수치해석

created : 2022-04-20T15:32:34+00:00
modified : 2022-06-16T01:30:04+00:00

Curves

Curves - Implicit Representation

  • Most curvest can be represented implicitly:
    • \[f(x, y) = 0\]
    • Examples:
      • \[ax + by + c = 0\]
      • \[x^2 + y^2 - r^2 = 0\]
    • In 3D, surfaces are represented as:
      • \[f(x, y, z) = 0\]

Curves - Parametric Representation

  • The value of each component (x, y, z) depends on an independent variable, u (the parameter).
  • \[p(u) = \begin{bmatrix} x(u) \\ y(u) \\ z(u) \end{bmatrix}\]
  • \[\frac{dp(u)}{du} = \begin{bmatrix} \frac{dx(u)}{du} \\ \frac{dy(u)}{y(u)} \\ \frac{dz(u)}{z(u)} \end{bmatrix}\]
    • Derivative : Direction = tangent, Magnitude = speed that curve changes with u

Criteria for choosing a representation

  • Parametric forms are not unique. Many different representations are possible for a single curve.
  • Criteria for a good representation:
    • Local control of shape
    • Smoothness and continuity
    • Ability to evaluate derivatives
    • Stability
    • Ease of Rendering
  • Parametric curves that are polynomials in u satisfy many of these criteria:
    • \[x(y) = c_0 + c_1 u + c_2 u^2 + \cdots + c_n u^n = \sum_{k=0}^n c_ku^k\]

Matrix Equation for parametric curve

  • We can write the 3 euqations for x(u), y(u), and z(u) in one matrix equation:
    • \[p(u) = \begin{bmatrix} x(u) \\ y(u) \\ z(u) \end{bmatrix} = \begin{bmatrix} \sum_{k=0}^n u^k c_{xk} \\ \sum_{k=0}^n u^k c_{yk} \\ \sum_{k=0}^n u^k c_{zk} \end{bmatrix} = \sum_{k=0}^n u^k \begin{bmatrix} c_{xk} \\ c_{yk} \\ c_{zk} \end{bmatrix} = \sum_{k=0}^n u^k c_k\]

Degree of the Polynomial

  • Tradeoff:
    • High degree can have rapid changes and lots of turns, but it requires more computation and curve may not be as smooth.
    • Low degree means a smoother curve, but it may not fit the data as well.
  • Compromise: Use low degree polynomials with short curve segments. Cubic polynomial work well.

Multiple curve segments

  • We usually want to define a curve segment between two endpoints. We define the curve between u=0 and u=1, so that:
    • $p(0) = p_0$ and $p(1) = p_1$
  • Longer curves are composed of multiple segments. We would like the connection between curves to be as smooth as possible.

Interpolation

  • If one point provides 3 equations and 12 unknowns, then 4 points provide 12 equations with 12 unknowns.
  • We can choose any value of u to correspond to the 4 points, so we will choose to divide the interval 0 to 1 evenly(0, 1/3, 2/3, 1).

PCA

Principal Components

  • All principal components(PCs) start at the origin
  • First PC is direction of maximum variance from origin
  • Subsequent PCs are orthogonal to 1st PC and describe maximum residual variance

Principal Component Analysis

  • For an arbitrary set of N vertices $P_1, P_2, …, P_N$
  • Mean position : $m = \frac{1}{N} \sum_{i=1}^N P_i$
  • 3x3 covariance matrix : $C= \frac{1}{N} \sum_{i=1}^N(P_i - m ) (P_i - m)^T$:
    • Represents the corrleation between each pair of the $x,y,$ and $z$ coordinates
  • Covariance matrix entries:
    • \[C_{11} = \frac{1}{N} \sum_{i=1}^N(x_i - m_x)^2, C_{12} = C_{21} = \frac{1}{N} \sum_{i=1}^N (x_i - m_x)(y_i - m_y)\]
    • \[C_{22} = \frac{1}{N} \sum_{i=1}^N(y_i - m_y)^2, C_{13} = C_{31} = \frac{1}{N} \sum_{i=1}^N (x_i - m_x)(z_i - m_z)\]
    • \[C_{33} = \frac{1}{N} \sum_{i=1}^N(z_i - m_z)^2, C_{23} = C_{32} = \frac{1}{N} \sum_{i=1}^N (y_i - m_y)(z_i - m_z)\]
  • An entry of zero : no corrleation
  • $C$ is diagonal matrix : three coordinates are uncorrelated

