MA5311. Take home exam

Posted on April 20, 2017 by Fabrice Baudoin

Exercise 1. Solve Exercise 44 in Chapter 1 of the book.

Exercise 2.  Solve Exercise 3 in Chapter 1 of the book.

Exercise 3.  Solve Exercise 39 in Chapter 1 of the book.

Exercise 4. The heat kernel on \mathbb{S}^1 is given by p(t,y) =\frac{1}{2\pi}\sum_{m \in \mathbb{Z}} e^{-m^2 t} e^{im y} =\frac{1}{\sqrt{4\pi t}} \sum_{k \in \mathbb{Z}} e^{-\frac{(y -2k\pi)^2}{4t} }.

  • By using the subordination identity e^{-\tau | \alpha | } =\frac{\tau}{2\sqrt{\pi}} \int_0^{+\infty} \frac{e^{-\frac{\tau^2}{4t}-t \alpha^2}}{t^{3/2}} dt, \quad \tau \neq 0, \alpha \in \mathbb{R}, show that for \tau > 0, \frac{1+e^{-2\pi \tau}}{1-e^{-2\pi \tau}} =\frac{1}{2\pi} \sum_{k \in \mathbb{Z}} \frac{2\tau}{\tau^2+n^2}
  • The Bernoulli numbers B_k are defined via the series expansion \frac{x}{e^x -1}=\sum_{k=0}^{+\infty} B_k \frac{x^k}{k!}. By using the previous identity show that for k \in \mathbb{N}, k \neq 0, \sum_{n=1}^{+\infty} \frac{1}{n^{2k}} =(-1)^{k-1} \frac{(2\pi)^{2k} B_{2k} }{2(2k)!}.

 

Exercise 5. Show that the heat kernel on the torus \mathbb{T}^n=\mathbb{R}^n / (2 \pi \mathbb{Z})^n is given by p(t,y) = \frac{1}{(4\pi t)^{n/2}} \sum_{k \in \mathbb{Z}^n} e^{-\frac{\|y+2k\pi\|^2}{4t} }=\frac{1}{(2\pi)^n} \sum_{l\in \mathbb{Z}^n} e^{i l \cdot y -\| l \|^2 t}.

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