Before unit 00 · readiness

Readiness: what this course assumes, tested

unit 00's slides · the curriculum

Checkpoint. Start unit 00 now if both of these hold. (1) Of the five Calculus fluency problems you finish at least four with the right final number or closed form, on paper, with no hint opened, in about two and a half hours for the five together. (2) In each of the other four groups you reach the answer to most parts with at most the first rung open. The calculus group decides: if (1) fails, spend the weeks on it before unit 00, whatever the rest says. If only (2) fails for some group, start anyway and fill that gap before the first unit that leans on it — each problem's exercises line names that unit.

Work each problem on paper first. Under it, Hints opens a ladder one rung at a time — each rung appears only once the one before it is open, and the last gives the first line of the working — and Worked solution has the whole thing: the idea, the working, then a check. Read the solution after your attempt, including when you got it right. F folds everything back; D switches dark and light.

Twelve problems, about four and three-quarter hours: roughly two and a half for calculus, an hour for linear algebra, forty-five minutes for analysis, and half an hour for the last two groups. Time yourself per group — the time is part of the result. A computer algebra system may check a final answer; it may not produce one.

What this sheet tests, and what it does not. The course assumes you read and write proofs, and if that is where you come from, it is not what is in doubt. Only two problems here are proofs from start to finish, and they are there to check the analysis, not your proof technique. The sheet tests whether computation will be slow. The course's worked solutions are long chains of substitution, parts, expansion and simplification, carried to a number and checked. Someone who can prove the mean value theorem but has not done a few hundred integrals follows every step and still takes several times as long, and over 41 units that adds up to months.

Reading the result, group by group. The times are typical for a mathematically mature reader working an hour or two a day. They are rough.

  • Calculus fluency went badly — you knew the method but got stuck in the algebra, or lost signs and factors, or took much longer than the guide. This is the gap that matters. The cure is volume, not reading: 15–25 integrals, expansions or series a day for three to six weeks, each carried to a checked answer. For drill, use Demidovich, Problems in Mathematical Analysis (the chapters on indefinite integrals, definite integrals and series: do the odd-numbered problems, and check the answers at the back). For the method behind the drill, use Spivak, Calculus (3rd or 4th ed.), chapter 19, Integration in elementary terms, and chapter 20, Approximation by polynomial functions, plus its chapter on infinite series for P4. All have plenty of problems. Stop when P1–P4 would take you an hour.
  • Real analysis went badly — uniform convergence, improper integrals or differentiating under the integral sign felt new. Allow three to four weeks with Abbott, Understanding Analysis (2nd ed.): chapter 6, Sequences and series of functions (uniform convergence, power series, Taylor's theorem), and chapter 7, The Riemann integral. The section on the factorial (Gamma) function in chapter 8 is where Abbott handles improper integrals and differentiating under the integral sign. Unit 01 is the first unit that needs this.
  • Linear algebra went badly — usually it is duals and bilinear forms, not eigenvalues. Allow two to four weeks with Axler, Linear Algebra Done Right, 4th edition (free online from the author): section 3F on duality; chapter 6 on inner products and chapter 7 for the spectral theorem; section 9A on bilinear and quadratic forms; and chapter 8, for Jordan form as a statement only. The third edition has no bilinear-form chapter. Axler does not dwell on signature, so read Sylvester's law of inertia and its proof in the chapter on bilinear forms in Hoffman and Kunze, Linear Algebra (2nd ed.). Unit 04 is the first to lean on it (the signature of a Hessian); units 07, 16, 26 and 31 need all of it, and unit 16 needs duals most.
  • Metric spaces went badly. Allow one to two weeks with Sutherland, Introduction to Metric and Topological Spaces (2nd ed.): its chapters on complete spaces and on compactness. Abbott's metric-space section in chapter 8 is a shorter alternative for the same two ideas. P11's solution lists what the contraction theorem needs; unit 01's first proof uses all of it.
  • Groups went badly. Allow a few days to a week with Armstrong, Groups and Symmetry: its opening chapters, up to and including the one on actions, orbits and stabilizers. Unit 13 is the first unit that leans on this; nothing further into group theory is assumed.

If everything went well, stop here and open unit 00. The sheet has done its job in an afternoon.

Calculus fluency

P1 Substitution, parts, partial fractions routine

exercises: substitution then integration by parts, partial fractions with an irreducible quadratic — partial fractions from unit 00 (quadratic drag); integration by parts heavily in 09, 11, 19

Evaluate exactly, then give each to five significant figures.

  1. \(\displaystyle \int_0^{\pi^2/4} \sin\sqrt x\;\mathrm dx\).
  2. \(\displaystyle \int_0^{1} \frac{\mathrm dx}{1+x^3}\).
Hints · 3 rungs
Rung 1

Both are plain real numbers. The first is an integer. The second is a logarithm plus a multiple of \(\pi\): the \(\pi\) comes from an arctangent.

Rung 2

(1) Remove the root with a substitution. What is left is a polynomial times a sine, which is a job for parts. (2) Factor \(1+x^3\) over the reals into a linear and an irreducible quadratic factor, and split into partial fractions. Then split the quadratic's numerator into a multiple of the denominator's derivative plus a constant.

Rung 3

(1) \(x = t^2\), \(\mathrm dx = 2t\,\mathrm dt\), \(t\) from \(0\) to \(\pi/2\): the integral is \(2\int_0^{\pi/2} t\sin t\,\mathrm dt\). (2) \(1+x^3 = (1+x)(x^2-x+1)\), and \(\dfrac1{1+x^3} = \dfrac{A}{1+x} + \dfrac{Bx+C}{x^2-x+1}\); put \(x=-1\) to get \(A\) at once.

Worked solution

Idea. Each integral takes two moves in sequence. The skill being tested is doing both without losing a factor of 2 or a sign.

1. With \(x = t^2\), \(\mathrm dx = 2t\,\mathrm dt\), limits \(0\to\pi/2\): \[ \int_0^{\pi^2/4}\sin\sqrt x\,\mathrm dx = 2\int_0^{\pi/2} t\sin t\,\mathrm dt = 2\Big[-t\cos t\Big]_0^{\pi/2} + 2\int_0^{\pi/2}\cos t\,\mathrm dt = 0 + 2\Big[\sin t\Big]_0^{\pi/2} = \mathbf 2 . \]

2. Here \(1+x^3 = (1+x)(x^2-x+1)\), and \(x^2-x+1 = (x-\tfrac12)^2+\tfrac34\) has no real root. Putting \(x=-1\) in \(1 = A(x^2-x+1) + (Bx+C)(1+x)\) gives \(3A = 1\). Comparing the \(x^2\) coefficients gives \(A + B = 0\), and the constant terms give \(A + C = 1\). So \[ \frac1{1+x^3} = \frac{1/3}{1+x} - \frac13\,\frac{x-2}{x^2-x+1}. \] Write \(x - 2 = \tfrac12(2x-1) - \tfrac32\). The first piece integrates to a logarithm, and the second to an arctangent, because \(\int \mathrm dx/((x-\frac12)^2+\frac34) = \frac2{\sqrt3}\arctan\frac{2x-1}{\sqrt3}\): \[ \int\frac{\mathrm dx}{1+x^3} = \tfrac13\ln(1+x) - \tfrac16\ln(x^2-x+1) + \tfrac1{\sqrt3}\arctan\frac{2x-1}{\sqrt3} . \] From \(0\) to \(1\), the middle term is \(0\) at both ends and the arctangent runs from \(-\pi/6\) to \(\pi/6\): \[ \int_0^1\frac{\mathrm dx}{1+x^3} = \frac{\ln 2}{3} + \frac{\pi}{3\sqrt3} \approx 0.23105 + 0.60460 = \mathbf{0.83565} . \]

Check. (1) On \([0,\pi^2/4]\) the integrand rises from \(0\) to \(1\), and the interval has length \(\pi^2/4\approx 2.47\). An answer of \(2\) means an average height of about \(0.81\), which is plausible for a concave rise to 1. (2) On \([0,1]\) the integrand falls from \(1\) to \(\tfrac12\). Simpson's rule with the midpoint value \(1/1.125\) gives \((1 + 4\cdot0.8889 + 0.5)/6 = 0.8426\), close to \(0.8357\). Differentiating the antiderivative gives back \(1/(1+x^3)\) (a CAS may confirm this).

