The Ringing That Never Dies
Round sine waves are layered to build a sharp square wave.
More terms press the flat parts onto the wave. But the ringing beside each jump does not shrink. No number of terms removes that bump.
For example, zoom into a π’ building in a π· photo. Smaller pixels sharpen the shape. Yet the jagged marks along the edges stay to the end.
S N ( x ) = 4 Ο β k = 1 N sin β‘ ( ( 2 k β 1 ) x ) 2 k β 1 S_N(x) = \frac{4}{\pi}\sum_{k=1}^{N} \frac{\sin((2k-1)x)}{2k-1} S N β ( x ) = Ο 4 β k = 1 β N β 2 k β 1 sin (( 2 k β 1 ) x ) β Examples & Exercises
Compute the sine-series coefficients b 2 k β 1 = 4 Ο ( 2 k β 1 ) b_{2k-1} = \frac{4}{\pi(2k-1)} b 2 k β 1 β = Ο ( 2 k β 1 ) 4 β
b n = 2 Ο β« 0 Ο sin β‘ ( n x ) β d x = 2 ( 1 β cos β‘ n Ο ) n Ο b_n = \frac{2}{\pi}\int_0^{\pi} \sin(nx)\,dx = \frac{2(1-\cos n\pi)}{n\pi} b n β = Ο 2 β β« 0 Ο β sin ( n x ) d x = nΟ 2 ( 1 β cos nΟ ) β N = 5 N = 5 N = 5 Show proof
The square wave is odd on ( β Ο , Ο ) (-\pi, \pi) ( β Ο , Ο ) b n = 1 Ο β« β Ο Ο f ( x ) sin β‘ ( n x ) β d x = 2 Ο β« 0 Ο sin β‘ ( n x ) β d x b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx = \frac{2}{\pi}\int_0^{\pi}\sin(nx)\,dx b n β = Ο 1 β β« β Ο Ο β f ( x ) sin ( n x ) d x = Ο 2 β β« 0 Ο β sin ( n x ) d x
Integrating gives 2 n Ο ( 1 β cos β‘ n Ο ) \frac{2}{n\pi}(1 - \cos n\pi) nΟ 2 β ( 1 β cos nΟ ) 0 0 0 n n n 4 n Ο \frac{4}{n\pi} nΟ 4 β n n n
So b 2 k β 1 = 4 Ο ( 2 k β 1 ) b_{2k-1} = \frac{4}{\pi(2k-1)} b 2 k β 1 β = Ο ( 2 k β 1 ) 4 β b 2 k = 0 b_{2k} = 0 b 2 k β = 0
Verify numerically that the peak near the jump tends to β 1.179 \approx 1.179 β 1.179
(Numerical check)
max β‘ S N β 2 β S i ( Ο ) Ο β 1.17898 \max S_N \to \frac{2\,\mathrm{Si}(\pi)}{\pi} \approx 1.17898 max S N β β Ο 2 Si ( Ο ) β β 1.17898 N = 50 N = 50 N = 50 β 1.179 \approx 1.179 β 1.179 0.179 0.179 0.179 Show proof
Measuring the peak just right of the jump at N = 50 N = 50 N = 50 β 1.179 \approx 1.179 β 1.179 2 β S i ( Ο ) Ο β 1.17898 \frac{2\,\mathrm{Si}(\pi)}{\pi} \approx 1.17898 Ο 2 Si ( Ο ) β β 1.17898
The jump size is 2 2 2 0.179 0.179 0.179 1 1 1 8.95 % 8.95\% 8.95%
This fraction does not go to 0 as N N N
Show algebraically that S N S_N S N β x = Ο / 2 x = \pi/2 x = Ο /2
S N ββ£ ( Ο 2 ) = 4 Ο β k = 1 N ( β 1 ) k + 1 2 k β 1 β 1 S_N\!\left(\tfrac{\pi}{2}\right) = \frac{4}{\pi}\sum_{k=1}^{N}\frac{(-1)^{k+1}}{2k-1} \to 1 S N β ( 2 Ο β ) = Ο 4 β k = 1 β N β 2 k β 1 ( β 1 ) k + 1 β β 1 At x = Ο / 2 x = \pi/2 x = Ο /2 4 Ο \frac{4}{\pi} Ο 4 β Show proof
Since sin β‘ ( ( 2 k β 1 ) Ο 2 ) = ( β 1 ) k + 1 \sin\left((2k-1)\frac{\pi}{2}\right) = (-1)^{k+1} sin ( ( 2 k β 1 ) 2 Ο β ) = ( β 1 ) k + 1 S N ( Ο / 2 ) = 4 Ο β k = 1 N ( β 1 ) k + 1 2 k β 1 S_N(\pi/2) = \frac{4}{\pi}\sum_{k=1}^{N}\frac{(-1)^{k+1}}{2k-1} S N β ( Ο /2 ) = Ο 4 β β k = 1 N β 2 k β 1 ( β 1 ) k + 1 β
As N β β N \to \infty N β β Ο 4 \frac{\pi}{4} 4 Ο β 1 2 k β 1 \frac{1}{2k-1} 2 k β 1 1 β 0 0 0 Ο 4 \frac{\pi}{4} 4 Ο β arctan β‘ \arctan arctan x = 1 x = 1 x = 1
Hence S N ( Ο / 2 ) β 4 Ο β
Ο 4 = 1 S_N(\pi/2) \to \frac{4}{\pi}\cdot\frac{\pi}{4} = 1 S N β ( Ο /2 ) β Ο 4 β β
4 Ο β = 1
9. Series of Functions & Fourier