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==== Question 2 ==== | ==== Question 2 ==== | ||
+ | We apply the Cauchy-Riemann criterion: | ||
+ | \begin{align} | ||
+ | \partial_x\cos x&=-\sin x&\, \partial_x(-\sinh y)&=0\\ | ||
+ | \partial_y\cos x&=0&\, \partial_y(-\sinh y)&=-\cosh y | ||
+ | \end{align} | ||
+ | |||
+ | While one condition ($\partial_x v=-\partial_y u$) is always | ||
+ | satisfied, the other ($\partial_x u=\partial_y v$) is only for points | ||
+ | where both the cosine and the hyperbolic sine are unity, since one is | ||
+ | less than one and the other larger than one otherwise. The function is | ||
+ | therefore derivable at points $\pi/2+2k\pi$ for $k\in\mathbf{Z}$ on | ||
+ | the real line. | ||
==== Question 3 ==== | ==== Question 3 ==== |
¿Hey, whatcha doin on this page?
It's just where I put stuff that I'm experimenting on for possible f¯uture use.
Contents |
Calculate:
$$\int_{-\infty}^\infty\frac{dx}{1+x^2+x^4}\,.$$
The poles are easily find by noticing that $(1+x^2+x^4)(1-x^2)=1-x^6$, i.e., they are the sixth root of unity which are not solutions of $x^2=1$, which leaves:
\begin{equation} e^{ik\pi/3}\quad\mathrm{for\ }k\in\{1,2,4,5\} \end{equation}
We can now calculate the integral by computing the residues and summing those on the upper half-plane (times $2i\pi$):
\begin{multline} \mathrm{Res}_{k=1}\frac{1}{(z-e^{i\pi/3})(z-e^{i2\pi/3})(z-e^{-i\pi/3})(z-e^{-i2\pi/3})} =\\ \frac{1}{(e^{i\pi/3}-e^{i2\pi/3})(e^{i\pi/3}-e^{-i\pi/3})(e^{i\pi/3}-e^{-i2\pi/3})}\,. \end{multline}
The denominator is calculated easily by dealing with the geometry of the unit circle (one is a sine, the other is after computing a product, the translated sum of a number with its opposite). One thus finds the residue as:
$$1/(2i\sqrt{3}e^{i\pi/3})$$
The other residue is found similarly as
$$1/(2i\sqrt{3}e^{-i\pi/3})$$
so that the integral is finally, since $1/e^{-i\pi/3}+1/e^{i\pi/3}=e^{-i\pi/3}+e^{i\pi/3}=1$, simply:
$$\frac{\pi}{\sqrt3}$$
We apply the Cauchy-Riemann criterion:
\begin{align} \partial_x\cos x&=-\sin x&\, \partial_x(-\sinh y)&=0\\ \partial_y\cos x&=0&\, \partial_y(-\sinh y)&=-\cosh y \end{align}
While one condition ($\partial_x v=-\partial_y u$) is always satisfied, the other ($\partial_x u=\partial_y v$) is only for points where both the cosine and the hyperbolic sine are unity, since one is less than one and the other larger than one otherwise. The function is therefore derivable at points $\pi/2+2k\pi$ for $k\in\mathbf{Z}$ on the real line.
We write $\sin(z)=\sin(x+iy)$ with $x$ and $y$ real, and expand trigonometrically:
\begin{equation} \sin(x+iy)=\sin x\cosh y+i\cos x\sinh y \end{equation}
so that
\begin{align} |\sin(x+iy)|^2&=\sin^2x\cosh^2y+\cos^2x\sinh^2y\\ &=\sin^2x\cosh^2y+\sin^2x\sinh^2y-\sin^2x\sinh^2y+\cos^2x\sinh^2y\\ &=\sin^2x+\sinh^2y \end{align}
here we have added and removed $\sin^2x\sinh^2y$ and used $\sin^2+\cos^2=1$ and $\cosh^2-\sinh^2=1$.
Now $\sinh(y)$ is zero only for $y=0$ (this is the only solution to $e^y=e^{-y}$), therefore all the zeros of $\sin(z)$ are real, and are given by the zeros of $\sin(x)$, i.e., $z=\pi\mathbf{Z}$.
