Wikipedia:Reference desk/Archives/Mathematics/2015 October 26

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October 26

Series of sum of previous reciprocals

Let A_1=1, let A_(n+1) = sum(k=1...n) of 1/(A_k).

So the series goes 1, 1, 2, 2.5, 2.9, ~3.244827586 , etc. I'm trying to figure out it this is logarithmic and if so how. If I want to find A_1000000, does anyone have any better idea than an excel spreadsheet? (A1 set to 1, B1 and below set to =1/A1 , A2 and below =sum($B$1:$B1) ) Naraht (talk) 05:01, 26 October 2015 (UTC)

Python program

res=1.0
tel=1 3
while tel<=1000000:
   res=res+(1.0/res)
   tel=tel+1
print(res)

3000 -> 77.49662419442322 77.4708101697002

10000 -> 141.44372894635987 141.42958798733721

30000 -> 244.96301286490154 244.9548481629191

100000 -> 447.22195789401366 447.2174858081414

300000 -> 774.6018518681202 774.5992698900433

1000000 -> 1414.21661385363 1414.2151996420582

3000000 -> 2449.49161668406 2449.4908001878994

10000000 -> 4472.1370486818205 4472.136601468301 Ssscienccce (talk) 06:39, 26 October 2015 (UTC) edited 15:51, 26 October 2015 (UTC)

With a bit of heuristics I got A(n) ~ √(2n) which seems to agree with the previous results. I suspect this is the first term of an asymptotic series. --RDBury (talk) 13:44, 26 October 2015 (UTC)

PS. I think the python program is actually computing A(n+2); at least when I computed the values using Excel I get A(3002) is 77.49662419 which matches the 3000 value above. In any case I get the empirical formula:

$A(n)\approx {\sqrt {2n+.5\ln n}}$

which correct for n=10000002 to 3 digits. --RDBury (talk) 14:28, 26 October 2015 (UTC)

Oops, you're right of course, it starts with A(3) . Changed it. Ssscienccce (talk) 15:51, 26 October 2015 (UTC)

OP here. I hadn't thought of re-expressing it as A_(n+1)=A_n + 1/A_n, that makes calculation easier....

Flip the sign (Chaotic?)

Given that the previous one could be expressed as A_(n+1)=A_n + 1/A_n , I'm now wondering about the behavior of A_(n+1)=A_n - 1/A_n. Presuming that A_1 is not equal to 1 or -1, this should never go to 0 (and then infinity), but the behavior seems Chaotic (and I mean that in the mathematical sense). I set 10 as the first value, and it decreases gradually until it drops below 1 and jumps to something with a much larger absolute value, decends again and jumps back out. Any feelings on this?

So A(n) is obtained by iterating the rational function f(x) = x + 1/x (or x - 1/x). Then you can find the corresponding Julia set which are almost always fractals and the action of f on the Julia set is almost always chaotic. The approach you use with quadratic polynomials is to conjugate with some invertable function to get the number of parameters down to 1, giving f(x) = x²+c, and then vary the parameter to get a Mandelbrot set. In this case though you don't have a quadratic polynomial but a biquadratic rational so it's not as well studied (afaik). The transformation z→iz conjugates x+1/x to x-1/x so they have the same behavior (just rotated 90°). It seems more convenient to work with x-1/x because the Julia set in this case seems to be real axis or some subset. Another variation is f(x) = c(x + 1/x) where c takes on various values. For c=1/2 this is Newton's method applied to the polynomial x²-1 and this is equivalent to the f(x) = x² so it's behavior is well known. --RDBury (talk) 04:01, 27 October 2015 (UTC)

PS2. I've done a bit of experimentation in Excel and have come up with the following scheme. For any real number r write a string, finite or infinite, of 0's and 1's as follows.

if r = 0 then write 1 and stop

if r<0 then write 0 and set r=f(r)=r-1/r

if r>0 then write 1 and set r=f(r)=r-1/r

repeat

So:

0→1

1→11

-1→01

φ→111

1/√2→101010...

etc. Now interpret the string as the binary digits of a number between 0 and 1. This seems to create an increasing, continuous but non-differentiable function from the reals to (0, 1) similar to Minkowski's question mark function. Assuming this is true then call it ??(x). Then ??(f(x))=2??(x) mod 1 and f is equivalent to a binary left shift and is therefore chaotic. For f(x)=(x-1/x)/2 then the corresponding ?? function is ??(x)=-π arccot(x) (or some permutation). There is probably similar behavior for f(x)=(x-1/x)/c for c between 1 and 2 but for other values the situation is likely to be more complicated. When you get into chaos everything starts getting messy and hard to prove so one tends to resort to making pictures and guessing what's going on in them. --RDBury (talk) 05:42, 27 October 2015 (UTC)

I've read this three times and I think I'm up to about 80% understanding. The one piece I don't get is that you have r<0 twice, should one be r>0?Naraht (talk) 14:19, 28 October 2015 (UTC)

Corrected, thanks. There are probably books that explain this stuff better than I can; one I learned a lot from was The Beauty of Fractals. I'm not exactly an expert either, I've just spent a lot of time with various fractal generating programs. --RDBury (talk) 03:42, 29 October 2015 (UTC)

bounding derivatives in L^p space

Hello,

I'm currently stuck with an analysis exercise. If $u\in C^{\infty }({\bar {U}})$, $1<p<\infty$ (with U open, bounded and with $C^{1}$ boundary) and $\epsilon >0$, I'm meant to show the existence of a constant C such that $||Du||_{L^{p}(U)}\leq \epsilon ||D^{2}u||_{L^{p}(U)}+C||u||_{L^{p}(U)}$.

I originally thought this would be a typical integration by parts question, but I can't see how to do it in this case since u doesn't necessarily have compact support. Aside from Poincare's inequality, I also don't know of any other inequalities relating a function with its derivative, so I'm lost at how to attack it. Also the fact that there are two constants on the right hand side, one of which we can control makes me think I'll need to use Cauchy's or Young's inequality with $\epsilon$, but cannot see how at this stage it'd be useful.

Any advise/help?

Cheers,

Neuroxic (talk) 07:44, 26 October 2015 (UTC)

I only have a minute to reply, so apologies for brevity/errors. The goal is to exploit compactness of the Kondrachov embedding of $W^{2,p}$ in $W^{1,p}$. So assume it's false. Normalize so that $\|Du\|=1$. So there is a sequence $u_{k}$ such that

$1\geq \epsilon \|D^{2}u_{k}\|+k\|u_{k}\|.$ (*)

Pass to a $W^{1,p}$ convergent subsequence $u_{k_{j}}$. In the (weaker) $L^{p}$ topology, we have $u_{k_{j}}\to 0$ because it follows from (*) that$\|u_{k}\|<1/k$. Hence the limit in the $W^{1,p}$ (stronger) topology must also be zero. But the limit function must also satisfy the constraint $\|Du\|=1$, which is a contradiction. Sławomir
Biały 00:15, 27 October 2015 (UTC)