From 61b48fd8baae321daf294ebf670ef4906240d260 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Wed, 5 May 2021 19:03:09 -0400 Subject: homeworks and attendance quizzes for kang --- kang/hw11.tex | 175 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 175 insertions(+) create mode 100644 kang/hw11.tex (limited to 'kang/hw11.tex') diff --git a/kang/hw11.tex b/kang/hw11.tex new file mode 100644 index 0000000..d55a9e8 --- /dev/null +++ b/kang/hw11.tex @@ -0,0 +1,175 @@ +\def\Res{\mathop{\rm Res}} +\def\Re{\mathop{\rm Re}\nolimits} +\def\Im{\mathop{\rm Im}\nolimits} +\let\rule\hrule +\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak} + +%page 253 + +\noindent{\bf 4.} + +\noindent{\it (a)} + +$$(\sec z)^{-1} = \cos z = q(z).$$ +$$p(z) = z.$$ +$$p(\pi/2+n\pi) = \pi/2 + n\pi = z_n\neq 0, \qquad q(\pi/2+n\pi) = 0, +\qquad q'(\pi/2+n\pi) = -\sin(\pi/2+n\pi) = (-1)^{n+1} \neq 0,$$ +so this is a simple pole with residue +$$\Res_{z=z_n}(z\sec z) = (-1)^nz_n.$$ + +\noindent{\it (b)} + +Similarly, +$$(\tanh(z)) = {\sinh z\over\cosh z},$$ +giving $p(z) = \sinh z$ and $q(z) = \cosh z,$ +which forms a simple pole because +$$p(z_n) = -\sin(\pi/2 + n\pi) = (-1)^n, \qquad q(z_n) = \cos(\pi/2 + +n\pi) = 0, \qquad q'(z_n) = \sin(\pi/2 + n\pi) = (-1)^n.$$ + +$$p(z_n)/q'(z_n) = (-1)^n/(-1)^n = 1 = \Res_{z=z_n}\tanh z$$ + +\noindent{\bf 6.} + +The value of this integral is equal to $2\pi i$ times the sum of the +residues of $$f(z) = {1\over z^2\sin z}$$ within the square. +These are $z^2 = 0 \to z = 0$ and $\sin z = 0 \to z = z_n = \pi n,$ +where $-N\leq n\leq N.$ +This set includes $z = 0.$ + +$$f(z) = {p(z)\over q(z)}$$ +where $p(z) = 1$ and $q(z) = z^2\sin z.$ $p(z_n)\neq 1,$ $q(z_n) = 0,$ +and $q(z_n) = z_n^2\cos z_n + 2z_n\sin z_n = z_n^2(-1)^n \neq 0$ (except +at $z_n = 0.$) +The sum of residues within the range of $-N\leq n\leq N$ and $n\neq 0$ +is $$\sum_{n=1}^N (-1)^nz_n^{-2}+z_{-n}^{-2} = \sum_{n=1}^N {2(-1)^n\over(\pi +n)^2}.$$ + +At $z_n = 0,$ +$$f(z) = {\phi(z)\over z^3},$$ +where $\phi(z) = {z\over\sin z},$ +This gives a residue at 0 of ${\phi''(0)\over 3!}$. +$$\phi''(z) = -{\cos z\over(\sin z)^2} - {\cos z - z\sin z\over(\sin z)^2} ++ {2z(\cos z)^2\over(\sin z)^3} = +{-2\cos z\sin z + z(\sin z)^2 + 2z(\cos z)^2 \over (\sin z)^3}.$$ +$$\phi''(0) = 1,$$ +found by noting that, near 0, $\sin z\to z$ and $\cos z\to 1.$ +This gives a 1/6 residue. + +Together, this makes +$$\int_{C_N} {dz\over z^2\sin z} = 2\pi i\left[{1\over 6}+2\sum_{n=1}^N +{(-1)^n\over\pi^2 n^2}\right].$$ + +If the integral tends to 0 as $N\to\infty,$ +$$0 = {1\over 6}+2\sum_{n=1}^N{(-1)^n\over \pi^2 n^2} \to +-{\pi^2\over 12} = \sum_{n=1}^N{(-1)^n\over n^2},$$ +meaning the series tends to $\pi^2/12.$ + +\noindent{\bf 10.} + +Because $p(z)/q(z)$ has a pole of order $m,$ +$${p(z)\over q(z)} = {\phi(z)\over (z-z_0)^m},$$ +where $\phi(z)$ is analytic and non-zero at $z_0,$ further giving +$$q(z) = {p(z)\over\phi(z)}(z-z_0)^m,$$ +and because $\phi(z)$ is nonzero, $p(z)/\phi(z)$ is analytic, and this +tells us that $q(z)$ has a zero of order $m$ because $q(z) = +g(z)(z-z_0)^m$ where $g(z)$ is a nonzero analytic function. + +\hrule +%page 264 + +\noindent{\bf 1.} + +$$z^2+1 = (z-i)(z+i).$$ +The only singularity above the real axis is $z = i.$ +This point is a simple pole, so its residue is $1/(i+i) = -i/2.$ +This gives the value of the integral as $\pi i*(-i/2) = \pi/2$ (since +the integrand is even and $\pi R/(R^2-1)$ vanishes to 0 as $R$ tends to +infinity). + +\noindent{\bf 4.