Exercises For The Feynman Lectures On Physics By Richard .
Exercises for the Feynman Lectures onPhysics by Richard Feynman, Et Al. Chapter36 Fourier Analysis of Wavesβ DetailedWork by James Pate Williams, Jr. BA, BS,MSwE, PhDFrom Exercises for the Feynman Lectures on Physics by Richard Feynman, Robert Leighton, MatthewSands, et al. 36 Fourier Analysis of Waves. Refer to The Feynman Lectures on Physics Vol. I, Chapter 50.Fourier series over a period P where x is contained in the half-open interval [x0, P). π02πππ₯2πππ₯π(π₯) [ππ cos () ππ sin ()]2πππ 0π₯0 π22πππ₯ππ π(π₯) cos () ππ₯πππ₯0π₯0 π22πππ₯ππ π(π₯) sin () ππ₯πππ₯0Eulerβs equation used to derive trigonometric identities.π π(π π) cos(π π) π sin(π π) π ππ π ππ (cos π π sin π)(cos π π sin π) cos π cos π sin π sin π π(cos π sin π sin π cos π)cos(π π) cos π cos π sin π sin πsin(π π) cos π sin π sin π cos π2cos π cos π cos(π π) cos(π π)2sin π cos π sin(π π) sin(π π)π ππ₯ π ππ₯ π 0 1 (cos π₯ π sin π₯)(cos π₯ π sin π₯) (cos π₯)2 (sin π₯)2π 2ππ₯ cos 2π₯ π sin 2π₯ π ππ₯ π ππ₯ (cos π₯ π sin π₯)(cos π₯ π sin π₯) (cos π₯)2 (sin π₯)2 π(cos π₯ sin π₯ sin π₯ cos π₯)cos 2π₯ (cos π₯)2 (sin π₯)2 1 (sin π₯)2 (sin π₯)2 1 2(sin π₯)21
1(sin π₯)2 (1 cos 2π₯)2sin 2π₯ 2 cos π₯ sin π₯36.1 (a) y(x) const. (b) y(x) sin x 0 x 2 * pi.(a)π¦(π₯) πΆ π₯ [0, π]π₯0 π2πΆ2πππ₯ππ cos () ππ₯πππ₯0ππ sin [2πΆπ2πππ₯ π₯ π₯0 ππΆ2ππ(π₯0 π)πΆ2πππ₯0 sin ()] sin [] sin ()π 2ππππππππππ₯ π₯02ππ(π₯0 π)2πππ₯02πππ₯02πππ₯0] cos () sin(2ππ) sin () cos(2ππ) sin ()ππππππ πΆ2πππ₯02πππ₯0) sin ()] 0[sin (πππππ₯0 π2πΆ2πππ₯ππ sin () ππ₯πππ₯0ππ 2πΆπ2πππ₯ π₯ π₯0 ππΆ2ππ(π₯0 π)2πππ₯0 cos ()] ] cos ()}{cos [π 2ππππππππ₯ 0 ) cos(2ππ) sin () sin(2ππ) cos ()][cos οΏ½οΏ½0 ) cos ()] 0[cos (πππππ0 πΆ π0 2πΆ2(b)π¦(π₯) sin π₯ π₯ [0,2π] π₯0 0, π 2π2π1111π₯ 2ππ0 sin π₯ ππ₯ cos π₯]π₯ 0 [cos(2π) cos(0)] (1 1) 0ππππ02
