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====== Differences ====== This shows you the differences between two versions of the page.

--- gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/15 12:46]
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+++ gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/17 10:23] (current)
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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*}
+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 $N$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.

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