FortyFive › Revision notes › IB Maths: Analysis & Approaches › IB Maths: Analysis & Approaches: Number and algebra
IB Maths: Analysis & Approaches: Number and algebra Sequences, series, logs, binomial theorem, proof. 16 subtopics, full free lesson below.
In this topic: 1.1 · 1.2 · 1.3 · 1.4 · 1.5 · 1.6 · 1.7 · 1.8 · 1.9 · 1.10 · 1.11 · 1.12 · 1.13 · 1.14 · 1.15 · 1.16
1.1 Scientific notationVery large and very small numbers are written in standard form so they are easy to compare and calculate with.
\[a\times10^{k},\qquad 1\le a\lt 10,\ k\in\mathbb{Z}\]
Large numbers have a positive power: \(4\,500\,000=4.5\times10^{6}\). Small numbers have a negative power: \(0.0032=3.2\times10^{-3}\). To multiply, multiply the \(a\)-values and add the powers; to divide, divide and subtract. Then adjust so that \(1\le a\lt10\). Worked example Calculate \((3\times10^{5})(4\times10^{-2})\) in standard form.
Multiply the numbers: \(3\times4=12\). Add the powers: \(10^{5}\times10^{-2}=10^{3}\), giving \(12\times10^{3}\). Adjust: \(12\times10^{3}=\mathbf{1.2\times10^{4}}\). 1.2 Arithmetic sequences and seriesIn an arithmetic sequence you add the same common difference \(d\) each time, e.g. \(3, 7, 11, 15, \dots\) with \(d=4\).
\[u_n=u_1+(n-1)d\]
\[S_n=\frac{n}{2}\big(2u_1+(n-1)d\big)=\frac{n}{2}\,(u_1+u_n)\]
\(d=u_{n+1}-u_n\). If \(d\lt0\) the sequence decreases. Sigma notation : \(\displaystyle\sum_{k=1}^{n}u_k\) means \(u_1+u_2+\dots+u_n\).Real contexts: simple interest, seats in rows, savings that rise by a fixed amount each month. Worked example An arithmetic sequence has \(u_3=11\) and \(u_7=27\). Find \(S_{20}\).
\(u_1+2d=11\) and \(u_1+6d=27\). Subtract: \(4d=16\), so \(d=4\) and \(u_1=3\). \(S_{20}=\frac{20}{2}\big(2(3)+19(4)\big)=10(82)=\mathbf{820}\). 1.3 Geometric sequences and seriesIn a geometric sequence you multiply by the same common ratio \(r\) each time, e.g. \(5, 10, 20, 40, \dots\) with \(r=2\).
\[u_n=u_1\,r^{\,n-1}\]
\[S_n=\frac{u_1\left(r^{n}-1\right)}{r-1}=\frac{u_1\left(1-r^{n}\right)}{1-r},\qquad r\ne1\]
\(r=\dfrac{u_{n+1}}{u_n}\). A negative \(r\) makes the terms alternate in sign. To find \(n\), solve \(u_1r^{n-1}=k\) with logarithms or a GDC table. Worked example Find the sum of the first 8 terms of \(3, 6, 12, \dots\)
\(u_1=3,\ r=2,\ n=8\). \(S_8=\dfrac{3\left(2^{8}-1\right)}{2-1}=3\times255\). \(S_8=\mathbf{765}\). 1.4 Financial applications of geometric sequencesCompound interest is a geometric sequence: each period the balance is multiplied by the same factor.
\[FV=PV\times\left(1+\frac{r}{100k}\right)^{kn}\]
\[\text{Depreciation: }V=V_0\left(1-\frac{r}{100}\right)^{n}\]
\(r\) = annual rate (%), \(k\) = compounding periods per year, \(n\) = years. GDC finance solver: \(N=kn\), \(I\%=r\), \(PV\) negative (money paid in), \(P/Y=C/Y=k\). Give money answers to 2 decimal places. Worked example €5000 is invested at 4% per year, compounded quarterly. Find the value after 3 years.
\(PV=5000,\ r=4,\ k=4,\ n=3\). \(FV=5000\left(1+\frac{4}{400}\right)^{12}=5000\times1.01^{12}\). \(FV=\mathbf{\text{€}5634.13}\). 1.5 Laws of exponents and introduction to logarithms\[a^{m}\times a^{n}=a^{m+n}\qquad \frac{a^{m}}{a^{n}}=a^{m-n}\qquad \left(a^{m}\right)^{n}=a^{mn}\]
\[a^{0}=1\qquad a^{-n}=\frac{1}{a^{n}}\]
\[a^{x}=b\iff x=\log_a b\]
\(\log x\) means \(\log_{10}x\); \(\ln x\) means \(\log_e x\), where \(e\approx2.718\). Logarithms and exponentials undo each other: \(\ln\left(e^{x}\right)=x\) and \(e^{\ln x}=x\). Worked example Solve \(10^{x}=50\).
