Suppose I have a dynamic table supporting $Insert$ procedure, which sets an input value after the tail of the dynamic table. If the underlying table is already full, we multiply its size by $q > 1$. (I have proof that that arrangement leads to amortized constant time for $Insert$: see this post.)
My best attempt so far is: $\Phi_i = \alpha n_i + \beta N_i$, where $n_i$ is the number of elements in the dynamic table after the $i$th operation, and $N_i$ is the total capacity of the underlying array after the $i$th operation.
The above leads to:
(1) If there is more room for a new value, $n_i = n_{i - 1} + 1$, and $N_i = N_{i - 1}$: \begin{aligned} \hat{c_i} &= c_i + \Phi_i - \Phi_{i - 1} \\ &= 1 + (\alpha n_i + \beta N_i) - (\alpha n_{i - 1} + \beta N_{i - 1}) \\ &= 1 + (\alpha n_{i - 1} + \alpha + \beta N_{i - 1}) - (\alpha n_{i - 1} + \beta N_{i - 1}) \\ &= 1 + \alpha \\ &= \Theta(1). \end{aligned}
(2) However, if there is no more room for the new value, we need to make the underlying array larger; in our case, we multiply its capacity by a constant $q > 1$:
Now, we have: $n_i = n_{i - 1} + 1$ and $N_i = qN_{i - 1}$: \begin{aligned} \hat{c}_i &= c_i + \Phi_i - \Phi_{i - 1} \\ &= 1 + (\alpha n_i + \beta N_i) - (\alpha n_{i - 1} + \beta N_{i - 1}) \\ &= 1 + (\alpha n_{i - 1} + \alpha + \beta qN_{i - 1}) - (\alpha n_{i - 1} + \beta N_{i - 1}) \\ &= 1 + \alpha + (q - 1) \beta N_{i - 1}. \end{aligned} ... and at this point, I am stuck.
