(New page: We want to calculate <math>\int_0^\pi \sin x\ dx</math> three days before we learn the Fundamental Theorem of Calculus, so our only tool is the limit of a Riemann sum. So <math>\int_0^...) |
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<math>e^{i\theta}=\cos\theta + i\sin\theta.</math> | <math>e^{i\theta}=\cos\theta + i\sin\theta.</math> | ||
| − | Hence, that Riemann sum is the imaginary part of | + | Hence, that Riemann sum <math>R</math> is the imaginary part of |
| − | <math>(\pi/N)\sum_{n=1}^N e^{ | + | <math>(\pi/N)\sum_{n=1}^N e^{in\pi/N}.</math> |
| − | But <math>e^{ | + | But |
| + | |||
| + | <math>e^{in\pi/N}=\left(e^{i\pi/N}\right)^n,</math> | ||
| + | |||
| + | so <math>R</math> is just the imaginary part of a geometric sum. | ||
| + | |||
| + | The formula | ||
| + | |||
| + | <math>1+r+r^2+\dots+r^N = \frac{1-r^{N+1}}{1-r}</math> | ||
| + | |||
| + | lets us calculate that <math>R</math> is the imaginary part of | ||
| + | |||
| + | <math>\frac{1-\left(e^{i\pi/N}\right)^{N+1}}{1-e^{i\pi/N}}.</math> | ||
Revision as of 08:15, 1 September 2011
We want to calculate
$ \int_0^\pi \sin x\ dx $
three days before we learn the Fundamental Theorem of Calculus, so our only tool is the limit of a Riemann sum.
So
$ \int_0^\pi \sin x\ dx\approx \sum_{n=1}^N \sin(n\pi/N)(\pi/N) $
when $ N $ is large.
Recall Euler's identity,
$ e^{i\theta}=\cos\theta + i\sin\theta. $
Hence, that Riemann sum $ R $ is the imaginary part of
$ (\pi/N)\sum_{n=1}^N e^{in\pi/N}. $
But
$ e^{in\pi/N}=\left(e^{i\pi/N}\right)^n, $
so $ R $ is just the imaginary part of a geometric sum.
The formula
$ 1+r+r^2+\dots+r^N = \frac{1-r^{N+1}}{1-r} $
lets us calculate that $ R $ is the imaginary part of
$ \frac{1-\left(e^{i\pi/N}\right)^{N+1}}{1-e^{i\pi/N}}. $