2π2π2π11ππ sin π₯ cos ππ₯ ππ₯ [ sin(π₯ ππ₯)ππ₯ sin(π₯ ππ₯)ππ₯ ]π2π0002π 2π1{ sin[(1 π)π₯] ππ₯ sin[(1 π)π₯] ππ₯ }2π00π₯ 2ππ₯ 2ππ₯ 2ππ₯ 2π 1cos[(1 2π)π₯]cos[(1 2π)π₯]] ]{ } 2π1 2π1 2ππ₯ 0π₯ 0 1cos[(1 2π)π₯]cos[(1 2π)π₯]] ]{ } 02π1 2π2π 1π₯ 0π₯ 02π1ππ sin π₯ sin ππ₯ ππ₯ 0 π 1π02π2π0011π1 (sin π₯)2 ππ₯ (1 cos 2π₯)ππ₯ 1π2π36.2π(π₯) {π 1 π₯ [0, π] 1 π₯ (π, 2π)2π11π₯ ππ₯ 2π )(sin ππ₯]π₯ 0ππ ( cos ππ₯ ππ₯ cos ππ₯ ππ₯ ) sin ππ₯]π₯ π 0πππ0ππ2π11π₯ ππ₯ 2π )(cos ππ₯]π₯ 0ππ ( sin ππ₯ ππ₯ sin ππ₯ ππ₯ ) cos ππ₯]π₯ ππππ0π1[cos ππ cos ππ 2] πππ2π 1 4( 1)2π 1 4( 1)2π 24 π(2π 1)π(2π 1) π(2π 1) 4sin[(2π 1)π₯]π(π₯) (2π 1)ππ 0(a) π4sin[(2π 1) π 2]π( ) (2π 1)2ππ 0sin[(2π 1) π 2] sin(ππ π 2) cos ππ sin(π 2) sin ππ cos(π 2) ( 1)π3
( 1)π41 (2π 1)ππ 0 π 0( 1)ππ (2π 1) 4(b) π 0π 0( 1)π( 1)ππ 2 π2 ( ) (2π 1)(2π 1)416 ( 1)π π11π2 (2π 1)(2π 1) 2(2π 1)2 16π 0 π 0π 0 1π2 (2π 1)28π 0(c) 4 π π 0 π 0π 0 111 4 43 114 π2 π2 π 4 (2π 1)2(2π 1)2 4π 3 86π 0 π 036.3π₯ π₯ [0, π)ππ(π₯) {π π₯1 π₯ [π, 1π0 π(π₯) ππ₯ 2 π₯ ππ₯ ππ₯ ππ₯ 2 π₯ ππ₯πππππ02π0π22113 2 (2π π) 2 [(2π)2 π 2 ] 2 111ππ π(π₯) cos ππ₯ ππ₯ 2 π₯ cos ππ₯ ππ₯ cos ππ₯ ππ₯ cos ππ₯ ππ₯ 2 π₯ cos ππ₯ ππ₯πππππ00π(π’π£) π’ππ£ π£ππ’ π(π’π£) π’π£ π’ππ£ π£ππ’4
π’ππ£ π’π£ π£ππ’π’(π₯) π₯ππ£ cos ππ₯ ππ₯ cos ππ₯ ππ₯ π1sin ππ₯πππ₯ ππ₯1111π₯ π π₯ cos ππ₯ ππ₯ sin ππ₯] sin ππ₯ ππ₯ 2 cos ππ₯]π₯ 0 2 (cos ππ 1) 2 [( 1)π 1]ππππππ₯ 0002 (2π 1)22π2ππ₯ 2ππ₯111π₯ 2π π₯ cos ππ₯ ππ₯ sin ππ₯] sin ππ₯ ππ₯ 2 cos ππ₯]π₯ π 2 (cos 2ππ cos ππ)πππππ₯ πππ12 2 [1 ( 1)π ] (2π 1)2ππ2π 1 2π224 2222(2π 1) π(2π 1) π(2π 1)2 π 2π2π2π2ππππ11111ππ π(π₯) sin ππ₯ ππ₯ 2 π₯ sin ππ₯ ππ₯ sin ππ₯ ππ₯ sin ππ₯ ππ₯ 2 π₯ sin ππ₯ ππ₯πππππ00π’(π₯) π₯ππ£ sin ππ₯ ππ₯1 sin ππ₯ ππ₯ cos ππ₯ππππ₯ ππ₯12π( 1)π π₯ sin ππ₯ ππ₯ cos ππ₯] cos ππ₯ ππ₯ ππππ₯ 0002π2ππππ₯ 2ππ₯12π 2π( 1)π π₯ sin ππ₯ ππ₯ cos ππ₯] cos ππ₯ ππ₯ πππππ₯ πππ 1212[( 1)π ( 1)π 2] [4( 1)π 4] [1 ( 1)π ](cos 2ππ cos ππ) 22πππππππππ2π 1 8442 (1 )(2π 1)π 2 (2π 1)π (2π 1)ππ π 0π 01442π(π₯) cos[(2π 1)π₯] (1 ) sin[(2π 1)π₯](2π 1)2 π 2(2π 1)π2π5
(a) 14π(0) π(2π) 0 (2π 1)2 π 22π 0 π 01π2 (2π 1)28(b)2 1π4 [ ](2π 1)264π 0(1 11111111 )(1 )32 52 72 9232 52 7 2 92111111111 1 2 2 2 2 2 (1 2 2 2 2 )3579335791111111111 2 (1 2 2 2 2 ) 2 (1 2 2 2 2 )5357973579 4111111π 2 (1 2 2 2 2 ) π 4(2π 1)9357964π 0 π 0π 2[1π4 π (2π 1)4 6411111 111 ( ) ]3 2 52 72 9232 52 72 92π 0.50732667704348711π4 π 1.522017047406288081819380198261 0.5073266770434871164 1.014690370362800971819380198261π4 1.014690370362800971819380198261πΏπΏ 95.998832628296223472948144984649πΏ 96(1) 1π4 (2π 1)4 96π 0 π 0π 0π 0111 444(2π 1)(2π 2)π6