Script error
$$ \begin{align*} \tag{1} a\ket{n}&=\sqrt{n}\ket{n-1}\,,&\bra{n}\,&a=\bra{n+1}\sqrt{n+1}\,,\\ \ud{a}\ket{n}&=\sqrt{n+1}\ket{n+1}\,,&\bra{n}\,&\ud{a}=\bra{n-1}\sqrt{n}\,, \end{align*} $$
$$ \begin{align*} \kern-1cm{(\mathrm{for}~i\le n+j)}\kern1cm a^i{\ud{a}}^j\ket{n}&={(n+j)!\over\sqrt{n!}\sqrt{(n+j-i)!}}\ket{n+j-i}\,,\\ \kern-1cm{(\mathrm{for}~i\le n)}\kern1cm a^{\dagger j}a^i\ket{n}&={\sqrt{n!}\sqrt{(n+j-i)!}\over(n-i)!}\ket{n+j-i}\,. \end{align*} $$
arial font algerian font bookman font braggadocio font courier font desdemona font garamond font modern font symbol font (These are pretty silly.) wingdings font (As are these.)
Blog:Sandbox File:laussy-jornada-divulgacion.ppt
$$f(z) = \left( \prod_{j=1}^n \frac{z - z_j}{1 - \overline{z_j}z} \right) \left( \prod_{j=1}^m \frac{z - w_j}{1 - \overline{w_j}z} \right)^{-1} g(z)$$
$$ \newcommand{\Re}{\mathrm{Re}\,} \newcommand{\pFq}[5]{{}_{#1}\mathrm{F}_{#2} \left( \genfrac{}{}{0pt}{}{#3}{#4} \bigg| {#5} \right)} $$
We consider, for various values of $s$, the $n$-dimensional integral \begin{align} \tag{2} W_n (s) &:= \int_{[0, 1]^n} \left| \sum_{k = 1}^n \mathrm{e}^{2 \pi \mathrm{i} \, x_k} \right|^s \mathrm{d}\boldsymbol{x} \end{align} % which occurs in the theory of uniform random walk integrals in the plane, where at each step a unit-step is taken in a random direction. As such, the integral (2) expresses the $s$-th moment of the distance to the origin after $n$ steps.
By experimentation and some sketchy arguments we quickly conjectured and strongly believed that, for $k$ a nonnegative integer \begin{align} \tag{3} W_3(k) &= \Re \, \pFq32{\frac12, -\frac k2, -\frac k2}{1, 1}{4}. \end{align} Appropriately defined, (3) also holds for negative odd integers. The reason for (3) was long a mystery, but it will be explained at the end of the paper.
\[ \begin{aligned} \label{def:1} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned} \]
\begin{aligned} \tag{4} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}
<google1 style="2"></google1>
Do you know this formula of mine <m>\frac{2\pi^2}{q}\int_0^\infty f(r)J_1(qr)rdr</m>?
21, May (2010) 17, August (2010)
<plot> set pm3d at s solid set palette rgb -6,-15,-7 unset colorbox set ticslevel 0 unset ztics unset surface set samples 70 set isosamples 70,70 complex(x,y)=x*{1,0}+y*{0,1} mandel(x,y,z,n) = (abs(z)>2.0 || n>=1000)? log(n): mandel(x,y,z*z+complex(x,y),n+1) a=-0.38 b=-0.612 set multiplot set origin 0,0 set size 0.55,0.77 splot [-0.5:0.5][-0.5:0.5] mandel(a,b,complex(x,y),0) set origin 0.35,-0.15 set size 0.7,0.96 set view 0,0,,, splot [-0.5:0.5][-0.5:0.5] mandel(a,b,complex(x,y),0) </plot>
<music>
\relative c' { e16-.->a(b gis)a-.->c(d b)c-.->e(f dis)e-.->a(b a) gis(b e)e,(gis b)b,(e gis)gis,(b e)e,(gis? b e) }
</music>
<music> \new Pianostaff << \new Staff { \time 2/2 \clef violin \key cis \minor \relative c \context Staff << \new Voice { \voiceOne
r4 cis8 dis e4 fis gis8 fis gis a gis fis e gis fis e fis gis fis e dis fis e dis e fis e d cis e d cis d e d cis b d cis b cis d cis b a cis b a b cis b a gis b a2 r cis2.
