5
$\begingroup$

The following is not exactly a research question (it was originated from manufacturing of exercises for calculus), and has no other motivation than explaining a phenomenon. I apologize if it is inappropriate (and will quickly remove it).

Consider the sequence of real numbers defined recursively as follows $$u_0:=\lambda > -1\\ ,$$ $$u_{n+1}=\frac{2u_n}{1+\sqrt{1+2^{-n}u_n}}\\ .$$ Numerical evidence suggests that $u_n$ always converges to $\log(1+\lambda)$. I imagine there should be a simple explanation. How can one prove it?

$\endgroup$
1
  • $\begingroup$ u_(n+1) is very close to the harmonic mean of 2^n and u_n when u_n/2^n is small. Perhaps that helps for another derivation? Gerhard "Ask Me About System Design" Paseman, 2012.10.26 $\endgroup$ Oct 26, 2012 at 22:51

1 Answer 1

7
$\begingroup$

Multiplying top and bottom by the conjugate and simplifying (assuming $u_n \neq 0$) we get: $$u_{n+1}=2^{n+1}(\sqrt{1+2^{-n}u_n}-1).$$ Calling $v_n:=\frac{u_n}{2^n}+1$ we have the recursion: $v_{n+1}=\sqrt{v_n}$ and therefore $v_n=v_0^{2^{-n}}$. Going back to $u$´s we have $u_n=2^n[(1+\lambda)^{2^{-n}}-1]$. Finally: $$\lim_{n\to\infty} u_n=\lim_{h\to 0}\frac{(1+\lambda)^h-(1+\lambda)^0}{h}= \ln (1+\lambda).$$

$\endgroup$
4
  • $\begingroup$ Your first display is valid even when $u_n=0$, and this case occurs if and only if $\lambda=0$. $\endgroup$
    – GH from MO
    Oct 26, 2012 at 22:20
  • $\begingroup$ that's very nice $\endgroup$ Oct 26, 2012 at 22:37
  • 1
    $\begingroup$ So $x=u_n$ is a solution of $(1+\frac{x}{N})^N=1+\lambda$ (with $N:=2^n$), which we may see as an approximated problem for $e^x=1+\lambda$... $\endgroup$ Oct 27, 2012 at 7:47
  • $\begingroup$ @Pietro: that´s a nice way to look at it. $\endgroup$ Oct 27, 2012 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.