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Unformatted text preview: ACM 95/100b Problem Set 5 Solutions
Faisal Amlani February 17, 2009
Problem 1 Consider the function given by f (x) = (1 − x2 /L2 ) 0 < x < L. (a) Develop the Fourier sine series over the interval 0 < x < L. (b) Develop the Fourier cosine series over the interval 0 < x < L . (c) Develop the fully periodic Fourier series over the interval 0 < x < L. (N.B. note the interval). (d) What is the value of each of the series at x = 0? Explain in each case why you get a given value. SOLUTION (a) : For this interval, we are looking for a series of the form
∞ f (x) =
n=1 An sin nπx . L Multiplying both sides by sin
L mπx L and integrating over the interval gives
∞ L f (x) sin
0 mπx dx = L An
n=1 0 sin mπx nπx sin dx. L L L mπx dx = Am . L 2 But by the orthogonality of the sine series, mπx = Am f (x) sin L 0 And so for all n = 1, 2, ..., An =
L L sin2
0 2L nπx dx f (x) sin L0 L nπx 2L dx (1 − x2 /L2 ) sin = L0 L 2 −2L cos(nπ ) + Ln2 π 2 + 2L = L n3 π 3 2 [2 + n2 π 2 − 2(−1)n ] = n3 π 3 1 so that
∞ f (x) =
n=1 2 n3 π 3 [2 + n2 π 2 − 2(−1)n ] sin nπx . L SOLUTION (b) : Similarly, we are looking for a series of the form
∞ f (x) =
n=0 Bn cos nπx . L Multiplying both sides by cos
L 0 mπx L and integrating over the interval gives
∞ L mπx dx = f (x) cos L Bn
n=0 0 cos mπx nπx cos dx. L L But by the orthogonality of the sine series, B0 =
L 1 L
L L f (x)dx
0 f (x) cos
0 mπx dx = Bm L
L cos
0 L mπx nπx cos dx = Bm . L L 2 And so for n = 0,1,2,... 1 B0 = L (1 − x2 /L2 )dx =
0 2 3 Bn = 2L nπx dx f (x) cos L0 L 2L nπx = dx (1 − x2 /L2 ) cos L0 L 2 −2Lnπ cos(nπ ) = L n3 π 3 −4(−1)n = n2 π 2
∞ so that nπx −4(−1)n 2 . cos f (x) = + 2π2 3 n=1 n L SOLUTION (c) : 2 To develop the fully periodic series, we want these functions to be periodic on 0 < x < L. That is, we look for
∞ f (x) =
n=0 Bn cos 2nπx L ∞ +
n=1 An sin 2nπx L . Multiplying both sides by cos orthogonality again yields B0 = 2mπx L and integrating over 0 < x < L and using 2 (1 − x2 /L2 )dx = , 3 1 L L 0 Bn = 2nπx 2L dx f (x) cos L0 L 2L 2nπx = (1 − x2 /L2 ) cos L0 L 2 −2nπ cos(2nπ ) = L L 4n3 π 3 −1 . = n2 π 2
2mπx L dx Similarly, multiplying both sides by sin using orthogonality again yields
L and integrating over 0 < x < L and 2mπx L f (x) sin
0 2mπx L L = Am
0 sin2 dx = Am L/2. And so for all n = 1, 2, ..., An = 2L f (x) sin L0 1 = . nπ 2nπx L dx Thus, the fully periodic series is given by 2 2nπx (−1) f (x) = + cos 2π2 3 n=1 n L SOLUTION (d) : Clearly by the above results, at x = 0, f (0) = 0 for the sine series, 3
∞ ∞ +
n=1 1 sin nπ 2nπx L . The sine series approximates an odd (periodic) extension of (1 − x2 /L2 ) to (−L, 0). Thus, from the left the value is 1 and from the right it is +1, and the average of the two is 0 (as expected from the theorem on uniform convergence). For the cosine series, at x = 0, we have (recalling series approximating π , e.g. Riemann Zeta function),