  • We want to transform points so that the covariance matrix is diagonal
    • \[\begin{aligned}C' &= \frac{1}{N} \sum_{i=1}^N(AP_i - Am)(AP_i - Am)^T \\ &= \frac{1}{N} \sum_{i=1}^N A(P_i - m)(P_i - m)^T A^T \\ &= ACA^T \end{aligned}\]
  • Find $A$ using eigenvectors:
    • Rows of $A$ are unit eigenvectors sorted by eigenvalues in decreasing order

Higher Diemnsions

  • Suppose data poitns are N-dimensional:
    • Same procedure applies :$C’ = ACA^T$
    • The eigenvectors define a new coordinates:
      • eigenvector with largest eigenvalue captures the most variation among training vectors
      • eigenvector with smallest eigenvalue has least variation
    • We can compress the data by only using the top few eigenvectors:
      • corresponds to choosing a “linear subspace”

Decomposition

Trianglular Systems

  • Lower triangular matrix $L$:
    • Square matrix for which $L_{ij} = 0$ when $i < j$
  • Linear system $Lx = r$
  • \[\begin{bmatrix} L_{11} & 0 & \cdots & 0 \\ L_{21} & L_{22} & \cdots 0 \\ \vdots & \vdots & \ddots & \vdots \\ L_{n1} & L_{n2} & \cdots & L_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{bmatrix} = \begin{bmatrix} r_1 \\ r_2 \\ \vdots \\ r_n \end{bmatrix}\]
  • Forward substitution:
    • \[x_i = \frac{1}{L_{ii}} ( r_i - \sum_{k=1}^{i-1} L_{ik}x_k)\]
  • Upper triangle
    • Square matrix for which $U_{ij} = 0$ when $i > j$
  • \[\begin{bmatrix} U_{11} & U_{12} & \cdots & U_{1n} \\ 0 & U_{22} & \cdots U_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & U_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{bmatrix} = \begin{bmatrix} r_1 \\ r_2 \\ \vdots \\ r_n \end{bmatrix}\]
  • Backward substitution:
    • \[x_i = \frac{1}{U_{ii}} ( r_i - \sum_{k=i+1}^{n} U_{ik}x_k)\]

LU Decomposition

  • $LU = M$
  • \[\begin{bmatrix} L_{11} & 0 & \cdots & 0 \\ L_{21} & L_{22} & \cdots 0 \\ \vdots & \vdots & \ddots & \vdots \\ L_{n1} & L_{n2} & \cdots & L_{nn} \end{bmatrix} \begin{bmatrix} U_{11} & U_{12} & \cdots & U_{1n} \\ 0 & U_{22} & \cdots U_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & U_{nn} \end{bmatrix} = \begin{bmatrix} M_{11} & M_{12} & \cdots & M_{1n} \\ M_{21} & M_{22} & \cdots & M_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ M_{n1} & M_{n2} & \cdots & M_{nn} \end{bmatrix}\]
  • \[M_{ij} = \sum_{k=1}^i L_{ik} U_{kj}, \text{ if } i \le j\]
  • \[M_{ij} = \sum_{k=1}^j L_{ik} U_{kj}, \text{ if } i \ge j\]
  • $Mx = r \Rightarrow LUx = r$:
    • Let $Ux = y$
    • Solve $Ly = r$
    • Solve $Ux = y$
  • Doolittle’s Method:
    • $L_{ii} \equiv 1, i =1, 2, …, n$
  • We can express L and U in one matrix:
    • \[D = \begin{bmatrix} U_{11} & U_{12} & U_{13} & \cdots & U_{1n} \\ L_{21} & U_{22} & U_{23} & \cdots & U_{2n} \\ L_{31} & L_{32} & U_{33} & \cdots & U_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ L_{n1} & L_{n2} & L_{n3} & \cdots & U_{nn} \end{bmatrix}\]
  • Solve U and L for each column j for top to bottom:
    • \[U_{1j} = M_{1j}; L_{i1} = \frac{M_i1}{U_{11}}\]
    • \[U_{ij} = M_{ij} - \sum_{k=1}^{i-1} L_{ik} U_{kj}, \text{ if } i > 1\]
    • \[L_{ij} = \frac{1}{U_{jj}} ( M_{ij} - \sum_{k=1}^{j - 1} L_{ik} U_{kj}), \text{ if } j > 1\]

