Author Archive | yuliy

## problems for review

1. Consider the language $$L$$ consisting of all words in alphabet a,b,c having no subwords aaa. Find the generating function
$$f(x)=\sum |L_n| z^n,$$
where $$L_n$$ is the set of words of lengths $$n$$ in $$L$$. Estimate how fast $$\# ## midterm 2 \(\def\pd{\partial} \def\Real{\mathbb{R}}$$

1. Consider the closed cycle $$\gamma:[0,1]\to\Real^2$$ shown on the left (we assume that $$\gamma(0)=\gamma(1)$$ is the dot on the positive $$x$$-axis). Define the functions
$I_-(t)=\int_0^t \left(\frac{(x+1)dy -ydx}{(x+1)^2+y^2}.\dot{\gamma}(t)\right)dt,\quad I_+(t)=\int_0^t \left(\frac{(x-1)dy -ydx}{(x-1)^2+y^2}.\dot{\gamma}(t)\right)dt,$
given as partial integrals over the

## Stokes and relatives

• The (general) Stokes Theorem: this unique theorem puts
Green’s/Gauss/Stokes Theorems (from intermediate calculus)
under one umbrella. Let $$C$$ be a polyhedron with boundary $$\partial C$$. Then
$${\int_{\partial C} \omega = \int_C d\omega }$$
• Example: integrate the form $$\omega = • ## solutions for problems on vector analysis and differential forms Solutions Exercise: Assume that 4 vectors in \(\mathbb{R}^3$$ satisfy $$A+B+C+D$$=0.
Simplify
$$A\times B-B\times C+C \times D-D\times A$$

Solution: By the anti-symmetry $$-B\times C = C\times B$$
and $$-D\times A = A\times D$$ we can rewrite the expression as

## 2.28 operators in Hermitian spaces and their spectra.

$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}} \def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}} \def\eye{\left(\begin{array}{cc}1&amp;0\\0&amp;1\end{array}\right)} \def\bra#1{\langle #1|}\def\ket#1{|#1\rangle}\def\j{\mathbf{j}}\def\dim{\mathrm{dim}} \def\ker{\mathbf{ker}}\def\im{\mathbf{im}} \def\tr{\mathrm{tr\,}} \def\braket#1#2{\langle #1|#2\rangle}$$

1.
Given a bilinear form $$Q(\cdot,\cdot)$$, or, equivalently, a mapping $$Q:U\to U^*$$, one can bake easily new bilinear forms from linear operators $$U\to U$$: just take $$Q_A(u,v):=Q(u,Av)$$.

This opens …