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--- gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/17 10:14]
gibson
+++ gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/17 10:23] (current)
gibson
@@ Line 15: / Line 15: @@
 We can also do the sum with a ''for'' loop. To see how to build the ''for'' loop, it's helpful to think of the series as a sequence of //partial sums//
 \begin{eqnarray*}
-P_1 &= 1 \\
+P_1 = 1
-P_2 &= 1 + \frac{1}{2^2} \\
-P_3 &= 1 + \frac{1}{2^2} + \frac{1}{3^2} \\
 \end{eqnarray*}
-The Nth partial sum $P_N$ is
 \begin{eqnarray*}
-P_N = \sum_{n=1}^{N} \frac{1}{n^2}
+P_2 = 1 + \frac{1}{2^2}
 \end{eqnarray*}
-Note that the difference between successive partial sums is a single term.
+\begin{eqnarray*}
+P_3 = 1 + \frac{1}{2^2} + \frac{1}{3^2}
+\end{eqnarray*}
+etc. Note that the difference between successive partial sums is a single term.
 \begin{eqnarray*}
 P_n = P_{n-1} + \frac{1}{n^2}
 \end{eqnarray*}
-So we can compute the $N$ partial sum $P_N$ by successively adding the term $1/n^2$ for n going from 1 to N.
+So we can compute the $N$th partial sum $P_N$ by successively adding the term $1/n^2$ for n going from 1 to N.
 That's exactly we do when we compute the sum with a ''for'' loop.

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