∞ n=1 −4(−1)n 4 4 = 2 (1 + 1/32 + 1/52 + ...) − 2 (1/22 + 1/42 + 1/62 + ...) n2 π 2 π π = 4 1 42 (π /8) − 2 (π 2 /24) = 2 π π 3 = 1, i.e. so that f (0) = 2 3 + ∞ −4(−1)n n=1 n2 π 2 f (0) = 1 for the cosine series. The cosine series approximates an even (periodic) extension of (1−x2 /L2 ) to (−L, 0). Thus, from the left the value is 1 and from the right it is still 1 hence giving the average value of 1. This is the correct value because a continuous extension of (1−x2 /L2 ) to (−L, 0) is naturally even. For the fully periodic series, at x = 0, we have that (using Riemann Zeta with para2 mater 2) f (0) = 3 − ∞ n21π2 = 2/3 − 1/π 2 (π 2 /6) = 1/2, i.e. n=1 f (0) = 1/2 for the fully periodic series. The full series makes (1 − x2 /L2 ) become periodic with a period of length L (e.g. a sawtooth shape). Thus from the left the value is zero and from the right +1 – we get the average value of 1/2. Problem 2 Compute the Fourier sine series of the function f (x) = cos x − 1 + 2x π 0<x<π How fast for the Fourier coefﬁcients decrease with increasing n? Explain this rate of decrease. SOLUTION : For a sine series, we will need the odd extension of f (x) to (−π, π ). For the odd extension, we want −f (−x) on (−π, 0). Then the fourier series of this extended function gives the values f (x) on (0, π ). That is,
∞ f (x) =
n=1 bn sin(nx)dx 4 where bn = = = = = 1 π 1 π 1 π 1 π 2 π
π f (x) sin(nx)dx
−π π 0 f (x) sin(nx)dx −
0 π f (−x) sin(nx)dx
−π −π f (x) sin(nx)dx +
0 π 0 π f (−x) sin(nx)dx f (x) sin(nx)dx
0 f (x) sin(nx)dx +
0 π f (x) sin(nx)dx,
0 where the fourth line follows from a change of variables x = −x =⇒ dx = −dx and noting that sin(−nx) = − sin(nx). So, we can compute bn by considering ﬁrst bn = 2 π
π π cos(x) sin(nx)dx −
0 0 sin(nx)dx + 2 π π x sin(nx)dx .
0 From double integration by parts, we have that for n = 1,
π π cos(x) sin(x)dx
0 π = =⇒ = sin(x) sin(x) 0. π 0 −
0 sin(x) cos(x)dx 2
0 cos(x) sin(x)dx For n > 1, the ﬁrst integral is
π π cos(x) sin(nx)dx
0 = = = sin(nx) sin(x) π 0 −n
0 π 0 sin(x) cos(nx)dx
π −n − cos(nx) cos(x) −n cos(nπ ) + 1 − n −n
0 π cos(x) sin(nx)dx cos(x) sin(nx)dx
0 π =⇒ cos(x) sin(nx)dx = =⇒ cos(x) sin(nx)dx = n(cos(nπ ) + 1) . n2 − 1 −n[cos(nπ ) + 1] (1 − n2 )
0 π 0 We also have that for all n ≥ 1,
π 0 1 sin(nx)dx = − cos(nx) n 5 π 0 = 1 − cos(nπ ) n and by a single integration by parts 2 π
π 0 x sin(nx)dx = x cos(nx) π 1 π 2 + − cos(nx)dx 0 π n n0 π cos(nπ ) 1 2 π − + 2 sin(nx) 0 = π n n 2 cos(nπ ) . =− n Putting it all together we ﬁnd that b1 = bn 2 [0 − 1 + cos(π ) − 2 cos(π )] = 0 π 2 n(cos(nπ ) + 1) cos(nπ ) 1 2 cos(nπ ) = − −− π n2 − 1 n n n 2 1 + cos(nπ ) = π n(n2 − 1) for n ≥ 2. Thus, the sine series is given by 2 f (x) = π
