The relation $(1+\lambda \Delta )^{-1}y_n=(a-1)a^{-n}\sum_{k=1}^{n-1}a^ky_k+Aa^{-n}$.

Question

I want to prove the above relation for $\lambda=a/(a-1)$ where $a\ne 1$, and $A$ is an arbitrary constant, without induction on $n$.

And $\Delta z_n = z_{n+1}-z_n$ for some sequence $z_n$.

So I tried to geometrically expand this term and I get:

$$(1+\lambda \Delta)^{-1}y_n = (1-\lambda \Delta+\lambda^2\Delta^2-\ldots)y_n$$

I don't see how to get this sum in the RHS, I know that $\Delta^{-1}y_n = \sum_{k=1}^{n-1}y_k+B$ where $B$ is some arbitrary constant.

I don't see how to relate this to the above relation.

Any hints? it should be straightforward, shouldn't it?

Answer 1

The idea is to invoke $ \left( 1 + \lambda \Delta \right) $ on the RHS.

Denote $ S_n = \sum_{k=1}^{n}{a^k y_k} $

then $ \mathrm{RHS} = (a-1)a^{-n}\cdot S_{n-1} + A \cdot a^{-n} $

Note that $ \Delta \left( a^{-n} \cdot S_{n-1} \right) = a^{-\left(n+1\right)} \cdot S_n - a^{-n}\cdot S_{n-1} $ and therefore

$$ \begin{split} (1 + \lambda \cdot \Delta )\left( a^{-n} \cdot S_{n-1} \right) = a^{-n} \cdot S_{n-1} + \frac{a}{a-1}\left(a^{-\left(n+1\right)} \cdot S_n - a^{-n}\cdot S_{n-1}\right) = \\ a^{-n} S_{n-1} \cdot \left(1- \frac{a}{a-1}\right) + a^{-n}\cdot S_n \cdot \frac{1}{a-1} = \\ a^{-n} \cdot \frac{S_n - S_{n-1}}{a-1} = a^{-n} \cdot \frac{{a^n \cdot y_n }}{a-1} = \frac{y_n}{a-1} \end{split} $$

Therefore we have $ \left( 1 + \lambda \Delta \right) \left( \left( a -1\right) a^{-n} \cdot S_{n - 1} \right) = y_n $

Denote $ z_n = \left( 1 + \lambda \Delta \right)^{-1} y_n $. Then $ z_n = \left( a -1\right) a^{-n} \cdot S_{n - 1} + w_n $ where $ w_n $ belongs to the kernel of $ 1 + \lambda \Delta $, i.e.

$$ \begin{split} w_n + \lambda \cdot \left( w_{n+1} - w_n \right) = 0 \\ \iff \\ w_{n+1} = -\frac{1}{\lambda} \cdot w_n + w_n = \left( \frac{1-a}{a} + 1 \right) \cdot w_n \\ \iff \\ w_{n+1} = a^{-1} w_{n} \end{split} $$

which means that $ w_{n} = A \cdot a^{-n} $ where $ A $ is a constant.