Math 5C, Summer 2011 Take-Home Midterm Due August 23, 2011 in Lecture Name 1 Pe rm No. 1 Name 2 Perm No. 2 Style Substance Total Directions: 1. Attach this page as a coversheet to your paper. 2. Y ou may work in groups of 2 and use course materi als—including th e T A and mysel f, but no other resources are allowed. Especially the internet! 3. There are 30 points on this exam. 4. Substance: Eac h part of each problem is worth 1 point for corr ectne ss—20 poin ts possible in total . 5. Sty le: There are 10 point s for ove rall presen tation; see belo w. 6. First and foremost your writin g shoul d be in fully develope d sent ences ; a mathematical formula should be treated as part of a sentence. Bad: calculus fact: fg = f g − g f→ f= 0 Good: From the product rule ( fg) = fg + g fwe can deduce that fg = f g − g f, which is known as ‘integration by parts.’ Our remarks from before imply that fis identically 0. 7. It is encouraged that you craft your midterm in to a coherent piece of prose, though you may simply answ er parts (a), (b), etc. if you wish. 8. Final ly , I’m unashamedly a sucker for mathema tica l aesthetic. Your final product should b e on crisp beautiful paper with nicely drawn symbols and handwriting (or computer typesetting if you’re super-cool). Remember the old saying: If momma [the grader] ain’t happy, ain’t nobody happy.
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Math 5C, Summer 2011
Take-Home Midterm
Due August 23, 2011 in Lecture
Name 1
Perm No. 1
Name 2
Perm No. 2
Style
Substance
Total
Directions:
1. Attach this page as a coversheet to your paper.
2. You may work in groups of 2 and use course materials—including the TA and myself, but no other resourcesare allowed. Especially the internet!
3. There are 30 points on this exam.
4. Substance: Each part of each problem is worth 1 point for correctness—20 points possible in total.
5. Style: There are 10 points for overall presentation; see below.
6. First and foremost your writing should be in fully developed sentences; a mathematical formula should betreated as part of a sentence.
Bad:
calculus fact:
f g = f g −
gf → f = 0
Good: From the product rule (f g) = f g + gf we can deduce that f g = f g −
gf,
which is known as ‘integration by parts.’ Our remarks from before imply that f is identically 0.
7. It is encouraged that you craft your midterm into a coherent piece of prose, though you may simply answerparts (a), (b), etc. if you wish.
8. Finally, I’m unashamedly a sucker for mathematical aesthetic. Your final product should be on crispbeautiful paper with nicely drawn symbols and handwriting (or computer typesetting if you’re super-cool).Remember the old saying: If momma [the grader] ain’t happy, ain’t nobody happy.
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Problem 1
In this problem we’re going to understand ∆ = · = ∂ 2x + ∂ 2y a bit better. Let D be the unit disk in R2
centered at the origin.
1. Use derivative rules to compute ∆(ln(x2 + y2)).
Something should look wrong; the last three questions seem to contradict each other. The problem is thatln(x2 + y2) has a singularity at the origin, so our goal is to understand what’s going on. For this we’regoing to need a smooth ‘test function’ φ. Assume that φ is zero on and near ∂D; in particular, φ and allof its derivatives are zero on the boundary of D.
As seen in the picture, B(a) is the annulus with outer radius 1. We can think of the annulus as a unit diskD with a smaller disc A(a), who’s boundary is C(a), taken out of it. Thus B(a) = D - A(a). Thus, if wethink about the boundary of B(a), we can say: ∂ B(a) = ∂ D - ∂ A(a) = ∂ D - C(a).
The subtraction of C(a) comes from the orientation of the inner circle and the normal pointing in theopposite direction of the outer boundary.
8. As a → 0 the continuity of φ means that φ ≈ φ(0, 0) (a constant) all along C (a). On the other hand ∂φ/∂nchanges values wildly along C (a), but for some constant M we have |∂φ/∂n| < M . Use these facts toevaluate
lima→0
C (a)
φ
∂
∂nln(x2 + y2)− ln(x2 + y2)
∂
∂nφ
ds.
To make the computations easier, notice that for a circle centered at the origin ∂/∂n = ∂/∂r, the derivativewith respect to the polar coordinate r.
10. The Dirac delta function δ is not really a function; nonetheless we pretend like it is a function and defineit by the property that for any function f ,
This shows that ∆ ln(x2 + y2) acts similarly to the Dirac delta function: when taking the integral over Dof a function multiplied by ∆ ln(x2 + y2), we get that function evaluated at (0,0) multiplied by 4 π. We canthen relate ∆ ln(x2 + y2) and δ as:
∆ln(x2 + y2) = 4πδ
Checking that this makes sense, we reevaluate the previous integral using the properties of the Dirac deltafunction:
Using the property of the Dirac delta function described in problem 10, we know: B
∆(1/
x2 + y2 + z2)dV =
B
CδdV = C (0, 0) = C
Since C is a constant function, finding C at any point will tell us what C is at all points. Thus we solve theintegral to find C. We use divergence theorem to change the integral from B to the boundary.