Error Reduction

  • Suppose we solve a linear system $Mx = r$:
    • If we obtained a solution $x= x_0$, $x_0$ usually slightly different from the true solution due to round-off error
    • Thus $Mx_0 = r_0$:
      • $M(x + \Delta x) = r + \Delta r$, where $\Deltax = x_0 - x, \Deltar = r_0 - r$
      • $M \Delta x = \Delta r$
      • $M \Delta = M x_0 - r$
      • solve for the error $\Delta x$ and improve the solution : $x = x_0 - \Delta x$

Matrix Inversion

  • Matrix inverse can be computed in a column-by-column method
  • \[MM^{-1} = I\]

Iterative

Solving Large Lienar Systems

  • Global illumination (Radiosity):
    • Solve $B = (I - \rho F)^{-1} E$
    • Need to solve for a large matrix
    • It takes too much time with conventional methods

Jacobi Method

  • Iterative or approximate method
  • Solve a linear system $AX=B$
  • General formula:
    • \[x_i^{(k+1)} = \frac{1}{a_{ii}} ( b_i - \sum_{j \not = i} a_{ij} x_{j}^{(k)}), i=1,2, ...,n\]
  • Convergence check:
    • $\epsilon_i = \vert \frac{x_i^j x_i^{j-1}}{x_i^j} \vert < \epsilon_s$

Gauss-Seidel Method

  • Speed up the convergence
  • Use better approximations when possible
  • $x_{k+1}$ is a better approximation to $x$ than $x_k$.
  • $x_i^{(k+1)} = \frac{1}{a_{ii}} (b_i - \sum_{j < i} a_{ij} x_j^{(k+1)} - \sum_{j > i} a_{ij} x_j^{(k)}), i = 1,2,…,n$

Convergence

  • Sufficient condition for convergence:
    • Method always converge if the matrix $A$ is diagonally dominant
    • Row diagonal dominance: For each row, the absolution value of the diagonal term is greater than the sum of absolute values of other terms
    • $\vert a_{ii} \vert > \sum_{i \not = j} \vert a_{ij} \vert$

Relaxation in Gauss-Seidel

  • Slight modification for improving convergence
  • $x_i^{(k+1)} = \lambda x_i^{(k+1)} + (1 - \lambda) x_i^{(k)}$, $0 < \lambda < 2$:
    • If $\lambda = 1$, no changes
    • If $0 < \lambda < 1$, underrelaxation helps to make a nonconvergent system converge
    • If $1 < \lambda < 2$, overrelaxation to accelerate the converge

Interpolation

Interpolation

  • Iterpolation:
    • A process of finding a function that passes through a given set of points $(x_i, y_i)$ (data fitting or curve fitting)
    • It can be used to estimate variable $y$ corresponding to unknown $x \in [a, b] - { x_i }$
  • Extrapolation:
    • Estimate variable $y$ corresponding to $x < x_0$ or $x > x_n$
  • Purpose of interpolation:
    1. Replace a set of data points ${(x_i, y_i)}$ with a function given analytically.
    2. Approximate functions with simpler ones, usually polynomials or ‘piecewise polynomials’.

Linear Interpolation

  • Interpolation with straight line:
    • Let two data poitns $(x_0, y_0)$ and $(x_1, y_1)$ be given. There is a unique straight line passing through these points. We can write the formula for a straight line as:
      • \[P_1(x) = y_0 + (\frac{y_1 - y_0}{x_1 - x_0})(x - x_0)\]

