{
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 {
		   "cell_type": "markdown",
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	   "source": [
       "# Chapter 11: Numerical Differentiation"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.1: First_order_Forward_Difference.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 11_01\n",
"//First order forward difference\n",
"//Pg No. 348\n",
"clear ;close ;clc ;\n",
"\n",
"x = poly(0,'x');\n",
"deff('F = f(x)','F = x^2');\n",
"deff('DF = df(x,h)','DF = (f(x+h)-f(x))/h');\n",
"dfactual = derivat(f(x));\n",
"h = [0.2 ; 0.1 ; 0.05 ; 0.01 ]\n",
"for i = 1:4\n",
"    y(i) = df(1,h(i));\n",
"    err(i) = y(i) - horner(dfactual,1)\n",
"end\n",
"table = [h y err];\n",
"disp(table)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.2: Three_Point_Formula.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 11_02\n",
"//Three-Point Formula\n",
"//Pg No. 350\n",
"clear ;close ;clc ;\n",
"\n",
"x = poly(0,'x');\n",
"deff('F = f(x)','F = x^2');\n",
"deff('DF = df(x,h)','DF = (f(x+h)-f(x-h))/(2*h)');\n",
"dfactual = derivat(f(x));\n",
"h = [0.2 ; 0.1 ; 0.05 ; 0.01 ]\n",
"for i = 1:4\n",
"    y(i) = df(1,h(i));\n",
"    err(i) = y(i) - horner(dfactual,1)\n",
"end\n",
"table = [h y err];\n",
"disp(table)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.3: Error_Analysis.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 11_03\n",
"//Pg No. 353\n",
"close ;clear ;clc ;\n",
"\n",
"x = 0.45;\n",
"deff('F = f(x)','F = sin(x)');\n",
"deff('DF = df(x,h)','DF = (f(x+h) - f(x))/h');\n",
"dfactual =  cos(x);\n",
"h = 0.01:0.005:0.04;\n",
"n = length(h);\n",
"for i = 1:n\n",
"    y(i) = df(x,h(i))\n",
"    err(i) = y(i) - dfactual ;\n",
"end\n",
"table = [ h' y err];\n",
"disp(table)\n",
"//scilab uses 16 significant digits so the bound for roundoff error is 0.5*10^(-16)\n",
"e = 0.5*10^(-16)\n",
"M2 = max(sin(x+h));\n",
"hopt = 2*sqrt(e/M2);\n",
"disp(hopt,'hopt = ',M2,'M2 = ')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.4: Approximate_Second_Derivative.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 11_04\n",
"//Approximate second derivative\n",
"//Pg No. 354\n",
"clear ;close ;clc ;\n",
"\n",
"x = 0.75;\n",
"h = 0.01;\n",
"deff('F = f(x)','F = cos(x)');\n",
"deff('D2F = d2f(x,h)','D2F = ( f(x+h) - 2*f(x) + f(x-h) )/h^2 ');\n",
"y = d2f(0.75,0.01);\n",
"d2fexact = -cos(0.75)\n",
"err = d2fexact - y ;\n",
"disp(err,'err = ',d2fexact,'d2fexact = ',y,'y = ')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.5: Differentiation_of_Tabulated_Data.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 11_05\n",
"//Differentiation of tabulated data\n",
"//Pg No. 358\n",
"clear ;close ;clc ;\n",
"\n",
"T = 5:9 ;\n",
"s = [10  14.5  19.5  25.5  32 ];\n",
"h = T(2)-T(1);\n",
"n = length(T)\n",
"function V = v(t)\n",
"    if find(T == t) == 1 then\n",