P2 The integral every orbit needs medium

exercises: trigonometric integrals, the half-angle substitution, differentiating with respect to a parameter — used in 08, 09, 10, 15

Let \(0 \le e \lt 1\).

  1. Show that \(\displaystyle\int_0^{2\pi}\frac{\mathrm d\theta}{1+e\cos\theta} = \frac{2\pi}{\sqrt{1-e^2}}\).
  2. Deduce \(\displaystyle\int_0^{2\pi}\frac{\mathrm d\theta}{(1+e\cos\theta)^2}\) in closed form. Evaluate both integrals at \(e = \tfrac12\) to four decimals.
  3. What happens to both as \(e\to1^-\), and why is that expected from the integrand?

(In unit 08 the second integral is, up to constants, the period of an elliptical orbit of eccentricity \(e\).)

Hints · 3 rungs
Rung 1

The answer to part 2 is \(2\pi\) times a power of \(1-e^2\). Parts 1 and 2 are the same integral: the second comes from the first by one differentiation.

Rung 2

Part 1: substitute \(t = \tan(\theta/2)\). But first move the range to \((-\pi,\pi)\) by periodicity, because \(\tan(\theta/2)\) blows up at \(\theta=\pi\), in the middle of \([0,2\pi]\). Part 2: add a parameter, \(I(a) = \int_0^{2\pi}\mathrm d\theta/(a + e\cos\theta)\) for \(a \gt e\), and differentiate in \(a\).

Rung 3

With \(t=\tan(\theta/2)\): \(\cos\theta = \dfrac{1-t^2}{1+t^2}\), \(\mathrm d\theta = \dfrac{2\,\mathrm dt}{1+t^2}\), so \(a + e\cos\theta = \dfrac{(a+e)+(a-e)t^2}{1+t^2}\) and the integral becomes \(\displaystyle\int_{-\infty}^{\infty}\frac{2\,\mathrm dt}{(a+e)+(a-e)t^2}\).

Worked solution

Idea. Rationalise with the half-angle substitution to get an arctangent integral. Then note that the squared denominator is what a derivative with respect to the constant term produces.

1. For \(a \gt e \ge 0\) put \(I(a) = \int_0^{2\pi}\mathrm d\theta/(a+e\cos\theta)\). The integrand has period \(2\pi\), so integrate over \((-\pi,\pi)\) instead, where \(t = \tan(\theta/2)\) is a smooth increasing bijection onto \(\mathbb R\). Rung 3's substitution gives \[ I(a) = \int_{-\infty}^{\infty}\frac{2\,\mathrm dt}{(a+e)+(a-e)t^2} = \frac{2}{\sqrt{(a+e)(a-e)}}\Big[\arctan\Big(\sqrt{\tfrac{a-e}{a+e}}\,t\Big)\Big]_{-\infty}^{\infty} = \frac{2\pi}{\sqrt{a^2-e^2}} . \] At \(a=1\) this is \(2\pi/\sqrt{1-e^2}\).

2. For \(a\) near \(1\), the integrand \((a+e\cos\theta)^{-1}\) and its \(a\)-derivative are continuous on \([a_0,a_1]\times[0,2\pi]\) with \(a_0 \gt e\). That is a compact set, so differentiating under the integral sign is allowed: \[ -\int_0^{2\pi}\frac{\mathrm d\theta}{(a+e\cos\theta)^2} = I'(a) = -\frac{2\pi a}{(a^2-e^2)^{3/2}} \quad\Longrightarrow\quad \int_0^{2\pi}\frac{\mathrm d\theta}{(1+e\cos\theta)^2} = \frac{2\pi}{(1-e^2)^{3/2}} . \] At \(e = \tfrac12\), \(1-e^2 = \tfrac34\), so the two integrals are \(2\pi/\sqrt{0.75} = \mathbf{7.2552}\) and \(2\pi/0.75^{3/2} = \mathbf{9.6736}\).

3. Both diverge, like \((1-e)^{-1/2}\) and \((1-e)^{-3/2}\). At \(e=1\) the denominator \(1+\cos\theta\) vanishes to second order at \(\theta=\pi\), since \(1+\cos\theta\approx\tfrac12(\theta-\pi)^2\). So the integrand behaves like \((\theta-\pi)^{-2}\) and \((\theta-\pi)^{-4}\), and neither is integrable there. (In unit 08, \(e\to1\) is the orbit opening into a parabola, and the period becoming infinite.)

Check. At \(e=0\) both integrals are \(\int_0^{2\pi}\mathrm d\theta = 2\pi\), and both formulas give \(2\pi\) ✓. Numerical quadrature at \(e=\tfrac12\) gives \(7.25520\) and \(9.67360\) ✓. The second integral exceeds the first, as it must: Cauchy–Schwarz gives \(\int f^2 \ge \big(\int f\big)^2/2\pi\), and here \(\int f \ge 2\pi\), so the right side is already at least \(\int f\). In numbers, \(7.2552^2/2\pi = 8.378\), which lies between \(7.2552\) and \(9.6736\) ✓.

Another way

Anyone who knows residues can do part 1 in one line with \(z = e^{i\theta}\). The course reaches that in unit 20 and does not assume it. The half-angle route needs nothing but the arctangent.

P3 Fourth order, with the remainder pinned down medium

exercises: Taylor's theorem with the Lagrange remainder, expanding a composite function to several orders — used from unit 00; heavily in 06, 21, 27, 37

Let \(\gamma(\beta) = (1-\beta^2)^{-1/2}\) for \(0\le\beta\lt1\).

  1. Show \(\gamma(\beta) = 1 + \tfrac12\beta^2 + \tfrac38\beta^4 + R(\beta)\). Using the Lagrange form of the remainder, find constants that give \(c\,\beta^6 \le R(\beta) \le C(\beta)\,\beta^6\), where \(C(\beta)\) is explicit.
  2. At \(\beta = 0.1\), give the interval your bound puts \(R\) in, and check it against \(\gamma(0.1)\) computed directly.
  3. Expand \(\ln\cos x\) about \(0\) up to and including the \(x^4\) term, and check the result at \(x = 0.1\).
Hints · 3 rungs
Rung 1

Expand in \(u = \beta^2\), not in \(\beta\). Then "fourth order in \(\beta\)" is second order in \(u\), and the remainder is a third derivative in \(u\). In part 3 the answer is \(-\tfrac12x^2\) plus a negative multiple of \(x^4\).