∞ n=2 1 + cos(nπ ) n(n2 − 1) sin(nx)dx, x ∈ (0, π ). Clearly, the coefﬁcients decay like 1/n3 . This makes sense because the odd extension x of f (x) = cos(x) − 1 + 2π must have a discontinuous n − 1 = 2 derivative since integration by parts can was repeated at most twice. This is easily conﬁrmed by noting that at x = 0,
x→0+ lim f (2) (x) = lim − cos(x) = −1, +
x→ 0 x→0− lim −f (2) (−x) = lim cos(x) = 1. −
x→ 0 Problem 3 (a) For what values of x does the Fourier series π2 (−1)n +4 cos nx = x2 ? 3 n2 n=1 (b) What is the value of the Fourier series for all x? (c) From the relation above, show the identities
∞ ∞ n=1 ∞ 1 π2 = , n2 6 n=1 π2 (−1)n+1 = . n2 12 6 SOLUTION (a) : We have a cosine series which requires an even extension. Since x2 is already even, 2 we have that f (x) = x2 on some interval [−L, L] whose cosine series gives π3 + n 2 4 ∞ (−1) cos nx. Clearly in this series, a0 = π3 . But n=1 n2 a0 = 1 2L
L x2 dx =
−L 1 x3 2L 3 L −L = L3 − (−L)3 L2 = . 3 3 So by the uniqueness of the cosine series for an even f (x), we have that L = π , i.e. the series converges to x2 for −π ≤ x ≤ π. SOLUTION (b) : By the 2π periodic nature of cos(nx), any x outside of the interval [−π, π ] will return a value equal to the value returned by an x0 = x − 2πm ∈ [−π, π ] for some m = 0, ±1, ±2, ... That is, the value of the series is (x − 2πm)2 for m ∈ Z such that (2m − 1)π ≤ x ≤ (2m + 1)π. SOLUTION (c) : If we let x = π , then cos(nπ ) = (−1)n for n = 1, 2, ... and so from part (a),
π2 3 +4 +4 4 ∞ (−1)2n n=1 n2 = π2 =⇒
π2 3 ∞ 1 n=1 n2 = π2
2π 2 3 =⇒
∞ 1 n=1 n2 = =⇒
∞ n=1 1 π2 =. n2 6 7 If we let x = 0, then cos(n0) = 0 for n = 1, 2, ... and so from part (a),
π2 3 +4 ∞ (−1)n n=1 n2 =0
2 =⇒ 4 −4 4
∞ ∞ (−1)n+1 n=1 n2 = − π3 = =
π2 3 π2 3 =⇒
∞ (−1)n n=1 n2 =⇒
∞ (−1)n+1 n=1 n2 =⇒ π2 (−1)n+1 =. n2 12 n=1 Problem 4 (a) Determine the Fourier cosine series of the function f (x) = cos(νx) where ν is an arbitrary real number. (b) From this series, deduce that for ν = n π sin πν = 1 + (−1)n ν n=1
∞ ∞ 0≤x≤π 1 1 + ν −n ν+n . , 1 π cot πν = + ν n=1 1 1 + ν −n ν+n (c) Integrate the last formula with respect to ν from ν = 0 to ν = θ with 0 < θ < 1, to show that ∞ sin πθ θ2 = 1− 2 . πθ n n=1 Note that the identities in part (b) represent meromorphic functions (functions that have only pole singularities) as sums over the poles. The identity of part (c) represents an entire function as a product of factors that indicate where the zeroes of the function are. These types of representations are known as MittagLefﬂer expansions. SOLUTION (a) : For a cosine series, we’d like an even extension of f (x) = cos(νx). Since f (x) is already even, we have that f (x) = cos(νx) on −π ≤ x ≤ π and so for
