Numerical Analysis
Part of a project to develop a solution to the general problem f(x) =0. Following roughly the order of Ralston's [i]A First Course In Numercal Analysis[/i], worksheets examining the related problems of [math]\;\;\;[/math] Interpolation [math]\;\;\;[/math] Numerical differentiation and quadrature [math]\;\;\;[/math] Numerical solution of differential equations [math]\;\;\;[/math] The zeros of a polynomial [math]\;\;\;[/math] Approximation error
Table of Contents
- Interpolation
- Piecewise Interpolation
- Lagrange Interpolation: Equal Spacing
- Lagrange Polynomial: Even/Odd Sums
- Cubic Spline
- Example: Adpating A Bezier Curve
- Bezier Approximation, 1
- Bezier Approximation, 2
- Bezier Circle
- Quadrature, 1
- Quadrature: Approximate Osculating Circles
- The Arc Tangent
- Bezier Curve Arc Length: Polynomial Approximation
- Zeros of a Polynomial
- Newton's Method
- The Cubic Equation
- Cubic Equation, 1
- Cubic Equation: bounds
- Roots of the Cubic Equation: Solution
- Roots of the Cubic Equation: Solution 1 (Iteration)
- Quadratic approximation
- One root as a continued fraction
- Trigonometric sums
- Approximate the function cos(x)
- Trigonometric Polynomial, 1
- Trigonometric Polynomial, 2
- Tools
- Tool: Cubic Spline (Construction)
Piecewise Interpolation
Piecewise Interpolation, using f(x), f'(x) at the two nearest tabular points.
Piecewise Interpolation
Bezier Approximation, 1
An order n Bezier curve (a vector function) can represent smooth curves not possible with an order n polynomial. Are they a practical choice for function approximation? Consider the order 3 Bezier curve, defined by two points and the tangent vectors at those points:
Bezier Approximation, 1
Quadrature: Approximate Osculating Circles
Before getting into the textbook, let me consider some ways that one might approximate the arc length of a curve. [br][br]Firstly, I can always use linear approximation: lay out points along the curve, draw straight lines between adjacent points, and add up the distances. But perhaps I can do better than this. [br][br]The lenght of the circular arc through three consecutive points on the curve responds to both first and second derivatives, and should be a much better approximation. Let me try it.
Quadrature: Approximate Osculating Circles
Newton's Method
Consider a sequence of x-values:[br][math]\;\;\;[/math]Begin with [math] {\small x_0} [/math], chosen at pleasure.[br][math]\;\;\;[/math]Draw the tangent to the curve at x0:[br] [math]\;\;\;\;\;\; {\small y - f(x_0) = f'(x_0) (x-x_0) }. [/math][br][math]\;\;\;[/math]The x-intercept of this tangent line, is the next number in the sequence:[br] [math]\;\;\;\;\;\; {\small x_1 = x_0+ \frac{- f(x_0)}{ f'(x_0)} }. [/math]
Newton's Method
Cubic Equation, 1
Working through the problem of solving the cubic, the question arises, [i]suppose I remove the offset?[/i] That is, I have [br][math]\;\;\; f(x) = x³+ c_2x² + c_1 x + c_0,[/math][br]which is the same same as[br][math]\;\;\; g(x) = x³+ d_2x² + d_1 x,[/math][br][br]...shifted to the right or left by some unknown number of units.[br]Writing [math]{\small g(x) = x(x² + d_2x + d_1),} [/math], the first root is 0, and the last two roots are given by the quadratic equation. [br][br]If we write g(x) = f(x-t), is it possible to solve for the mystery offset [i]t[/i] and the new mystery coefficients?
Cubic Equation, 1
Approximate the function cos(x)
NOTES:[br][list=1][br][*]The red arc is the 2nd order Bezier curve: p(t)= A + 2(B-A)t + (C-2B+A)t², 0≤t≤1. [br]To find the vertex, focus, and parabola equation: [url]http://www.geogebratube.org/material/show/id/37814[/url][br][br][br][*] To convert this to an equation in x: [br] [math]\;\;\;[/math]Solve [math]{\small x= 2rt + (1-r) t²}\; [/math] for t.[br] [math]\;\;\;[/math]Then [math] {\small y(x) = 1- t(x)².}[/math][br]The conversion can be put off as long as possible. A, B,C, are sufficient to retrieve both the original function and calculate an approximation. [br][br][*]The analysis graphs don't prove the results. However, it should be clear that we only need to establish[br] [math]\;\;\;[/math]Does the parabola arc cross cos(x)?[br] in the neighborhood of a point where they are tangent. [br][/list] [br]________[br]{The zeros of a polynomial}
Tool: Cubic Spline (Construction)
Spline[<Points>, n, f'(a), f'(b)][br][br]gives the cubic spline passing through n equally spaced points, and with slopes f'(a), f'(b) at the endpoints.