What is it to solve an equation forward? I'm reading a book in Monetary Economics and I don't understand a step.
I have this expression:
$$ \dfrac{\lambda_{t}}{P_{t}} = \beta \left( \dfrac{\lambda_{t+1} + \mu_{t+1}}{P_{t+1}} \right) $$
And then it says "solving this equation forward implies that":
$$ \dfrac{\lambda_{t}}{P_{t}} = \sum_{i=1}^{\infty} \beta^{i} \left( \dfrac{\mu_{t+i}}{P_{t+i}} \right) $$
I dont't know what they're doing. What is it to solve an equation forward?
 A: Assuming $\beta$ is in the first equation too, you can make use of the forward operator: $F x_k = x_{k+1}$ To define
$$ \frac{\lambda_t}{P_t} = \beta F \left(\frac{\lambda_t + \mu_t}{P_t} \right) \implies (1-\beta F) \frac{\lambda_t}{P_t} = \beta F \frac{\mu_t}{P_t}. $$
Inverting this operator (skipping details on justification) $(1-\beta F)^{-1} = \sum_{k=0}^\infty (\beta F)^k$ and so
$$ \frac{\lambda_t}{P_t} = \sum_{k=1}^\infty (\beta F)^k \frac{\mu_t}{P_t} = \sum_{k=1}^\infty \beta^k \frac{\mu_{t+k}}{P_{t+k}} $$
A: We can begin to solve this equation in two ways, an idea that generalizes to every time we have a relationship between the $t^{\text{th}}$ and $(t+1)^{\text{th}}$ term of a sequence.
One way is to start with the equation for $\frac{\lambda_{t}}{P_{t}}$, and use substitution to eliminate $\frac{\lambda_{t+1}}{P_{t+1}}$, then $\frac{\lambda_{t+2}}{P_{t+2}}$, and so on:
\begin{align}
\frac{\lambda_t}{P_t} &= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta \frac{\lambda_{t+1}}{P_{t+1}} \\
  &= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta \left(\beta \frac{\mu_{t+2}}{P_{t+2}} + \beta \frac{\lambda_{t+2}}{P_{t+2}}\right)  \\
  &=  \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta^2 \frac{\mu_{t+2}}{P_{t+2}} + \beta^2 \frac{\lambda_{t+2}}{P_{t+2}} \\
  &= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta^2 \frac{\mu_{t+2}}{P_{t+2}} + \beta^2 \left(\beta \frac{\mu_{t+3}}{P_{t+3}} + \beta \frac{\lambda_{t+3}}{P_{t+3}}\right) \\
  &= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta^2 \frac{\mu_{t+2}}{P_{t+2}} + \beta^3 \frac{\mu_{t+3}}{P_{t+3}} + \beta^3 \frac{\lambda_{t+3}}{P_{t+3}} \\
 &= \dots
\end{align}
The other way is to rewrite this equation as $\frac{\lambda_{t+1}}{P_{t+1}} = \beta^{-1} \frac{\lambda_t}{P_t} - \frac{\mu_{t+1}}{P_{t+1}}$ or better yet $\frac{\lambda_t}{P_t} = \beta^{-1} \frac{\lambda_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t}$, and then use substitution to eliminate $\frac{\lambda_{t-1}}{P_{t-1}}$, then $\frac{\lambda_{t-2}}{P_{t-2}}$, and so on:
\begin{align}
\frac{\lambda_t}{P_t} &= \beta^{-1} \frac{\lambda_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
  &= \beta^{-1} \left(\beta^{-1} \frac{\lambda_{t-2}}{P_{t-2}} - \frac{\mu_{t-1}}{P_{t-1}}\right)- \frac{\mu_t}{P_t} \\
  &= \beta^{-2} \frac{\lambda_{t-2}}{P_{t-2}} - \beta^{-1}\frac{\mu_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
  &= \beta^{-2} \left(\beta^{-1} \frac{\lambda_{t-3}}{P_{t-3}} - \frac{\mu_{t-2}}{P_{t-2}}\right)- \beta^{-1}\frac{\mu_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
  &= \beta^{-3} \frac{\lambda_{t-3}}{P_{t-3}} - \beta^{-2}\frac{\mu_{t-2}}{P_{t-2}}- \beta^{-1}\frac{\mu_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
  &= \dots
\end{align}
I feel like it makes sense to say that in the first method, we're solving the equation "forward", since we're getting higher and higher indices on $\lambda$. In the second method, we're solving the equation "backward", getting lower and lower indices.
In the first method, assuming that $\beta^i \frac{\lambda_{t+i}}{P_{t+i}} \to 0$ as $i \to \infty$, we're eventually left with just an infinite sum with no $\lambda$'s in it. In the second method, we're left with a finite sum with no $\lambda$'s in it, assuming that we have an initial condition for $\frac{\lambda_0}{P_0}$.