Rung 2

Part 1: \(f(u) = (1-u)^{-1/2}\) has derivatives \(f^{(k)}(u) = c_k(1-u)^{-1/2-k}\), all positive and increasing on \([0,1)\). So the remainder \(\frac{f'''(\xi)}{3!}u^3\) with \(0\lt\xi\lt u\) is bracketed by its values at \(\xi=0\) and \(\xi=u\). Part 3: compose two known series, \(\cos x = 1 - w\) with \(w = \tfrac{x^2}2 - \tfrac{x^4}{24}+\dots\), and \(\ln(1-w) = -w - \tfrac{w^2}2 - \dots\).

Rung 3

\(f' = \tfrac12(1-u)^{-3/2}\), \(f'' = \tfrac34(1-u)^{-5/2}\), \(f''' = \tfrac{15}8(1-u)^{-7/2}\), so \(R = \tfrac{15}{8}\cdot\tfrac16(1-\xi)^{-7/2}u^3 = \tfrac5{16}(1-\xi)^{-7/2}\beta^6\). For part 3, keep only the terms of \(-w - w^2/2\) up to \(x^4\): \(w^2 = x^4/4 + O(x^6)\).

Worked solution

Idea. Taylor's theorem does more than list coefficients. With the remainder in Lagrange form, it says how wrong the truncation is, and a monotone derivative turns that into a two-sided bound.

1. Put \(u = \beta^2\in[0,1)\) and \(f(u) = (1-u)^{-1/2}\). Then \(f(0)=1\), \(f'(0) = \tfrac12\), \(f''(0) = \tfrac34\), and \(f'''(u) = \tfrac{15}{8}(1-u)^{-7/2}\). Taylor's theorem to second order in \(u\), with the Lagrange remainder, gives, for some \(\xi\in(0,u)\), \[ f(u) = 1 + \tfrac12u + \tfrac{3}{8}u^2 + \frac{f'''(\xi)}{6}u^3 \quad\Longrightarrow\quad \gamma(\beta) = 1 + \tfrac12\beta^2 + \tfrac38\beta^4 + \tfrac5{16}(1-\xi)^{-7/2}\beta^6 . \] Since \((1-\xi)^{-7/2}\) increases in \(\xi\) and \(0\lt\xi\lt\beta^2\), \[ \tfrac5{16}\,\beta^6 \;\lt\; R(\beta) \;\lt\; \tfrac5{16}\,(1-\beta^2)^{-7/2}\,\beta^6 . \] There is no \(\beta^5\) term because \(\gamma\) is even. The expansion is fourth order in \(\beta\) with an error of order \(\beta^6\).

2. At \(\beta=0.1\): the lower bound is \(\tfrac5{16}\times10^{-6} = 3.125\times10^{-7}\), and the upper bound is \(3.125\times10^{-7}\times0.99^{-3.5} = 3.237\times10^{-7}\). The truncation gives \(1 + 0.005 + 0.0000375 = 1.0050375\). Directly, \(\gamma(0.1) = 1/\sqrt{0.99} = 1.00503781526\), so \(R = 3.153\times10^{-7}\), inside \([3.125, 3.237]\times10^{-7}\) ✓.

3. \(\cos x = 1 - w\) with \(w = \tfrac{x^2}{2} - \tfrac{x^4}{24} + O(x^6)\), and \(w^2 = \tfrac{x^4}4 + O(x^6)\). So \[ \ln\cos x = -w - \tfrac12w^2 + O(w^3) = -\tfrac{x^2}2 + \tfrac{x^4}{24} - \tfrac{x^4}{8} + O(x^6) = -\frac{x^2}{2} - \frac{x^4}{12} + O(x^6) . \] At \(x=0.1\): the expansion gives \(-0.005 - 0.00000833 = -0.00500833\). Directly, \(\ln\cos 0.1 = -0.00500836\), and the difference, \(2.2\times10^{-8}\), is the size of an \(x^6\) term (the next one is \(-x^6/45 = -2.2\times10^{-8}\)) ✓.

Check. Differentiate: \((\ln\cos x)' = -\tan x = -x - \tfrac{x^3}3 - \dots\), and integrating term by term gives \(-\tfrac{x^2}2 - \tfrac{x^4}{12}\) ✓. For part 1, \(\gamma = 1 + \tfrac12\beta^2+\dots\) is the binomial series \((1-u)^{-1/2} = \sum\binom{2k}{k}(u/4)^k\), and \(k=2\) gives \(6/16 = 3/8\) ✓.

P4 Sum it, or decide it medium

exercises: summing series in closed form, comparison and asymptotic tests, radius of convergence — used in 09, 19, 20, 21

  1. Sum \(\displaystyle\sum_{n\ge1}\frac{n}{2^n}\) and \(\displaystyle\sum_{n\ge1}\frac{1}{n(n+1)(n+2)}\).
  2. Decide convergence, with a reason, for \(\displaystyle\sum_{n\ge1}\Big(1-\cos\frac1n\Big)\), \(\displaystyle\sum_{n\ge1}\big(n^{1/n}-1\big)\) and \(\displaystyle\sum_{n\ge2}\frac{(-1)^n}{\ln n}\). For the last one, say whether it converges absolutely.
  3. Find the radius of convergence of \(\displaystyle\sum_{n\ge1}\frac{n!}{n^n}\,x^n\).
Hints · 3 rungs
Rung 1

Two rationals, then a verdict per series (one converges, one diverges, and one converges only conditionally), then a famous constant.

Rung 2

For \(\sum n/2^n\), differentiate the geometric series. For the second sum, look for a telescoping difference of \(1/(n(n+1))\)-type terms. For part 2, find the size of each term for large \(n\): \(1-\cos h\) and \(e^h-1\) for small \(h\). For part 3, use the ratio test.

Rung 3

\(\sum_{n\ge1} n x^{n} = x\,\dfrac{\mathrm d}{\mathrm dx}\dfrac{1}{1-x} = \dfrac{x}{(1-x)^2}\) for \(|x|\lt1\). And \(\dfrac{1}{n(n+1)(n+2)} = \dfrac12\Big[\dfrac1{n(n+1)} - \dfrac1{(n+1)(n+2)}\Big]\).

Worked solution

Idea. Closed forms come from a known series by differentiating or telescoping. Convergence questions are settled by the size of the \(n\)-th term, found by a one-term expansion.

1. A power series can be differentiated term by term inside its radius, so for \(|x|\lt1\) we have \(\sum n x^{n} = x/(1-x)^2\). At \(x=\tfrac12\) this is \(\tfrac12/\tfrac14 = \mathbf 2\). For the second sum, the partial fraction in rung 3 telescopes: \(\sum_{n=1}^{N} = \tfrac12\big[\tfrac1{1\cdot2} - \tfrac1{(N+1)(N+2)}\big]\to\tfrac14\). So the sum is \(\mathbf{\tfrac14}\).

2.

  • \(1-\cos\frac1n = 2\sin^2\frac1{2n} \le 2\big(\frac1{2n}\big)^2 = \frac{1}{2n^2}\). The terms are positive and bounded by a convergent \(p\)-series, so the series converges (its sum is about \(0.7788\)).
  • \(n^{1/n}-1 = e^{(\ln n)/n}-1 \ge \frac{\ln n}{n} \ge \frac1n\) for \(n\ge3\), since \(e^h-1\ge h\). Comparison with the harmonic series shows it diverges.
  • \(1/\ln n\) decreases to \(0\), so the alternating series test gives convergence. It is not absolute: \(1/\ln n \gt 1/n\) for \(n\ge2\).

