Survey of Mathematics
Unit 15

Optimisation, dynamics and computation

Finding the best choice, what happens when a rule is repeated forever, and how computers do mathematics.

1 chapter written · 11 definitions · 12 theorems · builds on Unit 1, Unit 2, Unit 5, Unit 7

The story

Chapters

  1. 15.1Numerical analysisA computer can only add, multiply and round, so it replaces hard functions by polynomials through sample points. The error of that replacement has an exact formula, equally spaced points can make it explode (Runge), Chebyshev points tame it, and rounding adds a second, separate kind of error.

The theorem chain

Every theorem in the unit, in the order it is proved. Each one is used by the ones after it. The label says whether it is on the 375/376 exams (T), an Apostol extra (A), or beyond both (B).

15.1 Numerical analysis

  1. Theorem · A Existence and uniqueness of the interpolating polynomial For distinct nodes there is exactly one polynomial of degree at most nn through the given n+1n+1 points, and it is p(x)=∑jyj ℓj(x)p(x) = \sum_j y_j\,\ell_j(x).
  2. Theorem · A The Vandermonde determinant det⁡V=∏0≤i<j≤n(xj−xi)\det V = \prod_{0 \le i < j \le n} (x_j - x_i). So VV is invertible exactly when the nodes are distinct.
  3. Theorem · A Interpolation error formula If ff has n+1n+1 continuous derivatives on [a,b][a, b] and pnp_n interpolates it at distinct nodes in [a,b][a, b], then for each xx in [a,b][a, b] there is a point ξ\xi in (a,b)(a, b) with f(x)−pn(x)=f(n+1)(ξ)(n+1)!∏j=0n(x−xj)f(x) - p_n(x) = \dfrac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{j=0}^{n}(x - x_j).
  4. Corollary · A Error bound for equally spaced nodes With nodes xj=a+jhx_j = a + jh, h=(b−a)/nh = (b-a)/n, we have ∣w(x)∣≤14hn+1n!|w(x)| \le \tfrac14 h^{n+1} n! on [a,b][a, b], so ∣f(x)−pn(x)∣≤Mn+1 hn+14(n+1)|f(x) - p_n(x)| \le \dfrac{M_{n+1}\,h^{n+1}}{4(n+1)}.
  5. Theorem · A Properties of the Chebyshev polynomials Tn(cos⁡θ)=cos⁡nθT_n(\cos\theta) = \cos n\theta; TnT_n has degree nn and leading coefficient 2n−12^{n-1} (for n≥1n \ge 1); it has nn distinct roots cos⁡(2k+1)π2n\cos\frac{(2k+1)\pi}{2n} in (−1,1)(-1, 1); ∣Tn∣≤1|T_n| \le 1 on [−1,1][-1, 1] and Tn(cos⁡kπn)=(−1)kT_n(\cos\frac{k\pi}n) = (-1)^k.
  6. Theorem · A Chebyshev's minimal property Among all monic polynomials of degree m≥1m \ge 1, none has a smaller largest absolute value on [−1,1][-1, 1] than 21−m2^{1-m}, and Tm/2m−1T_m/2^{m-1} achieves it.
  7. Corollary · A Error bound for Chebyshev nodes With the Chebyshev nodes on [−1,1][-1, 1], ∣f(x)−pn(x)∣≤Mn+12n (n+1)!|f(x) - p_n(x)| \le \dfrac{M_{n+1}}{2^n\,(n+1)!}, and no choice of n+1n+1 nodes makes the factor max⁡∣w∣\max|w| smaller than 2−n2^{-n}.
  8. Theorem · A Newton's divided-difference formula With pkp_k the interpolant through the first k+1k+1 nodes, pn(x)=∑k=0nf[x0,…,xk] (x−x0)⋯(x−xk−1)p_n(x) = \sum_{k=0}^{n} f[x_0, \dots, x_k]\,(x - x_0)\cdots(x - x_{k-1}), and the divided differences satisfy the quotient recursion in the definition.
  9. Lemma · A Cauchy's mean value theorem If g,kg, k are continuous on [0,h][0, h], differentiable on (0,h)(0, h) and k′≠0k' \ne 0 there, then g(h)−g(0)k(h)−k(0)=g′(τ)k′(τ)\dfrac{g(h) - g(0)}{k(h) - k(0)} = \dfrac{g'(\tau)}{k'(\tau)} for some τ∈(0,h)\tau \in (0, h).
  10. Theorem · A Error of the trapezoidal and Simpson rules On an interval of width b−ab - a, ∫abf−T=−(b−a)312f′′(η)\int_a^b f - T = -\dfrac{(b-a)^3}{12}f''(\eta) and ∫abf−S=−(b−a)52880f(4)(η)\int_a^b f - S = -\dfrac{(b-a)^5}{2880}f^{(4)}(\eta) for some η∈(a,b)\eta \in (a, b) (a different η\eta in each).
  11. Theorem · A Composite trapezoidal and Simpson rules With NN pieces of width hh: TN=h(12f0+f1+⋯+fN−1+12fN)T_N = h\big(\tfrac12 f_0 + f_1 + \cdots + f_{N-1} + \tfrac12 f_N\big) has error at most (b−a)h212M2\dfrac{(b-a)h^2}{12}M_2, and (for NN even) SN=h3(f0+4f1+2f2+4f3+⋯+4fN−1+fN)S_N = \dfrac h3\big(f_0 + 4f_1 + 2f_2 + 4f_3 + \cdots + 4f_{N-1} + f_N\big) has error at most (b−a)h4180M4\dfrac{(b-a)h^4}{180}M_4.
  12. Theorem · A Newton's method converges quadratically If f(r)=0f(r) = 0, f′(xk)≠0f'(x_k) \ne 0 and ff is twice differentiable, then xk+1−r=f′′(ξk)2f′(xk) (xk−r)2x_{k+1} - r = \dfrac{f''(\xi_k)}{2f'(x_k)}\,(x_k - r)^2 for some ξk\xi_k between xkx_k and rr.

Terms this unit defines

Unit review

A mixed set of 5 quick checks drawn at random from the 5 in this unit's chapters. Each one says which chapter it comes from, so a miss tells you what to reread. Press "New set" for another mix.

Where it leads

This is the last unit of the volume.