Binomial Expansion with Derivatives Question:
We have a function $f(t)$ that is defined for all $t \in [0,T]$. From our data, we can estimate two important parameters $\theta \in \mathbb{Z} \backslash 0$ and $n \in \mathbb{N}$. Given $n$, we can also estimate a function $g(n)$. Our estimations are fairly accurate.
Now it appears that there is a pattern. For example, for $\theta > 0$,
we have
\begin{align}
&n=1,\quad g(1)= \frac{d f(t)}{dt}+\theta f(t), \\
&n=2,\quad g(2)= \frac{d^2 f(t)}{dt^2}+\theta^2 f(t)+2\theta \frac{d f(t)}{dt},
\end{align}
and so on; therefore, I think we can write it as an ordinary differential equation $g(n)=(d_t+θ)^n f(t)$
where $d_t$ is a time-derivative operator. Now is there a way to explicitly determine $f(t)$?
My Attempt 1)
I tried to use Kovacic algorithm but it seems very complicated and I couldn't get anywhere with it.
My Attempt 2)
Although we have a nonhomogeneous equation due to $g(n)$, if for every $n$, we differentiate both sides, we have:
\begin{align}
& n=1,\quad f'' + \theta f'= 0, \\
& n=2,\quad f'''+2 \theta f''+ \theta^2 f'=0,
\end{align}
and so on. If we look closely, the characteristics equation for every $n$
emits two roots, $0$ and $-\theta$ so let us define $C_{1n}$ and $C_{2n}$ two constants and $f_n(t)$, given the value of $n$, then we can write
$$f_n(t) = C_{1n} + C_{2n}e^{- \theta t}.$$
But the problem is that the only condition we have is $f_n(T)=0.$
Any suggestions or comment would be very appreciated. I'm more interested in knowing if there is a solution for $f$ and the method to get there rather than the final solution.
Edit:
I think I have made a mistake: since the root $-\theta$ is a repeated root in the second set of ODEs, I think the solution to $f$ is:
$$f_n(t) = C_{0n} + e^{-\theta t}\sum_{k = 1}^{n} C_{kn} t^{k - 1}.$$
 A: Assuming that you are setting $g(n)=0,$ I think the solution can be found by an intuitive application of variation of parameters:
$$f(t)=e^{-\theta t}\sum_{i=0}^n a_it^i.$$
To get this, I reasoned that $C e^{-\theta t}$ is a solution of $(d_t+\theta)f(t).$ If I raise the power of the operator, I merely raise the order of the root. In the variation of parameters as applied to simple exponential solutions, you just get a higher and higher order polynomial multiplying the exponential. Now we prove that this solution solves the ODE:
\begin{align*}
(d_t+\theta)^nf(t)
&=(d_t+\theta)^n\!\left[e^{-\theta t}\sum_{i=0}^n a_it^i\right]\\
&=\sum_{i=0}^na_i\,(d_t+\theta)^n\left[e^{-\theta t}\,t^i\right].
\end{align*}
Now all we need do is prove that if $n\ge i,$ then
$$(d_t+\theta)^n\left[e^{-\theta t}\,t^i\right]=0.$$
Let us induct on $i,$ here. Suppose $i=0,$ and $n\ge i.$ Then we have
\begin{align*}
(d_t+\theta)^n\left[e^{-\theta t}\right]
&=(d_t+\theta)^{n-1}(d_t+\theta)\left[e^{-\theta t}\right]\\
&=(d_t+\theta)^{n-1}(d_t\left[e^{-\theta t}\right]+\theta\left[e^{-\theta t}\right])\\
&=(d_t+\theta)^{n-1}(0)\\
&=0.
\end{align*}
Now suppose the claim is true for arbitrary $i$ and $n\ge i.$ We show it is true for $i+1,$ assuming $n\ge i+1$ as well. We have
\begin{align*}
(d_t+\theta)^n\left[e^{-\theta t}\,t^{i+1}\right]
&=(d_t+\theta)^{n-1}(d_t+\theta)\left[e^{-\theta t}\,t^{i+1}\right]\\
&=(d_t+\theta)^{n-1}(d_t\left[e^{-\theta t}\,t^{i+1}\right]+\theta\left[e^{-\theta t}\,t^{i+1}\right])\\
&=(d_t+\theta)^{n-1}(-\theta e^{-\theta t}\,t^{i+1}+(i+1)e^{-\theta t}\,t^i+\theta\left[e^{-\theta t}\,t^{i+1}\right])\\
&=(i+1)(d_t+\theta)^{n-1}(e^{-\theta t}\,t^i).
\end{align*}
Since $n\ge i+1,$ we must also have $n-1\ge i,$ forcing $n\ge i,$ and this expression must also be zero. QED.
