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====== Math 445 lecture 10: more ''for'' ====== The following example problem spells out in great detail how to translate a formula in summation notation into a Matlab ''for'' loop. Recall this classic formula for $\pi^2/6$ due to Euler: \begin{eqnarray*} \frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \ldots \end{eqnarray*} We can sum the first N terms of this series with the Matlab one-liner <code matlab> N=100; sum((1:N).^(-2)) </code> 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 \end{eqnarray*} \begin{eqnarray*} P_2 = 1 + \frac{1}{2^2} \end{eqnarray*} \begin{eqnarray*} P_3 = 1 + \frac{1}{2^2} + \frac{1}{3^2} \end{eqnarray*} 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 $N$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. <code matlab> N=100; P=0; for n=1:N P = P + 1/n^2; end </code> At each step in the ''for'' loop, we compute $1/n^2$ for the current value of $n$, add it to the previously computed partial sum $P_{n-1}$, and then store the result into $P_n$. But, since we are only interested in the final value $P_N$, we just store the current value of the partial sum in the variable P and write over it with the next value each time we step through the loop.
