Max Bathman
Math 381
The Second Midterm
1. x = c sinh α sin β cos ϕ; y = c sinh α sin β sin ϕ; z = c cosh αDcos β. E D
E D
E
∂
∂
∂
∂
∂
∂
For these coordinates to be orthogonal, it must be true that ∂α
= ∂β
= ∂α
= 0, since
, ∂β
, ∂ϕ
, ∂ϕ
the inner product is symmetric. I will now compute these inner products:
∂
∂x ∂
∂y ∂
∂z ∂
=
+
+
= (c cosh α sin β cos ϕ, c cosh α sin β sin ϕ, c sinh α cos β)
∂α
∂α ∂x ∂α ∂y ∂α ∂z
∂
= (c sinh α cos β cos ϕ, c sinh α cos β sin ϕ, −c cosh α sin β)
∂β
∂
= (−c sinh α sin β sin ϕ, c sinh α sin β cos ϕ, 0)
∂ϕ
∂ ∂
,
= c2 sinh α cosh α sin β cos β cos2 β + c2 cosh α sinh α sin β cos β sin2 ϕ − c2 sinh α cosh α sin β cos β
∂α ∂β
= c2 sinh α cosh α sin β cos β(cos2 ϕ + sin2 ϕ − 1) = 0 X
∂ ∂
,
= −c2 cosh α sinh α sin ϕ cos ϕ(sin2 ϕ − sin2 ϕ − 0) = 0 X
∂α ∂ϕ
∂ ∂
,
= −c2 sinh2 α cos β sin β cos ϕ sin ϕ + c2 sinh2 α sin β cos β sin ϕ cos ϕ + 0 = 0 X
∂β ∂ϕ
The coordinates are indeed orthogonal. Now I find ds2 :
ds2 = h2α (dα)2 + h2β (dβ)2 + h2ϕ (dϕ)2
s
q
∂ ∂
hα =
,
= c2 cosh2 α sin2 β cos2 ϕ + c2 cosh2 α sin2 β sin2 ϕ + c2 sinh2 α cos2 β
∂α ∂α
q
q
= c2 cosh2 α sin2 β(cos2 ϕ + sin2 ϕ) + c2 sinh2 α cos2 β = c2 cosh2 α(1 − cos2 β) + c2 sinh2 α cos2 β
q
q
= c2 cosh2 α − c2 cos2 β(cosh2 α − sinh2 α) = c cosh2 α − cos2 β
q
q
hβ = c2 sinh2 α cos2 β(cos2 ϕ + sin2 ϕ) + c2 cosh2 α sin2 β = c2 sinh2 α cos2 β + c2 cosh2 α(1 − cos2 β)
q
q
= c2 cos2 β(sinh2 α − cosh2 α) + c2 cosh2 α = c cosh2 α − cos2 β = hα
q
q
hϕ = c2 sinh2 α sin2 β sin2 ϕ + c2 sinh2 α sin2 β cos2 ϕ = c2 sinh2 α sin2 β(sin2 ϕ + cos2 ϕ) = c sinh α sin β
⇒ ds2 = c2 (cosh2 α − cos2 β)((dα)2 + (dβ)2 ) + c2 sinh2 α sin2 β(dϕ)2
2. Find ∆u.
In class we found that the Laplacian is given by
∆u =
1
∂
hβ hϕ ∂
∂
hα hϕ ∂
∂
hβ hα ∂
u +
u +
u .
hα hβ hϕ ∂α
hα ∂α
∂β
hβ ∂β
∂ϕ
hϕ ∂ϕ
1
In the previous problem, I found that hα = hβ , so ∆u reduces to
2
∂
∂
∂
∂
∂
hα ∂
1
hϕ
u +
hϕ u +
u
∆u =
hα hβ hϕ ∂α
∂α
∂β
∂β
∂ϕ hϕ ∂ϕ
1
∂
∂
∂
∂
= 3
c
sinh
α
sin
β
u
+
c
sinh
α
sin
β
u
∂α
∂β
∂β
c sinh α sin β(cosh2 α − cos2 β) ∂α
2
2
2
∂
c (cosh α − cos β) ∂
+
u .
∂ϕ
c sinh α sin β
∂ϕ
Distributing the first term and removing independent terms from the partial derivatives gives
∂
∂
c sinh α
∂
∂
c sin β
sinh
α
u
+
sin
β
u
∆u = 3
∂α
∂β
c sinh α sin β(cosh2 α − cos2 β) ∂α
c3 sinh α sin β(cosh2 α − cos2 β) ∂β
+
∆u =
c(cosh2 α − cos2 β)
∂2u
2
2
c3 (cosh α − cos2 β) sinh α sin2 β ∂ϕ2
∂
1
c2 sinh α(cosh2 α − cos2 β) ∂α
3. Un (x) =
sinh α
∂
∂2u
∂
1
∂
1
u + 2
sin
β
u
+ 2
2
2
2
2
∂α
∂β
c sin β(cosh α − cos β) ∂β
c sinh α sin β ∂ϕ2
sin((n + 1) arccos x)
√
.
1 − x2
i
n+1 i
1 h i(n+1)θ
1 h iθ n+1
e
e
− e−i(n+1)θ =
− e−iθ
2i
2i
i
1 h
n+1
n+1
(cos θ + i sin θ)
− (cos(−θ) + i sin(−θ))
=
2i
i
1 h
n+1
n+1
(cos θ + i sin θ)
− (cos θ − i sin θ)
=
2i
Using the formula for binomial expansion, we get that
" n+1
#
n+1
X
1 X
(n + 1)!
(n + 1)!
k
k
n+1−k
k
n+1−k
k k
sin((n + 1)θ) =
cos
θi sin θ −
cos
θ(−1) i sin θ .
2i
k!(n + 1 − k)!
k!(n + 1 − k)!
Let θ = arccos x =⇒ sin((n + 1)θ) =
k=0
k=0
Using the fact that θ = arccos x, we can construct a right triangle with one angle equal to θ so that we may
find sin θ:
1 √
1 − x2
θ
x
cos θ = cos(arccos x) = x
√
sin θ = sin(arccos x) = 1 − x2
We now have that
sin((n + 1) arccos x)
√
Un (x) =
1 − x2
" n+1
#
n+1
2 k/2
2 k/2
X
1 X
(n + 1)!
(1
−
x
)
(n
+
1)!
(1
−
x
)
=
xn+1−k ik
−
xn+1−k (−1)k ik
2 )1/2
2i
k!(n + 1 − k)!
k!(n + 1 − k)!
(1
−
x
(1 − x2 )1/2
k=0
k=0
" n+1
#
n+1
X
1 X
(n + 1)!
(n + 1)!
n+1−k k
2 (k−1)/2
n+1−k
k k
2 (k−1)/2
=
x
i (1 − x )
−
x
(−1) i (1 − x )
.
2i
k!(n + 1 − k)!
k!(n + 1 − k)!
k=0
k=0
2
Notice that when k = 0 the left and right sums are equal, so their difference is equal to 0, so we sum from
k = 1 to ∞. Moreover, when k is even, (−1)k = 1, and the left and right sums are again equal. When k is
odd, the right sum is (−1) times the left sum, so their difference is equal to twice either sum. Thus, we have
that
n+1
(n + 1)!
1 X
xn+1−k ik (1 − x2 )(k−1)/2 .
Un (x) =
i
k!(n + 1 − k)!
k=1, odd
k
(k−1)/2
When k is odd, i = i(−1)
, so we get that
Un (x) =
n+1
X
k=1,odd
(n + 1)!
xn+1−k (x2 − 1)(k−1)/2 .
k!(n + 1 − k)!
(1)
2
(k−1)/2
will be
Because k is always odd, k−1
2 is always a positive integer or zero, so the expanded (x − 1)
0
comprised of positive integer powers of x or constants (i.e., multiples of x ). Also note that n + 1 − k ≥ 0
and is an integer for all k in the series. Hence, Un (x) is a polynomial in x.
I now investigate the sum Un−1 + Un+1 using the definition of Un given in this problem.
1
[sin((n + 2) arccos x) + sin(n arccos x)] .
Un−1 + Un+1 = √
1 − x2
Using the identity sin(α + β) = sin α cos β + sin β cos α, we get that
Un−1 + Un+1 = √
1
[sin(n arccos x) cos(2 arccos x) + sin(2 arccos x) cos(n arccos x) + sin(n arccos x)] .
1 − x2
Using the identities sin(2θ) = 2 sin θ cos θ and cos(2θ) = cos2 θ − sin2 θ, this expression becomes
1
√
sin(n arccos x)(cos2 (arccos x) − sin2 (arccos x)) + 2 cos(n arccos x) sin(arccos x) cos(arccos x)
2
1−x
+ sin(n arccos x)]
√
I found earlier that sin(arccos x) = 1 − x2 , so I substitute this into my expression to get
h
p
1
Un−1 + Un+1 = √
x2 sin(n arccos x) − (1 − x2 ) sin(n arccos x) + 2x 1 − x2 cos(n arccos x)
1 − x2
i
+ sin(n arccos x)
i
h
p
1
=√
2x2 sin(n arccos x) − sin(n arccos x) + 2x 1 − x2 cos(n arccos x) + sin(n arccos x)
1 − x2
i
h
p
1
=√
2x2 sin(n arccos x) + 2x 1 − x2 cos(n arccos x)
1 − x2
h
i
p
2x
=√
x sin(n arccos x) + 1 − x2 cos(n arccos x)
1 − x2
√
Notice that x = cos(arccos x) and 1 − x2 = sin(arccos x), so we have that
2x
[cos(arccos x) sin(n arccos x) + sin(arccos x) cos(n arccos x)]
1 − x2
2x
2x
=√
[sin(arccos x + n arccos x)] = √
[sin((n + 1) arccos x)]
2
1−x
1 − x2
= 2xUn (x)
Un−1 + Un+1 = √
Hence, I have established the recurrence relation
Un+1 (x) = 2xUn (x) − Un−1 (x).
3
4. Find U5 & U6 .
I will use the series definition of Un (x) which I found in the previous problem:
U5 =
6
X
k=1,odd
6!
6! 3 2
6!
6!
x6−k (x2 − 1)(k−1)/2 = x5 +
x (x − 1) + x(x2 − 1)2
k!(6 − k)!
5!
3!3!
5!
= 6x5 + 20x5 − 20x3 + 6x5 − 12x3 + 6x
U5 = 32x5 − 32x3 + 6x
U6 =
7
X
k=1,odd
7!
7!
7! 2 2
7! 2 2
x7−k (x2 − 1)(k−1)/2 = x6 +
x (x − 1) +
x (x − 1)2 + (x2 − 1)3
k!(7 − k)!
6!
3!4!
5!2!
= 7x6 + 35x6 − 35x4 + 21x6 − 42x4 + 21x2 + x6 − 3x4 + 3x2 − 1
U6 = 64x6 − 80x4 + 24x2 − 1
5. Obtain the generating function relation
∞
X
1
Un (x)tn .
=
1 − 2xt + t2
n=0
6. Find Un (0) & Un (1).
Using the series definition of Un (x) which I found in the third problem, we see that all of the terms of the
series are equal to zero except when n + 1 − k = 0, for this makes xn+1−k = x0 = 1. When n is odd, k will
never equal n + 1 because n = 1 is even and the series only has odd k. So for k odd, all of the terms in the
series are equal to zero (when x = 0), so Un (0) = 0 when n is odd. When n = 0 or n is even, k = n + 1 is
the last term in the series, and it is given by
Un (0) =
(n + 1)! 0 2
x (x − 1)n/2 = (x2 − 1)n/2
(n + 1)!0!
(when n is even or n = 0).
When x = 0, we get that
Un (0) = (−1)n/2 .
Hence, I have found that


1
Un (0) = (−1)n/2


0
if n = 0
if n is even
if n is odd
Now I investigate Un (1). From the series definition, we can see that when x = 1, (x2 − 1)(k−1)/2 = 0 for all
terms in the series except the k = 1 term. Thus, Un (1) is given by the k = 1 term, or that
Un (1) =
(n + 1)! n
(n + 1)n!
1 =
n!
n!
Un (1) = n + 1
7. Establish the orthogonality relation
Z 1
(1 − x2 )1/2 Un (x)Uk (x)dx = 0 when k 6= n.
−1
Find the integral when k = n.
4
R1
I will investigate the integral −1 (1 − x2 )1/2 Un (x)Uk (x)dx by making the substitution x = cos θ (and
thus dx = − sin θdθ) and using the definition given in problem 3:
Z arccos 1
Z 1
1
(1 − cos2 θ)1/2
(1 − x2 )1/2 Un (x)Uk (x)dx = −
sin((n + 1)θ) sin((k + 1)θ) sin θdθ
2θ
1
−
cos
arccos(−1)
−1
Z 0
sin2 θ
=−
2 sin((n + 1)θ) sin((k + 1)θ)dθ
−π sin θ
Z π
sin((n + 1)θ) sin((k + 1)θ)dθ
=
0
Using the product of sines identity sin α sin β = 21 [cos(α − β) − cos(α + β)], the integral is now in the
convenient form
Z 1
Z
Z
1 π
1 π
2 1/2
(1 − x ) Un (x)Uk (x)dx =
cos((n − k)θ)dθ −
cos((n + k + 2)θ)dθ
2 0
2 0
−1
1
1
π
π
=
[sin(n − k)θ]0 −
[sin(n + k + 2)θ]0
2(n − k)
2(n + k + 2)
1
1
[sin(n − k)π − sin 0] −
[sin(n + k + 2)π − sin 0]
=
2(n − k)
2(n + k + 2)
Because k and n are integers, sin(n − k)π = sin(n + k + 2)π = 0. Thus, every term in the above equation
equals 0 when n 6= k. Note that the denominator of the left term in the above equation equals zero when
k = n and the denominator in the right term equals zero when k = −2 − k, neither of which will occur for
n 6= k, n, k ≥ 0. Hence, I have verified that the orthogonality relation is given by
Z 1
(1 − x2 )1/2 Un (x)Uk (x)dx = 0 for n 6= k.
−1
I will now investigate the integral when k = n. The answer that I found for k 6= n had no attached
supposition that n 6= k, but n − k is in the denominator of the first term in the equation
Z 1
1
1
(1 − x2 )1/2 Un (x)Uk (x)dx =
[sin(n − k)π − sin 0] −
[sin(n + k + 2)π − sin 0]
2(n
−
k)
2(n
+
k + 2)
−1
which I found previously. All of the terms except the first are equal to zero when k = n without problems,
but I will take the limit of the first term to find the integral when n = k:
Z 1
0
sin((n − k)π)
=
(1 − x2 )1/2 Un (x)Uk (x)dx = lim
n→k
2(n
−
k)
0
−1
Using L’hôspital’s Rule, we get that
lim
n→k
sin((n − k)π)
cos((n − k)π)
π
= lim π
=
n→k
2(n − k)
2
2
Hence, I have found that
Z
1
(1 − x2 )1/2 Un (x)Uk (x)dx =
−1
π
for n = k.
2
8. Find the Fourier transform of f (t) = 1 − |t| if |t| ≤ 1, f (t) = 0 otherwise.
fˆ(t) =
Z
∞
e−iλt f (t)dt
−∞
Note that f is even, so that we only integrate against cos λt:
Z ∞
cos λtf (t)dt.
fˆ(t) =
−∞
5
Moreover, because f (and cos λt) is even, there is symmetry about x = 0, so fˆ(t) becomes
Z 1
Z 1
Z 1
ˆ
t cos λtdt
cos λtdt − 2
cos λt(1 − t)dt = 2
f (t) = 2
0
0
0
1
We integrate the right integral by parts: let u = t ⇒ du = dt, dv = cos λtdt ⇒ v = sin λt
λ
1
Z
2
2
2 1
1
=
sin λtdt
sin λt − [t sin λt]0 +
λ
λ
λ 0
0
2
2
2
1
= sin λ − sin λ − 2 [cos λt]0
λ
λ
λ
2(cos λ − 1)
2(1 − cos λ)
=−
=
2
λ
λ2
9. Find the Fourier transform of f (t) = e−t if t ≥ 0, f (t) = −et if t < 0.
Z ∞
Z 0
Z ∞
fˆ(λ) =
e−iλt f (t)dt = −
et(1−iλ) dt +
e−t(1+iλ) dt
−∞
= −
−∞
1
et(1−iλ)
1 − iλ
0
0
1
−
e−t(1+iλ)
1 + iλ
−∞
∞
0
1
1
−(1 + iλ) + 1 − iλ
=−
+0−0+
=
1 − iλ
1 + iλ
(1 − iλ)(1 + iλ)
−2iλ
fˆ(λ) =
1 + λ2
6