Class Exercise

BGU Physics Dept. Introduction to Mathematical Methods in Physics
Class Exercise
Summary
1. Use de Moiver’s theorem to prove that
tan 5θ =
t5 − 10t3 + 5t
5t4 − 10t2 + 1
where t = tan θ. Deduce the values of tan(nπ/10) for n = 1, 2, 3, 4.
Solution
cos(5θ) + i sin(5θ) = (cos θ + i sin θ)5
= cos5 θ + 5i cos4 θ sin θ − 10 cos3 θ sin2 θ − 10i cos2 θ sin3 θ + 5 cos θ sin4 θ + i sin5 θ
sin(5θ) = 5 cos4 θ sin θ − 10 cos2 θ sin3 θ + sin5 θ
cos(5θ) = cos θ5 − 10 cos3 θ sin2 θ + 5 cos θ sin4 θ
tan(5θ) =
sin(5θ)
=
cos(5θ)
sin(5θ)
cos5 θ
cos(5θ)
cos5 θ
=
tan5 θ − 10 tan3 θ + 5 tan θ
5 tan4 θ − 10 tan2 θ + 1
Finding the values of tan(nπ/10):
For n = 1 we set θ = π/10 so tan(5θ) = tan(π/2) → ∞ which means that the denominator is
zero
s
r
8
5t4 − 10t2 + 1 = 0 ⇒ t = ± 1 ±
10
we are interested in the positive solution (since both sin θ and cos θ is positive for 0 < θ < π/2,
and in t < 1 since sin θ is smaller then cosθ for 0 < θ < π/4. So
s
r
8
tan(π/10) = 1 −
.
10
For n = 3 we set θ = 3π/10 so tan(5θ) = tan(3π/2) → ∞ which means that the denominator
is zero again, only this time we expect t > 1 since pi/4 < θ < π/2, hence
s
r
8
tan(3π/10) = 1 +
.
10
For n = 2 and n = 4 we have tan(5θ) = 0 so now the nominator is zero
q
√
5
3
t − 10t + 5t = 0 ⇒ t = 0, t = ± 5 ± 2 5
So
q
√
tan(2π/10) = 5 − 2 5
q
√
tan(4π/10) = 5 + 2 5
1
2. By writing z = tan w in terms of exponentials show that
1
1 + iz
−1
tan z =
ln
2i
1 − iz
Solution
1 e2iw − 1
sin w
1 eiw − e−iw
=
=
cos w
i eiw + e−iw
i e2iw + 1
1 + iz
1
1 + iz
2iw
2iw
2iw
⇒ (e − 1)iz = (e + 1) ⇒ e
=
⇒w=
ln
1 − iz
2i
1 − iz
z = tan w =
3. Identify the series
∞
X
(−1)n+1 x2n
n=1
(2n − 1)!
,
and then, by integration and differntiantion, deduce the values S of the following series:
(a)
(b)
(c)
(d)
∞
X
(−1)n+1 n2
n=1
∞
X
n=1
∞
X
n=1
∞
X
n=1
(2n)!
(−1)n+1 n
(2n + 1)!
(−1)n+1 nπ 2n
4n (2n − 1)!
(−1)n (n + 1)π 2n
(2n)!
Solution
f (x) =
∞
X
(−1)n+1 x2n
n=1
∞
X
=x
n=1
(2n − 1)!
(−1)n+1 x2n−1
(2n − 1)!
x3 x5
=x x−
+
+ ...
3!
5!
= x sin(x)
2
(a) S =
∞
X
(−1)n+1 n2
n=1
(2n)!
0
f (x) =
f 00 (x) =
=
∞
X
(−1)n+1 2nx2n−1
n=1
∞
X
n=1
∞
X
n=1
(2n − 1)!
(−1)n+1 2n(2n − 1)x2n−2
(2n − 1)!
∞
(−1)n+1 4n2 x2n−2 X (−1)n+1 2nx2n−2
−
(2n − 1)!
(2n − 1)!
n=1
f 00 (1) + f 0 (1)
f 00 (1) = 4S − f 0 (1) ⇒ S =
4
f 00 (x) + f 0 (x)
1
f (x) = x sin x ⇒
= (2 cos x + x cos x + sin x − x sin x)
4
4
3 cos(1)
f 00 (1) + f 0 (1)
=
S=
4
4
(b) S =
∞
X
(−1)n+1 n
n=1
(2n + 1)!
0
f (x) =
∞
X
(−1)n+1 2nx2n−1
n=1
∞
X
=2
=2
n0 =0
∞
X
(2n − 1)!
0
0
(−1)n +2 (n0 + 1)x2n +1
(2n0 + 1)!
0
∞
0
0
0
0
(−1)n +1 n0 x2n +1
= −2
n0 =0
∞
X
n0 =1
(2n0 + 1)!
n =0
∞
X
+2
n=1
0
0
(−1)n +1 n0 x2n +1
(2n0 + 1)!
(−1)n+1 x2n−1
(2n − 1)!
+ 2f (x)
f 0 (1) = −2S + 2f (1)
S=
(c) S =
1
2f (1) − f 0 (1)
= (− cos(1) + sin(1)))
2
2
∞
X
(−1)n+1 nπ 2n
n=1
4n (2n − 1)!
f 0 (x) =
1 0
xf (x) =
2
S=
0
X (−1)n +2 x2n +1
(−1)n +2 n0 x2n +1
+
2
(2n0 + 1)!
(2n0 + 1)!
0
n0 =0
∞
X
= −2
= n = n0 + 1
∞
X
(−1)n+1 2nx2n−1
n=1
∞
X
n=1
∞
X
n=1
(2n − 1)!