3. With \(a_n = n!/n^n\), \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} = \Big(\frac{n}{n+1}\Big)^n = \Big(1+\frac1n\Big)^{-n}\to\frac1e , \] so the radius is \(R = \mathbf e \approx 2.71828\).

Check. For \(\sum n/2^n\), the partial sums \(0.5, 1, 1.375, 1.625, 1.78, \dots\) rise towards 2. For the telescoping sum, the first term alone is \(1/6 = 0.1667\), and adding \(1/24\) and \(1/60\) gives \(0.225\), approaching \(0.25\). The radius \(e\) agrees with Stirling, \(n!/n^n\approx\sqrt{2\pi n}\,e^{-n}\), whose \(n\)-th root tends to \(1/e\).

P5 A long chain to one number hard

exercises: factoring a quartic over the reals, partial fractions, improper limits, keeping a long calculation clean — partial fractions from unit 00; chains this long are what the worked solutions in every unit look like, and unit 20's residues are the other way to do integrals like this one

Evaluate \(\displaystyle\int_0^\infty\frac{\mathrm dx}{1+x^4}\) exactly by partial fractions, and give it to five significant figures. Then find a second, shorter route that uses the substitution \(x\mapsto1/x\).

Hints · 3 rungs
Rung 1

The answer, a little over 1, is \(\pi\) divided by an algebraic number. The logarithmic parts of the antiderivative cancel in the limit, and the arctangents supply everything.

Rung 2

\(1+x^4\) has no real root, but it is a difference of squares once you add and subtract \(2x^2\). For the second route, compare \(\int_0^\infty\mathrm dx/(1+x^4)\) with what \(x\mapsto1/x\) turns it into, add the two, and look for the derivative of \(x - 1/x\).

Rung 3

\(1+x^4 = (1+x^2)^2 - 2x^2 = (x^2+\sqrt2\,x+1)(x^2-\sqrt2\,x+1)\), and by the symmetry \(x\mapsto-x\) the decomposition has the form \(\dfrac{Ax+B}{x^2+\sqrt2x+1} + \dfrac{-Ax+B}{x^2-\sqrt2x+1}\).

Worked solution

Idea. Factor into two real quadratics, split, and integrate each piece as logarithm plus arctangent. Nothing here is deep. The difficulty is the length of the chain, which is exactly what the course's solutions look like.

Factor and split. Rung 3's form gives \(1 = (Ax+B)(x^2-\sqrt2x+1) + (-Ax+B)(x^2+\sqrt2x+1)\). The \(x^2\) terms give \(-2\sqrt2A + 2B = 0\), and the constants give \(2B = 1\). So \(B = \tfrac12\) and \(A = \tfrac1{2\sqrt2}\), and \[ \frac1{1+x^4} = \frac1{2\sqrt2}\left[\frac{x+\sqrt2}{x^2+\sqrt2x+1} - \frac{x-\sqrt2}{x^2-\sqrt2x+1}\right] . \]

Integrate each piece. Write \(x\pm\sqrt2 = \tfrac12(2x\pm\sqrt2) \pm \tfrac{\sqrt2}{2}\) and \(x^2\pm\sqrt2x+1 = (x\pm\tfrac{\sqrt2}2)^2 + \tfrac12\). Since \(\int\mathrm dx/((x\pm\frac{\sqrt2}2)^2+\frac12) = \sqrt2\arctan(\sqrt2x\pm1)\), this gives \[ \int\frac{\mathrm dx}{1+x^4} = \frac1{4\sqrt2}\ln\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1} + \frac1{2\sqrt2}\Big[\arctan(\sqrt2x+1) + \arctan(\sqrt2x-1)\Big] . \]

Limits. As \(x\to\infty\) the log's argument tends to \(1\), and each arctangent tends to \(\pi/2\). At \(x=0\) the log is \(\ln1 = 0\) and the arctangents are \(\pi/4\) and \(-\pi/4\). So \[ \int_0^\infty\frac{\mathrm dx}{1+x^4} = \frac{1}{2\sqrt2}\cdot\pi = \frac{\pi}{2\sqrt2} \approx \mathbf{1.1107} . \]

Check. Differentiating the antiderivative returns \(1/(1+x^4)\). A CAS confirms this, and by hand the arctangent pair differentiates to \(\frac{\sqrt2}{1+(\sqrt2x+1)^2} + \frac{\sqrt2}{1+(\sqrt2x-1)^2}\). For size: the integral lies between \(\int_0^1\tfrac12\,\mathrm dx + \int_1^\infty\tfrac{\mathrm dx}{2x^4} = \tfrac12+\tfrac16\) and \(1 + \tfrac13\), and \(1.1107\) is in \([0.667, 1.333]\) ✓. Numerical quadrature gives \(1.11072\) ✓.

Another way

With \(x\mapsto1/x\), \(I = \int_0^\infty\frac{\mathrm dx}{1+x^4} = \int_0^\infty\frac{x^2\,\mathrm dx}{1+x^4}\). Adding the two and dividing the numerator and denominator by \(x^2\), \[ 2I = \int_0^\infty\frac{1+x^{-2}}{x^2+x^{-2}}\,\mathrm dx = \int_{-\infty}^{\infty}\frac{\mathrm du}{u^2+2} = \frac{\pi}{\sqrt2}, \qquad u = x - \frac1x , \] since \(x^2+x^{-2} = u^2+2\), \(\mathrm du = (1+x^{-2})\,\mathrm dx\), and \(u\) runs increasingly over \(\mathbb R\). So \(I = \pi/(2\sqrt2)\) again ✓.

Real analysis

P6 A limit you may not pass under the integral medium

exercises: pointwise versus uniform convergence, interchanging limit and integral — used in 01 (Picard iteration), 09, 19

Let \(f_n(x) = n\,x\,e^{-n x^2}\) on \([0,1]\).

  1. Find the pointwise limit \(f\).
  2. Compute \(\sup_{[0,1]}|f_n - f|\), and decide whether \(f_n\to f\) uniformly on \([0,1]\).
  3. Compute \(\lim_n\int_0^1 f_n\) and \(\int_0^1 f\). Which theorem would have made them equal, and which of its hypotheses fails?
  4. Is the convergence uniform on \([\delta,1]\) for a fixed \(\delta\gt0\)?
Hints · 3 rungs
Rung 1

Part 1's answer is a function; part 2's is a formula in \(n\) followed by a yes or no; part 3's is two numbers and the name of a theorem. Sketch \(f_n\) for a few values of \(n\) before computing anything.

Rung 2

Find where \(f_n\) is maximal by differentiating. The bump moves towards \(0\) and grows taller. The integral has an elementary antiderivative, because \(x\,e^{-nx^2}\) is almost a derivative.

Rung 3

\(f_n'(x) = n e^{-nx^2}(1 - 2nx^2)\), which vanishes at \(x_n = 1/\sqrt{2n}\in(0,1]\). And \(\int_0^1 n x e^{-nx^2}\,\mathrm dx = \big[-\tfrac12 e^{-nx^2}\big]_0^1\).

Worked solution

Idea. A bump that narrows and rises can keep its area while vanishing at every fixed point. Pointwise convergence sees each point; the integral sees the bump.

1. \(f_n(0) = 0\). For \(x\gt0\), \(n e^{-nx^2}\to0\) because the exponential beats the linear factor. So \(f\equiv0\).