Quadratic Interpolation

  • We want to find a polynomial:
    • $P_2(x) = a_0 + a_1 x + a_2 x^2$
    • which satisfies:
      • $P_2(x_i) = y_i, i = 0, 1, 2$
      • for given data points $(x_0, y_0), (x_1, y_1), (x_2, y_2)$
  • Lagrange interpolation:
    • $P_2(x) = y_0 L_0(x) + y_1 L_1(x) + y_2 L_2(x)$
    • with:
      • $L_0(x) = \frac{(x - x_1)(x - x_2)}{x_0 - x_1)(x_0 - x_2)}$
      • $L_1(x) = \frac{(x - x_0)(x - x_2)}{x_1 - x_0)(x_1 - x_2)}$
      • $L_2(x) = \frac{(x - x_0)(x - x_1)}{x_2 - x_0)(x_2 - x_1)}$
  • Uniqueness:
    • Can there be another polynomial, call it $Q(x)$, for which:
      • $deg(Q) \le 2$
      • Let $R(x) = P_2(x) - Q(x)$
      • From the properties of $P_2$ and $Q$, we have $deg(R) \le 2$
      • $R(x_i) = P_2(x_i ) - Q(x_i) = y_i - y_i = 0$
      • Three zeros for quadratic function
      • So, $R(x) = 0$ for all x, $Q(x) = P_2(x)$ for all x

Higher Degree Interpolation

  • Polynomail function of degree $n$:
    • $deg(P_n) \le n$
    • $P_n(x_i) = y_i, i = 0, 1, …, n$
    • with data points $(x_0, y_0), …, (x_n, y_n)$
  • Lagrange interpolation:
    • $P_n(x) = y_0 L_0(x) + y_1 L_1(x) + \cdots + y_n L_n(x)$
    • $L_k(x) = \frac{(x - x_0) … (x - x_{k-1})(x - x_{k+1}) … (x - x_n)}{(x_k - x_0) … (x_k - x_{k-1})(x_k - x_{k+1})…(x_k - x_n)}$

Newton Polynomials

  • It is sometimes useful to find $P_1(x), P_2(x), …, P_N(x)$
  • Lagrange polynomials:
    • No constructive relationship between $P_{N-1}(x)$ and $P_N(x)$
  • Newton polynomials:
    • Recursive pattern
    • $P_N(x) = P_{N-1}(x) + g_N(x)$
    • $g_N(x) = a_N(x - x_0)(x - x_1) … (x - x_{N-1})$
  • To find $a_k$ for all polynomials $P_1(x), P_2(x), …, P_N(x)$ that approximate a given function $f(x)$
  • $P_k(x)$ is based on the centers $x_0, x_1, …, x_k$
  • $a_1$ : slope of the line between $(x_0, f(x_0))$ and $(x_1, f(x_1))$

General Divided Differnce

  • Given $n + 1$ distinct points $x_0, …, x_n$, with $n \ge 2$, define:
    • $$f[x_0, …, x_n] = \frac{f[x_1, …, x_n] - f[x_0, …, x_{n-1}]}{x_n - x_0}$
    • $$f[x_0, …l, x_n] = \frac{1}{n!} f^{(n)}(xc)$
    • for some $c$ intermediate to the points ${x_0, …, x_n }$

Spherical Linear Interpolation

  • \[q(t) = \frac{sin \theta (1 - t)}{sin \theta} q_1 + \frac{sing \theta t}{sin \theta} q_2\]

Bilinear Interopolation

  • \[f(R_1) \approx \frac{x_2 - x}{x_2 - x_1} f(Q_{11}) + \frac{x - x_1}{x_2 - x_1} f(Q_{21})\]
  • \[f(R_2) \approx \frac{x_2 - x}{x_2 - x_1} f(Q_{12}) + \frac{x - x_1}{x_2 - x_1} f(Q_{22})\]
  • \[f(P) \approx \frac{y_2 - y}{y_2 - y_1} f(R_1) + \frac{y - y_1}{y_2 - y_1} f(R_2)\]

Least Squares

Least-Squares Line

  • Data points $(x_1, y_1), …, (x_N, y_N), {x_k }$ are distinct
  • Numerical method is to determine $y=f(x)$
  • Lienar approximation: $y = f(x) = Ax + B$
  • If there is an error in the data points:
    • \(f(x_k) = y_k + e_k\),
    • where $e_k$ is the error

Errors

  • Erros: $e_k = f(x_k) - y_k$ for $1 \le k \le N$
  • Norms to measure how far the curve $y=f(x)$ lies from the data
  • Maximum error: $E_{\infty} (f) = {max}_{1 \le k \le N} { \vert f(x_k) - y_k }$
  • Average error: $E_1 (f) = \frac{1}{N} \sum_{k=1}^N \vert f(x_k) - y_k \vert$
  • Root-mean-squre error: $E_2 (f) = ( \frac{1}{N]} \sum_{k=1}^N \vert f(x_k) - y_k \vert ^2)^{1/2}$
  • Maximum error is sensitive to extreme data
  • Average is often used since it’s easy to compute
  • RMS is often used when statistical nature is used

