2. Further Pure Mathematics 2

At $x=0$:\sinh0=0,\quad\cosh0=1,\quad\tanh0=0,\quad\operatorname{sech}0=1,$$while cosech 0 and coth 0 are undefined.$$

$$From the definitions,$$\cosh^2x-\sinh^2x=\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{4}=1.Divide by $\cosh^2x$ or $\sinh^2x$ to obtain1-\tanh^2x=\operatorname{sech}^2x,\qquad \coth^2x-\operatorname{cosech}^2x=1.$$$$

$$$$\sinh(x\pm y)=\sinh x\cosh y\pm\cosh x\sinh y,$$$$\cosh(x\pm y)=\cosh x\cosh y\pm\sinh x\sinh y,$$where the paired signs correspond.$$

Setting $y=x$ gives\sinh2x=2\sinh x\cosh x,$$$$\cosh2x=\cosh^2x+\sinh^2x=2\cosh^2x-1=1+2\sinh^2x,$$$$\tanh2x=\frac{2\tanh x}{1+\tanh^2x}. $$$$

Let $u=\tanh^{-1}y$, soy=\frac{e^u-e^{-u}}{e^u+e^{-u}}=\frac{e^{2u}-1}{e^{2u}+1}.$$Then$$e^{2u}(1-y)=1+y,and because $|y|<1$, both sides of the logarithm are positive:u=\frac12\ln\left(\frac{1+y}{1-y}\right). $$$$

$$The system$$a_{11}x+a_{12}y+a_{13}z=b_1,\quad a_{21}x+a_{22}y+a_{23}z=b_2,\quad a_{31}x+a_{32}y+a_{33}z=b_3$$is exactly$$A\mathbf x=\mathbf b,\qquad A=(a_{ij}),\quad\mathbf x=\begin{pmatrix}x\y\z\end{pmatrix},\quad\mathbf b=\begin{pmatrix}b_1\b_2\b_3\end{pmatrix}. $$$$

$$$$\begin{aligned}2x-y+z&=4\x+3y-2z&=5\3x+y+z&=6\end{aligned}\quad\Longleftrightarrow\quad \begin{pmatrix}2&-1&1\1&3&-2\3&1&1\end{pmatrix}\begin{pmatrix}x\y\z\end{pmatrix}=\begin{pmatrix}4\5\6\end{pmatrix}. $$$$

$$The equations$$x+y+z=1,\quad2x+2y+2z=2,\quad x-y+z=0have one dependent row and two independent constraints, so they are consistent with one free parameter: three planes share a line. Replacing the second right side by 3 creates $0=1$ after elimination, so there is no common point.

If $A\mathbf e=\lambda\mathbf e$, thenA^2\mathbf e=A(\lambda\mathbf e)=\lambda A\mathbf e=\lambda^2\mathbf e.$$Repeating (or inducting) gives$$A^n\mathbf e=\lambda^n\mathbf e,so e is an eigenvector of $A^n$ with eigenvalue $\lambda^n$.

$$For$$A=\begin{pmatrix}1&1&0\0&2&1\0&0&3\end{pmatrix},$$$$\det(A-\lambda I)=(1-\lambda)(2-\lambda)(3-\lambda),$$so the real distinct eigenvalues are 1, 2 and 3.$$

If $A\mathbf q_i=\lambda_i\mathbf q_i$ and the eigenvectors form a basis, setQ=(\mathbf q_1\ \mathbf q_2\ \cdots),\qquad D=\operatorname{diag}(\lambda_1,\lambda_2,\ldots).Then $AQ=QD$, soA=QDQ^{-1}. $$$$

$$Adjacent factors cancel:$$(QDQ^{-1})^n=QD^nQ^{-1},and $D^n$ is obtained by raising each diagonal eigenvalue to n. The order of columns in Q must match the order of entries in D.

$$For the previous matrix,$$Q=\begin{pmatrix}1&1&1\0&1&2\0&0&2\end{pmatrix},\quad D=\operatorname{diag}(1,2,3),\quad Q^{-1}=\begin{pmatrix}1&-1&1/2\0&1&-1\0&0&1/2\end{pmatrix}. $$Hence$$A^n=\begin{pmatrix}1&2^n-1&(3^n-2\cdot2^n+1)/2\0&2^n&3^n-2^n\0&0&3^n\end{pmatrix}. $$$$

$$Cayley-Hamilton states: if$$p(\lambda)=\lambda^m+c_{m-1}\lambda^{m-1}+\cdots+c_1\lambda+c_0$$is the characteristic polynomial of A, then$$p(A)=A^m+c_{m-1}A^{m-1}+\cdots+c_1A+c_0I=0. $$$$

$$For$$A=\begin{pmatrix}2&1\1&1\end{pmatrix},$$$$p(\lambda)=\lambda^2-3\lambda+1,$$so$$A^2-3A+I=0. $$Therefore$$A^3=3A^2-A=8A-3I,and multiplying the identity by $A^{-1}$ givesA^{-1}=3I-A=\begin{pmatrix}1&-1\-1&2\end{pmatrix}. $$$$

$$For a composite, multiply by the inner derivative. For example,$$\frac{d}{dx}\left[\cosh^{-1}(2x)\right]=\frac{2}{\sqrt{(2x)^2-1}}=\frac{2}{\sqrt{4x^2-1}},\qquad x>\tfrac12.$$$$

