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$$ φ(n)=5 φ\left(\frac{n}{2}\right)-6 φ\left(\frac{n}{4}\right)+n $$ where $$ \varphi (1) = 2 \\ \text{and} \\ \varphi (2) = 1 $$

With $n=2^x$, I have the following equation. Am I wrong in this translation?

$$ Q(x) = 5Q(x-1) - 6Q(x-2) + 2^x $$

To obtain its particular solution, I try the following.

$$ A \cdot 2^x=5 \cdot A \cdot 2^{(x-1)}-6 \cdot A \cdot 2^{(x-2)}+2^x $$ which yields the following unexpected(wrong) thing $$ 2^x=0 $$

Where is my wrong? Can you help and explain please?

Soner from The Ottoman Empire
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  • Isn't $x$ meant to be $\log_2 n$? So shouldn't the last term be $x$, not $2^x$? But, I don't understand the definition here. What is $\varphi (1)$? Is there any non-zero value for $n$ for which you know $\varphi(n)$? Not that $\varphi(0)$ is clear either, but at least it doesn't endlessly regress. – lulu Dec 11 '23 at 19:49
  • @lulu Why, can you please look at this link? https://math.stackexchange.com/q/615586 – Soner from The Ottoman Empire Dec 11 '23 at 19:56
  • @lulu edited initial values. – Soner from The Ottoman Empire Dec 11 '23 at 20:00
  • Yeah, ok...I agree on the last term. Why are you so sure the expression has solution $A2^n$? Why not work out the first five or six values to see if that looks persuasive. – lulu Dec 11 '23 at 20:06
  • @lulu I just think it should be something that satisfies the first recursive equality. That's why I try $A2^n$. Would you mind writing an answer for education? Why not work out the first five or six values to see if that looks persuasive. Even I don't understand precisely what you mean about the expansion. – Soner from The Ottoman Empire Dec 11 '23 at 20:16
  • I don't know the answer. I'd first compute a lot of terms so I could test things. It often happens that, when you have the sort of recursion for which $A2^n$ can be a solution, it's worth trying $Bn2^n$ as well. Just as a suggestion. – lulu Dec 11 '23 at 20:19
  • Yeah, since $2$ is a root of the characteristic equation, you'll want a $k2^k$ term, as in the near duplicate you sent. – lulu Dec 11 '23 at 20:38
  • @lulu 3 is also root isn't it? why is it $k2^k$ I don't get. – Soner from The Ottoman Empire Dec 11 '23 at 20:47
  • From the homogeneous piece you get a $2^k$ term and a $3^k$ term. So, adding an additional $2^k$ terms changes nothing. In that situation you look for polynomials times $2^k$, here just $k2^k$. – lulu Dec 11 '23 at 20:53
  • Similarly, the homogeneous problem $a_n=6a_{n-1}-9a_{n-2}$ has $3^n$ and $n3^n$ as fundamental solutions, The characteristic equation, $x^2=6x-9$ has $3$ as a double root. – lulu Dec 11 '23 at 20:55

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