I have following recurrence: $f(n) = 2 \cdot f(n-1) + f(n-2)$ with $f(1) = 1$ and $f(2) = 2$.
How can I find an upper bound with the form $f(n) \leq c\alpha^n$?
I know that in this task the value of $c$ does not have to be determined.
Asked
Active
Viewed 95 times
2
-
this is a linear difference equation – vidyarthi Oct 30 '18 at 11:42
-
you can solve this difference equation by the ansatz $$f(n)=q^n$$ – Dr. Sonnhard Graubner Oct 30 '18 at 11:45
-
Do you know that these are Pell numbers, and thus they grow on the order of $(\sqrt2+1)^n$? – Parcly Taxel Oct 30 '18 at 12:18
-
Sorry, but that doesn't really help me :( – Oct 30 '18 at 18:43
1 Answers
0
As it was suggested in the comments, there is a technique to solve this type of problems (here is an example, you will find a lot more on MSE). By noting $a_n=f(n)$, we have a homogeneous recurrence $$a_{n}=2a_{n-1}+a_{n-2}$$ which can be solved using characteristic polynomial technique. In this case, the polynomial is $$x^2-2x-1=0$$ It has $1-\sqrt{2}$ and $1+\sqrt{2}$ as solutions, thus the general term of the recurrence is $$a_n=A(1-\sqrt{2})^n+B(1+\sqrt{2})^n \tag{1}$$ Given initial conditions $a_1=1, a_2=2$ we have $$1=A(1-\sqrt{2})+B(1+\sqrt{2})$$ $$2=A(1-\sqrt{2})^2+B(1+\sqrt{2})^2$$ leading to $A=-\frac{\sqrt{2}}{4}$ and $B=\frac{\sqrt{2}}{4}$ or $$a_n=\frac{\sqrt{2}}{4}\left(1+\sqrt{2}\right)^n-\frac{\sqrt{2}}{4}\left(1-\sqrt{2}\right)^n \tag{2}$$ As a result $$f(n)<\frac{\sqrt{2}}{2}\left(1+\sqrt{2}\right)^n$$ because $|1-\sqrt{2}|<1$ and $\left(1-\sqrt{2}\right)^n \rightarrow 0$ when $n\rightarrow\infty$.
rtybase
- 16,907