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 March 18th, 2015, 07:47 AM #1 Newbie   Joined: Sep 2009 From: Greece Posts: 12 Thanks: 0 Math Focus: Calculus Recurrence Hello all! I have some problem in figuring out how to solve the recurrence $a_n= 2 a_{n-1} sqrt{1-a_{n-1}^2}$ for n > 0 with $a_0= \frac{1}{2}$ and with $a_0= \frac{1}{3}$. In particular, what I want to ask, is some good transformation in order to change variables. Trying $b_n= n a_n$ or $b_n= log_2 a_n$, I find that can not do the trick. If anyone has encountered this recurrence or knows some useful transformation that can be used, I'd appreciate it very much. Thanks in advance! Last edited by Cloud8; March 18th, 2015 at 08:28 AM.
 March 18th, 2015, 08:22 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms This will be of the form $a_n=2^n\cdot a_0.$ Thanks from Cloud8
 March 18th, 2015, 12:14 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 Consider $b_n = \sin^{-1}(a_n)$. Thanks from zylo
 March 18th, 2015, 12:54 PM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Please don't edit your post to ask a different question. Thanks from topsquark
March 18th, 2015, 03:21 PM   #5
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by CRGreathouse Please don't edit your post to ask a different question.
So is the original post correct for the discussion after? I'm seeing a really cute simplification in skipjack's post (coolness!) about the listed OP but I'm not understanding where CRGreathouse's solution comes from.

-Dan

Last edited by skipjack; March 18th, 2015 at 09:11 PM.

 March 18th, 2015, 08:59 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 The recurrence was given as $a_n = 2a_{n-1}$ when the first reply was posted. My reply related to the recurrence currently shown. Thanks from topsquark
 March 19th, 2015, 08:30 AM #7 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timey-wimey stuff. Makes sense now, thanks. -Dan
April 6th, 2015, 09:51 AM   #8
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Quote:
 Originally Posted by skipjack Consider $b_n = \sin^{-1}(a_n)$.
Took me a while to see this so I pass it on.

an=sinbn

substitute into OP:

sinbn=sin2bn-1

bn=2bn-1

bn=(2^n)b0

an=sin(2^n)b0

a0=sinb0

Sorry, I don"t know how to do subscripts. I could do it in Word and copy/paste if you like (and if it works).

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