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    <p>A "nicer" way to do the trapezoid rule with equally-spaced points...</p> <p>Let <code>dx = pi/n</code> be the width of the interval. Also, let f(i) be sin(i*dx) to shorten some expressions below. Then interval i (in <code>range(1,n)</code>) contributes:</p> <pre><code>dA = 0.5*dx*( f(i) + f(i-1) ) </code></pre> <p>...to the sum (which is an area, so I'm using dA for "delta area"). Factoring out the 0.5*dx, makes the whole some look like:</p> <pre><code>A = 0.5*dx * ( (f(0) + f(1)) + (f(1) + f(2)) + .... + (f(n-1) + f(n)) ) </code></pre> <p>Notice that there are two f(1) terms, two f(2) terms, on up to two f(n-1) terms. Combine those to get:</p> <pre><code>A = 0.5*dx * ( f(0) + 2*f(1) + 2*f(2) + ... + 2*f(n-1) + f(n) ) </code></pre> <p>The 0.5 and 2 factors cancel except in the first and last terms:</p> <pre><code>A = 0.5*dx(f(0) + f(n)) + dx*(f(1) + f(2) + ... + f(n-1)) </code></pre> <p>Finally, you can factor dx out entirely to do just one multiplication at the end. Converting back to sin() calls, then:</p> <pre><code>def TrapArea(n): dx = pi/n asum = 0.5*(sin(0) + sin(pi)) # this is 0 for this problem, but not others for i in range(1, n-1): asum += sin(i*dx) return sum*dx </code></pre> <p>That changed "sum" to "asum", or maybe "area" would be better. That's mostly because sum() is a built-in function, which I'll use below the line.</p> <hr> <p>Extra credit: The loop part of the sum can be done in one step with a generator expression and the sum builtin function:</p> <pre><code>def TrapArea2(n): dx = pi/n asum = 0.5*(sin(0) + sin(pi)) asum += sum(sin(i*dx) for i in range(1,n-1)) return asum*dx </code></pre> <p>Testing both of those:</p> <pre><code>&gt;&gt;&gt; for n in [1, 10, 100, 1000, 10000]: print n, TrapArea(n), TrapArea2(n) 1 1.92367069372e-16 1.92367069372e-16 10 1.88644298557 1.88644298557 100 1.99884870579 1.99884870579 1000 1.99998848548 1.99998848548 10000 1.99999988485 1.99999988485 </code></pre> <p>That first line is a "numerical zero", since math.sin(math.pi) evaluates to about 1.2e-16 instead of exactly zero. Draw the single interval from 0 to pi and the endpoints are indeed both 0 (or nearly so.)</p>
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