Question

Is it possible to transform a sequence defined by a recursive formula into a regular sequence?

Answer 1

Sometimes, but not always.

Answer 2

Not always, for example the Fibonnacci sequence

1, 1, 2, 3, 5, 8, 13, 21, ...

cannot be converted into a regular sequence

Answer 3

Ok, thank you!

Answer 4

can't we use characteristic equation for fibonnacci?

Answer 5

That is not true. Recursive formulas can be put into closed form, which is a formula based on n, the number of the term, rather than the previous terms. The Fibonacci Sequence is no different. Instead of:
\[F_{n+2} = F_{n+1}+F_{n},F_1 = 1,F_2=1\]we get:\[F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)\]
Which is the closed formula for the nth term of the Fibonacci Sequence.

Answer 6

Yes, but I do not think this is what was meant be regular??

Answer 7

What else would be a regular formula? You want an equation that takes in the number of the term, and spits out the term. Thats what i view as regular.

Answer 8

Yes that is what I meant!.

Answer 9

But is there a method to do that @joemath314159

Answer 10

Yes. Using Linear Algebra. Most people know the shortcuts, but the whole process uses a lot of Linear Algebra. Being able to Diagonalize a matrix and whatnot.

Answer 11

The short way is to create the charateristic equation:\[F_{n+2}=F_{n+1}+F_n\iff F_{n+1}-F_{n+1}-F_n=0\]\[\Longrightarrow \lambda^2-\lambda-1 = 0\]This gives roots of:\[\lambda_{1,2}=\frac{1\pm\sqrt{5}}{2}\]Now we know the general form of the closed form
is:\[F_n = c_1\left(\frac{1+\sqrt{5}}{2}\right)^n+c_2\left(\frac{1-\sqrt{5}}{2}\right)^n\]and we use n = 1 F_1 = 1, n = 2 F_2 = 1 (the initial conditions) to see what c_1 and c_2 are. In this case, we obtain:\[c_1 = \frac{1}{\sqrt{5}},c_2 = -\frac{1}{\sqrt{5}}\]Which gives what i posted above.