1. Further Pure Mathematics 1

For $A x^4+B x^3+C x^2+D x+E=0$ with roots $\alpha,\beta,\gamma,\delta$:\sum \alpha=-\frac{B}{A},\qquad \sum_{i<j}\alpha_i\alpha_j=\frac{C}{A},\qquad \sum_{i<j<k}\alpha_i\alpha_j\alpha_k=-\frac{D}{A},\qquad \alpha\beta\gamma\delta=\frac{E}{A}.$$The same alternating pattern truncates for degrees 2 and 3.$$

If $e_1=\sum\alpha$, $e_2=\sum_{i<j}\alpha_i\alpha_j$ and $e_3=\sum_{i<j<k}\alpha_i\alpha_j\alpha_k$, then\sum\alpha^2=e_1^2-2e_2,\qquad \sum\alpha^3=e_1^3-3e_1e_2+3e_3.$$$$

For reciprocal roots $y=1/x$, substitute $x=1/y$. Iff(x)=2x^3-3x^2+5x-7,$$then$$y^3f(1/y)=2-3y+5y^2-7y^3=0,$$so an equation for the reciprocal roots is$$7y^3-5y^2+3y-2=0.$$$$

After cancellation, $Q(a)=0$ gives a vertical asymptote $x=a$ when the numerator is non-zero there. If $\deg P=\deg Q$, the horizontal asymptote is the ratio of leading coefficients. If $\deg P=\deg Q+1$, division gives\frac{P(x)}{Q(x)}=m x+c+\frac{R(x)}{Q(x)},so $y=m x+c$ is the oblique asymptote.

$$For$$y=\frac{x^2-2x+3}{x^2+1},$$the denominator is always positive and$$(y-1)x^2+2x+(y-3)=0.$$Hence$$\Delta=4-4(y-1)(y-3)\ge0\iff y^2-4y+2\le0,$$so the range is$$2-\sqrt2\le y\le2+\sqrt2.The horizontal asymptote $y=1$ is crossed at $x=1$; an asymptote is limiting behaviour, not automatically a forbidden y-value.

$$$$\sum_{r=1}^{n}r=\frac{n(n+1)}2,\qquad \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}6,\qquad \sum_{r=1}^{n}r^3=\left[\frac{n(n+1)}2\right]^2.$$$$

$$$$\sum_{r=1}^{n}(2r-3)(r+4)=\sum_{r=1}^{n}(2r^2+5r-12)$$$$=\frac{n(n+1)(2n+1)}3+\frac{5n(n+1)}2-12n=\frac{n(4n^2+21n-55)}6.$$$$

$$$$\frac1{(r+1)(r+3)}=\frac12\left(\frac1{r+1}-\frac1{r+3}\right).$$$$

$$$$\sum_{r=1}^{n}\frac1{(r+1)(r+3)}=\frac12\left(\frac12+\frac13-\frac1{n+2}-\frac1{n+3}\right).$$The shift is 2, so two initial and two final fractions survive.$$

For $S_n=\sum_{r=1}^{n}a_r$, the infinite series converges to $L$ exactly when\lim_{n\to\infty}S_n=Lexists and is finite. Then $\sum_{r=1}^{\infty}a_r=L$.

$$From the finite result$$S_n=\frac12\left(\frac12+\frac13-\frac1{n+2}-\frac1{n+3}\right),$$both final fractions tend to 0, so$$\sum_{r=1}^{\infty}\frac1{(r+1)(r+3)}=\frac12\left(\frac12+\frac13\right)=\frac5{12}.$$$$

$$$$A=\begin{pmatrix}1&2&0\-1&3&1\end{pmatrix},\quad B=\begin{pmatrix}2&1\0&-2\4&3\end{pmatrix}$$$$AB=\begin{pmatrix}1(2)+2(0)+0(4)&1(1)+2(-2)+0(3)\-1(2)+3(0)+1(4)&-1(1)+3(-2)+1(3)\end{pmatrix}=\begin{pmatrix}2&-3\2&-4\end{pmatrix}. $$$$

$$For$$M=\begin{pmatrix}a&b\c&d\end{pmatrix},\qquad \det M=ad-bc,and, when $ad-bc\ne0$,M^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\-c&a\end{pmatrix}. $$$$

$$For$$M=\begin{pmatrix}1&1&0\0&1&1\0&0&2\end{pmatrix},\quad \det M=2,$$so M is non-singular. Row reduction gives$$M^{-1}=\begin{pmatrix}1&-1&1/2\0&1&-1/2\0&0&1/2\end{pmatrix},and direct multiplication gives $MM^{-1}=I$.

$$For non-singular A and B,$$(AB)^{-1}=B^{-1}A^{-1},$$because$$(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=I.$$More generally,$$(A_1A_2\cdots A_k)^{-1}=A_k^{-1}\cdots A_2^{-1}A_1^{-1}. $$$$

$$Let$$A=\begin{pmatrix}1&2\0&1\end{pmatrix},\quad B=\begin{pmatrix}3&0\0&1\end{pmatrix},\quad AB=\begin{pmatrix}3&2\0&1\end{pmatrix}. $$Then$$B^{-1}A^{-1}=\begin{pmatrix}1/3&0\0&1\end{pmatrix}\begin{pmatrix}1&-2\0&1\end{pmatrix}=\begin{pmatrix}1/3&-2/3\0&1\end{pmatrix}=(AB)^{-1}. $$$$