} \new Voice { \voiceTwo
e,8 gis a b cis dis bis cis dis4 r r2 r1 r1 r4 fis, b b b a8 gis a2 gis1~ gis8 gis fis eis fis2 gis2.
} \new Voice { \voiceThree \stemDown
s1 s s s s2. fis4 eis2 fis
} >> } \new Staff { \clef bass \time 2/2 \key cis \minor \relative c' \context Staff << \new Voice { \voiceOne
s1 r4 gis cis cis cis bis8 ais bis2 cis1 b2. s4 s1 b2 cis~ cis~ cis8 cis b a gis2.
} \new Voice { \voiceTwo
\stemUp cis,1 bis2 e dis1 \stemDown cis4 e a a a gis8 fis gis2~ \stemUp gis fis gis1 a2 fis~ fis8 fis e dis e4
} \new Voice { \voiceThree
\stemDown cis4 b a2 gis4 r4 g2\rest e1\rest e1\rest e1\rest r4 cis' fis fis fis eis8 dis eis2 fis r r
} >> } >> </music>
http://www.youtube.com/watch?v=MZAKjKC7Gho http://www.youtube.com/watch?v=2_HXUhShhmY
# | Municipality | Autonomous community |
Pop. (2009) |
---|---|---|---|
1 | Madrid | 3,255,944 | |
2 | Barcelona | 1,621,537 | |
3 | Valencia | 852,208 | |
4 | Seville | 703,206 | |
5 | Zaragoza | 674,317 | |
6 | Málaga | 568,305 | |
7 | Murcia | 436,870 | |
8 | Palma | 401,270 | |
9 | Las Palmas | 381,847 | |
10 | Bilbao | 354,860 | |
11 | Alicante | 334,757 | |
12 | Córdoba | 328,428 | |
13 | Valladolid | 317,864 | |
14 | Vigo | 297,332 | |
15 | Gijón | 277,554 | |
16 | L'Hospitalet de Llobregat | 257,038 | |
17 | A Coruña | 246,056 | |
18 | Vitoria-Gasteiz | 235,661 | |
19 | Granada | 234,325 | |
20 | Elche | 230,112 | |
21 | Oviedo | 224,005 | |
22 | Santa Cruz de Tenerife | 222,417 | |
23 | Badalona | 219,547 | |
24 | Cartagena | 211,996 | |
25 | Terrassa | 210,941 | |
26 | Jerez de la Frontera | 207,532 | |
27 | Sabadell | 206,493 | |
28 | Móstoles | 206,478 | |
29 | Alcalá de Henares | 204,574 | |
30 | Pamplona | 198,491 | |
31 | Fuenlabrada | 197,836 | |
32 | Almería | 188,810 | |
33 | Leganés | 186,066 | |
34 | Donostia-San Sebastián | 186,066 | |
35 | Santander | 182,700 | |
36 | Castellón de la Plana | 180,005 | |
37 | Burgos | 178,966 | |
38 | Albacete | 169,716 | |
39 | Alcorcón | 167,967 | |
40 | Getafe | 167,164 | |
41 | Salamanca | 155,619 | |
42 | Logroño | 152,107 | |
43 | San Cristóbal de La Laguna | 150,661 | |
44 | Huelva | 148,806 | |
45 | Badajoz | 148,334 | |
46 | Tarragona | 140,323 | |
47 | Lleida | 138.416 | |
48 | Marbella | 134,623 | |
49 | León | 134,305 | |
50 | Cádiz | 126,766 |
Template:MunicipalitiesinMurcia
File:AbedularCanencia-27Oct2013.gpx
File:AbedularCanencia-27Oct2013.gpx
Download the gpx file for this track.
Download the gpx file for this track.
Download the gpx file for this track.
$$ax^2+bx+c$$
$$\color{red}{ax^2}+bx+c$$