111 4(2π 2)(π 1)416π 0π 0 1ππ₯1ππ¦1 16 (π₯ 1)4 16 π¦ 4 4801 π 01π4 π‘π4 96 π‘ 11 0.067645202106928815(π 1)416π 0(2)π4π4 π‘ 9690π‘ π4 π4 1.082323233711138191516003696541290 96 1.0146780316041920545462534655073 0.06764520210694613696975023103382π 4 16 π 4 16 π 4 96 156 15 90 π 0π 1111 111111π(4)6π 4π41 π4π‘ ( ) (π 1)4 16 14 24 34 441616π4168640 1440 16 90 π(4) π49036.4 Evaluate the following integral: π₯ 3 ππ₯π₯ 3 π π₯ ππ₯ π₯ π 11 π π₯00 π 0π 01 1 π π₯ π 2π₯ π 3π₯ (π π₯ )π π ππ₯1 π π₯ π₯ 3 π π₯ ππ₯ π₯ 3 π (π 1)π₯ ππ₯1 π π₯0π 0 0π’(π₯) π₯ 3ππ£ π (π 1)π₯ ππ₯π£(π₯) π (π 1)π₯ ππ₯ 7π (π 1)π₯π 1
3 (π 1)π₯ π₯ π03ππ₯ π₯ 2 π (π 1)π₯ ππ₯π 10π’(π₯) π₯ 2ππ£ π (π 1)π₯ ππ₯π£(π₯) π (π 1)π₯π (π 1)π₯ππ₯ π 1 2 π₯ 2 π (π 1)π₯ ππ₯ π₯ π (π 1)π₯ ππ₯π 100π’(π₯) π₯ππ£ π (π 1)π₯ ππ₯π£(π₯) π (π 1)π₯ ππ₯ π (π 1)π₯π 1 π₯π (π 1)π₯011ππ₯ π (π 1)π₯ ππ₯ (π 1)2π 10 π₯ 3 π π₯ ππ₯16π 4 π 4 6 6π(4) (π 1)41 π π₯9015π 0036.5 ( 1)π 1(2π 1)ππ₯(2π 1)ππ‘8βπ¦(π₯, π‘) 2 sin [] cos []2(2π 1)π22π 1 ( 1)π 1(2π 1)π( 1)π 1 ( 1)π 1 ( 1)π 12π8β8βπ¦ (1, ) 2 sin [] cos[(2π 1)π] 2 (2π 1)2(2π 1)2ππ2ππ 1 ( 1)3π 3 π 1( 1)π 18β8β8β1118β π 2β 2 2 (1 2 2 2 ) 2 222(2π 1)(2π 1)πππ357π4ππ 1π 1 ( 1)π 1(2π 1)π( 1)π 1 ( 1)π 18βπ¦(1,0) 2 sin[] (2π 1)2(2π 1)2π2π 1π 1 π 1π 1( 1)2π 2 8β8β18β111 2 2 2 (1 2 2 2 ) 22(2π 1)(2π 1)πππ357π΄1π΄2π΄311 1, 0, 2 π΄0π΄0π΄0 3936.68
β(π₯) 2ππ₯ π₯ [0,2π)2π2π2π00π₯ 2π11π₯11π₯ 2πππ β(π₯) cos ππ₯ ππ₯ 2 π₯ cos ππ₯ ππ₯ sin ππ₯] sin ππ₯ ππ₯ 2 cos ππ₯]π₯ 0π2πππππ₯ 00 02π2π0011ππ β(π₯) sin ππ₯ ππ₯ 2 π₯ sin ππ₯ ππ₯π2π2π2π00π₯ 2ππ₯12π π₯ sin ππ₯ ππ₯ cos ππ₯] cos ππ₯ ππ₯ ππππ₯ 0ππ 2π2π001ππ1111π₯ 2ππ0 β(π₯) ππ₯ 2 π₯ ππ₯ 2 π₯ 2 ]π₯ 0 1π2π2π2 1 1sin ππ₯β(π₯) 2 πππ 1Conjecture:ππ sin ( 2 ) 1 1 ( 1)π 1 1 1π1 11 1 11β( ) (1 ) 0.25 (2π 1) 2 π22 ππ2 π3 5 74π 1π 1 ( 1)π 1 ππ π tan 1 1(2π 1) 42 4π 136.7π 2π¦ 2π¦ π 0 π₯ 2 π‘ 2π π‘πππ πππ πππ π πππ π ππππ ππ‘π¦π₯ π₯ [0, π₯π ]π₯ππ¦(π₯, 0) π₯ π₯ππ΄0 (1 ) π₯ (π₯π , πΏ]πΏ π₯π{π΄0π¦Μ (π₯, 0) 0π΄0 ππππππ‘π’ππ, πΏ π‘βπ πππππ‘β ππ π‘βπ π π‘ππππ9