B
∆(1/ x2 + y2 + z2)dV = B
· ((1/ x2 + y2 + z2))dV = ∂B
(1/ x2 + y2 + z2) · d A
=
∂B
−x
(x2 + y2 + z2)3/2,
−y
(x2 + y2 + z2)3/2,
−z
(x2 + y2 + z2)3/2
· (x, y, z)dσ
=
∂B
−x2 + y2 + z2
(x2 + y2 + z2)3/2dσ
Converting to polar coordinates, and knowing on the boundary of B, r = 1, we solve the integral.
What we’ve done in this problem is shown that (1/2π) ln
x2 + y2 is the electrostatic potential of a point chargein R2. In general if the density of charge in space is ρ, then the electrostatic potential V solves ∆V = ρ. TheDirac delta is simply the charge density of a point charge. In R
3 the potential of a point charge is proportionalto 1/
x2 + y2 + z2; if we write that instead as 1/r we get the (probably familiar) formula
V =kQ
r,
where k and Q are just a constant and unit of charge respectively. This is the beginning of electrostatics as wellas the mathematical subject of potential theory.
Problem 2
In this problem we’re going to use series to solve a problem in statistical physics and thermodynamics. Anyobject whose temperature is above absolute zero will emit thermal radiation—that is, electromagnetic radiationgenerated by the internal thermal energy of the object. This phenomenon is probably familiar; any object whichis raised to a high enough temperature will glow ‘red-hot’ (in fact, other colors are possible too). Around theturn of the 20th century, understanding the underlying physics of thermal radiation from known principles wasa big problem. When Lord Rayleigh attempted to derive the amount of electromagnetic radiation a ‘blackbody’would emit at a certain frequency, a so-called ultraviolet catastrophe occurred: the rate at which the body emits
energy would beP ∝
∞
0
dλ
λ4= ∞
(a nice instance of a divergent integral!). By looking at experimental data, Max Planck was able to deduce amore accurate law. Though he amazingly guessed a complicated formula from the data, he later derived it byassuming energy came in discrete increments—this was the first time energy quantization and Planck’s constanth appeared in physics. Accepting something as crazy as energy quantization requires some convincing. We’d liketo compute the same rate of energy flow that caused Rayleigh’s law to fall apart. In doing so, we’ll derive theso-called Stefan-Boltzmann law of blackbody radiation.
1. Let’s cut to the chase. After a derivation starting from Planck’s law, the rate of energy flow (or ‘power’)via thermal radiation for a blackbody is given by the formula
P = 2πhAc2
∞
0
ν 3
exp(hν/kT )− 1dν,
where A is the surface area of the object, c is the speed of light, k is Boltzmann’s constant, and T isthe temperature of the object (recall the notation exp t = et). Let’s remove all physical content from theintegral by substituting x = hν/kT . From here deduce the Stefan-Boltzmann law: how does P depend onT ? Your expression for P will still contain the dimensionless integral
At this point we could say that we’re done. However, a great test of Planck’s theory comes from evaluatingthe integral and comparing numbers against experimental results. We’ll stop the story of the physicshere and focus on evaluating the integral. Finding an antiderivative is out of the question and a numerictreatment would be arduous, but we can use series to evaluate the integral explicitly!
Assuming the monotone convergence theorem, we can say that:
gn(x) = f 1(x) + f 2(x) + ... + f m(x) =mi=1
f i(x)
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Where all f i(x) are continuous and defined on the interval [a,b].We will assume that all f i(x) ≥ 0 for any x ∈ [a, b].Since gn(x) is monotonically increasing:
Let’s think of the p series as a function of p; define the Riemann zeta function
ζ ( p) =
∞n=1
1
n p.
This function is ubiquitous in number theory and seemingly random areas of science such as quantummechanics. We’ll only need one important fact about it.
7. We already know that ζ (1) diverges; now let’s find two more values of ζ . Euler came up with the following
formula for the sine function:
x−x3
6+
x5
120+ · · · = sin x = x
1−
x2
π2
1−
x2
4π2
1−
x2
9π2
· · ·
Compare the coefficients of x3 and x5 in these expressions to evaluate ζ (2) and ζ (4).
in terms of the gamma function and the Riemann zeta function. Be sure to check the hypotheses of thetheorem! What is the value of the integral in the Stefan-Boltzmann law?
I am morally obligated to mention the Riemann hypothesis, which is math’s biggest unsolved problem. Usingformulas such as
ζ ( p) = 2(2π) p−1 sin(πp/2)Γ(1− p)ζ (1− p)
ζ can be defined for any complex number—except 1 of course. For technical reasons, people care about the zeroesof ζ . Here is the Riemann hypothesis: for complex numbers a + bi with 0 < a < 1 is it TRUE or FALSE that
ζ (a + bi) = 0 =⇒ a = 1/2
Millions of dollars and eternal glory await the person who can solve this problem.