If $x^2+xy+y^2=7$, then2x+y+(x+2y)y'=0.At $(1,2)$, $y'=-4/5$. Differentiating again gives2+2y'+2(y')^2+(x+2y)y''=0,$$so$$y''=-\frac{2+2(-4/5)+2(16/25)}{5}=-\frac{42}{125}.$$$$

When $x=x(t)$ and $y=y(t)$ with $dx/dt\ne0$,\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \frac{d^2y}{dx^2}=\frac{1}{dx/dt}\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{x'y''-y'x''}{(x')^3},where primes in the final fraction mean derivatives with respect to $t$.

For $x=t^2+1$ and $y=t^3$,\frac{dy}{dx}=\frac{3t^2}{2t}=\frac32t,$$hence$$\frac{d^2y}{dx^2}=\frac{1}{2t}\frac{d}{dt}\left(\frac32t\right)=\frac{3}{4t}.At $t=1$, the second derivative is $3/4$.

$$The first terms of the Maclaurin series are$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots.To derive terms through $x^n$, find only the derivatives through order $n$, evaluate each at zero, and divide by its factorial. A general term is not required here.

For $y=\tan x$, start from $y'=1+y^2$. At $x=0$, $y=0$ and $y'=1$. Theny''=2yy'\Rightarrow y''(0)=0,$$$$y'''=2(y')^2+2yy''\Rightarrow y'''(0)=2.$$Therefore$$\tan x=x+\frac{x^3}{3}+\cdots.$$This is successive implicit differentiation: no formula for the general derivative was needed.$$

A truncated series can simplify a nearby calculation. From $e^{-x^2}=1-x^2+\cdots$,\int_0^{1/5}e^{-x^2},dx\approx\int_0^{1/5}(1-x^2),dx=\left[x-\frac{x^3}{3}\right]_0^{1/5}=\frac{74}{375}.$$$$

Complete the square before choosing. For $x>0$, set $u=x+1$:\int\frac{dx}{\sqrt{x^2+2x}}=\int\frac{du}{\sqrt{u^2-1}}=\cosh^{-1}u+C=\cosh^{-1}(x+1)+C.$$$$

$$Let$$I_n=\int_0^{\pi/2}\sin^n x,dx,\qquad n\ge2.Integrate by parts with $u=\sin^{n-1}x$ and $dv=\sin x\,dx$.

The boundary term $[-\sin^{n-1}x\cos x]_0^{\pi/2}$ is zero, soI_n=(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x,dx=(n-1)(I_{n-2}-I_n).$$Hence$$I_n=\frac{n-1}{n}I_{n-2}. $$$$

The base cases are $I_0=\pi/2$ and $I_1=1$. ThereforeI_6=\frac56I_4=\frac56\cdot\frac34I_2=\frac56\cdot\frac34\cdot\frac12I_0=\frac{5\pi}{32}. Even indices end at $I_0$; odd indices end at $I_1$.

For decreasing $f(x)=1/x$, unit-width rectangles give\int_1^{n+1}\frac{dx}{x}<\sum_{r=1}^{n}\frac1r<1+\int_1^n\frac{dx}{x},$$hence$$\ln(n+1)<\sum_{r=1}^{n}\frac1r<1+\ln n.$$The extra endpoint rectangle explains the isolated 1.$$

With $n$ rectangles on $[0,1]$, $\Delta x=1/n$:\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\frac1n\sum_{r=1}^{n}\frac{1}{1+(r/n)^2}\longrightarrow\int_0^1\frac{dx}{1+x^2}=\frac{\pi}{4}. $$$$

For $x=t^2$, $y=\tfrac23t^3$, $0\le t\le1$,L=\int_0^1\sqrt{(2t)^2+(2t^2)^2},dt=\int_0^1 2t\sqrt{1+t^2},dt=\frac23(2\sqrt2-1). $$$$

For $y=x^2$, $0\le x\le1$, rotated about the x-axis, the radius is $y=x^2$ and $ds=\sqrt{1+4x^2}\,dx$. ThusS=2\pi\int_0^1x^2\sqrt{1+4x^2},dx.$$The limits must trace the required arc once.$$

Writing $\operatorname{cis}\theta=\cos\theta+i\sin\theta$, de Moivre's theorem for an integer $n$ is[r\operatorname{cis}\theta]^n=r^n\operatorname{cis}(n\theta).Arguments differing by $2\pi$ describe the same point.

