Mathematics II

$y = x^{3/2}$, so $\dfrac{dy}{dx} = \dfrac32 x^{1/2}$ and $1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + \dfrac{9x}{4}$.

$L = \int_0^4 \sqrt{1 + \frac{9x}{4}}\,dx$

Let $u = 1 + \dfrac{9x}{4}$, so $dx = \dfrac49\,du$; $x = 0 \Rightarrow u = 1$, $x = 4 \Rightarrow u = 10$.

$L = \frac49\int_1^{10} u^{1/2}\,du = \frac49\cdot\frac23\left[u^{3/2}\right]_1^{10} = \frac{8}{27}\left(10\sqrt{10} - 1\right)$

$L \approx 9.07 \text{ units}$

For $f(x)=\dfrac1{1-x}=(1-x)^{-1}$, the derivatives are

$f'(x)=(1-x)^{-2},\quad f''(x)=2(1-x)^{-3},\quad f'''(x)=3!\,(1-x)^{-4},\quad \ldots,\quad f^{(n)}(x)=n!\,(1-x)^{-(n+1)}$

so $f^{(n)}(0)=n!$.

Taylor's series about $x=0$:

$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^{\infty}x^n=1+x+x^2+x^3+\cdots$

By the ratio test, the series converges for $|x|<1$, so

$\frac{1}{1-x}=1+x+x^2+x^3+\cdots,\qquad |x|<1$

Definitions. Vectors $v_1, v_2, \dots, v_n$ are linearly dependent if there exist scalars $c_1, \dots, c_n$, not all zero, such that $c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0$. They are linearly independent if the equation holds only when $c_1 = c_2 = \cdots = c_n = 0$.

The given set. Form the matrix whose rows are the vectors and find its determinant: $\begin{vmatrix}1&2&3\\3&4&5\\1&-1&1\end{vmatrix} = 1(4+5) - 2(3-5) + 3(-3-4) = 9 + 4 - 21 = -8.$

Since the determinant is non-zero, the equation $c_1(1,2,3) + c_2(3,4,5) + c_3(1,-1,1) = (0,0,0)$, i.e. $c_1 + 3c_2 + c_3 = 0,\quad 2c_1 + 4c_2 - c_3 = 0,\quad 3c_1 + 5c_2 + c_3 = 0,$ has only the trivial solution $c_1 = c_2 = c_3 = 0$.

Hence the vectors $(1,2,3), (3,4,5), (1,-1,1)$ are linearly independent.

Let $v_1 = (0,2,-4),\ v_2 = (1,-2,-1),\ v_3 = (1,-4,3)$. Form the determinant of the matrix with these rows:

$\begin{vmatrix} 0 & 2 & -4\\ 1 & -2 & -1\\ 1 & -4 & 3\end{vmatrix} = 0 - 2(3 + 1) + (-4)(-4 + 2) = -8 + 8 = 0$

The determinant is zero, so the vectors are linearly dependent.

Explicitly, solving $a v_1 + b v_2 + c v_3 = 0$ gives $b + c = 0$, $2a - 2b - 4c = 0$, $-4a - b + 3c = 0$, with solution $a = -b,\ c = -b$. Taking $b = 1$:

$-v_1 + v_2 - v_3 = 0, \quad\text{i.e. } v_2 = v_1 + v_3$

Check: $(0,2,-4) + (1,-4,3) = (1,-2,-1)$.

The Taylor series of $f$ about $x=0$ is

$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$

For $f(x)=e^{-x}$ the derivatives are

$f'(x)=-e^{-x},\quad f''(x)=e^{-x},\quad \ldots,\quad f^{(n)}(x)=(-1)^n e^{-x}$

so $f^{(n)}(0)=(-1)^n$. Therefore

$e^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{n!}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\cdots$

By the ratio test the radius of convergence is infinite, so the series is valid for all real $x$.

