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 for n > 0 with and with . In particular, what I want to ask, is some good transformation in order to change variables. Trying or , 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 $a_n=2^n\cdot a_0.$

March 18th, 2015, 12:14 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210 
Consider $b_n = \sin^{1}(a_n)$.

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.

March 18th, 2015, 03:21 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey stuff.  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: 20,942 Thanks: 2210 
The recurrence was given as $a_n = 2a_{n1}$ when the first reply was posted. My reply related to the recurrence currently shown.

March 19th, 2015, 08:30 AM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey stuff. 
Makes sense now, thanks. Dan 
April 6th, 2015, 09:51 AM  #8 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126  Took me a while to see this so I pass it on. an=sinbn substitute into OP: sinbn=sin2bn1 bn=2bn1 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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