2. By rung 3, \(f_n\) increases on \([0,x_n]\) and decreases after, so \[ \sup_{[0,1]}|f_n| = f_n\big(1/\sqrt{2n}\big) = \frac{n}{\sqrt{2n}}\,e^{-1/2} = \sqrt{\frac{n}{2e}}\;\to\;\infty . \] Not only not uniform: the sup is unbounded (\(4.29\) at \(n=100\)).

3. \(\int_0^1 f_n = \tfrac12(1-e^{-n})\to\tfrac12\), while \(\int_0^1 f = 0\). The theorem is: if \(f_n\to f\) uniformly on \([a,b]\) and each \(f_n\) is Riemann-integrable, then \(\int f_n\to\int f\). Its proof is one line, \(|\int(f_n-f)|\le(b-a)\sup|f_n-f|\). That is exactly the hypothesis part 2 destroyed.

4. Yes. Once \(x_n = 1/\sqrt{2n}\lt\delta\), \(f_n\) is decreasing on \([\delta,1]\), so \(\sup_{[\delta,1]}f_n = n\delta e^{-n\delta^2}\to0\). The failure is concentrated at \(0\).

Check. At \(n=1\): the sup is \(\sqrt{1/(2e)} = 0.4289\), attained at \(x=0.7071\), where \(0.7071\,e^{-0.5} = 0.4289\) ✓, and the integral is \(\tfrac12(1-e^{-1}) = 0.3161\). The integrals \(0.3161, 0.4323, 0.4751, \dots\) approach \(\tfrac12\) fast (\(0.49998\) at \(n = 10\)), since the bump is nearly all inside \([0,1]\) after a few steps.

P7 Proof: the Dirichlet integral, by differentiating a parameter hard

exercises: improper integrals, conditional convergence, differentiating under the integral sign and justifying it, a uniform tail bound — used in 11, 20, 21

  1. Show that \(\int_0^\infty\frac{\sin x}{x}\,\mathrm dx\) converges as an improper integral, but \(\int_0^\infty\frac{|\sin x|}{x}\,\mathrm dx\) does not.
  2. For \(a\gt0\) let \(I(a) = \int_0^\infty e^{-ax}\,\frac{\sin x}{x}\,\mathrm dx\). Justify differentiating under the integral sign for \(a\ge\delta\gt0\), compute \(I'(a)\), and find \(I(a)\) in closed form. Give \(I(1)\) to five decimals.
  3. Prove \(I(a)\to\int_0^\infty\frac{\sin x}{x}\,\mathrm dx\) as \(a\to0^+\), and conclude its value.
Hints · 3 rungs
Rung 1

\(I(a)\) is an elementary function of \(a\), an inverse trigonometric one. The difficulty in part 3 is that the integral is only conditionally convergent, so it has no integrable dominating function near \(a=0\). You need a tail bound that is uniform in \(a\).

Rung 2

Part 1: integrate by parts on \([1,R]\) to trade \(\sin x/x\) for \(\cos x/x^2\), and bound \(\int_{k\pi}^{(k+1)\pi}|\sin x|/x\) below. Part 2: \(|\partial_a(\cdot)|\le e^{-\delta x}\), and \(I(a)\to0\) as \(a\to\infty\) fixes the constant. Part 3: \(g(x) = e^{-ax}/x\) is positive and decreasing for every \(a\ge0\).

Rung 3

For \(g\ge0\) decreasing and \(0\lt R\lt S\), integrating by parts gives \(\int_R^S g\sin = \big[-g\cos\big]_R^S + \int_R^S g'\cos\), so \(\big|\int_R^S g\sin\big|\le g(R)+g(S)+\big(g(R)-g(S)\big) = 2g(R)\).

Worked solution

Idea. Insert a damping factor that makes everything absolutely convergent. Solve the damped problem by differentiating in the damping, then remove the damping. That last step needs care, because the undamped integral converges only conditionally.

1. Near \(0\), \(\sin x/x\to1\), so \([0,1]\) causes no trouble. On \([1,R]\), \(\int_1^R\frac{\sin x}{x} = \big[-\frac{\cos x}{x}\big]_1^R - \int_1^R\frac{\cos x}{x^2}\). The bracket converges as \(R\to\infty\), and the last integral converges absolutely because \(|\cos x/x^2|\le1/x^2\). For the absolute value, \(\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{x}\,\mathrm dx\ge\frac{1}{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\sin x|\,\mathrm dx = \frac{2}{(k+1)\pi}\), and \(\sum 1/(k+1)\) diverges.

2. For \(a\ge\delta\gt0\), \(\big|\partial_a\big(e^{-ax}\sin x/x\big)\big| = e^{-ax}|\sin x| \le e^{-\delta x}\), which is integrable on \([0,\infty)\) and independent of \(a\). So the differentiated integral converges uniformly on \([\delta,\infty)\) (the Weierstrass test for integrals), and differentiation under the integral sign is licensed there. Cited: the Leibniz rule for improper integrals under a uniform dominating bound, as in Abbott's Gamma-function section. Then \[ I'(a) = -\int_0^\infty e^{-ax}\sin x\,\mathrm dx = -\frac{1}{1+a^2}, \] where integrating by parts twice gives \(a^2J = 1 - J\) for \(J = \int_0^\infty e^{-ax}\sin x\,\mathrm dx\). So \(I(a) = C - \arctan a\). Since \(|\sin x/x|\le1\), \(|I(a)|\le\int_0^\infty e^{-ax} = 1/a\to0\) as \(a\to\infty\). That forces \(C = \pi/2\), so \[ I(a) = \frac\pi2 - \arctan a = \arctan\frac1a, \qquad I(1) = \frac\pi4 = \mathbf{0.78540} . \]

3. Let \(I(0)\) denote the Dirichlet integral, which exists by part 1. For every \(a\ge0\), \(g(x) = e^{-ax}/x\) is positive and decreasing, and \(g(S)\to0\). So rung 3 gives the tail bound \(\big|\int_R^\infty e^{-ax}\frac{\sin x}{x}\big|\le\frac{2}{R}\), uniformly in \(a\ge0\). Now, for any \(R\gt0\), using \(|1-e^{-ax}|\le ax\) and \(|\sin x/x|\le1\): \[ |I(a) - I(0)| \le \int_0^R (1-e^{-ax})\,\mathrm dx\;+\;\frac2R+\frac2R \le \frac{aR^2}{2} + \frac4R . \] Given \(\varepsilon\), take \(R = 8/\varepsilon\), and then \(a \lt \varepsilon/R^2\). So \(I(a)\to I(0)\), and \(\int_0^\infty\frac{\sin x}{x}\,\mathrm dx = \lim_{a\to0^+}\arctan\frac1a = \frac\pi2\).

Check. Numerical quadrature: \(I(1) = 0.785398\) ✓ and \(I(\tfrac12) = 1.107149 = \arctan 2\) ✓. The Dirichlet integral, by oscillatory quadrature, gives \(1.570796\) ✓. As \(a\to\infty\), \(\arctan(1/a)\approx1/a\), which matches the crude bound \(|I(a)|\le1/a\).

Linear algebra

P8 Dual basis, and a bilinear form is not a linear map medium

exercises: dual spaces and dual bases, components of vectors and covectors under a change of basis, the matrix of a bilinear form — used in 16 (tensors as multilinear maps), 26, 31

In \(\mathbb R^3\) with standard basis \(e_1,e_2,e_3\) and standard dual basis \(e^1,e^2,e^3\), let \(b_1 = (1,0,0)\), \(b_2 = (1,1,0)\), \(b_3 = (1,1,1)\).

