Unit 15
Optimisation, dynamics and computation
Finding the best choice, what happens when a rule is repeated forever, and how computers do mathematics.
The story
Chapters
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
- Theorem · A Existence and uniqueness of the interpolating polynomial For distinct nodes there is exactly one polynomial of degree at most through the given points, and it is .
- Theorem · A The Vandermonde determinant . So is invertible exactly when the nodes are distinct.
- Theorem · A Interpolation error formula If has continuous derivatives on and interpolates it at distinct nodes in , then for each in there is a point in with .
- Corollary · A Error bound for equally spaced nodes With nodes , , we have on , so .
- Theorem · A Properties of the Chebyshev polynomials ; has degree and leading coefficient (for ); it has distinct roots in ; on and .
- Theorem · A Chebyshev's minimal property Among all monic polynomials of degree , none has a smaller largest absolute value on than , and achieves it.
- Corollary · A Error bound for Chebyshev nodes With the Chebyshev nodes on , , and no choice of nodes makes the factor smaller than .
- Theorem · A Newton's divided-difference formula With the interpolant through the first nodes, , and the divided differences satisfy the quotient recursion in the definition.
- Lemma · A Cauchy's mean value theorem If are continuous on , differentiable on and there, then for some .
- Theorem · A Error of the trapezoidal and Simpson rules On an interval of width , and for some (a different in each).
- Theorem · A Composite trapezoidal and Simpson rules With pieces of width : has error at most , and (for even) has error at most .
- Theorem · A Newton's method converges quadratically If , and is twice differentiable, then for some between and .
Terms this unit defines
Interpolating polynomialLagrange basis polynomialVandermonde matrixChebyshev polynomialChebyshev nodesDivided differenceQuadrature ruleTrapezoidal ruleSimpson's ruleFloating-point numberCatastrophic cancellation
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.
from 15.1 Numerical analysis
Your answerWhat is the determinant of the Vandermonde matrix for the nodes 0, 2 and 5?
Multiply every difference x_j − x_i with i < j.
The product of the differences: . Check with the rows , , : expanding along the first row gives .
from 15.1 Numerical analysis
Your answerFor the data (1, 2), (2, 3), (4, 9), what is the divided difference f[1, 2, 4]?
Divide the difference of the two chord slopes by x₂ − x₀.
and , so . It is also the leading coefficient of .
from 15.1 Numerical analysis
Your answerSimpson's rule with one piece (b − a = 2, nodes 0, 1, 2) for the integral of x⁴ from 0 to 2. What value does it give? (The exact value is 6.4.)
Use (b − a)/6 times f(a) + 4 f(m) + f(b).
. The error is , and the formula predicts , with constant: they agree exactly.
from 15.1 Numerical analysis
Your answerx = 1.0000001 and y = 1 are stored with relative error at most u = 1.1×10⁻¹⁶ each. The bound on the relative error of x − y is u(|x| + |y|)/|x − y|. What is it, measured in units of 10⁻⁹? (For example, 3 means 3×10⁻⁹.)
Plug in: u times 2 divided by 1e-7.
and , so the bound is , which is in units of . Seven of the sixteen digits are gone.
from 15.1 Numerical analysis
Your answerNewton's method for x² − 2 has e_(k+1) = e_k²/(2 x_k). If x_k is close to √2 and e_k = 10⁻³, about how large is e_(k+1), measured in units of 10⁻⁷? (For example, 3 means 3×10⁻⁷.)
Square the error to get 10⁻⁶, divide by 2 times the root (about 2.83), and then count in units of 10⁻⁷.
, which is in units of .
Where it leads
This is the last unit of the volume.