The function $f(x)=x$ on $(-\pi,\pi)$ has period $2\pi$ and is odd, so $a_0=0$ and $a_n=0$. Only the sine terms remain:

$f(x)=\sum_{n=1}^{\infty} b_n\sin nx,\qquad b_n=\frac{2}{\pi}\int_0^{\pi}x\sin nx\,dx$

Integrating by parts:

$\int_0^{\pi}x\sin nx\,dx=\left[-\frac{x\cos nx}{n}\right]_0^{\pi}+\frac1n\int_0^{\pi}\cos nx\,dx=-\frac{\pi\cos n\pi}{n}+\frac{1}{n^2}\big[\sin nx\big]_0^{\pi}=\frac{\pi(-1)^{n+1}}{n}$

Hence

$b_n=\frac{2}{\pi}\cdot\frac{\pi(-1)^{n+1}}{n}=\frac{2(-1)^{n+1}}{n}$

$x=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nx=2\left(\sin x-\frac{\sin 2x}{2}+\frac{\sin 3x}{3}-\cdots\right),\quad -\pi<x<\pi$

The series is geometric with first term $a=1$ and common ratio $r=\dfrac{1/2}{1}=\dfrac12$.

The $n$th partial sum is

$S_n=\frac{a(1-r^n)}{1-r}=\frac{1-(1/2)^n}{1-\tfrac12}=2\left(1-\frac{1}{2^n}\right)$

As $n\to\infty$, $\dfrac1{2^n}\to0$, so $S_n\to2$. The partial sums have a finite limit, hence the series is convergent (also because $|r|=\tfrac12<1$).

$\text{Sum}=\frac{a}{1-r}=\frac{1}{1-\tfrac12}=2$

The function is defined on $(0,2\pi)$, so take period $2\pi$ with

$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\big(a_n\cos nx+b_n\sin nx\big)$

where $f=-1$ on $(0,\tfrac\pi2)$ and $f=0$ on $(\tfrac\pi2,2\pi)$.

Coefficient $a_0$.

$a_0=\frac1\pi\int_0^{2\pi}f(x)\,dx=\frac1\pi\int_0^{\pi/2}(-1)\,dx=-\frac12$

Coefficient $a_n$.

$a_n=\frac1\pi\int_0^{\pi/2}(-1)\cos nx\,dx=-\frac1{n\pi}\Big[\sin nx\Big]_0^{\pi/2}=-\frac{1}{n\pi}\sin\frac{n\pi}{2}$

Coefficient $b_n$.

$b_n=\frac1\pi\int_0^{\pi/2}(-1)\sin nx\,dx=\frac1{n\pi}\Big[\cos nx\Big]_0^{\pi/2}=\frac{1}{n\pi}\Big(\cos\frac{n\pi}{2}-1\Big)$

Series.

$f(x)=-\frac14-\frac1\pi\sum_{n=1}^{\infty}\frac1n\left[\sin\frac{n\pi}{2}\cos nx+\Big(1-\cos\frac{n\pi}{2}\Big)\sin nx\right]$

Written out, the first terms are

$f(x)=-\frac14-\frac1\pi\Big[(\cos x+\sin x)+\sin 2x+\tfrac13(\sin 3x-\cos 3x)+\cdots\Big]$

since for $n=1$: $\sin\frac\pi2=1$, $1-\cos\frac\pi2=1$; for $n=2$: $\sin\pi=0$, $1-\cos\pi=2$ (giving $\tfrac12\cdot2\sin2x$); for $n=3$: $\sin\frac{3\pi}2=-1$, $1-\cos\frac{3\pi}2=1$.

Definition. A geometric series $a+ar+ar^2+\cdots$ is said to converge if the sequence of its partial sums $S_n=\dfrac{a(1-r^n)}{1-r}$ tends to a finite limit as $n\to\infty$. This happens exactly when $|r|<1$, and then the sum is $S=\dfrac{a}{1-r}$. If $|r|\ge1$ (with $a\ne0$) the series diverges.

The given series. $1+\dfrac13+\dfrac19+\dfrac1{27}+\cdots$ is geometric with $a=1$ and $r=\dfrac13$.

$S_n=\frac{1-(1/3)^n}{1-\tfrac13}=\frac32\left(1-\frac1{3^n}\right)$

As $n\to\infty$, $\dfrac1{3^n}\to0$, so $S_n\to\dfrac32$, a finite limit. Also $|r|=\tfrac13<1$. Hence the series is convergent.

