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June 28

Closed Form Formula

Working on a problem, I have come across this recursively defined sequence,
$0,2,{\frac {9}{2}},{\frac {44}{6}},{\frac {250}{24}},{\frac {1644}{120}},\cdots$ .
The denominators are clearly the factorials but is there a closed form formula to produce the numerators. The recursive relation (if it helps) is
$a_{0}=0,\,a_{k}=(a_{k-1}+1){\frac {k+1}{k}}$
Thanks!--08:24, 28 June 2008 (UTC)

The numerators are sequence A052881 at Sloane's. The fractions themselves are

a_{n}=(n+1)\sum _{i=1}^{n}{\frac {1}{i}}.

Algebraist 10:20, 28 June 2008 (UTC)[reply]

Wow, this is exactly what I was looking for. But how do you know what the explicit form is? How can one derive it (from the recursive relation or otherwise)? Thanks--A Real Kaiser (talk) 21:58, 28 June 2008 (UTC)[reply]

I derived it by looking at Sloane's (that's what it's there for!). As a check, I proved it by induction. Of course, there's still that summation in there, so it's not as closed a form as one might like. I suppose if you needed to approximate terms cheaply, you'd replace the sum by log(n) or log(n)+γ. Algebraist 22:18, 28 June 2008 (UTC)[reply]

Actually, believe it or not, this IS the form I needed. The summation is exactly what was needed. So this form is not bad at all for what I was doing, it actually made everything easier (even the (n+1) canceled out).--A Real Kaiser (talk) 02:14, 29 June 2008 (UTC)[reply]

Question about Floor and ceiling functions and INT(x) in computers etc

for reals, floor(x) returns int(x) except when x is already an integer - in which case I get x-0.5

ie floor (2.78)=2 floor (2.1)=2 floor(2)=1.5

But if I use f(x)=floor(floor(x)+n) where 0.5<n<1 I get f(x)=int(x) for all reals..

Floor function can be expressed as a fourier series or infinte polynomial, and so I imagine that with great difficulty I could express floor(floor(x)+n) as an infinite series too...

Is this series known? requesting links to look where to start, or explaining why it doesn't work.. Thanks.87.102.86.73 (talk) 10:53, 28 June 2008 (UTC)[reply]

Huh ? Which language or program are you using here ? I have never seen an implementation of floor which gives floor(2) = 1.5. There is sometimes a confusion between floor and int for negative reals, for which int may round up in some languages, but neither floor nor int should ever give a non-integer result. Gandalf61 (talk) 11:17, 28 June 2008 (UTC)[reply]

silly gandalf - for floor(x) I mean the mathematical version ie

\lfloor x\rfloor =x-{\frac {1}{2}}+{\frac {1}{\pi }}\sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}.

(copied from the wiki page)

and by 'INT(x)' I meant the computer implementations.. I was trying to get floor(x) (as a power or fourier series) to match INT(x).87.102.86.73 (talk) 11:37, 28 June 2008 (UTC)[reply]

Oh yes, silly me, apology.. I confused floor(x) and its power series...

What I meant was a power or fourier series etc that matches floor(x) or INT(x) for all real finite x including x=2,3,etc.. thanks.87.102.86.73 (talk) 11:52, 28 June 2008 (UTC)[reply]

Power series are far too well behaved to match a discontinuous function like floor. My memories of the ways of Fourier analysis are hazy, but I'm reasonably sure that the series you've got (with the value at discontinuities being the average of the left and right limits, rather than the correct value) is the best possible. Algebraist 12:25, 28 June 2008 (UTC)[reply]

And as a matter of interest, why do you want to do this? Algebraist 12:28, 28 June 2008 (UTC)[reply]

Question - I was wondering if I tried to constructed a power series from f(f(x)+n) (n=0.7?) what it would actually do? (ie f(x) is the power series or fourier series closest equivalent to int(x) )

I could easily get the first few terms of xⁿ.. there definately looks like there is a (set of) series...

I'm tentatively in agreement with you that I won't be able to get a discontinuity that goes from say 1 to 2 when x goes from 1.99... to 2.000...0001 , so what does happen? anyone know?