} + +The singularities of this integrand are, like in the example, the sixth +roots of $-1,$ and they are simple poles, but the residues $B_k$ +evaluate instead to $$B_k = \Res_{z=c_k} {z^2\over z^6+1} = {c_k^2\over +6c_k^5} = {c_k^3\over 6c_k^6} = -{c_k^3\over 6}.$$ +$$c_k = e^{i(\pi/6+k\pi/3)} \to c_k^3 = e^{i(\pi/2+k\pi)} = i(-1)^k.$$ +Since $k\in\{0,1,2\},$ the residues sum to $-i/6,$ giving the value of +the integral to be $\pi i*(-i/6) = -\pi/6,$ (again valid because the +integrand is even and $\pi R^3/(R^6-1)$ vanishes to 0 as $R$ tends to +infinity). + +\noindent{\bf 8.} + +$$q(z)=(z^2+1)(z^2+2z+2) = (z+i)(z-i)(z+(1+i))(z+(1-i)).$$ +The singularities above the real axis are simple poles $z = i, -1+i.$ +At these singularities, the residues $p(z)/q'(z)$ are respectively +$${i\over (2i)(1+2i)(1)} = {1-2i\over10}$$ and $${-1+i\over +(-1+2i)(-1)(2i)} = {-1+3i\over 10}.$$ + +$M_R = {R\over (R^2-1)(R^2-2)} \to R\pi M_R = {R^2\over +(R^2-1)(R^2-2)},$ +so $\int_{C_R}f(x)dx = 0$ as $R$ tends to infinity. +This allows us to determine the Cauchy principal value of the integral +to be $2\pi i\cdot i/10 = -\pi/5.$ + +\hrule +%page 273 + +\noindent{\bf 1.} + +$$\int_{-\infty}^\infty {\cos x\thinspace dx\over (x^2+a^2)(x^2+b^2)} = +\Re \int_{-\infty}^\infty {e^{ix}dx\over (x^2+a^2)(x^2+b^2)}.$$ +The singularities in the upper half of the plane are simple poles +$x=ai, x=bi.$ +These have residues $${e^{-a}\over (2ai)(-a^2+b^2)}$$ and $$e^{-b}\over +(2bi)(-b^2+a^2),$$ respectively. +$$\int_{-\infty}^\infty {e^{ix}dx\over (x^2+a^2)(x^2+b^2)} += 2\pi iB - \int_{C_R} f(z)e^{iz} dz.$$ +This last integral tends to $0$ by Jordan's lemma because $f(z)$ has +maximum value ${1\over (R^2-a^2)(R^2-b^2)},$ which tends to 0 as $R$ +tends to infinity where $R>a$ and $R>b.$ +$$2\pi iB = 2\pi i\left({e^{-a}/2a-e^{-b}/2b \over i(a^2-b^2)}\right) = +\pi({e^{-a}/a - e^{-b}/b \over (a^2-b^2)}),$$ +proving the integral formula. + +\noindent{\bf 5.} + +$$\int_{-\infty}^\infty {x^3\sin ax\over x^4+4}dx = \Im +\int_{-\infty}^\infty {x^3e^{iax}\over x^4+4}dx.$$ + +$$\int_{-\infty}^\infty {x^3e^{iax}\over x^4+4}dx = 2\pi iB - \int_{C_R} +f(x)e^{iax}dx.$$ +Where $|x|>R>\sqrt 2,$ the maximum value of $f(x)$ is +$$M_R<{R^3\over R^4-4},$$ +which tends to $0$ as $R\to\infty,$ telling us the integral evaluates to +$0.$ +The singularities in the upper half of the plane are $c_k = \sqrt +2e^{i(\pi/4+k\pi/2)},$ where $k\in\{0,1\}.$ +These are simple poles, so they evaluate to $$p(c_k)/q'(c_k) = {c_k^3 +e^{iac_k}\over 4c_k^3} = e^{iac_k}/4 = +e^{ia\sqrt2 e^{i(\pi/4+k\pi/2)}}/4 = +$$$$ e^{ia\sqrt2(\cos(\pi/4+k\pi/2)+i\sin(\pi/4+k\pi/2))}/4 += e^{ia((-1)^k+i)}/4 += e^{a(i(-1)^k-1)}/4 += e^{-a}(\cos(a(-1)^k)+i\sin(a(-1)^k))/4,$$ +summing to $2e^{-a}\cos(a)/4.$ +Therefore, the integral evaluates to $2\pi ie^{-a}\cos a/2,$ +giving imaginary component (and value of the original integral) +$\pi e^{-a}\cos a.$ + +\noindent{\bf 8.} + +$$\int_{-\infty}^\infty {\sin x dx\over x^2+4x+5} = +\Im\int_{-\infty}^\infty {e^{ix} dx\over (x+(2+i))(x+(2-i))}$$ + +The only singularity in the upper half of the plane is $-2+i,$ +with a residue of +$$e^{-2i-1}\over 2i,$$ +giving the integral a value +${2\pi ie^{-2i-1}\over 2i} = \pi e^{-1-2i} = {\pi\over e}(\cos(-2)+i\sin(-2)),$ +giving an imaginary component $-\pi\sin(2)/e.$ + +This answer is valid because $f(z)$ tends to zero on $C_R$ as +$R\to\infty,$ where $$f(z) = {1\over z^2+4z+5}.$$ + +\bye -- cgit