Least-Squares Line

  • A best-fitting line is minimizing the rror
  • RMS is the traditional choice
  • Coefficients of least-squares line $y=Ax + B$ are the solution of the followign linear system:
    • \[(\sum_{k=1}^N x_k^2) A + (\sum_{k=1}^N x_k) B = \sum_{k=1}^N x_k y_k\]
    • $$\sum_{k=1}^N x_k) A + NB = \sum_{k=1}^N y_k$
  • Eror is teh vertical distance between $(x_k, y_k)$ and $(x_k, Ax_k + B)$ and we want minimize their squared sum
  • $E(A, B)$ is minimum when the partical derivatives $\partial E / \partial A$ and $\partial E / \partial B$ equal to zero
  • \[\begin{aligned} \frac{\partial E(A, B)}{\partial A} &= \sum_{k=1}^N 2(A x_k + B - y_k)(x_k) \\ &= 2 \sum_{k=1}^N (A x_k^2 + B x_k - x_k y_k) \end{aligned}\]
    • \[0 = A \sum_{k=1}^N x_k^2 = B \sum_{k=1}^N x_k - \sum_{k=1}^N x_k y_k\]
  • $$\begin{aligned} \frac{\partial E(A, B)}{\partial B} &= \sum^N 2(Ax_k + B - y_k) \ &= 2 \sum_{k=1}^N (A x_k + B - y_k)\end{aligned}$
    • \[0 = A \sum_{k=1}^N x_k + NB - \sum_{k=1}^N y_k\]

Power Fit

  • Fitting $f(x) = Ax^M$, where $M$ is known
  • Least-squares power curve: minimize the RMS (Root-Mean-Square) error:
    • \[A = \sum_{k=1}^N x_k^M y_k) / (\sum_{k=1}^N x_k^{2M})\]
  • Least-Squares method: minimize $E(A)$:
    • \[E(A) = \sum (Ax_k^M - y_k)^2\]
  • Solve $E’(A) = 0$:
    • \[E'(A) = 2 \sum (A x_k^{2M} - x_k^M y_k)\]
    • \[0 = A \sum x_k^{2M} - \sum x_k^M y_k\]

Data Linearization

  • Data linearization : process to transform a non-linear relation into a linear relation

Matrix Formulation

  • Lienar system:
    • $F^TFC = F^TY$ for the coeeficient matrix $C$

Polynomial Fitting

  • Polynomial function:
    • \[f(x) = c-1 + c_2 x + c_3 x^2 + \cdots + c_{M+1} x^M\]
  • Least-squares parabola

Nonlinear Least-Squares Method

  • Nonlinear system can be solved with Newton’s Method:
    • Time consuming
    • Requires good starting values for $A$ and $C$
  • Minimize $E(A, C)$ directly using optimization methods

Optimization

  • Methods for locating extrema(maxima or minima) of functions

Functions of One Variable

  • The function $f$ has a local minimum value at $x=p$, if there exists an open interval $i$ containing $p$ s.t. $f(p) \le f(x)$ for all $ x \in I$
  • $f$ has a local maximum value at $x = p$, if there exists an open interval $i$ containing $p$ s.t. $f(x) \le f(p)$ for all $ x \in I$
  • $f$ has a local extremum value at $x=p$, if it has local minimum or maximum value at $x=p$.

  • Theorem 1:
    • Suppose that $f(x)$ is continuous on $i= [a,b]$ and is differentiable on $(a, b)$:
      • If $f’(x) > 0$ for all $x \in 9a, b)$, then $f(x)$ is increasing on $I$.
      • If $f’(x) < 0$ for all $x \in (a, b)$, then $f(x)$ is decreasing on $I$.
  • Theorem 3:
    • Suppose that $f(x)$ is continuous on $i= [a,b]$ and $f’(x)$ is defined for all $x \in (a, b)$, except possibly at $x=p$:
      • If $f’(x) < 0$ on $(a, p)$ and $f’(x) > 0$ on $(p, b)$, then $f(p)$ is a local minimum.
      • If $f’(x) > 0$ on $(a, p)$ and $f’(x) < 0$ on $(p, b)$, then $f(p)$ is a local maximum.
  • Theorem 4:
    • Suppose that $f(x)$ is continuous on $i= [a,b]$ and $f’,f’’$ is defined for all $x \in (a, b)$, Also, $p \in (a, b)$ is a critical point where $f’(p) = 0$:
      • If $f’‘(p) > 0$, then $f(p)$ is a local minimum of $f$.
      • If $f’‘(p) < 0$, then $f(p)$ is a local maximum of $f$.
      • If $f’‘(p) = 0$, then this test is inconclusive.

