Math 1321
Week 3 Lab Worksheet
Due Thursday 01/31
1. Warmup:
Find the radius of convergence and the interval of convergence of the
following series.
∞
X
xn
nn
n=1
Solution: By the ratio Test,
xn+1
nn
an+1
= lim
.
n→∞ (n + 1)n+1 xn
n→∞ an
x
1
= lim
.
n→∞ n + 1 (1 + 1 )n
n
lim
=0
for any x
The radius of convergence is ∞ and the interval of convergence is R.
2. Let
P∞
n=1
an and
P∞
n=1 bn
be convergent series of positive terms.
(a) Why is it true that an < 1 for large values of n?
(b) Using part(a). show that an bn < bn for large values of n.
P
(c) Show that ∞
n=1 an bn converges.
P
2
(d) Using part(c) show that ∞
n=1 an converges.
Solution:P
(a) Since ∞
n=1 an converges, an → 0 as n → ∞, so an < 1 for large values of n
(b) Since the terms are all positive, if an < 1, then an bn < bn .
P
P∞
(c) Since ∞
n=1 bn converges and 0 < an bn < bn ,
n=1 an bn converges
P∞
P∞
P
(d) Use the series n=1 an in place of n=1 bn in part (c), since ∞
n=1 an converges.
3. Radiation from the Stars: Any object emits radiation when heated. A blackbody is a
system that absorbs all the radiation that falls on it. For instance, a matte black surface
or a large cavity with a small hole in its wall (like a blastfurnace) is a blackbody and
emits blackbody radiation. Even the radiation from the sun is close to being blackbody
radiation.
Proposed in the late 19th century, the Rayleigh-Jeans Law expresses the energy density
of blackbody radiation of wavelength λ as
f (λ) =
8πkT
λ4
where λ is measured in meters, T is the temperature in kelvins (K), and k is Boltzmann’s
constant. The Rayleigh-Jeans Law agrees with experimental measurements for long
wavelengths but disagrees drastically for short wavelengths. [The law predicts that
f (λ) → ∞ as λ → 0+ but experiments have shown that f (λ) → 0.] This fact is known
as the ultraviolet catastrophe.
In 1900 Max Planck found a better model (known now as Planck’s Law) for blackbody
radiation:
f (λ) =
8πhcλ−5
ehc/(λkT ) − 1
where λ is measured in meters, T is the temperature (in kelvins), and
h = Planck’s Constant = 6.6262 × 10−34 J.s
c = Speed of light = 2.997925 × 108 m/s
k = Boltzmann’s constant = 1.3807 × 10−23 J/K
(a) Use Taylor’s polynomial to expand ex up to the 4th order. What values of x gives
an approximation error of at most 10−4 ? Assume 0 < x ≤ 1.
Expand ehc/(λkT ) by using the Taylor’s polynomial you just derived. What values
of λ will achieve the same accuracy? Use T = 5700K (the temperature of the sun)
(b) Use a Taylor polynomial to show that, for large wavelengths, Planck’s Law gives
approximately the same values as the Rayleigh-Jeans Law. (Hint: Use the Taylor’s
Polynomial approximation of ehc/(λkT ) )
(c) Graph f as given by both laws (preferably on the same plot) and comment on
the similarities and differences. Use T = 5700K (the temperature of the sun).
(You may want to change from meters to the more convenient unit of micrometers:
1µm = 10−6 m.)
(d) Use your graph in part(c) to estimate the value of λ for which f (λ) is a maximum
under Planck’s Law.
Solution:
(a) The Taylor’s polynomial representation for the exponential function is given by
the following:
x 2 x3 x4
+
+
+ . . . for all x
ex = 1 + x +
2!
3!
4!
Taylor’s inequality is given by:
| Rn (x) |≤
M
| x − a |n+1
(n + 1)!
for | f n+1 (x) |≤ M
We know that the nth derivative of ex is itself. So,
f n (x) = ex
Notice that ex is an increasing function in (0, 1]. So we can bound the derivative as,
f n (x) = ex ≤ e1
so, if we require the following then we know how accurate the approximation will be.
| R4 (x) |≤
e1
| x |5 < 10−4
5!
10−4 5!
e1
=⇒ | x |≤ 0.3380
=⇒ | x |5 ≤
Consider the function ehc/(λkT ) . Find the Taylor’s polynomial by substituting x =
hc/(λkT )
ehc/λkT = 1 + (hc/λkT ) +
(hc/λkT )2 (hc/λkT )3 (hc/λkT )4
+
+
+ ...
2!
3!
4!
Therefore,
hc
|≤ 0.3380
kT λ
(6.6262 × 10−34 )(2.997925 × 108 )
=⇒ λ ≥
(0.3380)(1.3807 × 10−23 )(5700)
=⇒ λ ≥ 7.4668µm
|
(b) We know that the Taylor’s polynomial representation for the exponential function
is given by the following:
ex = 1 + x +
x 2 x3 x4
+
+
+ ...
2!
3!
4!
for all x
Write the Taylor polynomial for the expression ehc/(λkT )
ehc/λkT = 1 + (hc/λkT ) +
(hc/λkT )2 (hc/λkT )3 (hc/λkT )4
+
+
+ ...
2!
3!
4!
Let’s only consider the polynomial up to the first term, we have:
8πhcλ−5
ehc/(λkT ) − 1
8πhcλ−5
≈
1 + (hc/λkT ) − 1
8πkT
≈
λ4
f (λ) =
(c)
−22
Rayleigh−Jeans Law
f(λ)
x 10
1
0
10
20
30
40
5
12
50
λ(m)
60
70
80
90
100
Planck’s Law
x 10
X: 5.15e−07
Y: 1.033e+06
10
f(λ)
8
6
4
2
0
0
0.5
(d) λmax = 0.515µm .
1
1.5
λ(m)
2
2.5
3
−6
x 10