Stating the problem algebraically
The information we've got is enough to find the derivatives \(E'_i(t)\) as functions of \(E_j(t) (0 \leq j < n)\). We need to calculate the difference in temperatures between the body and its surroundings (as Newton put it) for each point. First, the point \((0, 0)\) is surrounded only by \((1, 0)\). Similarly, the point \((n - 1, 0)\) is surrounded only by \((n - 2, 0)\). All other points \((i, 0)\) are surrounded by both \((i - 1, 0)\) and \((i + 1, 0)\). We can now find the derivatives.
We'll drop the \((t)\)'s for a while. Read \(E^{(j)}_i\) as \(E^{(j)}_i(t)\).
\[
\begin{eqnarray}
E'_0 & = & E_1 - E_0 = - E_0 + E_1\\
E'_i & = & (E_{i - 1} - E_i) + (E_{i + 1} - E_i) = E_{i - 1} - 2E_i + E_{i + 1} (1 \leq i < n - 1)\\
E'_{n-1} & = & E_{n - 2} - E_{n-1}
\end{eqnarray}
\]
Rearranging terms, we get
We'll drop the \((t)\)'s for a while. Read \(E^{(j)}_i\) as \(E^{(j)}_i(t)\).
\[
\begin{eqnarray}
E'_0 & = & E_1 - E_0 = - E_0 + E_1\\
E'_i & = & (E_{i - 1} - E_i) + (E_{i + 1} - E_i) = E_{i - 1} - 2E_i + E_{i + 1} (1 \leq i < n - 1)\\
E'_{n-1} & = & E_{n - 2} - E_{n-1}
\end{eqnarray}
\]
Rearranging terms, we get
\[
\begin{eqnarray}
E_1 & = & E_0 + E'_0\\
E_{i} & = & E'_{i - 1} - E_{i - 2} + 2E_{i - 1} (2 \leq i < n)
\end{eqnarray}
\]
We'll now deal with finding \(a_i (1 \leq i < n)\). This section, I tell you, is beautiful. Aside from the conceptual inversion of our system of equations, we will find the coefficients in Pascal's triangle.
Let's look at our progress so far. We know \(a_0 = (1,)\) and \(a_1 = (1, 1)\) (corresponding to \(E_0 = E_0\) and \(E_1 = E_0 + E'_0\), respectively). For any \(i \geq 2\), we can find \(a_i\) if we know \(a_{i - 1}\) and \(a_{i - 2}\). It is possible to find a recurrence relation for \(a_i\):
\[
a_{i, j} = a_{i - 1, j - 1} - a_{i - 2, j} + 2a_{i - 1, j}
\]
This recurrence relation is similar to that of Pascal's triangle (\( p_{i,j} = p_{i-1,j} + p_{i-1,j-1} \)). As it turns out, if we construct a triangle of \(a_{i,j}\) similarly to how Pascal's triangle is constructed from \(p_{i,j}\), our new triangle is half of Pascal's (see the figure below, where the orange segments form our new triangle when moved together). We find that \( a_{i, j} = \binom{i+j}{2j} \).
Because energy is conserved, \( \sum_{i = 0}^{n-1} E_i(t) = \sum_{i = 0}^{n-1} E_i(0) \). Call this constant quantity E. Using the expressions obtained above for \(E_i\), we'll get a non-homogenous linear differential equation.
\[\begin{eqnarray}
E_1 & = & E_0 + E'_0\\
E_{i} & = & E'_{i - 1} - E_{i - 2} + 2E_{i - 1} (2 \leq i < n)
\end{eqnarray}
\]
Combining the equations into one massive differential equation
Looking at the latest set of expressions, we see that \(E_1\) depends only on \(E_0\) and its derivative. \(E_2\) depends on \(E_0\) and \(E_1\) plus derivatives, but if we substitute our expression for \(E_1\), we get \(E_2\) as a linear combination of only \(E_0\) plus derivatives. Continuing in a similar fashion, we conclude that every \(E_i\) can be written as a linear combination of only \(E_0\) plus derivatives. Thus for every \(i\), there must exist a sequence \(a_i\) independent of \(t\), such that \( E_i = \sum_{j = 0} a_{i, j}E_0^{(j)} \).We'll now deal with finding \(a_i (1 \leq i < n)\). This section, I tell you, is beautiful. Aside from the conceptual inversion of our system of equations, we will find the coefficients in Pascal's triangle.
Let's look at our progress so far. We know \(a_0 = (1,)\) and \(a_1 = (1, 1)\) (corresponding to \(E_0 = E_0\) and \(E_1 = E_0 + E'_0\), respectively). For any \(i \geq 2\), we can find \(a_i\) if we know \(a_{i - 1}\) and \(a_{i - 2}\). It is possible to find a recurrence relation for \(a_i\):
\[
a_{i, j} = a_{i - 1, j - 1} - a_{i - 2, j} + 2a_{i - 1, j}
\]
This recurrence relation is similar to that of Pascal's triangle (\( p_{i,j} = p_{i-1,j} + p_{i-1,j-1} \)). As it turns out, if we construct a triangle of \(a_{i,j}\) similarly to how Pascal's triangle is constructed from \(p_{i,j}\), our new triangle is half of Pascal's (see the figure below, where the orange segments form our new triangle when moved together). We find that \( a_{i, j} = \binom{i+j}{2j} \).
Because energy is conserved, \( \sum_{i = 0}^{n-1} E_i(t) = \sum_{i = 0}^{n-1} E_i(0) \). Call this constant quantity E. Using the expressions obtained above for \(E_i\), we'll get a non-homogenous linear differential equation.
\begin{eqnarray}
E &=& \sum_{i = 0}^{n-1} E_i(t)\\
&=& \sum_{i = 0}^{n-1} \sum_{j = 0}^{n-1} a_{i, j}E_0^{(j)}\\
&=& \sum_{j = 0}^{n-1} (E_0^{(j)} \sum_{i = 0}^{n-1} a_{i, j})
\end{eqnarray}
\]
The rightmost sum may be evaluated by looking at its final term in Pascal's triangle and going down one to the right. Our final result is:
\[
E = \sum_{j = 0}^{n-1} \binom{j+n}{2j+1} E_0^{(j)}
\]
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