$\text{Sum}=\frac{a}{1-r}=\frac{1}{1-\tfrac13}=\frac32$

For $z = 3 - 4i$, the modulus is $r = \sqrt{3^2 + (-4)^2} = 5.$

The point lies in the fourth quadrant, so $\theta = -\tan^{-1}\left(\frac{4}{3}\right) \approx -53.13^\circ \ (-0.927\text{ rad}).$

Hence $3 - 4i = 5\left[\cos\left(-\tan^{-1}\tfrac{4}{3}\right) + i\sin\left(-\tan^{-1}\tfrac{4}{3}\right)\right] \approx 5\,(\cos 53.13^\circ - i\sin 53.13^\circ).$

The interval is $(-2, 2)$, so the half-period is $L = 2$ (period $4$): $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty\left(a_n\cos\frac{n\pi x}{2} + b_n\sin\frac{n\pi x}{2}\right).$

Since $f = 0$ on $(-2, 0)$, only $\int_0^2$ contributes.

Coefficient $a_0$. $a_0 = \frac12\int_0^2 x\,dx = \frac12\cdot 2 = 1,\qquad \frac{a_0}{2} = \frac12.$

Coefficient $a_n$. Let $k = \dfrac{n\pi}{2}$. $a_n = \frac12\int_0^2 x\cos kx\,dx = \frac12\left[\frac{x\sin kx}{k} + \frac{\cos kx}{k^2}\right]_0^2 = \frac12\cdot\frac{\cos n\pi - 1}{k^2} = \frac{2\left[(-1)^n - 1\right]}{n^2\pi^2}.$ So $a_n = -\dfrac{4}{n^2\pi^2}$ for odd $n$ and $a_n = 0$ for even $n$.

Coefficient $b_n$. $b_n = \frac12\int_0^2 x\sin kx\,dx = \frac12\left[-\frac{x\cos kx}{k} + \frac{\sin kx}{k^2}\right]_0^2 = \frac12\left(-\frac{2(-1)^n}{k}\right) = \frac{2(-1)^{n+1}}{n\pi}.$

Fourier series. $f(x) = \frac12 - \frac{4}{\pi^2}\sum_{n=1,3,5,\dots}\frac{1}{n^2}\cos\frac{n\pi x}{2} + \frac{2}{\pi}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sin\frac{n\pi x}{2}.$

Check at $x = 0$: the series gives $\tfrac12 - \tfrac{4}{\pi^2}\cdot\tfrac{\pi^2}{8} = 0 = f(0)$.

For $z = \sqrt3 - i$: $r = \sqrt{(\sqrt3)^2 + (-1)^2} = 2,\qquad \tan\theta = \frac{-1}{\sqrt3}.$

The point is in the fourth quadrant, so $\theta = -\dfrac{\pi}{6}$. Hence $\sqrt3 - i = 2\left[\cos\left(-\frac\pi6\right) + i\sin\left(-\frac\pi6\right)\right] = 2\left(\cos\frac\pi6 - i\sin\frac\pi6\right).$

The series is geometric with first term $a = 1$ and common ratio $r = \dfrac{1/3}{1} = \dfrac13$.

A geometric series converges when $|r| < 1$. Here $|r| = \tfrac13 < 1$, so the series is convergent.

Its sum to infinity is $S = \frac{a}{1 - r} = \frac{1}{1 - \frac13} = \frac{3}{2}.$

Formulae. For period $2\pi$: $f(x)=\dfrac{a_0}{2}+\sum_{n\ge1}(a_n\cos nx+b_n\sin nx)$ with $a_0=\dfrac1\pi\displaystyle\int_0^{2\pi}f\,dx$, $a_n=\dfrac1\pi\displaystyle\int_0^{2\pi}f\cos nx\,dx$, $b_n=\dfrac1\pi\displaystyle\int_0^{2\pi}f\sin nx\,dx$.

Take $f(x)$ on $(0,2\pi)$ and extend it periodically with period $2\pi$. Since $f=-1$ on $(0,\pi/2)$ and $f=0$ on $(\pi/2,2\pi)$, only $(0,\pi/2)$ contributes to each integral.

Coefficients.

$a_0=\frac1\pi\int_0^{\pi/2}(-1)\,dx=-\frac12,$

$a_n=-\frac1\pi\int_0^{\pi/2}\cos nx\,dx=-\frac{\sin(n\pi/2)}{n\pi},$

$b_n=-\frac1\pi\int_0^{\pi/2}\sin nx\,dx=-\frac1\pi\left[\frac{-\cos nx}{n}\right]_0^{\pi/2}=-\frac{1-\cos(n\pi/2)}{n\pi}.$

Fourier series.