∞ f (x) = a0 +
n=1 an cos(nx)dx 8 we have that since cos is even, a0 =
−π 1 cos(νx)dx 2π π 1π = cos(νx)dx π0 π 1 = sin(νx) 0 νπ sin(νπ ) = , νπ and an = = = = = = = = = = = 1π cos(νx) cos(nx)dx π −π 2π cos(νx) cos(nx)dx π0 π π 1 cos((ν − n)x)dx + cos((ν + n)x)dx π0 0 π π 1 1 1 sin((ν − n)x) + sin((ν + n)x) 0 0 π ν −n ν+n 1 1 1 sin((ν − n)π ) + sin((ν + n)π ) π ν −n ν +n 1 ν sin((ν − n)π ) + n sin((ν − n)π ) + ν sin((ν + n)π ) − n sin((ν + n)π ) π (ν − n)(ν + n) 2 ν [sin((ν − n)π ) + sin((ν + n)π )] − n[sin((ν + n)π ) − sin((ν − n)π )] π 2(ν − n)(ν + n) 2 ν sin(νπ ) cos(nπ ) − n cos(νπ ) sin(nπ ) π (ν − n)(ν + n) 2 ν sin(νπ ) cos(nπ ) π (ν − n)(ν + n) 1 1 2 ν sin(νπ )(−1)n + π 2ν (ν − n) 2ν (ν + n) sin(νπ ) 1 1 (−1)n + π ν −n ν+n 1 for n ≥ 1. Here we used the trig identities cos(u) cos(v ) = 2 [cos(u − v )+cos(u + v )], sin(u) cos(v ) = 1 [sin(u + v )+sin(u − v )] and cos(u) sin(v ) = 1 [sin(u + v ) − sin(u − 2 2 9 v )]. Thus, the cosine series is given by 1 1 sin(νπ ) sin(νπ ) cos(nx) + (−1)n + f (x) = νπ π ν −n ν+n n=1 = sin(νπ ) 1 1 1 cos(nx) . + (−1)n + π ν n=1 ν−n ν +n
∞ ∞ SOLUTION (b) : If we let x = 0, then f (0) = cos(0) = 1 and cos(nx) = 1 for all n ≥ 1 From the above representation we have 1 π sin(νπ ) = =⇒ = 1 1 sin(νπ ) 1 (−1)n + + π ν n=1 ν −n ν +n 1 1 1 . + (−1)n + ν n=1 ν−n ν +n
∞ ∞ If we let x = π , then f (π ) = cos(νπ ) and cos(nx) = (−1)n for all n ≥ 1 Then we have that cos(νπ ) π cos(νπ ) sin(νπ ) π cot(νπ ) = =⇒ = =⇒ = sin(νπ ) 1 1 1 (−1)n + (−1)n + π ν n=1 ν−n ν +n 1 1 1 + + ν n=1 ν − n ν + n 1 1 1 . + + ν n=1 ν − n ν + n
∞ ∞ ∞ SOLUTION (c) : Subtracting 1/ν from both sides and integrating, π cot πν − limǫ→0
θ −ǫ ǫ πν ) π cos(πν ) sin( 1 ν = ∞ n=1 1 ν −n + 1 ν +n 1 ν −n − 1 ν =⇒ dν = limǫ→0 θ −ǫ ǫ + 1 ν +n dν. 10 But by L’hopital’s rule, the left hand side is computed as
θ −ǫ lim ǫ→0 π
ǫ cos(πν ) 1 θ dν = lim [ln sin(πν ) − ln ν ]ǫ −ǫ − ǫ→0 sin(πν ) ν = = = = = sin(πν ) lim ln ǫ→0 ν ǫ sin(πǫ) sin(π (θ − ǫ)) − ln lim ln ǫ→0 θ−ǫ ǫ sin(πθ) π cos(πǫ) ln − ln lim ǫ→0 θ 1 sin(πθ) ln − ln π θ sin(πθ) . ln πθ since ν ∈ (0, 1) so that the right hand side
θ −ǫ 1 1 We have that ν −n + ν +n = is computed as θ −ǫ 2ν ν 2 −n2 = −2ν n2 −ν 2 lim ǫ→0 ǫ θ −ǫ 1 1 −2ν dν = lim + dν 2 − ν2 ǫ→0 ǫ ν−n ν +n n = lim[ln(n2 − ν 2 )]θ−ǫ ǫ ǫ→0 = lim[ln(n2 − (θ − ǫ)2 ) − ln(n2 − ǫ2 )]
ǫ→0 = ln(n2 − θ2 ) − ln(n2 ) n2 − θ2 = ln n2 θ2 = ln 1 − 2 . n We thus have that after integrating, sin(πθ) ln = πθ
∞ ln 1 −
n=1 θ2 n2 =⇒ sin(πθ) = exp πθ
∞ ∞ ln 1 −
n=1 θ2 n2 θ2 n2 =
n=1 ∞ exp ln 1 − 1−
n=1 = θ2 n2 . 11 ...
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This note was uploaded on 12/12/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.
 Winter '09
 NilesA.Pierce

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