like 1/12 or 239/362880. I'm not good with putting names with equations yet so if Eulers summation formula works along these lines then yes I have just recently started studying them.

Thanks for the interest river rat.  10. Hi omerta

So are you still interested in seeing how to derive that expansion?  11. Definately still want to learn. I havent responded since I've been reading through my gamma book. It has a pretty good section on the Euler-Mclaren summation. I'm assuming this is what you are talking about. It gives two examples and I have been following along as to try and get a firm grasp on it.  12. While looking for examples I did see this on mathforum where he uses matrix math. Really goes well with my type of learning since I can see whats going on. I'm still going to learn the euler summation though but I though this was pretty cool. Have you seen this method before? Is it easier, harder or just another way to do it?

*From the mathforum*
I'm still not very familiar with the integral-expression,
although in the books of G.H. Hardy and of K.Knopp (the latter
is online available, though only in german) it is used
frequently.
I express the Euler-summation in matrix-notation, which
can easily be seen to be equivalent to Hardy's and Knopp's
sum-notation.
With a rowvector E~ = [1,1,1,1,...] and a column-vector
A = [a0,a1,a2,...an,...]~ a summation of A would simply
be written as

s = E~ * A (1)

The Euler-summation uses a replacement of E~ by a product
of a (lower triangular) binomial-matrix P

P = [1 0 0 0 ... ]
[1 1 0 0 ... ]
[1 2 1 0... ]
[1 3 3 1 ... ]
[ ... ]
and a rowvector V(1/2)~ = [1, 1/2, 1/4, 1/8,... ]

such that

1/2*V(1/2)~ * P = [1, 1, 1, 1, ... ] = E~ (2)

Now the Euler-summation uses replacement of E~ in
(1) by (2), thus establishing

E~ * A =
1/2*V(1/2)~ * P * A = s (3)

Hth -

Gottfried Helms  13. BTW River rat, Since there are still other areas where I need to strengthen so that this all flows well, dont feel that you have to respond imediately to these. In the meantime I'm still studying so pretty much whenever is fine.  Bookmarks
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