If B acts first and A second, the composite matrix is AB. For a stretch parallel to the x-axis by 3 followed by reflection in $y=x$,A=\begin{pmatrix}0&1\1&0\end{pmatrix},\quad B=\begin{pmatrix}3&0\0&1\end{pmatrix},\quad AB=\begin{pmatrix}0&1\3&0\end{pmatrix}. $$$$

For $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and a candidate line $y=mx$, the direction $(1,m)^T$ maps to $(a+bm,c+dm)^T$. Invariance requires parallel directions:c+dm=m(a+bm). Also check a vertical line separately using direction $(0,1)^T$.

$$For$$M=\begin{pmatrix}4&-1\2&1\end{pmatrix},$$$$2+m=m(4-m)\iff m^2-3m+2=0,so $m=1$ or $m=2$. Thus $y=x$ and $y=2x$ are invariant lines through the origin; points on them are generally scaled, not fixed.

With the syllabus convention $r\ge0$,x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2,\qquad \tan\theta=\frac yx\ (x\ne0). $$$$

Polar to Cartesian: replace $r\cos\theta$ by x, $r\sin\theta$ by y and $r^2$ by $x^2+y^2$. Forr=2(\cos\theta+\sin\theta),$$multiply by r to obtain$$x^2+y^2=2x+2y,$$or$$(x-1)^2+(y-1)^2=2.$$$$

A thin sector of angle $d\theta$ has area approximately $\tfrac12r^2d\theta$. Therefore, when $r=f(\theta)$ traces the intended boundary once from $\alpha$ to $\beta$,A=\frac12\int_{\alpha}^{\beta}r^2,d\theta. $$$$

The upper half of $r=a(1+\cos\theta)$ is traced once for $0\le\theta\le\pi$. ThusA=\frac{a^2}{2}\int_0^\pi(1+\cos\theta)^2d\theta$$$$=\frac{a^2}{2}\int_0^\pi(1+2\cos\theta+\cos^2\theta)d\theta=\frac{3\pi a^2}{4}. $$$$

$$For$$\mathbf r=\begin{pmatrix}1\0\2\end{pmatrix}+\lambda\begin{pmatrix}1\1\0\end{pmatrix}+\mu\begin{pmatrix}0\1\1\end{pmatrix},$$$$\mathbf n=\begin{pmatrix}1\1\0\end{pmatrix}\times\begin{pmatrix}0\1\1\end{pmatrix}=\begin{pmatrix}1\-1\1\end{pmatrix},\qquad p=3.Hence $\mathbf r\cdot(1,-1,1)=3$, orx-y+z=3.$$$$

For the angle $0\le\theta\le\pi$ between non-zero vectors a and b,\mathbf a\times\mathbf b=|\mathbf a|,|\mathbf b|\sin\theta,\hat{\mathbf n},where $\hat{\mathbf n}$ is the unit normal selected by the right-hand rule. Thus $|\mathbf a\times\mathbf b|$ is the parallelogram area and half of it is the triangle area.

If $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$,\mathbf a\times\mathbf b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1). $$$$

For $\mathbf a=(1,2,-1)$ and $\mathbf b=(3,0,2)$,\mathbf a\times\mathbf b=(4,-5,-6).Check: $(4,-5,-6)\cdot(1,2,-1)=0$ and $(4,-5,-6)\cdot(3,0,2)=0$. Its magnitude is $\sqrt{77}$, so a unit normal is $\pm(4,-5,-6)/\sqrt{77}$.

For skew lines $L_1:\mathbf r=\mathbf a+\lambda\mathbf u$ and $L_2:\mathbf r=\mathbf b+\mu\mathbf v$, let $\mathbf w=\mathbf u\times\mathbf v$. Their shortest distance is\frac{|(\mathbf b-\mathbf a)\cdot\mathbf w|}{|\mathbf w|}. $$To find the common perpendicular itself, solve$$[\mathbf b+\mu\mathbf v-(\mathbf a+\lambda\mathbf u)]\cdot\mathbf u=0,$$$$[\mathbf b+\mu\mathbf v-(\mathbf a+\lambda\mathbf u)]\cdot\mathbf v=0.The resulting points P and Q determine $\mathbf r=\mathbf P+t(\mathbf Q-\mathbf P)$.

Claim: $24\mid(5^{2n}-1)$ for every positive integer n. Base: $5^2-1=24$. Assume $5^{2k}-1=24q$ for some integer q. Then5^{2(k+1)}-1=25\cdot5^{2k}-1=25(5^{2k}-1)+24=24(25q+1),$$so the k+1 expression is divisible by 24. Hence the claim follows for all positive integers n.$$

Base: $S_1=1=2!-1$. Assume $S_k=(k+1)!-1$. ThenS_{k+1}=S_k+(k+1)(k+1)!$$$$=(k+1)!-1+(k+1)(k+1)!=(k+2)(k+1)!-1=(k+2)!-1.$$Therefore the conjecture holds for all positive integers n by induction.$$