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Given the following recurrence equation:

$$ (b_{n+2})^2 - 7(b_{n+1})^2 + 12(b_n)^2 = (5n^2 + 3)4^n $$

Which after expanding the equation is equal to:

$$ (b_{n+2})^2 - 7(b_{n+1})^2 + 12(b_n)^2 = 5n^2(4^n) + 3(4^n) $$

Now since this is definitely not lineal and I have never worked with such types of recurrence equations before, I am not where where to begin in solving it.

Where do I start and how do I proceed?

UPDATE:

Following from the answer provided by @h-h-rugh:

$$ b_n^2 = 4^na_n \\ 4^{n+2}a_{n+2} - 7(4^{n+1}a_{n+1}) + 12(4^na_n) = (5n^2+3)4^n \\ (4^2a_{n+2}-7(4)(a_{n+1}) + 12a_n)4^n = (5n^2+3)4^n $$

Dividing by $4^n$:

$$ 16a_{n+2} - 28a_{n+1} + 12a_n = 5n^2+3 \rightarrow \text{ eq. 1} $$

I then separate the equation:

$$ a_n^{(h)} = 16a_{n+2} - 28a_{n+1} + 12a_n = 0 \\ a_n^{(p)} = an^2 + bn + c $$

Substituting a_n^{(p)} en eq. 1:

$$ 16[a(n+2)^2 + b(n+2) + c] - 28[a(n+1)^2 + b(n+1) + c] + 12[an^2 + bn + c] = 5n^2 + 3 \\ - \\ 16[a(n^2 + 4n + 4) + bn +2b + c] - 28[a(n^2 + 2n + 1) + bn + b + c] + 12[an^2 + bn + c] = 5n^2 + 3 \\ - \\ 16[an^2 + 4an + 4a + bn + 2b + c] - 28[an^2 + 2an + a + bn + b + c] + 12[an^2 + bn + c] = 5n^2 + 3 \\ - \\ 16an^2 + 64an + 64a + 16bn + 32b + 16c - 28an^2 - 56an - 28a - 28bn - 28b - 28c + 12an^2 + 12bn + 12c = 5n^2 + 3 \\ - \\ 16an^2 - 28an^2 + 12an^2 + 64an -56an + 64a -28a + 16bn - 28bn + 12bn + 32b - 28b + 16c - 28c + 12c = 5n^2 + 3 \\ - \\ 8an + 36a + 4b = 5n^2 + 3 $$

Would that be correct?

Omari Celestine
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2 Answers2

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This is a nonlinear equation. This can however be converted into linear form using the substitution $a_n = b_n^2$. Now the equation becomes: $$a_{n+2} - 7a_{n+1} + 12a_n = (5n^2+3)4^n$$

Hope you can take it from here.

Now, the solution to the homogenous part is: $a^{\{h\}} = c_1(3)^n + c_2(4)^n$. Note that: $4$ is a root of the characteristic equation with multiplicity $1$. Using the theorem presented in my answer here, then the particular solution takes the form: $$a^{\{p\}} = n(p_2n^2+p_1n + p_0)(4)^n$$

Substituting this into the recurrence relation we have: $$p_2[16(n+2)^3-28(n+1)^3+12n^3] + p_1[16(n+2)^2-28(n+1)^2+12n^2] + p_0[16(n+2)-28(n+1)+12n] = 5n^2+3$$ $$\implies n^2\color{red}{[12p_2]}+n\color{blue}{[108p_2+8p_1]}+\color{green}{[100p_2+36p_1+4p_0]} = \color{red}{5}n^2+\color{blue}{0}n + \color{green}{3}$$

  • For the most part, yes. But for what I am thinking, in separating it into the homogeneous and the polynomial parts, since it's a multiplication at the end, how do I do the separation? – Omari Celestine Dec 28 '17 at 11:55
  • My problem is I did not really understand that form of explanation. Do think you can extend your answer by providing a complete solution so that I have a better idea as to dealing with such problems? – Omari Celestine Dec 28 '17 at 12:35
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    @OmariCelestine Edited my answer. I have just left to you the final part of carefully calculating the $p_i$’s. –  Dec 28 '17 at 12:56
  • The thing is I am having problems understanding the equation you used and for that, I am unable to follow it. – Omari Celestine Dec 28 '17 at 16:00
  • @OmariCelestine This is the method you find in most discrete mathematics books to solve recurrence relations of this form, sir. Thoroughly edited my answer. –  Dec 29 '17 at 06:59
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You may simultaneously convert it into linear recurrence with a "pure" polynomial RHS by substituting: $b_n^2=4^n a_n$

H. H. Rugh
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  • Then eliminating the $4^n$ by dividing? – Omari Celestine Dec 28 '17 at 13:35
  • yes indeed and then proceed as usual: Find a polynomial particular solution and add the homogeneous solution – H. H. Rugh Dec 28 '17 at 14:04
  • Could you help me with this; After one of the substitutions I am getting the following equation: $8an+36a+4b=5n^2+3$. Does this look correct to you? – Omari Celestine Dec 28 '17 at 15:59
  • No, and I don't quite understand your letters? By direct substitution I rather get something like $16 a_{n+2} -28 a_{n+1} + 12 a_n = 5n^2 + 3$ which you may then divide by 4 (or 16) before solving. – H. H. Rugh Dec 28 '17 at 16:29
  • I updated the question with part of what I did. Could you tell me where I went wrong? – Omari Celestine Dec 28 '17 at 16:45
  • That is what I got but I mean after substituting $a_n^{(p)} = an^2 + bn + c$ into the equation $16a_{n+2} - 28a_{n+2} + 12a_n = 5n^2 + 3$ – Omari Celestine Dec 28 '17 at 16:54
  • I see, but in this case you need a polynomial of degree 3 (because $16-28+12=0$) – H. H. Rugh Dec 28 '17 at 17:32
  • And how do I determine that from such equations since it appeared to be a maximum of $ax^2$? – Omari Celestine Dec 28 '17 at 17:35
  • Not quite sure what you mean. The reason why there is a problem is that $1$ is a root of the characteristic polynomial on the LHS. The solution is that you try with a polynomial of the form $an^3+bn^2+cn$. The optimal path is to first solve for $a$, then $b$ and finally $c$. Coefficients are rather 'ugly' (if my computation is correct) – H. H. Rugh Dec 28 '17 at 18:56
  • I mean determine the maximum grade of polynomial equation? – Omari Celestine Dec 28 '17 at 18:58
  • Trial and error is probably the best answer. But when 1 is a root you need to go to a higher degree. If it is a double root you need to go to two degrees higher, e.g. for $a_{n+2}-2 a_{n+1}+a_n= 5n^2+3$ you would actually need a 4th degree polynomial. – H. H. Rugh Dec 28 '17 at 19:50