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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?

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  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Dec 11, 2023 at 19:49
  • $\begingroup$ @lulu Why, can you please look at this link? math.stackexchange.com/q/615586 $\endgroup$
    – Soner from The Ottoman Empire
    Commented Dec 11, 2023 at 19:56
  • $\begingroup$ @lulu edited initial values. $\endgroup$
    – Soner from The Ottoman Empire
    Commented Dec 11, 2023 at 20:00
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Dec 11, 2023 at 20:06
  • $\begingroup$ @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. $\endgroup$
    – Soner from The Ottoman Empire
    Commented Dec 11, 2023 at 20:16

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