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March 18th, 2015, 07:47 AM   #1
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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.
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March 18th, 2015, 08:22 AM   #2
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This will be of the form $a_n=2^n\cdot a_0.$
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March 18th, 2015, 12:14 PM   #3
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Consider $b_n = \sin^{-1}(a_n)$.
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March 18th, 2015, 12:54 PM   #4
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Please don't edit your post to ask a different question.
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March 18th, 2015, 03:21 PM   #5
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Quote:
Originally Posted by CRGreathouse View Post
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.
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March 18th, 2015, 08:59 PM   #6
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The recurrence was given as $a_n = 2a_{n-1}$ when the first reply was posted. My reply related to the recurrence currently shown.
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March 19th, 2015, 08:30 AM   #7
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Makes sense now, thanks.

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April 6th, 2015, 09:51 AM   #8
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Quote:
Originally Posted by skipjack View Post
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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