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Further cleanup of 07_differentiation notebook
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‎07_differentiation.ipynb

Lines changed: 120 additions & 28 deletions
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@@ -63,7 +63,10 @@
6363
"\n",
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"However, for our purposes here it's somewhat more convenient to use the *Newton Polynomial* Basis\n",
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"\n",
66-
"$$n_j(x) = \\prod^{j-1}_{i=0} (x - x_i)$$"
66+
"\\begin{align}\n",
67+
" n_0 &= 1\\\\\n",
68+
" n_j(x) &= \\prod^{j-1}_{i=0} (x - x_i)\\quad\\mathrm{for}\\quad j>0 \\\\\n",
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"\\end{align}\n"
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]
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},
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{
@@ -96,7 +99,7 @@
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"source": [
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"Note: The Newton polynomials have some features common to both the monomials and the Lagrange polynomials\n",
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"\n",
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"* Like the monomials they are of monic polynomials of increasing degree in $x$, i.e. $n_2(x)$ is quadratic\n",
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"* Like the monomials they are monic polynomials of increasing degree in $x$, i.e. $n_2(x)$ is quadratic\n",
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"* Like the Lagrange polynomials, some Newton polynomials vanish identically at the interpolation points $x_i$. In particular\n",
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"\n",
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"$$\n",
@@ -118,10 +121,10 @@
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"\n",
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"$$P_N(x) = \\sum^N_{j=0} a_j n_j(x)$$\n",
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"\n",
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"as the Linear system\n",
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"as the $(N+1)\\times (N+1)$ linear system\n",
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"\n",
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"$$\n",
124-
" P_N(x_i) = y_i\n",
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" P_N(x_i) = y_i\\quad\\mathrm{for}\\quad i=0,\\ldots,N\n",
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"$$"
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]
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},
@@ -279,12 +282,12 @@
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},
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"outputs": [],
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"source": [
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"\n",
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"# Calculate a polynomial in Newton Form\n",
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"data = numpy.array([[-2.0, 1.0], [-1.5, -1.0], [-0.5, -3.0], [0.0, -2.0], [1.0, 3.0], [2.0, 1.0]])\n",
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"x = numpy.linspace(-2.0, 2.0, 100)\n",
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"basis = newton_basis(x, data[:,0])\n",
287-
"P = P_newton(x, data)"
289+
"P = P_newton(x, data)\n",
290+
"N = data.shape[0] - 1"
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]
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},
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{
@@ -358,14 +361,23 @@
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "subslide"
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"slide_type": "fragment"
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}
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},
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"source": [
365-
"We know from Lagrange's Theorem that the remainder term looks like\n",
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"\n",
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"$$R_N(x) = (x - x_0)(x - x_1)\\cdots (x - x_{N})(x - x_{N+1}) \\frac{f^{(N+1)}(c)}{(N+1)!}$$\n",
368+
"And we know from Lagrange's Theorem that the remainder term looks like\n",
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"\n",
370+
"$$R_N(x) = (x - x_0)(x - x_1)\\cdots (x - x_{N})(x - x_{N+1}) \\frac{f^{(N+1)}(c)}{(N+1)!}$$"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "subslide"
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}
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},
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"source": [
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"noting that we need to require that $f(x) \\in C^{N+1}$ on the interval of interest. Taking the derivative of the interpolant $P_N(x)$ then leads to \n",
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"\n",
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"$$P_N'(x) = [y_0, y_1] + ((x - x_1) + (x - x_0)) [y_0, y_1, y_2] + \\cdots + \\left(\\sum^{N-1}_{i=0}\\left( \\prod^{N-1}_{j=0,~j\\neq i} (x - x_j) \\right )\\right ) [y_0, y_1, \\ldots, y_N]$$"
@@ -375,7 +387,7 @@
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "subslide"
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"slide_type": "fragment"
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}
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},
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"source": [
@@ -401,13 +413,26 @@
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "subslide"
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"slide_type": "fragment"
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}
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},
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"source": [