  1. Find the dual basis \(\beta^1,\beta^2,\beta^3\) of \(b_1,b_2,b_3\) in terms of \(e^1,e^2,e^3\).
  2. Find the components of \(v = (3,2,1)\) and of \(\omega = 2e^1 - e^2 + 3e^3\) in the new bases, and check \(\omega(v)\) both ways.
  3. Let \(B(x,y) = x_1y_2 + x_2y_1 + x_3y_3\), with matrix \(A\) in the standard basis. Find its matrix in the basis \(b_i\).
  4. Find the matrix, in the basis \(b_i\), of the linear map whose standard matrix is the same \(A\). Explain why it differs from part 3, and say when the two rules would agree.
Hints · 3 rungs
Rung 1

Each \(\beta^i\) is a difference of at most two \(e^j\). Vectors and covectors transform in opposite ways: one by \(P^{-1}\), the other by \(P\). The bilinear form's new matrix is symmetric, and the linear map's is not.

Rung 2

Let \(P\) be the matrix whose columns are the \(b_i\). The dual basis is defined by \(\beta^i(b_j) = \delta^i_j\), so the \(\beta^i\), written as rows, form \(P^{-1}\). The new matrix of \(B\) has entries \(B(b_i,b_j)\).

Rung 3

\(P = \begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix}\). Its inverse is bidiagonal: solve \(P\,c = v\) from the bottom row up. The form's new matrix is \(P^{\mathsf T}AP\), and the map's is \(P^{-1}AP\).

Worked solution

Idea. A basis change acts on vectors, on covectors and on bilinear forms by three different rules, all forced by the definitions. A bilinear form and a linear map can share a matrix in one basis and part company in the next.

1. \(P^{-1} = \begin{pmatrix}1&-1&0\\0&1&-1\\0&0&1\end{pmatrix}\), and its rows are the dual basis: \[ \beta^1 = e^1 - e^2,\qquad \beta^2 = e^2 - e^3,\qquad \beta^3 = e^3 . \] For example, \(\beta^1(b_2) = 1 - 1 = 0\) and \(\beta^2(b_2) = 1 - 0 = 1\) ✓.

2. Vector components are \(v^i = \beta^i(v)\), so \((3-2,\ 2-1,\ 1) = (1,1,1)\). Indeed \(b_1+b_2+b_3 = (3,2,1)\) ✓. Covector components are \(\omega_i = \omega(b_i)\): \(\omega(b_1) = 2\), \(\omega(b_2) = 2-1 = 1\), \(\omega(b_3) = 2-1+3 = 4\). This is the row vector \(\omega P = (2,1,4)\). In the standard basis \(\omega(v) = 6 - 2 + 3 = 7\), and in the new one \(2\cdot1+1\cdot1+4\cdot1 = 7\) ✓.

3. \(A = \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\), and the new matrix has entries \(B(b_i,b_j)\), that is \(P^{\mathsf T}AP\): \[ [B]_b = \begin{pmatrix}0&1&1\\1&2&2\\1&2&3\end{pmatrix} . \] For instance, \(B(b_2,b_3) = 1\cdot1 + 1\cdot1 + 0\cdot1 = 2\).

4. For a linear map, \(P^{-1}AP = \begin{pmatrix}-1&0&0\\1&1&0\\0&0&1\end{pmatrix}\). A linear map eats a vector and returns a vector, so the output must be converted back with \(P^{-1}\). A bilinear form eats two vectors and returns a number, so both inputs convert with \(P\) and nothing converts back. The two rules agree for every \(A\) exactly when \(P^{\mathsf T} = P^{-1}\), that is when the change is orthogonal. That is why the distinction is invisible in orthonormal bases, and why it matters in unit 26, where the natural bases are not orthonormal in the Euclidean sense.

Check. The invariants each rule must preserve are preserved. \([B]_b\) is symmetric, and \(\det[B]_b = \det A\,(\det P)^2 = -1\). Directly, \(0\cdot(6-4) - 1\cdot(3-2) + 1\cdot(2-2) = -1\) ✓. The map's matrix has trace \(-1+1+1 = 1 = \operatorname{tr}A\) ✓ and determinant \(-1 = \det A\) ✓.

P9 Signature, two ways routine

exercises: quadratic forms, completing squares, Sylvester's law of inertia, signature from eigenvalues — used in 04, 16, and above all 26 (Minkowski space is a quadratic form of signature (1,3))

  1. Find the signature of \(q(x,y,z) = xy + yz + zx\) by completing squares, and again from the eigenvalues of its symmetric matrix.
  2. Find the signature of \(p(x,y,z) = x^2 + 4xy + y^2 + 2z^2\).
  3. For \(q\), exhibit a 2-dimensional subspace on which \(q\) is negative definite, and prove that no 2-dimensional subspace makes \(q\) positive definite.
Hints · 3 rungs
Rung 1

A signature is a pair (number of positive squares, number of negative squares). Sylvester says it does not depend on how you diagonalise, so any route that ends in a sum of signed squares will do. Both forms are indefinite. Part 3 is about what the counts mean.

Rung 2

\(q\) has no square terms, so first make one: substitute \(x = X+Y\), \(y = X-Y\). For part 3, use \((x+y+z)^2 = x^2+y^2+z^2 + 2q\).

Rung 3

With \(x = X+Y\), \(y = X-Y\): \(xy = X^2-Y^2\) and \(z(x+y) = 2Xz\), so \(q = X^2 + 2Xz - Y^2\), and you can complete the square in \(X\).

Worked solution

Idea. Any diagonalisation gives the same counts of signs, so use whichever is quickest. The counts have an intrinsic meaning: the largest dimension of a subspace on which the form is positive (or negative) definite.

1. \(q = X^2 + 2Xz - Y^2 = (X+z)^2 - z^2 - Y^2\). The substitution \((x,y,z)\mapsto(X+z,\ Y,\ z)\) is invertible, so the signature is \(\mathbf{(1,2)}\). The matrix is \(\tfrac12\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}\), with eigenvalue \(1\) on \((1,1,1)\) and \(-\tfrac12\) twice on the plane \(x+y+z=0\), which again gives \((1,2)\) ✓.

2. \(p = (x+2y)^2 - 3y^2 + 2z^2\), so the signature is \(\mathbf{(2,1)}\). The eigenvalues are \(3\) and \(-1\) from the \(\begin{pmatrix}1&2\\2&1\end{pmatrix}\) block, and \(2\) ✓.

3. On \(W = \{x+y+z = 0\}\), the identity in rung 2 gives \(q = -\tfrac12(x^2+y^2+z^2)\), which is negative definite. Now suppose \(U\) is a 2-dimensional subspace with \(q\) positive definite on \(U\). Then \(\dim U + \dim W = 4 \gt 3\), so \(U\cap W\) contains some \(v\ne0\), and \(q(v)\) would be both \(\gt0\) and \(\lt0\), a contradiction. This is the whole proof of Sylvester's law in miniature.

Check. The signs of the determinant and trace are consistent. For (1,2), \(\det = 1\cdot(-\tfrac12)^2 = \tfrac14 \gt 0\) and the trace is \(0 = 1 - \tfrac12 - \tfrac12\) ✓. A spot value: \(q(1,1,-1) = 1 - 1 - 1 = -1 \lt 0\), and that vector is not in \(W\), so negative values occur off \(W\) too. The form is indefinite, as a signature of (1,2) says.

