1
$\begingroup$

Prove convergence of $a_{n+1}=\frac{c}{2}-\frac{ a_n^{2}}{2}$, $a_1=\frac{c}{2}$ for $0<c<1$

I managed to prove it using the contraction mapping theorem, but i want to prove it using more elementary method such as monotonicity + boundedness and such.

Checking for $ c=0.5 $, all the elements in the odd places are a descreasing subsequence and all elements in the even places are increasing subsequence.

Beyond that, didn't achieve much advancement.

$\endgroup$
5
  • 1
    $\begingroup$ Doesn't the standard iterated convergence theorem work? $\endgroup$
    – Calvin Lin
    Commented Jan 8, 2023 at 20:35
  • 1
    $\begingroup$ Hint: for "spiral convergence" study $f(f(x))-x$ where $f(x)=(c-x^2)/2$. See here and here. $\endgroup$
    – zwim
    Commented Jan 8, 2023 at 20:39
  • $\begingroup$ If you want to prove it without a fixed point theorem, then you could proceed as you suggested. Prove that the odd subsequence is decreasing and the even one is increasing. There are only polynomials involved that can be discussed by elementary means. The boundaries should come automatically with bounded values of c. $\endgroup$
    – Marius S.L.
    Commented Jan 8, 2023 at 21:25
  • $\begingroup$ @CalvinLin I was looking for a solution for a first semester student $\endgroup$
    – rotem aracky
    Commented Jan 9, 2023 at 14:05
  • $\begingroup$ @rotemaracky 1) Show that $ c/2 - c^2 /4 < a_n < c/2$. 2) Show that $ |a_{n+1} - a_n | < c |a_{n} - a_{n-1} |$. 3) Hence, the limit exists. $\quad$ This is the ratio test (though disguised), which I'd expect a first-semester analysis student to be able to understand. $\endgroup$
    – Calvin Lin
    Commented Jan 10, 2023 at 22:55

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .