Download Recurrence Relation - Math - Assignment Solutions and more Exercises Mathematics in PDF only on Docsity!
Math 334
Assignment 8 — Solutions
- (a) Evaluate I =
0
e−x
2 dx. (Hint: Consider I^2 and use double integration.)
(b) Show that Γ(
1 2 ) =^
π. (Hint: Use the result of part (a).)
(c) Show that Γ(n +
1 2
(2n)!
n! 2^2 n
π, for n ∈ N.
Solution
(a) Let I =
0
e
−x^2 dx, then
I
2
0
e
−x^2 dx
0
e
−y^2 dy
0
0
e
−x^2 dx
e
−y^2 dy
0
0
e
−(x^2 +y^2 ) dx
dy =
∫ π 2
0
0
e
−r^2 rdr
dθ
∫ π 2
0
e
−r^2
∞
0
dθ =
∫ π 2
0
dθ =
π
4
and therefore I =
0
e
−x^2 dx =
π
2
(b)
0
e
−t √ t
dt = 2
0
e
−x^2 dx = 2(
π/2) =
π.
(c) Proof #1 (direct). From the recurrence relation Γ(μ + 1) = μΓ(μ) we get
π, Γ(
π,
π, Γ(
π,
2 n + 1
2
2 n − 1
2
2 n − 1
2
(2n − 1) · (2n − 3) · · · 5 · 3 · 1
2 n
π
2 n · (2n − 1) · (2n − 2) · (2n − 3) · · · 5 · 4 · 3 · 2 · 1
[2n · (2n − 2) · (2n − 4) · · · 4 · 2] 2n
π
(2n)!
2 n[n · (n − 1) · (n − 2) · · · 2 · 1] 2n
π =
(2n)!
22 nn!
π
Proof #2 (by induction). For n = 0 we have Γ(
0! 2^0
π =
π.
Assume the formula holds for n = m, i.e. Γ(m +
1 2 ) =^
(2m)!
m! 2^2 m
π.
Examine the formula Γ(n +
1 2
) for n = m + 1:
Γ(m + 1 +
) = (m +
)Γ(m +
2 m + 1
2
(2m)!
m! 2^2 m
π =
2 m + 2
2(m + 1)
2 m + 1
2
(2m)!
m! 2^2 m
π
(2m + 2)!
(m + 1)m! 2^2 m+
π =
[2(m + 1)]!
(m + 1)! 22(m+1)
π.
Thus formula holds for n = m + 1 and hence, by induction, it holds for all n ∈ Z
.
- Show that Bessel functions of the first kind satisfy the following recurrence relations:
(a)
d
dx
(x
−μ Jμ(x)) = −x
−μ Jμ+1(x);
(b)
d
dx
(x
μ Jμ(x)) = x
μ Jμ− 1 (x);
(c)
2 μ
x
Jμ(x) = Jμ− 1 (x) + Jμ+1(x).
Solution
(a)
d
dx
(x
−μ Jμ(x)) =
d
dx
∑^ ∞
k=
(−1)kx^2 k
2 μ+2k^ k! Γ(μ + k + 1)
∑^ ∞
k=
(−1)k 2 k x^2 k−^1
2 μ+2k^ k! Γ(μ + k + 1)
∑^ ∞
k=
k 2 k x
2 k− 1
2 μ+2k^ k(k − 1)! Γ(μ + k + 1)
∑^ ∞
k=
k x
2 k− 1
2 μ+2k−^1 (k − 1)! Γ(μ + k + 1)
∑^ ∞
k=
(−1)k+1^ x^2 k+
2 μ+2k+1^ k! Γ(μ + k + 2)
= −x
−μ
∑^ ∞
k=
(−1)k^ x(μ+1)+2k
2 (μ+1)+2k^ k! Γ((μ + 1) + k + 1)
= −x
−μ Jμ+1(x).
(b)
d
dx
(x
μ Jμ(x)) =
d
dx
∑^ ∞
k=
(−1)kx2(μ+k)
2 μ+2k^ k! Γ(μ + k + 1)
∑^ ∞
k=
(−1)k2(μ + k) x2(μ+k)−^1
2 μ+2k^ k! Γ(μ + k + 1)
∑^ ∞
k=
k 2(μ + k) x
2(μ+k)− 1
2 μ+2k^ k! (μ + k)Γ(μ + k)
∑^ ∞
k=
k x
2(μ+k)− 1
2 μ+2k−^1 k! Γ(μ + k)
∑^ ∞
k=
(−1)k^ xμ+(μ−1)+2k
2 (μ−1)+2k^ k! Γ((μ − 1) + k + 1)
= x
μ
∑^ ∞
k=
(−1)k^ x(μ−1)+2k
2 (μ−1)+2k^ k! Γ((μ − 1) + k + 1)
= x
μ Jμ− 1 (x).
(c) From the results of parts (a) and (b) we have
Jμ− 1 (x) = x
−μ d
dx
(x
μ Jμ(x)) = x
−μ (x
μ J
′ μ(x) +^ μx
μ− 1 Jμ(x)) = J
′ μ(x) +^
μ
x
Jμ(x),
Jμ+1(x) = −x
μ d
dx
(x
−μ Jμ(x)) = −x
μ (x
−μ J
′ μ(x)^ −^ μx
−μ− 1 Jμ(x)) = −J
′ μ(x) +^
μ
x
Jμ(x).
Adding these equations gives the result: Jμ− 1 (x) + Jμ+1(x) =
2 μ
x
Jμ(x).
- Show that
(a) J 1 2
(x) =
πx
sin x; (b) J− 1 2
(x) =
πx
cos x; (c) J 3 2
(x) =
πx
sin x
x
− cos x
Solution
Now, going back to the original question, we get
0
xJ
2 n(^
αnm
ℓ
x) dx =
2
α^2 nm
∫ (^) αnm
0
sJ
2 n(s)^ ds^ =^
2
2 α^2 nm
[
s
2 J
′ 2 n (s)^ −^ (n
2 − s
2 )J
2 n(s)
]
αnm
0
ℓ^2
2 α^2 nm
[
α
2 nmJ
′ 2 n (αnm)^ −^ (n
2 − α
2 nm)J
2 n(αnm) +^ n
2 J
2 n(0)
]
2
J
′ 2 n (αnm)^ (since^ Jn(αnm) = 0,^ and^ Jn(0) = 0 for^ n^ >^ 1).
Using the formula
d
dx
(x
−μ Jμ(x)) = −x
−μ Jμ 1 (x) with μ = n results in
J
′ n(x) =^
n
x
Jn(x) − Jn+1(x), and, in particular J
′ n(αnm) =^ −Jn+1(αnm).
The result follows: ∫ (^) ℓ
xJ
2 n(^
αnm
ℓ
x) dx =
2
J
2 n+1(αnm).
