gibson:teaching:fall-2014:math445:lecture10-diary
====== Differences ====== This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
|
gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/15 13:12] gibson |
gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/17 10:23] (current) gibson |
||
|---|---|---|---|
| Line 1: | Line 1: | ||
| ====== Math 445 lecture 10: more ''for'' ====== | ====== Math 445 lecture 10: more ''for'' ====== | ||
| - | ===== example 1 ===== | + | 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: | Recall this classic formula for $\pi^2/6$ due to Euler: | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| - | \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \ldots | + | \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*} | \end{eqnarray*} | ||
| - | Previously, we showed how to sum the first $N$ terms of this series with the Matlab one-liner | + | We can sum the first N terms of this series with the Matlab one-liner |
| <code matlab> | <code matlab> | ||
| + | N=100; sum((1:N).^(-2)) | ||
| + | </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*} | \begin{eqnarray*} | ||
| - | \frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2} | + | P_1 = 1 |
| \end{eqnarray*} | \end{eqnarray*} | ||
| - | The Nth partial sum $P_N$ of the infinite series is | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| - | P_N = \sum_{n=1}^{N} \frac{1}{n^2} | + | P_2 = 1 + \frac{1}{2^2} |
| \end{eqnarray*} | \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 | ||
gibson/teaching/fall-2014/math445/lecture10-diary.1413403956.txt.gz · Last modified: 2014/10/15 13:12 (external edit)
Page Tools
Except where otherwise noted, content on this wiki is licensed under the following license: CC Attribution-Share Alike 3.0 Unported