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"If we let $\\Delta x = \\max_i |x_k - x_i|$ we then know that the remainder term will be $\\mathcal{O}(\\Delta x^N)$ as $\\Delta x \\rightarrow 0$ thus showing that this approach converges and we can find arbitrarily high order approximations (ignoring floating point error)."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "fragment"
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}
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},
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"source": [
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"### <font color='red'>Caution</font>\n",
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"\n",
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"High order does not necessarily imply high-accuracy!"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
@@ -486,6 +511,7 @@
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{
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"cell_type": "markdown",
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"metadata": {
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"hide_input": true,
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"slideshow": {
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"slide_type": "slide"
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}
@@ -720,7 +746,7 @@
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"source": [
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"$$\\begin{aligned}\n",
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" P_2'(x) &= [f(x_n), f(x_{n+1})] + ((x - x_n) + (x - x_{n+1})) [f(x_n), f(x_{n+1}), f(x_{n-1})] \\\\\n",
723-
" &= \\frac{f(x_{n+1}) - f(x_n)}{x_{n+1} - x_n} + ((x - x_n) + (x - x_{n+1})) \\left ( \\frac{f(x_{n-1}) - f(x_{n+1})}{(x_{n-1} - x_{n+1})(x_{n-1} - x_n)} - \\frac{f(x_{n+1}) - f(x_n)}{(x_{n+1} - x_n)(x_{n-1} - x_n)} \\right )\n",
749+
" &= \\frac{f(x_{n+1}) - f(x_n)}{x_{n+1} - x_n} + ((x - x_n) + (x - x_{n+1}))\\\\ &* \\left ( \\frac{f(x_{n-1}) - f(x_{n+1})}{(x_{n-1} - x_{n+1})(x_{n-1} - x_n)} - \\frac{f(x_{n+1}) - f(x_n)}{(x_{n+1} - x_n)(x_{n-1} - x_n)} \\right )\n",
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"\\end{aligned}$$"
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]
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},
@@ -777,8 +803,9 @@
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"hide_input": true,
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"slideshow": {
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"slide_type": "skip"
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"slide_type": "subslide"
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}
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},
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"outputs": [],
@@ -794,7 +821,7 @@
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"\n",
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"# Compute derivative\n",
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"f_prime_hat = numpy.empty(x_hat.shape)\n",
797-
"f_prime_hat[1:-1] = (f(x_hat[2:]) - f(x_hat[:-2])) / (2 * delta_x)\n",
824+
"f_prime_hat[1:-1] = (f(x_hat[2:]) - f(x_hat[:-2])) / (2. * delta_x)\n",
798825
"\n",
799826
"# Use first-order differences for points at edge of domain\n",
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"f_prime_hat[0] = (f(x_hat[1]) - f(x_hat[0])) / delta_x # Forward Difference at x_0\n",
@@ -807,7 +834,7 @@
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"hide_input": false,
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"hide_input": true,
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"slideshow": {
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"slide_type": "fragment"
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}
@@ -818,11 +845,14 @@
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"plt.rcdefaults()\n",
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"axes = fig.add_subplot(1, 1, 1)\n",
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"\n",
821-
"axes.plot(x, f_prime(x), 'k')\n",
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"axes.plot(x, f_prime(x), 'k',label=\"$f'(x)$\")\n",
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"axes.plot(x_hat, f_prime_hat, 'ro')\n",
850+
"axes.plot(x, f(x),'g--',label=\"$f(x)$\")\n",
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"axes.plot(x_hat, f(x_hat), 'go')\n",
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"axes.set_xlim((x[0], x[-1]))\n",
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"# axes.set_ylim((-1.1, 1.1))\n",
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"axes.grid()\n",
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"axes.legend()\n",
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"\n",
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"plt.show()"
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]
@@ -884,7 +914,7 @@
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{
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"cell_type": "markdown",
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"metadata": {
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"hide_input": false,
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"hide_input": true,
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"slideshow": {
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"slide_type": "subslide"
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}
@@ -907,16 +937,66 @@
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"source": [
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"Say we want to derive the second order accurate, first derivative approximation that we just did, this requires the values $(x_{n+1}, f(x_{n+1})$ and $(x_{n-1}, f(x_{n-1})$. We can express these values via our Taylor series approximation above as\n",
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"\n",