Take \(\log_{10}\) of both sides: \(x=\log 50\). \(x\approx\mathbf{1.70}\) (3 s.f.). 1.6 Simple deductive proofStart from one side (usually the left-hand side) and transform it, step by step, into the other side. Never work on both sides at once, and never start from the result you are trying to prove. Finish with a clear statement such as LHS = RHS. The symbol \(\equiv\) means “identically equal”. Worked example Prove that \((n+1)^{2}-(n-1)^{2}\equiv4n\).
LHS \(=\left(n^{2}+2n+1\right)-\left(n^{2}-2n+1\right)\) \(=n^{2}+2n+1-n^{2}+2n-1\) \(=4n=\) RHS. Proved. 1.7 Rational exponents and laws of logarithms\[a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^{m}\]
\[\log_a(xy)=\log_a x+\log_a y\qquad \log_a\!\left(\frac{x}{y}\right)=\log_a x-\log_a y\]
\[\log_a\!\left(x^{m}\right)=m\log_a x\qquad \log_a x=\frac{\log_b x}{\log_b a}\]
To solve \(a^{x}=b\), take logs of both sides. Every log argument must be positive, so check your solutions. Worked example Solve \(2^{x+1}=5^{x}\).
Take \(\ln\): \((x+1)\ln2=x\ln5\). \(x(\ln5-\ln2)=\ln2\). \(x=\dfrac{\ln2}{\ln2.5}\approx\mathbf{0.756}\). 1.8 Sum of an infinite geometric sequenceIf the terms get smaller and smaller, the series approaches a finite limit.
\[S_\infty=\frac{u_1}{1-r},\qquad |r|\lt1\]
If \(|r|\ge1\) the series diverges: it has no sum to infinity. Recurring decimals are infinite geometric series. Worked example Write \(0.\dot{4}\dot{5}=0.454545\ldots\) as a fraction.
\(0.45+0.0045+\dots\) has \(u_1=0.45\) and \(r=0.01\). \(S_\infty=\dfrac{0.45}{0.99}=\dfrac{45}{99}\). \(=\mathbf{\dfrac{5}{11}}\). 1.9 The binomial theorem\[(a+b)^{n}=\sum_{r=0}^{n}\binom{n}{r}a^{\,n-r}b^{\,r}\]
\[\binom{n}{r}={}^{n}C_r=\frac{n!}{r!\,(n-r)!}\]
The coefficients form the rows of Pascal's triangle. The general term \(\binom{n}{r}a^{n-r}b^{r}\) lets you find one term without expanding everything. Worked example Find the coefficient of \(x^{3}\) in \((2+x)^{5}\).
General term: \(\binom{5}{r}2^{5-r}x^{r}\), with \(r=3\). \(\binom{5}{3}\times2^{2}=10\times4\). Coefficient \(=\mathbf{40}\). 1.10 Counting principles and the extended binomial theorem HL \[{}^{n}P_r=\frac{n!}{(n-r)!}\ \ (\text{order matters})\qquad {}^{n}C_r=\frac{n!}{r!\,(n-r)!}\ \ (\text{order does not matter})\]
\[(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+\dots,\quad n\in\mathbb{Q},\ |x|\lt1\]
Arrangements with repeated items: divide by the factorial of each repeat. For \((a+bx)^{n}\) with rational \(n\), first write it as \(a^{n}\left(1+\frac{b}{a}x\right)^{n}\), valid for \(\left|\frac{b}{a}x\right|\lt1\). Worked example Expand \(\sqrt{1+x}\) up to the term in \(x^{2}\).
Use \(n=\tfrac12\): \(1+\tfrac12x+\dfrac{\frac12\left(-\frac12\right)}{2!}x^{2}\). \(=\mathbf{1+\tfrac{x}{2}-\tfrac{x^{2}}{8}}\), valid for \(|x|\lt1\). 1.11 Partial fractions HL \[\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}\]
Multiply through by the denominator, then substitute \(x=a\) and \(x=b\) to find \(A\) and \(B\) quickly. Used to integrate rational functions (see 5.15). Worked example Write \(\dfrac{5x+1}{(x+1)(x-1)}\) in partial fractions.
