I am stuck on this question:
Consider the delay-differential equation $\frac{d}{ dt}x(t) = x(t − 1)$, with $x(t) = 1$ for $t \in [−1, 0]$. The solution is given by $x(t) = \sum^{n}_{0}{\frac{[t − (k − 1)]^k}{k!}}$ for $n − 1 \leq t \leq n$, where $n$ is a non-negative integer. Use proof by induction (i.e., method of steps) to find the solution
I have tried to use the sum of a geometric series on the series and then differentiating this and setting it equals to $x(t-1)$ also in the geometric series however when i tried to the basis case $k=1$ I ended up with $\frac{1-t^n}{t-1}=\frac{1+t^2(t-1)^n-nt^3(t-1)^{n-1}}{t^2}$ which I am unable to prove is true, can anyone see where I have gone wrong or suggest a better method of approach this?
-edited so that n is the upper limit
Here is how you might formulate the answer: Let $x_n(t) = \sum_{k=0}^n \frac{(t-(k-1))^k}{k!}$. Then $x(t) = x_n(t)$ where $n - 1 < t \leq n$. Just to be explicit, $$ x(t) = \begin{cases} 1 & \text{if $0 < x \leq 1$} \\ 1+t & \text{if $1 < x \leq 2$} \\ 1+t+\frac{1}{2}(t-1)^2 & \text{if $2 < x \leq 3$} \\ 1+t+\frac{1}{2}(t-1)^2 + \frac{1}{6}(t-2)^3 & \text{if $3 < x \leq 4$} \\ \dots & \dots \end{cases} $$
You need to show $x'(t) = x(t-1)$ for all $t \geq 0$. On the interval $n-1 < t \leq n$, we have $x(t) = x_n(t)$ and $x(t-1) = x_{n-1}(t-1)$. So it suffices to show that $x_n'(t) = x_{n-1}(t-1)$ for each $n$.
For this, I'm not even sure you need induction. For \begin{align*} x_n'(t) &= \sum_{k=1}^n \frac{k(t-(k-1))^{k-1}}{k!} = \sum_{k=1}^n \frac{(t-(k-1))^{k-1}}{(k-1)!} = \sum_{k=0}^{n-1} \frac{(t-k)^{k}}{k!} \\ x_{n-1}(t-1) &= \sum_{k=0}^{n-1} \frac{((t-1)-(k-1))^{k}}{k!} = \sum_{k=0}^{n-1} \frac{(t-k)^{k}}{k!} \end{align*}