gibson:teaching:fall-2014:math445:lecture10-diary
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gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/15 12:53] gibson |
gibson:teaching:fall-2014:math445:lecture10-diary [2014/10/17 10:23] (current) gibson |
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| ====== Math 445 lecture 10: more ''for'' ====== | ====== Math 445 lecture 10: more ''for'' ====== | ||
| - | Here's a in finite series for $pi^2/6$ | + | 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*} | \begin{eqnarray*} | ||
| - | & & \frac{3}{4 \pi} \sqrt{4 \cdot x^2 12}\\ | + | \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 |
| - | & & \lim_{n \to \infty} | + | |
| - | \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}\\ | + | |
| - | & & {\it f}(x) = \frac{1}{\sqrt{x} x^2}\\ | + | |
| - | & & e^{i \pi} + 1 = 0\; | + | |
| \end{eqnarray*} | \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*} | \begin{eqnarray*} | ||
| - | \pi^2/6 = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + \ldots | + | P_1 = 1 |
| - | \end{eqnarry*} | + | \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. | ||
| + | |||
| + | |||
gibson/teaching/fall-2014/math445/lecture10-diary.1413402804.txt.gz · Last modified: 2014/10/15 12:53 by gibson
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