As to why I ask - pure curiosity - hoping I will become 'cleverer'...with your help

Could anyone explain why and if the fourier series above is the best possible..87.102.86.73 (talk) 13:01, 28 June 2008 (UTC)[reply]

I imagine this question boils down to "what happens/can go wrong with nested infinite power series" eg a similar expression would be e^(e^x) - which at first sight seems to be expressable as a power series.. does something 'go wrong' or not?87.102.86.73 (talk) 13:04, 28 June 2008 (UTC)[reply]

Power series are everywhere continuous, so can't work here. What do you mean when you say you can get the first few terms of xⁿ? Broadly speaking, if two functions are expressible as power series, then so is any combination of them. Thus

e^{e^{x}}

is fine, for example. See analytic function for more info. Algebraist 13:13, 28 June 2008 (UTC)[reply]

Mmmh, you're right - I spoke before I actually checked I can't get ::

\lfloor x\rfloor =x-{\frac {1}{2}}+{\frac {1}{\pi }}\sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}.

as a polynomial in x for reasons that are now obvious to me..87.102.86.73 (talk) 13:17, 28 June 2008 (UTC)[reply]

I think I made the assumption that because floor(x) can be expressed as a fourier series it would be an analytical function.. which it isn't.. is that right? (Thanks for your help I think I half understand now)87.102.86.73 (talk) 13:19, 28 June 2008 (UTC)[reply]

Yes. Fourier series are far more general than power series. For example, any function with finitely many discontinuities and local extrema has a Fourier series that converges to the right value (except maybe at discontinuities, as here), but almost all such functions are nowhere differentiable, and thus have no power series. Algebraist 13:57, 28 June 2008 (UTC)[reply]

Thanks. error in my reasoning now understood, a little wiser now.87.102.86.73 (talk) 15:01, 28 June 2008 (UTC)[reply]

Calculating infinite series fractions sum

Hi. This is not homework. I'm just curious as to the formulae used to calculate these, expecially with fractions, and if there is an article on this. I'm not talking about the more obvious ones like 1/3+1/30+1/300... or 1/10+1/200+1/1000... . I mean the other ones like 1/2+1/4+1/8+1/16... or 1/2+1/3+1/4+1/5... or 1/2+1/4+1/6+1/8+1/10... or 1/2+1/4+1/16+1/256... or 1/2+3/4+5/6+7/8... , etc. This is probably a silly or random or obvious question, but I'm just wondering how this is usually canculated in our base-10 number system. Thanks. ~A H 1^(TCU) 14:38, 28 June 2008 (UTC)[reply]

If a series converges, then you can calculate the partial sums (i.e. sums of the first n numbers) to get (better and better) approximations to the limit. You can probably also calculate how good these approximations are (depending on the series). For especially nice series, you can calculate the limit explicitly, and not have to use approximations. Of course, if the series diverges, then there's no limit at all. There are a number of techniques for determining whether a series converges (and if so, if there's a nice form for the limit); fortunately, your examples are all dead easy. The first ~~and fourth~~ are geometric series, for which the limit is easy to find, and the other three can easily be shown to diverge, due to their relation to the harmonic series. For a nice example of a series whose limit is not obvious, see Goldbach–Euler theorem. Algebraist 14:53, 28 June 2008 (UTC)[reply]

Very powerful methods to calculate a number of different series are using the Taylor series or the Fourier series. (Igny (talk) 18:24, 28 June 2008 (UTC))[reply]

I disagree with Algebraist. The fourth is not geometric. I believe that the sum of the fourth series is unknown, perhaps it cannot be expressed in terms of elementary functions. It is approximately 0.8164. The big picture is that there are no general methods to determine the sum of an infinite series exactly. For a few simple series, this is easy. In other cases (such as the series

1+1/4+1/9+1/16+1/25+\cdots

) it requires some insight or tricks to find the sum. For most series that you would write they either diverge or the sum cannot be expressed in simpler terms except for the infinite series itself. In some of these cases, which are important for one reason or another, a name is given to to sum of the series. (For example, Apery's constant.) Oded (talk) 20:42, 28 June 2008 (UTC)[reply]

Sorry, I misread the fourth series. Algebraist 21:57, 28 June 2008 (UTC)[reply]

It looks like series four has been studied somewhat, but no nice form is known. The decimal expansion is A007404 at Sloane's. Plouffe's has 20000 digits. Algebraist 22:29, 28 June 2008 (UTC)[reply]