PHZ 3113 Fall 2010
Homework #2, Due Friday, September 10
1. Write the first five terms of the series for sin[(1 + x)1/3 − 1]. Use logical reasoning or
common sense to decide what is the radius of convergence of this series expansion.
Start with the expansion of (1+x)p (see http://www.phys.ufl.edu/ fry/3113/series.pdf),
1
1
p(p − 1) x2 + p(p − 1)(p − 2) x3
2
6
1
1
p(p − 1)(p − 2)(p − 3) x4 +
p(p − 1)(p − 2)(p − 3)(p − 4) x5 + · · · .
+
24
120
(1 + x)p = 1 + p x +
Evaluated for p = 1/3, this gives
y = (1 + x)1/3 − 1 =
1
5 3
10 4
22 5
1
x − x2 +
x −
x +
x + ···.
3
9
81
243
729
Then, as a function of y,
sin y = y −
=
=
1 3
1 5
y +
y +···
6
120
1
1
5
10 4
22 5
x − x2 + x3 −
x +
x
3
9
81
243
729
1
−
6
5
1
1 2
1
5 3 3
1
x− x + x
x
+
3
9
81
120 3
1
1
1 3
17 4
721 5
x − x2 +
x −
x +
x + ···.
3
9
18
486
29160
The series for sin y converges for any argument, and so any failure to converge must come
from the series for (1 + x)1/3 , which converges for |x| < 1.
The figure (next page) shows the exact result (solid line), the linear approximation (dotted
line, red), the quadratic approximation (short-dash line, green), the cubic approximation
(long-dash line, blue), the quartic approximation (dot-dash line, cyan), and the quintic
approximation (short-long dash, magenta). Higher order approximations are better for |x| <
1, and successively worse for |x| > 1.
2. A function p(x) is defined as
3 sin x − x(3 + x2 ) cos x
.
x4
What is the behavior of p(x) as x → 0? As x → ∞? Sketch the function p(x).
p(x) =
For small x, use the series expansions
1
1 5
1
1
sin x = x − x3 +
x + ···,
cos x = 1 − x2 + x4 + · · · .
6
120
2
24
Then,
1
1 3
1 2
1 5
1 2
1 4
3
p(x) = 4 3 x − x +
x + · · · − 3x 1 − x + x − x 1 − x
6
120
2
24
2
x
1 2 5
2x
= 4
x +··· =
+ ···.
5
5
x
In the numerator, the linear terms cancel and the cubic terms cancel, leaving a function that
is well-behaved at the origin. For large x, you can’t do anything with sin x and cos x, but
you can with the factors of 1/x,
3
1
3
1
p(x) = 4 sin x − 3 cos x − cos x ≈ cos x.
x
x
x
x
3. A function f (x) has the series expansion
∞
X
n xn
f (x) =
.
(n + 1)
n=1
Write f (x) in closed form (sum the series).
This isn’t a “known” series, but it can be obtained systematically starting from the geometric
series
∞
X
1
2
xn .
= 1+x +x +··· =
1−x
n=0
To insert factors of n, differentiate with respect to x:
d
dx
1
1−x
∞
X
=
n xn−1 .
n=0
Factors of n in the denominator come from integration; but we want factors 1/(n + 1), which
comes from integrating xn , not xn−1 . So, first multiply by x and then integrate,
Z x
0
d
x0
dx0
1
1 − x0
dx0 =
∞
X
nxn+1
.
n+1
n=0
This has xn+1 , so divide by x:
Z ∞
X
1 x 0 d
1
n xn
0
x
dx
=
.
x 0
dx0 1 − x0
n+1
n=0
The derivative step gives
d
x
dx
1
1−x
=
x
,
(1 − x)2
and the integral step results in
Z
Z
1 x x0 dx0
1
1 x [(x0 − 1) + 1] dx0
1
+ ln(1 − x).
=
=
0
2
0
2
x 0 (1 − x )
x 0
1−x x
(x − 1)
Having obtained this result, it is easy to see that the logarithm term subtracts 1/(n+1) from
the coefficient 1 of xn in the geometric series. I suppose you can intuit this and then show
that it gives the right series, but that isn’t really a reliable procedure (“wait for inspiration”).
4. The functions sinh x and cosh x are defined as
1
sinh x = (ex − e−x ),
2
1
cosh x = (ex + e−x ).
2
Write the first four nonvanishing terms of the series expansions for sinh x and cosh x. Write
the series for 1/ cosh x. Use this obtain the first four terms of the series for tanh x.
The series for e±x are
1 x4 + · · · .
e±x = 1 ± x + 12 x2 ± 16 x3 + 24
The functions cosh x and sinh x involve the sum and difference of these. In the sum, the odd
terms cancel and the even terms reinforce; while in the difference the even terms cancel and
the odd terms reinforce, leaving
1
1
1 6
cosh x = 1 + x2 + x4 +
x ;
2
24
720
1
1 5
1 7
x +
x .
sinh x = x + x3 +
6
120
5040
The series for 1/ cosh x is
1
1
=
1
1
1 x6 + · · ·)
cosh x
1 + ( 2 x2 + 24 x4 + 720
1 x4 + 1 x6 ) + ( 1 x2 + 1 x4 )2 − ( 1 x2 )3 + · · ·
= 1 − ( 12 x2 + 24
720
2
24
2
5 x4 − 61 x6 + · · · .
= 1 − 12 x2 + 24
720
The series for tanh x is
1 x5 + 1 x7 × 1 − 1 x2 + 5 x4 − 61 x6
tanh x = x + 16 x3 + 120
5040
2
24
720
2
17 7
1
x +···
= x − x3 + x5 −
3
15
315
5. A rope with length one foot longer than the circumference of the earth is looped around
the equator and stretched tight from one point. What is the height of this point above the
earth’s surface? Take R⊕ = 4000 mi, take 1 mi = 5000 ft.
The taut rope lies mostly along the surface of the earth, but departs over a small arc ±θ
near the peak. The two segments of rope form legs of two right triangles, as in the figure.
The length of the two straight segments is 2R tan θ, the length along the surface is 2Rθ, and
so the excess length is
2R (tan θ − θ) = ∆s.
The excess height is
h=
R
− R.
cos θ
The solution is going to involve a small angle, so expand,
1
1 3
tan θ − θ = θ + θ + · · · − θ ≈ θ3 ,
3
3
θ3 =
3∆s
;
2R
h
i 1
3∆s 2/3
1
2
2
1
h ≈ R (1 + 2 θ + · · ·) − 1 ≈ R θ = R
2
2
2R
2/3
3(1 ft)
1
= (4000)(5000 ft)
2
2(4000)(5000 ft)
= (107 ft)(75 × 10−9)2/3 = (107 ft)(17.8 × 10−6 ) = 178 ft.
The distance from the high point to the points where the rope reaches the earth (the horizon
distance from that height) is about 17 miles.