The AIC for the piecewise function is indeed the sum of the AIC scores of each piece under some independence assumptions explained below, so yes, you can just add them together.

Let $X_j$ denote the set of points which "belong" to segment $j$ of your model. The AIC can be derived in the following way. We use $k$ to denote the number of parameters in the model, with $k_j$ being the number of parameters in segment $j$.

\begin{align} AIC &= 2k-2\log(P(X|\theta))\\ &= 2 \sum_j k_j - 2 \log \prod_j P(X_j | \theta_j) \\ &= \sum_j 2k_j - 2\log P(X_j|\theta_j)\\ &= \sum_j AIC_j \end{align}

For this proof to go through, there is the caveat that each group $X_j$ must be independent from all other groups. That is, the population growth in each era is independent from all other eras. For this particular problem, I suspect this is not a horrible assumption to make.

The AIC for the piecewise function is indeed the sum of the AIC scores of each piece, so yes, you can just add them together.

Let $X_j$ denote the set of points which "belong" to segment $j$ of your model. The AIC can be derived in the following way. We use $k$ to denote the number of parameters in the model, with $k_j$ being the number of parameters in segment $j$.

\begin{align} AIC &= 2k-2\log(P(X|\theta))\\ &= 2 \sum_j k_j - 2 \log \prod_j P(X_j | \theta_j) \\ &= \sum_j 2k_j - 2\log P(X_j|\theta_j)\\ &= \sum_j AIC_j \end{align}

For this proof to go through, there is the caveat that each group $X_j$ must be independent from all other groups in your statistical model. That is, the population growth in each era is independent from all other eras. This is true under a piecewise model of population growth that you are using.

The AIC for the piecewise function is indeed the sum of the AIC scores of each piece, so yes, you can just add them together.

Let $X_j$ denote the set of points which "belong" to segment $j$ of your model. The AIC can be derived in the following way. We use $k$ to denote the number of parameters in the model, with $k_j$ being the number of parameters in segment $j$.

\begin{align} AIC &= 2k-2\log(P(X|\theta))\\ &= 2 \sum_j k_j - 2 \log \prod_j P(X_j | \theta_j) \\ &= \sum_j 2k_j - 2\log P(X_j|\theta_j)\\ &= \sum_j AIC_j \end{align}

For this proof to go through, there is the caveat that each group $X_j$ must be independent from all other groups. That is, the population growth in each era is independent from all other eras. For this particular problem, I suspect this is not a horrible assumption to make.

The AIC for the piecewise function is indeed the sum of the AIC scores of each piece under some independence assumptions explained below, so yes, you can just add them together.

Let $X_j$ denote the set of points which "belong" to segment $j$ of your model. The AIC can be derived in the following way. We use $k$ to denote the number of parameters in the model, with $k_j$ being the number of parameters in segment $j$.

\begin{align} AIC &= 2k-2\log(P(X|\theta))\\ &= 2 \sum_j k_j - 2 \log \prod_j P(X_j | \theta_j) \\ &= \sum_j 2k_j - 2\log P(X_j|\theta_j)\\ &= \sum_j AIC_j \end{align}

For this proof to go through, there is the caveat that each group $X_j$ must be independent from all other groups. That is, the population growth in each era is independent from all other eras. For this particular problem, I suspect this is not a horrible assumption to make.