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

N=100; sum((1:N).^(-2))

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.

N=100;
P=0;
for n=1:N
  P = P + 1/n^2;
end

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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Math 445 lecture 10: more ''for''

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