mersenneforum.org Recurrence Equation
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2004-04-06, 12:15 #1 jinydu     Dec 2003 Hopefully Near M48 2·3·293 Posts Recurrence Equation c(n+2) = -8c(n+1) - 16c(n), c(1) = -1, c(2) = 8 Given the above recurrence equation and starting conditions, the goal is to find the solution (that is, an equation that calculates c(n) without reference to previous terms). The characteristic polynomial is: n^2 = -8n - 16 n^2 + 8n + 16 = 0 n = -4 (repeated root) What has made this problem difficult for me is the repeated root. Because of the repeated root, the solution given at http://mathworld.wolfram.com/LinearR...eEquation.html doesn't work because it involves dividing by zero. In case it helps, here are some more terms: c(0) = 0 c(1) = -1 c(2) = 8 c(3) = -48 c(4) = 256 c(5) = -1280 c(6) = 6144 c(7) = -28672 Also, the ratio between successive terms appears to be approaching some number close to -4. Thanks
 2004-04-06, 19:14 #2 FeLiNe     Dec 2003 23 Posts I'm probably just missing something obvious somewhere, but what is wrong with eq. (11) on the Mathworld page? In the situation you present, A=-8, x1=-1 and x2=8 so the first term that starts with (Ax1- x2) is identical to zero and all that remains is the second term. Last fiddled with by FeLiNe on 2004-04-06 at 19:20 Reason: I just learned how to do proper subscripts 'round'ere.
 2004-04-07, 01:32 #3 jinydu     Dec 2003 Hopefully Near M48 110110111102 Posts I meant Equation 13 (I do know of a way to adjust the formula to take account of different starting terms, but that doesn't solve the problem of dividing by zero).
2004-04-07, 02:41   #4
wblipp

"William"
May 2003
New Haven

23·103 Posts

Quote:
 Originally Posted by jinydu What has made this problem difficult for me is the repeated root.
Repeated roots in difference equations are similar to repeated roots in differential equations. In this case the general solutions is

A*(-4)n + B*n*(-4)n

You pick A and B to match the first two terms.

One way to derive this is to take the two root solution with alpha = beta + epsilon and figure out what happens in the limit as epsilon goes to zero. You have
[(beta+epsilon)n - betan]/(beta+epsilon-beta)

=[betan + n*epsilon* betan-1 + epsilon2*?? - betan]/epsilon

= n*betan-1 + epsilon*??

Absobing a factor of beta into the constant, in the limit this is proportional to

n * betan

William

 2004-04-07, 06:08 #5 jinydu     Dec 2003 Hopefully Near M48 2×3×293 Posts Ok, that's the right answer (proven by induction). Thanks
 2004-05-15, 11:59 #6 Qbert   May 2004 416 Posts Just for final clarification, c(n)=(-1)^n*4^(-1 + n)*n c(n)=(-1)**n*4**(-1 + n)*n Is TeX form or even better MathML form possible to post?
2004-05-15, 14:02   #7
wblipp

"William"
May 2003
New Haven

45018 Posts

Quote:
 Originally Posted by Qbert Just for final clarification, c(n)=(-1)^n*4^(-1 + n)*n c(n)=(-1)**n*4**(-1 + n)*n Is TeX form or even better MathML form possible to post?
HTML superscript codes are supported as [sup]n[/sup], so you could write this as

c(n) = (-1)n4n-1n

 Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Combinatorics & Combinatorial Number Theory 1 2017-12-21 00:42 jasong jasong 4 2012-02-20 03:33 flouran Math 7 2009-12-12 18:48 Unregistered Information & Answers 2 2007-01-18 17:13 Erasmus Factoring 3 2004-05-14 09:26

All times are UTC. The time now is 16:16.

Mon Jan 17 16:16:53 UTC 2022 up 178 days, 10:45, 0 users, load averages: 1.39, 1.30, 1.19

Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