"        V = [ -3*s( find(T == t) ) + 4*s( find(T==(t+h)) ) - s( find( T == (t+2*h) ) ) ]/(2*h) //Three point forward difference formula\n",
"    elseif find(T == t) == n\n",
"        V = [ 3*s( find(T == t) ) - 4*s( find(T==(t-h)) ) + s( find( T == (t-2*h) ) ) ]/(2*h) //Backward difference formula\n",
"    else\n",
"        V = [ s( find(T == (t+h)) ) - s( find(T == (t-h)) ) ]/(2*h) //Central difference formula\n",
"    end\n",
"endfunction\n",
"\n",
"v_5 = v(5)\n",
"v_7 = v(7)\n",
"v_9 = v(9)\n",
"\n",
"disp(v_9,'v(9) = ',v_7,'v(7) = ',v_5,'v(5) = ')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.6: Three_Point_Central_Difference_Formula.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 11_06\n",
"//Three Point Central Difference formula\n",
"//Pg No. 359\n",
"clear ;close ;clc ;\n",
"\n",
"T = 5:9 ;\n",
"s = [10  14.5  19.5  25.5  32 ];\n",
"h = T(2)-T(1);\n",
"deff('A = a(t)','A = [ s( find( T == (t+h) ) ) - 2*s( find( T == t) ) + s( find( T == (t-h) ) ) ]/h^2')\n",
"a_7 = a(7)\n",
"\n",
"disp(a_7,'a(7) = ')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.7: Second_order_Derivative.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 11_7\n",
"//Pg No. 359\n",
"clear ; close ; clc ;\n",
"\n",
"h = 0.25 ;\n",
"//y''(x) = e^(x^2)\n",
"//y(0) = 0 , y(1) = 0\n",
"// y''(x) = y(x+h) - 2*y(x) + y(x-h)/h^2  = e^(x^2)\n",
"//( y(x + 0.25) - 2*y(x) + y(x-0.25) )/0.0625 = e^(x^2)\n",
"//y(x+0.25) - 2*y(x) + y(x - 0.25) = 0.0624*e^(x^2)\n",
"//y(0.5) - 2*y(0.25) + y(0) = 0.0665\n",
"//y(0.75) - 2*y(0.5) + y(0.25) = 0.0803\n",
"//y(1) - 2*y(0.75) + y(0.5) = 0.1097\n",
"//given y(0) = y(1) = 0\n",
"//\n",
"//0 + y2 - 2y1 = 0.06665\n",
"//y3 - 2*y2 + y1 = 0.0803\n",
"//-2*y3 + y2 + 0 = 0.1097\n",
"//Therefore\n",
"A = [0 1 -2 ; 1 -2 1 ; -2 1 0 ]\n",
"B = [ 0.06665 ; 0.0803 ; 0.1097 ]\n",
"C = A\B \n",
"mprintf('solving the above equations we get \n\n y1 = y(0.25) = %f \n y2 = y(0.5) = %f \n y3 = y(0.75) = %f \n ',C(3),C(2),C(1))"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 11.8: Richardsons_Extrapolation_Technique.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Example No. 9_01\n",
"//Richardson's Extrapolation Technique\n",
"//Pg No. 362\n",
"clear ;close ;clc ;\n",
"\n",
"x = -0.5:0.25:1.5\n",
"h = 0.5 ;\n",
"r = 1/2 ;\n",
"\n",
"deff('F = f(x)','F = exp(x)');\n",
"deff('D = D(x,h)' ,'D = [ f(x + h) - f(x-h) ]/(2*h) ' );\n",
"deff('df = df(x,h,r)','df = [D(x,r*h) - r^2*D(x,h)]/(1-r^2)');\n",
"\n",
"df_05 = df(0.5,0.5,1/2);\n",
"disp(df_05,'richardsons technique -  df(0.5) = ',D(0.5,0.25),'D(rh) = D(0.25) = ',D(0.5,0.5),'D(0.5) = ')\n",
"dfexact = derivative(f,0.5)\n",
"disp(dfexact,'Exact df(0.5) = ')\n",
"disp('The result by richardsons technique is much better than other results')\n",
"\n",
"//r = 2\n",
"disp(df(0.5,0.5,2),'df(x) = ',D(0.5,2*0.5),'D(rh) = ','for r = 2')"
   ]
   }
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