Find Limit of an in Series with a0,a1 & n

[cdummie]

Homework Statement

If we have a number sequence such that: a₀, a₁ are given, and every other element is given as ##a_n=\frac{(a_{n-1} + a_{n-2})}{2} then express a_n in terms of a₀, a₁ and n , and fin the limit of a_n

Homework Equations

The Attempt at a Solution

If i try to express a₃ in terms of a₁ and a₀ and then a₄ in terms of a₃ and a₂ (using the a₂ i previously expressed in terms of a₁ and a₀ ) and going so on with next elements up to a_n element, that isn't actually a good idea, since i am supposed to express it only using a1 and a0, no a_n-1 and stuff. So if i try to find a2 - a1, then a3 - a2 all the way to the a_n - a_n-1 i would have something like this:

$a_2 - a_1= - \frac{a_1 - a_0}{2} \\ a_3 - a_2= \frac{a_1 - a_0}{2^2} \\ \vdots \\ a_n - a_{n-1} = (-1)^{n-1} \frac{a_1 - a_0}{2^{n-1}}$

If i sum up all of these, i actually have:
$a_n - a_1 = -\frac{a_1 - a_0}{2} + \frac{a_1 - a_0}{2^2} - \frac{a_1 - a_0}{2^3} + \cdots +(-1)^{n-1} \frac{a_1 - a_0}{2^{n-1}}$

Which is what i need to have, i should just "move" a_1 to the RHS but i don't know how to simplify that series so that i could find a limit of it.

 

[HallsofIvy]

The "a_1" on the left is not really important. What is important is that the right side is a geometric series, \sum_{i=0}^n ar^i with a= a_1- a_0 and common factor -\frac{1}{2}. After you have taken that sum, then add a_1 to both sides.

 

[BvU]

If i try to express a3 in terms of a₁ and a₀ and then...

But you want it in terms of a₁ and a₀ , not a₂ etc.
So:
$$a_0, \ a_1, \ {a_0 + a_1 \over 2}, \ { a_1 + {a_0 + a_1 \over 2} \over 2} , \ {{a_0 + a_1 \over 2} + \ { a_1 + {a_0 + a_1 \over 2} \over 2} \over 2 }. \ ... $$and that you can easily generalize, which is exactly what is asked from you.

I see Ivy has already spoiled the thing for you. Pity.

 

[cdummie]

The "a_1" on the left is not really important. What is important is that the right side is a geometric series, \sum_{i=0}^n ar^i with a= a_1- a_0 and common factor -\frac{1}{2}. After you have taken that sum, then add a_1 to both sides.

Well, i think i understand the idea, but the fact that sings are switching is confusing me, i mean, when i have something like this: $\sum\limits_{k=1}^n \frac{1}{2k}$ (which is almost exactly the same as in this case, except for the (-1)^k part) then it's simple:
$\sum\limits_{k=1}^{n-1} \frac{1}{2k} = \frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}$ So, what can i do about that $(-1)^k$ ?

 

[micromass]

Do you know matrix diagonalization?

 

[cdummie]

Do you know matrix diagonalization?

Matrix diagonalization? I know that if i have upper or lower triangular matrix that determinant of such matrix is product of elements on the diagonal, if that's what you meant.

 

[Ray Vickson]

Homework Statement

If we have a number sequence such that: a₀, a₁ are given, and every other element is given as ##a_n=\frac{(a_{n-1} + a_{n-2})}{2} then express a_n in terms of a₀, a₁ and n , and fin the limit of a_n

Homework Equations

The Attempt at a Solution

If i try to express a₃ in terms of a₁ and a₀ and then a₄ in terms of a₃ and a₂ (using the a₂ i previously expressed in terms of a₁ and a₀ ) and going so on with next elements up to a_n element, that isn't actually a good idea, since i am supposed to express it only using a1 and a0, no a_n-1 and stuff. So if i try to find a2 - a1, then a3 - a2 all the way to the a_n - a_n-1 i would have something like this:

$a_2 - a_1= - \frac{a_1 - a_0}{2} \\ a_3 - a_2= \frac{a_1 - a_0}{2^2} \\ \vdots \\ a_n - a_{n-1} = (-1)^{n-1} \frac{a_1 - a_0}{2^{n-1}}$

If i sum up all of these, i actually have:
$a_n - a_1 = -\frac{a_1 - a_0}{2} + \frac{a_1 - a_0}{2^2} - \frac{a_1 - a_0}{2^3} + \cdots +(-1)^{n-1} \frac{a_1 - a_0}{2^{n-1}}$

Which is what i need to have, i should just "move" a_1 to the RHS but i don't know how to simplify that series so that i could find a limit of it.

The solution of the recurrence can be obtained using a standard method for such a "constant coefficient" problem. If we assume a solution of the form $a_n = c r^n$ (with constant $c$), then the recursion $a_{n+2} = (a_{n+1} + a_n)/2$ yields $r^n \,r^2 = r^n \,(1+r)/2$, hence $r^2 = (1+r)/2$ and so $r = 1,-1/2$. That implies $a_n = A 1^n + B(-1/2)^n = A + B\,(-1/2)^n$. Find $A$ and $B$ by using the given conditions at $n = 0$ and $n = 1$.

Note: you can easily check that for any constants $A,B$ the form $a_n = A + B \,(-1/2)^n$ satisfies the recursion.