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"$$\\begin{aligned}\n",
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"\\begin{aligned}\n",
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" f(x_{n+1}) &= f(x_n) + (x_{n+1} - x_n) f'(x_n) + \\frac{(x_{n+1} - x_n)^2}{2!} f''(x_n) + \\frac{(x_{n+1} - x_n)^3}{3!} f'''(x_n) + \\mathcal{O}((x_{n+1} - x_n)^4) \\\\\n",
912-
" &= f(x_n) + \\Delta x f'(x_n) + \\frac{\\Delta x^2}{2!} f''(x_n) + \\frac{\\Delta x^3}{3!} f'''(x_n) + \\mathcal{O}(\\Delta x^4)\n",
913-
"\\end{aligned}$$\n",
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"\n",
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"\\end{aligned}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "fragment"
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}
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},
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"source": [
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"or\n",
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"\\begin{aligned}\n",
955+
"&= f(x_n) + \\Delta x f'(x_n) + \\frac{\\Delta x^2}{2!} f''(x_n) + \\frac{\\Delta x^3}{3!} f'''(x_n) + \\mathcal{O}(\\Delta x^4)\n",
956+
"\\end{aligned}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "subslide"
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}
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},
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"source": [
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"and\n",
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"\n",
917-
"$$f(x_{n-1}) = f(x_n) + (x_{n-1} - x_n) f'(x_n) + \\frac{(x_{n-1} - x_n)^2}{2!} f''(x_n) + \\frac{(x_{n-1} - x_n)^3}{3!} f'''(x_n) + \\mathcal{O}((x_{n-1} - x_n)^4) $$\n",
918-
"\n",
919-
"$$ = f(x_n) - \\Delta x f'(x_n) + \\frac{\\Delta x^2}{2!} f''(x_n) - \\frac{\\Delta x^3}{3!} f'''(x_n) + \\mathcal{O}(\\Delta x^4) $$"
969+
"\\begin{align}\n",
970+
"f(x_{n-1}) &= f(x_n) + (x_{n-1} - x_n) f'(x_n) + \\frac{(x_{n-1} - x_n)^2}{2!} f''(x_n) + \\frac{(x_{n-1} - x_n)^3}{3!} f'''(x_n) + \\mathcal{O}((x_{n-1} - x_n)^4) \n",
971+
"\\end{align}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "fragment"
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}
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},
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"source": [
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"\\begin{align} \n",
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"&= f(x_n) - \\Delta x f'(x_n) + \\frac{\\Delta x^2}{2!} f''(x_n) - \\frac{\\Delta x^3}{3!} f'''(x_n) + \\mathcal{O}(\\Delta x^4)\n",
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"\\end{align}"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "subslide"
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}
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},
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"source": [
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"Or all together (for regularly spaced points),\n",
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"\\begin{align} \n",
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"f(x_{n+1}) &= f(x_n) + \\Delta x f'(x_n) + \\frac{\\Delta x^2}{2!} f''(x_n) + \\frac{\\Delta x^3}{3!} f'''(x_n) + \\mathcal{O}(\\Delta x^4)\\\\\n",
998+
"f(x_{n-1})&= f(x_n) - \\Delta x f'(x_n) + \\frac{\\Delta x^2}{2!} f''(x_n) - \\frac{\\Delta x^3}{3!} f'''(x_n) + \\mathcal{O}(\\Delta x^4)\n",
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"\\end{align}"
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]
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},
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{
@@ -933,7 +1013,18 @@
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" f'(x_n) + R(x_n) = A f(x_{n+1}) + B f(x_n) + C f(x_{n-1})\n",
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"$$\n",
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"\n",
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"where $R(x_n)$ is our error. Plugging in the Taylor series approximations we find\n",
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"where $R(x_n)$ is our error. "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "fragment"
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}
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},
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"source": [
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"Plugging in the Taylor series approximations we find\n",
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"\n",
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"$$\\begin{aligned}\n",
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" f'(x_n) + R(x_n) &= A \\left ( f(x_n) + \\Delta x f'(x_n) + \\frac{\\Delta x^2}{2!} f''(x_n) + \\frac{\\Delta x^3}{3!} f'''(x_n) + \\mathcal{O}(\\Delta x^4)\\right ) \\\\\n",
@@ -985,7 +1076,7 @@
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"cell_type": "markdown",
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"metadata": {
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"slideshow": {
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"slide_type": "subslide"
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"slide_type": "skip"
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}
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},
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"source": [
@@ -1072,8 +1163,9 @@
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"hide_input": true,
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"slideshow": {
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"slide_type": "skip"
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"slide_type": "fragment"
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}
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},
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"outputs": [],

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