οΏ½οΏ½οΏ½, π‘) sin () [ππ sin () ππ cos ()]πΏπΏπΏπ 1 π¦Μ (π₯, π‘) π οΏ½οΏ½οΏ½π‘sin () [ππ cos () ππ sin ()]πΏπΏπΏπΏππ πππ 0πππ₯π2πΏ2 sin ()πΏππ π΄0 2 2π π π₯π (πΏ π₯π )πππ₯ππΏ2 sin ()πππ₯ππππ‘πΏπ¦(π₯, π‘) 2π΄0 2 2sin () cos ()πΏπΏπ π π₯π (πΏ π₯π ) π 1πππ₯ππΏ2 sin ( πΏ )πππ₯π¦(π₯, 0) 2π΄0 2 2sin ()πΏπ π π₯π (πΏ π₯π ) π 1π π¦(π₯, π‘) 8π΄0 π 12πΏπ 2πΏ ππ1π2 π 2πππ₯ππππ‘sin () cos ()πΏπΏ π¦(π₯, 0) οΏ½οΏ½οΏ½ 2 sin () 2 [sin ( ) sin () sin () ]2πππΏππΏ4πΏ9πΏπ 1π πΏ 2 π οΏ½π¦ (π₯, ) 8π΄0 2 2 sin () cos(ππ) 2 [ sin ( ) sin () sin () ]ππ ππΏππΏ4πΏ9πΏπ 1 ( 1)π 1πΏ πΏ8π΄0π¦( , ) 2 2 πππ2π 136.8π(π₯) sin π₯ π₯ [0, π) π0π(π₯) [ππ cos(2ππ₯) ππ sin(2ππ₯)]2π 110
π2224π₯ ππ0 sin π₯ ππ₯ cos π₯}π₯ 0 ( 1 1) πππ22ππ sin π₯ cos(2ππ₯) ππ₯ { sin[(1 2π)π₯] ππ₯ sin[(1 π)π₯] ππ₯ }ππ00π₯ π0π₯ π1cos[(1 2π)π₯]cos[(1 2π)π₯] { ] ]} π1 2π1 2ππ₯ 0π₯ 0π₯ ππ₯ π1cos[(1 2π)π₯]cos[(1 2π)π₯] { ] ]}π1 2π2π 1π₯ 0π₯ 0π1 π2 π2 1 cos(3π) 11 21 2 64[ cos π 1] ( 2) ( ) π33π 3π 3 33π 0.42441318157838756205035670232671 cos(5π) 1 cos(3π) 12 1 12 354[ ] ( ) ( ) π5533π 5 3π 15 1515π 0.084882636315677512410071340465341 cos(7π) 1 cos(5π) 12 1 12 574[ ] ( ) ( ) π7755π 7 5π 35 3535π 0.03637827270671893389003057448515π2ππ sin π₯ sin(2ππ₯) ππ₯π0 sin π₯ π 0π( 1)π π₯ 2π 1π₯3 π₯5 π₯ (2π 1)!3! 5!πππ11 sin π₯ sin(2ππ₯) ππ₯ π₯ sin(2ππ₯) ππ₯ π₯ 3 sin(2ππ₯) ππ₯ π₯ 5 sin(2ππ₯) ππ₯ 3!5!0000π’(π₯) π₯ππ£ sin(2ππ₯) ππ₯π£(π₯) 1cos(2ππ₯)2ππππ₯ ππ₯ ππ₯11 π₯ sin(2ππ₯) ππ₯ cos(2ππ₯)] cos(2ππ₯) ππ₯ 2 sin(2ππ₯)] 02π2πππ₯ 0π₯ 000π’(π₯) π₯ 3ππ£ sin(2ππ₯) ππ₯11
π£(π₯) 1cos(2ππ₯)2ππππ₯ ππ₯33 π₯ 3 sin(2ππ₯) ππ₯ cos(2ππ₯)] π₯ 2 cos(2ππ₯) ππ₯2π2ππ₯ 000π’(π₯) π₯ 2ππ£ cos(2ππ₯) ππ₯π£(π₯) 1sin(2ππ₯)2πππ₯ πππ₯21 π₯ cos(2ππ₯) ππ₯ sin(2ππ₯)] π₯ sin(2ππ₯) ππ₯ 02πππ₯ 0200Conjecture:π π₯ 2π 1 sin(2ππ₯) ππ₯ 0 ππ 00(a)π 1π028 16 1112π0111 π(π₯) 2 ππ₯ ππ2 2 2 ( ) π02 ( )π2ππ 9 225 1225π9 225 12250π 1(b)π2 4π0 15π15Fourier coefficients and graph for Exercise 36.1 (a) for f(x) 1 for all x in the half-open interval [x0, P).12