For $n<0$, division explains the result. For example,(2\operatorname{cis}(\pi/6))^{-2}=\frac{1}{(2\operatorname{cis}(\pi/6))^2}=\frac14\operatorname{cis}(-\pi/3).$$The modulus stays positive while its reciprocal power shrinks the point toward the origin.$$

For every positive integer $n$, the claim isP(n):\quad(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).$$$$

Assume $P(k)$ is true. Then\begin{aligned}(\cos\theta+i\sin\theta)^{k+1}&=\cos(k\theta)+i\sin(k\theta)\&=\cos((k+1)\theta)+i\sin((k+1)\theta),\end{aligned}because the real and imaginary parts are exactly the cosine and sine addition formulae. Thus $P(k)\Rightarrow P(k+1)$.

For a multiple-angle identity, expand $(\cos\theta+i\sin\theta)^n$ and equate real or imaginary parts. For example,\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta.Dividing matching real and imaginary expressions by a power of $\cos\theta$ can produce identities in $\tan\theta$.

To express powers in multiple angles, set $z=\cos\theta+i\sin\theta$. Since $z+z^{-1}=2\cos\theta$ and $z^m+z^{-m}=2\cos m\theta$, expanding $(z+z^{-1})^4$ gives\cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3).Use $z-z^{-1}=2i\sin\theta$ for sine powers.

$$For a trigonometric series, combine it as$$C+iS=\sum_{r=0}^{n}a_r(\cos r\theta+i\sin r\theta)=\sum_{r=0}^{n}a_rz^r,\qquad z=e^{i\theta}.Evaluate the resulting algebraic or geometric series, rationalise if needed, then take its real part for $C$ and imaginary part for $S$.

If $w^n=R\operatorname{cis}\phi$, all roots arew_k=R^{1/n}\operatorname{cis}\left(\frac{\phi+2\pi k}{n}\right),\qquad k=0,1,\ldots,n-1.For the nth roots of unity, $R=1$ and $\phi=0$, so the roots are equally spaced and their arguments differ by $2\pi/n$.

$$First write$$\frac{dy}{dx}+P(x)y=Q(x).$$Then$$\mu(x)=e^{\int P(x),dx},\qquad \frac{d}{dx}(\mu y)=\mu Q,$$so$$y=\frac{1}{\mu}\left(\int\mu Q,dx+C\right).Divide by the original coefficient of $dy/dx$ before identifying $P$.

For $x>0$, solvey'+y\coth x=\cosh x.Since $\mu=e^{\int\coth x\,dx}=\sinh x$,\frac{d}{dx}(y\sinh x)=\sinh x\cosh x.$$Hence$$y\sinh x=\tfrac12\sinh^2x+C,\qquad y=\tfrac12\sinh x+C\operatorname{cosech}x.$$$$

$$For$$y''-3y'+2y=e^{3x},the homogeneous roots are 1 and 2, so $y_c=C_1e^x+C_2e^{2x}$. Trying $y_p=Ae^{3x}$ gives $(9-9+2)Ae^{3x}=e^{3x}$, hence $A=1/2$. Thereforey=C_1e^x+C_2e^{2x}+\tfrac12e^{3x}. $$$$

For a constant-coefficient homogeneous equation, try $y=e^{mx}$. A first-order equation $ay'+by=0$ gives $am+b=0$. A second-order equationay''+by'+cy=0$$gives the auxiliary equation$$am^2+bm+c=0.$$$$

$$For$$y''-4y'+13y=0,$$$$m^2-4m+13=0\quad\Rightarrow\quad m=2\pm3i.$$Hence$$y_c=e^{2x}(C_1\cos3x+C_2\sin3x).$$$$

$$For the syllabus case$$y''+4y=\sin2x,sine and cosine of 2x already belong to the CF, so use the supplied form $y_p=kx\cos2x$. Theny_p''+4y_p=-4k\sin2x.Matching $\sin2x$ givesk=-\tfrac14,so one PI is $y_p=-\tfrac14x\cos2x$.

For the given substitution $x=e^t$ (so $t=\ln x$, $x>0$),x\frac{dy}{dx}=\frac{dy}{dt},\qquad x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}.$$Therefore$$ax^2y''+bxy'+cy=F(x)$$becomes$$a\ddot y+(b-a)\dot y+cy=F(e^t),$$a constant-coefficient equation when the transformed right side permits it.$$

For a given homogeneous substitution $y=ux$,\frac{dy}{dx}=u+x\frac{du}{dx}.$$In$$\frac{dy}{dx}=\frac{x+y}{x-y},$$this gives$$u+xu'=\frac{1+u}{1-u},\qquad x\frac{du}{dx}=\frac{1+u^2}{1-u}. $$$$

$$Now separate and integrate:$$\frac{1-u}{1+u^2},du=\frac{dx}{x},$$$$\tan^{-1}u-\tfrac12\ln(1+u^2)=\ln|x|+C.Finally substitute $u=y/x$ and retain any domain restrictions or separately lost constant solutions.

$$Suppose$$y=3+C_1e^{-t}+C_2e^{-2t},\qquad y(0)=5,\quad y'(0)=-6.$$Then$$C_1+C_2=2,\qquad -C_1-2C_2=-6,so $C_1=-2$ and $C_2=4$. The particular solution isy=3-2e^{-t}+4e^{-2t}. $$$$