P10 Diagonalise a chain, and meet one you cannot medium

exercises: the spectral theorem for real symmetric matrices, orthonormal eigenbases, Jordan form as a statement, powers of a Jordan block — used in 06 (critical damping), 07 (normal modes), 16, 22

  1. Let \(K = \begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix}\). Find an orthogonal \(Q\) with \(Q^{\mathsf T}KQ\) diagonal. Give the eigenvalues exactly and to four decimals.
  2. Say in one sentence why the eigenvectors of a real symmetric matrix belonging to different eigenvalues are orthogonal, and prove it.
  3. Show that \(J = \begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}\) is not diagonalisable, and find \(J^n\) in closed form for \(n\ge1\).
Hints · 3 rungs
Rung 1

The three eigenvalues are symmetric about \(2\): \(2\) and \(2\pm\) a surd. The eigenvectors are symmetric or antisymmetric under reversing the order of the coordinates. \(J^n\) has one entry that is not a power of \(\lambda\): a factor of \(n\) appears.

Rung 2

\(K = 2I - N\), where \(N\) has ones next to the diagonal, so find the eigenvalues of \(N\). Try \((1,0,-1)\) and \((1,c,1)\). For part 3, write \(J = \lambda I + E\) with \(E^2 = 0\).

Rung 3

\(N(1,c,1)^{\mathsf T} = (c,\,2,\,c)^{\mathsf T}\), which is a multiple \(\mu(1,c,1)^{\mathsf T}\) when \(c = \mu\) and \(2 = \mu c\), that is \(\mu^2 = 2\). And \(\lambda I\) commutes with \(E\), so the binomial theorem applies to \((\lambda I + E)^n\).

Worked solution

Idea. Guess eigenvectors from the matrix's symmetry, then normalise. A matrix that is not diagonalisable is not exotic: the smallest example already has a factor of \(n\) in its powers, and that factor is where the \(t\,e^{\lambda t}\) of unit 06's critical damping comes from.

1. \(N(1,0,-1)^{\mathsf T} = (0,0,0)^{\mathsf T}\), so \(K\) has eigenvalue \(2\) on \((1,0,-1)\). By rung 3, \(N\) has eigenvalue \(\pm\sqrt2\) on \((1,\pm\sqrt2,1)\), so \(K\) has \(2\mp\sqrt2\) on those vectors. Normalising, \[ Q = \begin{pmatrix}\tfrac12&\tfrac1{\sqrt2}&\tfrac12\\ \tfrac{\sqrt2}{2}&0&-\tfrac{\sqrt2}2\\ \tfrac12&-\tfrac1{\sqrt2}&\tfrac12\end{pmatrix}, \qquad Q^{\mathsf T}KQ = \operatorname{diag}\big(2-\sqrt2,\ 2,\ 2+\sqrt2\big) = \operatorname{diag}(0.5858,\ 2,\ 3.4142) . \]

2. Symmetry makes \(K\) self-adjoint for the dot product, and a self-adjoint map cannot mix eigenvalues. If \(Ku = \lambda u\) and \(Kv = \mu v\), then \(\lambda\,u\cdot v = (Ku)\cdot v = u\cdot(Kv) = \mu\,u\cdot v\). So \((\lambda-\mu)\,u\cdot v = 0\), and \(u\cdot v = 0\) when \(\lambda\ne\mu\).

3. \(J\)'s only eigenvalue is \(\lambda\), and \(J - \lambda I = E\) has rank 1, so the eigenspace is 1-dimensional. There is no basis of eigenvectors. (Jordan's theorem says every complex matrix is similar to a block-diagonal matrix of blocks like \(J\). The course states and uses it and does not prove it.) Since \(\lambda I\) and \(E\) commute and \(E^2 = 0\), the binomial expansion of \((\lambda I+E)^n\) stops after two terms: \[ J^n = \lambda^nI + n\lambda^{n-1}E = \begin{pmatrix}\lambda^n&n\lambda^{n-1}\\0&\lambda^n\end{pmatrix} . \] (Summing \(t^nJ^n/n!\) turns the \(n\lambda^{n-1}\) into \(t\,e^{\lambda t}\). That sum is the matrix exponential, which unit 07 builds; nothing here assumes it.)

Check. The trace of \(K\) is \(6\) and the eigenvalues sum to \((2-\sqrt2)+2+(2+\sqrt2) = 6\) ✓. \(\det K = 2(4-1) - (-1)(-2) = 4\) and the product is \(2(4-2) = 4\) ✓. \(Q^{\mathsf T}Q = I\): the first column has length \(\tfrac14+\tfrac12+\tfrac14 = 1\) ✓. For part 3, \(n = 1\) gives back \(J\) ✓, and \(J^{n+1} = J\,J^n\) has top-right entry \(\lambda\cdot n\lambda^{n-1} + 1\cdot\lambda^n = (n+1)\lambda^n\) ✓. At \(\lambda = 2\), \(n = 5\) the formula gives \(5\cdot2^4 = 80\), and multiplying out \(J^5\) gives the same ✓.

Metric spaces

P11 Proof: a contraction and its fixed point medium

exercises: completeness, the Banach fixed-point theorem and its error bounds, which hypothesis does what — used in 01 (the first proof of the course), 04 (inverse function theorem), 09 (Kepler's equation by iteration)

  1. Show that \(T(x) = \cos x\) maps \([0,1]\) into itself and is a contraction there. Give its constant \(q\), and conclude that \(\cos x = x\) has exactly one solution \(x^*\) in \([0,1]\).
  2. Starting from \(x_0 = 1\), use the a priori bound \(|x_n - x^*|\le\frac{q^n}{1-q}|x_1-x_0|\) to find how many iterations guarantee \(|x_n-x^*|\lt10^{-6}\). Then iterate (a calculator is fine) and find the number actually needed. Explain the gap.
  3. \(S(x) = x + 1/x\) on \([1,\infty)\) satisfies \(|Sx - Sy| \lt |x-y|\) for \(x\ne y\), and the space is complete. Show that \(S\) has no fixed point, and say which hypothesis of Banach's theorem fails.
Hints · 3 rungs
Rung 1

The constant in part 1 is the largest value of \(|T'|\) on the interval. The fixed point is about \(0.739\). The a priori bound asks for more than twice the iterations that are really needed. In part 3 the missing ingredient is a single \(q\lt1\) that works for all pairs.

Rung 2

Use the mean value theorem, \(|\cos x - \cos y| = |\sin\xi|\,|x-y|\). \([0,1]\) is closed in \(\mathbb R\), hence complete. For the gap: near \(x^*\) the local rate is \(|T'(x^*)|\), not the worst case over the interval.

Rung 3

\(x_1 = \cos1 = 0.5403\), \(|x_1-x_0| = 0.4597\), \(1-q = 0.1585\). Solve \(q^n\cdot0.4597/0.1585\lt10^{-6}\) for \(n\) by taking logarithms, and remember \(\ln q\lt0\) flips the inequality. For part 3, compute \(|Sx - Sy|/|x-y|\) exactly.

Worked solution

Idea. Banach's theorem needs three things: a complete space, a self-map, and a uniform constant \(q\lt1\). Check each. The a priori bound uses only \(q\), which is why it is safe but pessimistic.