$f(x)=-\frac14-\frac1\pi\sum_{n=1}^{\infty}\frac1n\left[\sin\frac{n\pi}{2}\cos nx+\left(1-\cos\frac{n\pi}{2}\right)\sin nx\right].$

The first terms are

$f(x)=-\frac14-\frac1\pi\left[\cos x+\sin x+\sin 2x-\frac{\cos 3x}{3}+\frac{\sin 3x}{3}+\cdots\right].$

Check at $x=\pi$, where $f=0$: the sine terms vanish and the series gives $-\dfrac14-\dfrac1\pi\sum_{n\ \mathrm{odd}}\dfrac{(-1)^n(-1)^{(n-1)/2}}{n}=-\dfrac14+\dfrac1\pi\cdot\dfrac\pi4=0$.

From $y=(x/2)^{2/3}$ we get $x=2y^{3/2}$. At $x=0$, $y=0$; at $x=2$, $y=1$. Because $dy/dx$ is infinite at $x=0$, integrate with respect to $y$:

$\frac{dx}{dy}=3y^{1/2}.$

$L=\int_0^1\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy=\int_0^1\sqrt{1+9y}\,dy=\frac19\cdot\frac23\Big[(1+9y)^{3/2}\Big]_0^1$

$L=\frac{2}{27}\left(10^{3/2}-1\right)=\frac{2}{27}\left(10\sqrt{10}-1\right)\approx 2.27 \text{ units}.$

Let $z = 1 + i$, so $x=1$, $y=1$.

Modulus: $r = \sqrt{x^2+y^2} = \sqrt{1+1} = \sqrt2$.

Argument: $\theta = \tan^{-1}\!\left(\dfrac{y}{x}\right) = \tan^{-1}(1) = \dfrac{\pi}{4}$ (the point lies in the first quadrant).

$1 + i = r(\cos\theta + i\sin\theta) = \sqrt2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

Check: $\sqrt2\cdot\frac{1}{\sqrt2}=1$ for both the real and imaginary parts.

This is an improper integral. Let $u = 1 + x^2$, so $du = 2x\,dx$.

$\int_2^{\infty} \frac{2x\,dx}{(1+x^2)^{3/2}} = \lim_{b\to\infty}\left[-\frac{2}{\sqrt{1+x^2}}\right]_2^{b} = \lim_{b\to\infty}\left(-\frac{2}{\sqrt{1+b^2}} + \frac{2}{\sqrt 5}\right)$

Since $\dfrac{2}{\sqrt{1+b^2}} \to 0$,

$\int_2^{\infty} \frac{2x\,dx}{(1+x^2)^{3/2}} = \frac{2}{\sqrt5}$

The integral converges.

Equilibrium: $D(x) = S(x)$ gives $25 - x^2 = x + 5$, so $x^2 + x - 20 = 0$, i.e. $(x+5)(x-4) = 0$. Hence $x_0 = 4$ and $p_0 = 4 + 5 = 9$.

Consumer's surplus

$CS = \int_0^4 D(x)\,dx - x_0p_0 = \int_0^4 (25 - x^2)\,dx - 36 = \left[25x - \frac{x^3}{3}\right]_0^4 - 36 = 100 - \frac{64}{3} - 36 = \frac{128}{3}$

Producer's surplus

$PS = x_0p_0 - \int_0^4 S(x)\,dx = 36 - \left[\frac{x^2}{2} + 5x\right]_0^4 = 36 - 28 = 8$

So $CS = \dfrac{128}{3} \approx 42.67$ and $PS = 8$.

Vector space. A non-empty set $V$ with addition and scalar multiplication (over a field $F$, e.g. $\mathbb R$) is a vector space if it is closed under both operations and satisfies: addition is commutative and associative, there is a zero vector, every $u$ has an additive inverse $-u$, and for scalars $a,b$: $a(u+v) = au + av$, $(a+b)u = au + bu$, $(ab)u = a(bu)$, $1\cdot u = u$.

Subspace test. A non-empty subset $W\subseteq V$ is a subspace if $u + v\in W$ and $ku\in W$ for all $u,v\in W$ and all scalars $k$.

Here $W = \{(x,y,z): 2x + y + 2z = 0\}$.