Bracketing Search Methos

  • Evaluate the function many times and search for a local minimum/maximum
  • To reduce the number of function evaluations, a good strategy is needed for determining where the function is to be evaluated
  • Definition:
    • The function $f(x)$ is unimodal on $I = [a, b]$, if there exists a unique number $p \in I$ such that:
      • $f(x)$ is decreasing on $[a, p]$
      • $f(x)$ is increasing on $[p, b]$
  • If $f(x)$ is unimodal on $[a,b]$, then it is possible to replace the interval with a subinterval on which $f(x)$ takes on its miniumum value
  • Select two interior points $c < d \Rightarrow a < c < d < b$
  • $f(c), f(d) < max(f(a), f(b))$:
    • When $f(c) \le f(d)$:
      • The minimum must occur in $[a,d]$
      • New subinterval $[a, d]$
    • When $f(d) < f(c)$:
      • The minimum must occur in $[c, d]$
      • New subinterval $[c, b]$
  • $c = ra + (1 - r)b$
  • $d = (1 - r)a + rb$
  • $r = \frac{-1 + \sqrt 5}{2}$
  • The value r is not constant on each subinterval
  • Number of subintervals (iterations) is predetermined and based on the specified tolerance
  • The interior poitns $c_k$ and $d_k$ of the k-th subinterval $[a_k, b_k]$ are:
    • $c_k = a_k + (1 - \frac{F_{n-k-1}{F_{n-k}}})(b_k - a_k)$
    • $d_k = a_k + \frac{F_{n-k-1}}{F_{n-k}}(b_k - a_k)$
  • For a tolerance $\epsilon$, find the smallest value of $n$ such that:
    • \[\frac{b_0 - a_0}{F_n} < \epsilon, F_n > \frac{b_0 - a_0}{\epsilon}\]

Multidimensional Optimization

  • Find the extremum of a function of several variables:
    • Direct method (No derivatives)
    • Gradient method (Use derivatives)
  • Based on evaluation of the fuction randomly at selected values of the independent variables.
  • If a sufficient number of samples are ocnducted, the optimum will be eventually located
  • Advantages:
    • Works even for discontiuous and nondifferentiable functions.
  • Disadvantages:
    • As the number of independent variables grows, the task can become onerous.
    • Not efficient, it does not account for the behavior of underlying function
  • More efficient than random search and still doesn’t require derivative evaluation
  • The baisc strategy:
    • Change one variable at a time while the other variables are held constant.
    • Problem is reduced to a sequence of one-dimensional searches that can be solved by variety of methods
    • The search becomes less efficient as you approach the maximum.

Gradient

  • Graident of a function $f$, $\nabla f$, tells us:
    • $$\nabla f = \begin{bmatrix} \frac{\partial f}{\partial x_1} \ \vdots \ \frac{\partial f}{\partial x_n} \end{bmatrix}$

Hessian Matrix (of Hessian of $f$)

  • $$H = \begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partialx_1 \partial x} & \cdots & \frac{\partial^2 f}{\partial x_1 \partial x_n} \ \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2} & \cdots & \frac{\partial^2 f}{\partial x_2 \partial x_n} \ \vdots & \vdots & \ddots & \vdots \ \frac{\partial^2 f}{\partial x_n \partial x_1} & \frac{\partial^2 f}{\partial x_n \partial x_2} & \cdots & \frac{\partial^2 f}{\partialx_n^2} \end{bmatrix}$
  • Also known as the matrix of second partial derivatives.
  • It provides a way to discern if a funciton has reached an optimum or not.

Solution of Nonlinear Equation

Basis of Bisection Method

  • An equation $f(x) = 0$, where $f(x)$ is a real continuous function, has at least one root between $x_l$ and $x_u$ if $f(x_l)f(x_u) < 0$.
  • If