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,

n+1 n an+1 x n lim = lim n+1 . n→∞ an n→∞ (n + 1) xn x 1 = lim . n→∞ 1 n n + 1 (1 + n ) = 0 for any x

The radius of convergence is ∞ and the interval of convergence is R.

P∞ P∞ 2. Let n=1 an and 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 anbn < bn for large values of n. P∞ (c) Show that n=1 anbn 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 anbn < bn. P∞ P∞ (c) Since n=1 bn converges and 0 < anbn < bn, n=1 anbn 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

8πkT f(λ) = 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:

8πhcλ−5 f(λ) = ehc/(λkT ) − 1 where λ is measured in meters, T is the temperature (in kelvins), and

h = Planck’s Constant = 6.6262 × 10−34J.s c = Speed of light = 2.997925 × 108m/s k = Boltzmann’s constant = 1.3807 × 10−23J/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 = 10−6m.) (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: x2 x3 x4 ex = 1 + x + + + + ... for all x 2! 3! 4! Taylor’s inequality is given by: M | R (x) |≤ | x − a |n+1 for | f n+1(x) |≤ M n (n + 1)!

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.

e1 | R (x) |≤ | x |5< 10−4 4 5! 10−45! =⇒ | x |5≤ e1 =⇒ | x |≤ 0.3380

Consider the function ehc/(λkT ). Find the Taylor’s polynomial by substituting x = hc/(λkT )

(hc/λkT )2 (hc/λkT )3 (hc/λkT )4 ehc/λkT = 1 + (hc/λkT ) + + + + ... 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:

x2 x3 x4 ex = 1 + x + + + + ... for all x 2! 3! 4!

Write the Taylor polynomial for the expression ehc/(λkT )

(hc/λkT )2 (hc/λkT )3 (hc/λkT )4 ehc/λkT = 1 + (hc/λkT ) + + + + ... 2! 3! 4!

Let’s only consider the polynomial up to the ﬁrst term, we have:

8πhcλ−5 f(λ) = ehc/(λkT ) − 1 8πhcλ−5 ≈ 1 + (hc/λkT ) − 1 8πkT ≈ λ4

(c)

−22 x 10 Rayleigh−Jeans Law )

λ 1 f(

0 10 20 30 40 50 60 70 80 90 100 λ(m)

5 x 10 Planck’s Law 12 X: 5.15e−07 Y: 1.033e+06 10

8 )

λ 6 f(

0 0 0.5 1 1.5 2 2.5 3 λ −6 (m) x 10

(d) λmax = 0.515µm .