1. Non-empty: $(0,0,0)\in W$ since $0 = 0$.
2. Closed under addition: let $u = (x_1,y_1,z_1)$ and $v = (x_2,y_2,z_2)$ be in $W$, so $2x_1 + y_1 + 2z_1 = 0$ and $2x_2 + y_2 + 2z_2 = 0$. Then $u + v = (x_1+x_2,\ y_1+y_2,\ z_1+z_2)$ and $2(x_1+x_2) + (y_1+y_2) + 2(z_1+z_2) = (2x_1 + y_1 + 2z_1) + (2x_2 + y_2 + 2z_2) = 0$ so $u + v\in W$.
3. Closed under scalar multiplication: for $k\in\mathbb R$, $ku = (kx_1, ky_1, kz_1)$ and $2kx_1 + ky_1 + 2kz_1 = k(2x_1 + y_1 + 2z_1) = 0$ so $ku\in W$.

Hence $W$ is a subspace of $\mathbb R^3$.

The inner integral is with respect to $x$ from $3$ to $4$, and the outer with respect to $y$ from $0$ to $2$: $\int_0^2\int_3^4 xy\,dx\,dy = \int_0^2 y\left[\frac{x^2}{2}\right]_3^4 dy = \int_0^2 y\cdot\frac{16-9}{2}\,dy = \frac72\left[\frac{y^2}{2}\right]_0^2 = \frac72\cdot 2 = 7.$

A function $f$ has period $T$ if $f(x+T)=f(x)$ for all $x$, the smallest such $T>0$ being the period.

$\sin 2(x+T)=\sin(2x+2T)=\sin 2x \iff 2T=2\pi \text{ (smallest positive)}.$

So $T=\pi$. In general $\sin kx$ has period $\dfrac{2\pi}{k}$; with $k=2$ the period of $\sin 2x$ is $\pi$.

Integrate once:

$\frac{dy}{dx} = \frac{x^3}{3} + 3x + C_1$

Integrate again:

$y = \frac{x^4}{12} + \frac{3x^2}{2} + C_1x + C_2$

This is the general solution, with arbitrary constants $C_1, C_2$.

Particular solution. From $f'(2) = 10$: $\dfrac{8}{3} + 6 + C_1 = 10$, so $C_1 = \dfrac43$.

From $f(0) = 22$: $C_2 = 22$.

$y = \frac{x^4}{12} + \frac{3x^2}{2} + \frac{4x}{3} + 22$

Harmonic test. A function is harmonic if it satisfies Laplace's equation $u_{xx}+u_{yy}=0$.

$u_x = 2x,\quad u_{xx}=2,\qquad u_y=-2y-1,\quad u_{yy}=-2$

$u_{xx}+u_{yy}=2-2=0$

Hence $u=x^2-y^2-y$ is harmonic.

Harmonic conjugate. By the Cauchy-Riemann equations, $v_y=u_x$ and $v_x=-u_y$.

From $v_y=2x$, integrating with respect to $y$:

$v=2xy+g(x)$

Then $v_x=2y+g'(x)$, and this must equal $-u_y=2y+1$. So $g'(x)=1$, giving $g(x)=x+c$.

$v(x,y)=2xy+x+c$

Analytic function.

$f(z)=u+iv=(x^2-y^2+2ixy)+(ix-y)+ic$

Since $z^2=x^2-y^2+2ixy$ and $iz=ix-y$,

$f(z)=z^2+iz+ic$

where $c$ is a real constant (equivalently $f(z)=z^2+iz+C$).

Definition. Let $f(z)=u(x,y)+iv(x,y)$, $z=x+iy$. The Cauchy-Riemann (C-R) equations are

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\qquad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

They are necessary conditions for $f$ to be analytic (differentiable) at a point.

Case 1: $f(z)=z^2$.

$z^2=(x+iy)^2=(x^2-y^2)+i(2xy),\qquad u=x^2-y^2,\;v=2xy$

$u_x=2x,\quad v_y=2x\;\Rightarrow\;u_x=v_y$

$u_y=-2y,\quad v_x=2y\;\Rightarrow\;u_y=-v_x$

Both equations hold for all $z$, so C-R equations are satisfied for $f(z)=z^2$.

Case 2: $f(z)=|z|^2$.

$|z|^2=x^2+y^2,\qquad u=x^2+y^2,\;v=0$

$u_x=2x,\quad v_y=0;\qquad u_y=2y,\quad -v_x=0$

The first equation needs $2x=0$ and the second needs $2y=0$. Both hold only at $x=y=0$, i.e. $z=0$. Hence for $z\neq0$ the C-R equations are not satisfied, and $|z|^2$ is not analytic away from the origin.