1. \(\cos\) decreases on \([0,1]\), so \(T([0,1]) = [\cos1,1] = [0.5403,1]\subset[0,1]\). By the mean value theorem, \(|\cos x-\cos y| = |\sin\xi|\,|x-y|\le\sin1\,|x-y|\), since \(\sin\) increases on \([0,1]\). So \(q = \sin1 = 0.841471\lt1\). \([0,1]\) is a closed subset of the complete space \(\mathbb R\), so it is complete. Banach's theorem gives exactly one fixed point, \(x^* = 0.7390851\).

2. The bound requires \(q^n \lt 10^{-6}(1-q)/|x_1-x_0| = 10^{-6}\times0.158529/0.459698\), that is \(n \gt \ln(3.449\times10^{-7})/\ln(0.841471) = 86.2\). So 87 iterations are guaranteed to suffice. Iterating from \(1\) gives \(0.5403,\ 0.8576,\ 0.6543,\ 0.7935,\ 0.7014,\dots\), and \(|x_n-x^*|\) first drops below \(10^{-6}\) at \(n = \mathbf{32}\). The gap arises because the iterates soon live near \(x^*\), where the contraction rate is \(|T'(x^*)| = \sin x^* = 0.6736\), not \(0.8415\). The iterates alternate about \(x^*\) because \(T' \lt 0\).

3. A fixed point would need \(1/x = 0\), which is impossible. For \(x\ne y\), \(Sx - Sy = (x-y)\big(1 - \tfrac1{xy}\big)\), and \(0\le1-\tfrac1{xy}\lt1\) on \([1,\infty)\). So \(|Sx-Sy|\lt|x-y|\), but the ratio \(1 - 1/(xy)\) tends to \(1\) as \(x,y\to\infty\). There is no uniform \(q\lt1\), and that is the failed hypothesis. The iterates \(x_{n+1} = x_n + 1/x_n\) increase without bound, because \(x_{n+1}^2 \ge x_n^2 + 2\).

Check. Predict the actual count from the local rate: the error starts at \(|1 - x^*| = 0.2609\) and shrinks by about \(0.6736\) per step, so \(n\approx\ln(10^{-6}/0.2609)/\ln0.6736 = 31.6\). The next integer is \(32\) ✓. And \(\cos(0.7390851) = 0.7390851\) ✓.

Groups

P12 A group of hyperbolic rotations routine

exercises: groups, homomorphisms, group actions, orbits and stabilisers — used lightly in 13 (one-parameter groups), 16 (SO(3)) and 26 (this very group, as Lorentz boosts)

For \(\varphi\in\mathbb R\) let \(L(\varphi) = \begin{pmatrix}\cosh\varphi&\sinh\varphi\\ \sinh\varphi&\cosh\varphi\end{pmatrix}\), acting on column vectors \((t,x)^{\mathsf T}\in\mathbb R^2\).

  1. Show \(L(\varphi_1)L(\varphi_2) = L(\varphi_1+\varphi_2)\). Deduce that \(G = \{L(\varphi)\}\) is a group under matrix multiplication, isomorphic to \((\mathbb R,+)\).
  2. Show that every \(L(\varphi)\) preserves \(t^2 - x^2\).
  3. Describe the orbit and the stabiliser of \((1,0)\), and the orbit of \((1,1)\).
  4. Put \(v = \tanh\varphi\). Find the composition law for \(v\), and evaluate it at \(v_1 = v_2 = \tfrac12\) and at \(v_1 = v_2 = 0.9\).
Hints · 3 rungs
Rung 1

Part 1 is the addition formulas for \(\cosh\) and \(\sinh\). Each orbit lies in a level set of \(t^2-x^2\) (that is part 2), but need not be the whole of it. Composing two values of \(v\) below 1 gives a value still below 1.

Rung 2

A homomorphism that is injective is an isomorphism onto its image, and the image of a group under a homomorphism is a group. For part 2 check \(L^{\mathsf T}\eta L = \eta\) with \(\eta = \operatorname{diag}(1,-1)\). For part 4 use the addition formula for \(\tanh\).

Rung 3

\(\cosh(\varphi_1+\varphi_2) = \cosh\varphi_1\cosh\varphi_2 + \sinh\varphi_1\sinh\varphi_2\) and \(\sinh(\varphi_1+\varphi_2) = \sinh\varphi_1\cosh\varphi_2 + \cosh\varphi_1\sinh\varphi_2\). Also \(L(\varphi)(1,1)^{\mathsf T} = (\cosh\varphi+\sinh\varphi)(1,1)^{\mathsf T} = e^\varphi(1,1)^{\mathsf T}\).

Worked solution

Idea. This is a one-parameter group, like the rotations \(R(\theta)\), with the circle \(t^2+x^2\) replaced by the hyperbola \(t^2-x^2\). Everything follows from one addition formula.

1. Multiplying out and using rung 3's formulas, the product has diagonal \(\cosh(\varphi_1+\varphi_2)\) and off-diagonal \(\sinh(\varphi_1+\varphi_2)\). So \(h:\varphi\mapsto L(\varphi)\) is a homomorphism \((\mathbb R,+)\to GL(2,\mathbb R)\) (note \(\det L = \cosh^2-\sinh^2 = 1\)). Its image \(G\) is therefore a subgroup: it contains \(L(0) = I\), is closed under products, and has inverses \(L(\varphi)^{-1} = L(-\varphi)\). \(h\) is injective, since the off-diagonal entry \(\sinh\varphi\) determines \(\varphi\). So \(G\cong(\mathbb R,+)\).

2. \((t',x') = (t\cosh\varphi + x\sinh\varphi,\ t\sinh\varphi + x\cosh\varphi)\) gives \(t'^2 - x'^2 = (t^2-x^2)(\cosh^2\varphi-\sinh^2\varphi) = t^2-x^2\), because the cross terms \(2tx\cosh\varphi\sinh\varphi\) cancel.

3. The orbit of \((1,0)\) is \(\{(\cosh\varphi,\sinh\varphi)\}\), the whole branch \(t^2-x^2 = 1\), \(t\gt0\). Its stabiliser is \(\{I\}\), since \(\sinh\varphi = 0\) forces \(\varphi = 0\). The orbit of \((1,1)\) is the open ray \(\{s(1,1): s\gt0\}\), since \(e^\varphi\) takes every positive value. It lies on the level set \(t^2-x^2 = 0\), as part 2 requires, but it is only one of that level set's four rays: preserving the form does not make a level set a single orbit.

4. \(\tanh(\varphi_1+\varphi_2) = \dfrac{\tanh\varphi_1+\tanh\varphi_2}{1+\tanh\varphi_1\tanh\varphi_2}\), so \[ v = \frac{v_1+v_2}{1+v_1v_2}, \qquad v(\tfrac12,\tfrac12) = \frac{1}{1.25} = \mathbf{0.8},\qquad v(0.9,0.9) = \frac{1.8}{1.81} = \mathbf{0.99448} . \] In unit 26 this is the relativistic addition of velocities, in units where light's speed is 1.

Check. Through \(\varphi\): \(\operatorname{artanh}\tfrac12 = 0.549306\), doubled is \(1.098612 = \ln 3\), and \(\tanh(\ln3) = (9-1)/(9+1) = 0.8\) ✓. Limits: \(v_2 = 0\) gives \(v = v_1\) ✓, and \(v_1 = 1\) gives \(v = 1\) for every \(v_2\), so 1 cannot be exceeded ✓.