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Fourier coefficients and graph for Exercise 36.1 (b) f(x) sin x for all x in the half-open interval [0, 2Ο).15
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Fourier coefficients and graph for Exercise 36.2, the square wave.17
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Fourier coefficients and graph for Exercise 36.3, the triangle wave.22
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Fourier coefficients and graph for Exercise 36.6, the sawtooth wave.27
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Fourier coefficients and graph for Exercise 36.8, the rectified sine wave.29
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From Exercises for the Feynman Lectures on Physics by Richard Feynman, Robert Leighton, Matthew Sands, et al. 36 Fourier Analysis of Waves. Refer to The Feynman Lectures on Physics Vol
May 02, 2018Β Β· D. Program Evaluation ΝThe organization has provided a description of the framework for how each program will be evaluated. The framework should include all the elements below: ΝThe evaluation methods are cost-effective for the organization ΝQuantitative and qualitative data is being collected (at Basics tier, data collection must have begun)
Silat is a combative art of self-defense and survival rooted from Matay archipelago. It was traced at thΓ© early of Langkasuka Kingdom (2nd century CE) till thΓ© reign of Melaka (Malaysia) Sultanate era (13th century). Silat has now evolved to become part of social culture and tradition with thΓ© appearance of a fine physical and spiritual .
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The Feynman Lectures The lectures were published as the Feynman Lectures in Physics in three red volumes (list copied from Wikipedia): Volume 1: Mechanics, radiation and heat. Volume 2: Electromagnetism and matter. Volume 3: Quantum mechanics. Abbreviated edition
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