\(5x+1=A(x-1)+B(x+1)\). \(x=1\): \(6=2B\Rightarrow B=3\). \(x=-1\): \(-4=-2A\Rightarrow A=2\). \(=\mathbf{\dfrac{2}{x+1}+\dfrac{3}{x-1}}\). 1.12 Complex numbers: Cartesian form HL \[i^{2}=-1\qquad z=a+bi\qquad \operatorname{Re}(z)=a,\ \operatorname{Im}(z)=b\]
\[z^{*}=a-bi\qquad zz^{*}=a^{2}+b^{2}\]
Add and multiply like algebra, replacing \(i^{2}\) with \(-1\). Divide by multiplying numerator and denominator by the conjugate of the denominator. On an Argand diagram, \(a+bi\) is the point \((a,b)\). Worked example Write \(\dfrac{3+2i}{1-i}\) in the form \(a+bi\).
Multiply by \(\dfrac{1+i}{1+i}\). Numerator: \(3+3i+2i+2i^{2}=1+5i\). Denominator: \(1+1=2\). \(=\mathbf{\tfrac12+\tfrac52 i}\). 1.13 Modulus–argument (polar) and Euler form HL \[|z|=r=\sqrt{a^{2}+b^{2}}\qquad \arg z=\theta\]
\[z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta=re^{i\theta}\]
\[z_1z_2=r_1r_2\,e^{i(\theta_1+\theta_2)}\qquad \frac{z_1}{z_2}=\frac{r_1}{r_2}\,e^{i(\theta_1-\theta_2)}\]
Sketch the point first to choose the correct quadrant for \(\theta\). Give the principal argument, \(-\pi\lt\theta\le\pi\), unless told otherwise. Worked example Write \(z=1+i\) in Euler form.
\(r=\sqrt{1^{2}+1^{2}}=\sqrt2\). \(\theta=\arctan\frac11=\frac{\pi}{4}\) (first quadrant). \(z=\mathbf{\sqrt2\,e^{i\pi/4}}\). 1.14 Conjugate roots, De Moivre's theorem and roots of complex numbers HL \[\left[r(\cos\theta+i\sin\theta)\right]^{n}=r^{n}(\cos n\theta+i\sin n\theta)\]
\[z^{n}=re^{i\theta}\ \Rightarrow\ z=r^{\frac1n}e^{\,i\frac{\theta+2k\pi}{n}},\quad k=0,1,\dots,n-1\]
If a polynomial has real coefficients, its complex roots come in conjugate pairs. The \(n\) roots of a complex number are equally spaced around a circle of radius \(r^{1/n}\). Worked example Find \((1+i)^{8}\).
\(1+i=\sqrt2\operatorname{cis}\frac{\pi}{4}\). \((\sqrt2)^{8}\operatorname{cis}\left(8\cdot\frac{\pi}{4}\right)=16\operatorname{cis}2\pi\). \(=\mathbf{16}\). 1.15 Proof by induction, contradiction and counterexample HL Induction : (1) prove true for \(n=1\); (2) assume true for \(n=k\); (3) prove true for \(n=k+1\); (4) write the conclusion: “true for \(n=1\), and true for \(k\) implies true for \(k+1\), so true for all \(n\in\mathbb{Z}^{+}\)”.Contradiction : assume the opposite and reach something impossible (e.g. proving \(\sqrt2\) is irrational).Counterexample : one example that breaks a statement disproves it.Worked example Prove by induction that \(\displaystyle\sum_{r=1}^{n}r=\frac{n(n+1)}{2}\).
\(n=1\): LHS \(=1\), RHS \(=\frac{1\cdot2}{2}=1\). True. Assume true for \(n=k\): \(\displaystyle\sum_{r=1}^{k}r=\frac{k(k+1)}{2}\). For \(n=k+1\): \(\dfrac{k(k+1)}{2}+(k+1)=\dfrac{(k+1)(k+2)}{2}\), which is the formula with \(n=k+1\). True for \(n=1\) and true for \(k\Rightarrow\) true for \(k+1\), so true for all \(n\in\mathbb{Z}^{+}\) . 1.16 Systems of linear equations HL Up to 3 equations in 3 unknowns. Solve by elimination or with your GDC. Three outcomes: a unique solution ; no solution (a contradiction such as \(0=5\)); or infinitely many (a row \(0=0\)). For infinitely many solutions, write the general solution using a parameter, e.g. \(z=\lambda\). Worked example Solve \(x+y+z=6,\ \ x-y+z=2,\ \ 2x+y-z=1\).
Equation 1 minus equation 2: \(2y=4\Rightarrow y=2\). Then \(x+z=4\) and \(2x-z=-1\). Add: \(3x=3\Rightarrow x=1,\ z=3\). \(\mathbf{x=1,\ y=2,\ z=3}\). Check in equation 3: \(2+2-3=1\) ✓. Functions →
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