channelflow.org

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

--- gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/17 09:44]
gibson
+++ gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/17 10:23] (current)
gibson
@@ Line 13: / Line 13: @@
 </code>
-Rewrite this with summation notation
+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*}
-\frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}
+P_1 = 1
 \end{eqnarray*}
-The Nth partial sum $P_N$ of the infinite series is
 \begin{eqnarray*}
-P_N = \sum_{n=1}^{N} \frac{1}{n^2}
+P_2 = 1 + \frac{1}{2^2}
 \end{eqnarray*}
-Then $\pi^2/6 = \lim_{N \rightarrow \infty} P_N$.
+\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.
-Now

channelflow.org

User Tools

Site Tools

Page Tools