(−1)n+1 nx2n
(2n − 1)!
(−1)n+1 nπ 2n
1π 0
π
=
f (π/2) =
n
4 (2n − 1)!
22
4
3
(d)
∞
X
(−1)n (n + 1)π 2n
(2n)!
n=1
00
f (x) =
=
∞
X
(−1)n+1 2n(2n − 1)x2n−2
n=1
∞
X
(2n − 1)!
(−1)n+1 2nx2n−2 = n = n0 + 1
(2n − 2)!
n=1
∞
X
0
=2
n0 =0
(−1)n +2 (n0 + 1)x2n
(2n0 )!
0
0
0
∞
X
(−1)n +1 (n0 + 1)x2n
=2 x−
(2n0 )!
0
"
#
n =1
f 00 (π) = 2 [π − S]
f 00 (π)
=1+π
2
S=π−
4. By noting that for 0 ≤ η ≤ 1, η 1/2 ≥ η 3/4 ≥ η prove that
Z a
2
1
π
≤ 5/2
(a2 − x2 )3/4 dx ≤
3
4
a
0
Solution
1
Z
a
2
2 3/4
1
Z
1
2 3/4
Z
1
dx = {x = at} 5/2
a (1 − t ) adt =
(1 − t2 )3/4 dt
a5/2 0
a
0
0
Z 1
Z 1
Z 1
0 < t < 1 ⇒ (1 − t2 ) < 1 ⇒
(1 − t2 )dt <
(1 − t2 )3/4 dt <
(1 − t2 )1/2 dt
0
0
0
Z 1
1
2
(1 − t2 )dt = 1 − =
3
3
0
Z 1
Z π/2
Z π/2
π
2 1/2
2
1/2
(1 − t ) dt =
(1 − sin z) cos zdz =
cos2 zdz =
4
0
0
0
(a − x )
3/2
5. The gamma function Γ(n) is defined for all n > 1 by
Z ∞
Γ(n + 1) =
xn e−x dx
0
(a) Find a recurrence relation connecting Γ(n + 1) and Γ(n).
(b) Deduce the value of Γ(n + 1) when n is a non-negative integer.
√
(c) Find the value of Γ(7/2), given that Γ(1/2) = π.
(d) Taking factorial m for any m to be defined by m! = Γ(m + 1), evaluate (3/2)!
Solution
(a)
Z
Γ(n + 1) =
0
∞
∞
xn e−x dx = − xn e−x 0 +
4
Z
0
∞
nxn−1 e−x dx = nΓ(n)
(b)
Z
∞
n = 0 ⇒ Γ(1) =
e−x dx = 1
0
n = k ⇒ Γ(k + 1) = kΓ(k) = k(k − 1)Γ(k − 1) = k · (k − 1) · · · 3 · 2 · 1 · Γ(1) = k!
(c) Find the value of Γ(7/2), given that Γ(1/2) =
√
π.
(d) Taking factorial m for any m to be defined by m! = Γ(m + 1), evaluate (3/2)!
6. Evaluate the Laplacian of the function
ψ(x, y, z) =
x2
zx2
+ y2 + z2
(a) Directly in Cartesian coordinates.
(b) After changing to a spherical polar coordinate system.
(c) Verify that, as they must, the two methods give the same result.
Solution
(a) ∇2 f (x, y, z) =
2z (−4x2 +y 2 +z 2 )
(x2 +y 2 +z 2 )2
(b) f˜(r, θ, ϕ) = f (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ) = r cos θ sin2 θ cos2 ϕ
cos θ(2 cos2 θ−(3+5 cos 2ϕ) sin2 θ)
∇2 f˜(r, θ, ϕ) =
r
7. Maxwells equations for electromagnetism in free space can be written as
~ =0
(i)∇ · B
~ =0
(ii)∇ · E
~
~ = 1 ∂E
(iii)∇ × B
2
c ∂t
~
~ = − ∂B
(iv)∇ × E
∂t
~ is defined by B
~ = ∇ × A,
~ and a scalar ϕ by E
~ = −∇ · ϕ −
A vector A
condition
~
∂A
∂t .
Show that if the
~ + 1 ∂ϕ = 0
(v)∇ · A
c2 ∂t
~ and ϕ satisfy wave equations
is imposed (this is known as choosing the Lorentz gauge), then A
as follows:
1 ∂2ϕ
=0
c2 ∂t2
2~
~− 1 ∂ A =0
(vii)∇2 A
c2 ∂t2
(vi)∇2 ϕ −
5
!
~
∂
A
~ = 0 ⇒ ∇2 ϕ + ∇
(ii)∇ · E
∂t
∂
~ =0
⇒ ∇2 ϕ +
∇A
∂t
1 ∂2ϕ
⇒ (v) ⇒ ∇2 ϕ − 2 2 = 0
c ∂t
~
~
∂
E
1
~ =
~ = 1 ∂E
(iii)∇ × B
⇒ ∇ × (∇ × A)
2
2
c ∂t
c ∂t
!
~
∂
A
1
∂
~ + ∇(∇ · A)
~ =
−∇ϕ −
⇒ −∇2 A
c2 ∂t
∂t
!
~
∂ϕ
1
∂
∂
A
1
~−∇
= 2
−∇ϕ −
⇒ (v) ⇒ −∇2 A
c2 ∂t
c ∂t
∂t
~−
⇒ ∇2 A
~
1 ∂2A
=0
2
2
c ∂t
if any typos or corrections found please let me know: Ben Yellin [email protected]
6