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solutions to exercises 1.19

Exercise: do the polynomials \(p_0(x)=1,~~p_1(x)=(x-1),~~p_2(x)=(x-1)^2,~~p_3=(x-1)^3\) form a basis of the vector space of cubic polynomials (with real coefficients)? If so, express \(x^3\) in this basis.


Solution: use the binomial: \(x^n = \big((x-1)+1\big)^n = \sum_{m=0}^n {n\choose m}(x-1)^m\) to infer …

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1.19 Linear spaces

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

1. Linear (or vector) spaces (over a field \(k\) – which in our case will be almost always the complex numbers) are sets that satisfy the following standard list of properties:

  • There exists
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solutions to exercises 1.17

Exercise 1: describe the curve \(1/z\), where \(z=1+ti, ~~t\in (-\infty,\infty)\).


Solution: apply \(\frac{1}{z} = \frac{\bar z}{|z|^2}\) and get \( \frac{1}{1+ti} = \frac{1-ti}{1+t^2}\), which is a parametric curve  of the form \(p(t) = x(t)+iy(t)\) where \(x(t) = \frac{1}{1+t^2},~~~ y(t) = \frac{-t}{1+t^2}\).…

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1.17 Complex numbers

\(\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}}\def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}}
\def\eye{\left(\begin{array}{cc}1&0\\0&1\end{array}\right)}
\)

Complex numbers are expressions \(z=x+\i y\), where \(\i\) is the imaginary unit, defined by \(\i^2=-1\). Note the ambiguity (\(-\i\) would work just as well).

Complex numbers can be added subtracted, multiplied the usual way, with the …

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