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diff --git a/75/CH4/EX4.2/ex_1.sce b/75/CH4/EX4.2/ex_1.sce
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+        //    PG (200)
+
+deff('[y]=f(x)','y=exp(x)')
+
+xset('window',0);
+x=-1:.01:1;                // defining the range of x.
+y=feval(x,f);
+ 
+a=gca(); 
+ 
+a.y_location = "origin";
+ 
+a.x_location = "origin"; 
+plot(x,y)                // instruction to plot the graph
+
+
+
+//    possible approximation
+//        y = q1(x)
+
+//    Let e(x) = exp(x) - [a0+a1*x]
+//    q1(x) & exp(x) must be equal at two points in [-1,1], say at x1 & x2
+//    sigma1 = max(abs(e(x)))
+//    e(x1) = e(x2) = 0.
+//    By another argument based on shifting the graph of y = q1(x),
+//    we conclude that the maximum error sigma1 is attained at exactly 3 points.
+//    e(-1) = sigma1
+//    e(1) = sigma1
+//    e(x3) = -sigma1
+//    x1 < x3 < x2
+//    Since e(x) has a relative minimum at x3, we have e'(x) = 0
+//    Combining these 4 equations, we have..
+//    exp(-1) - [a0-a1] = sigma1 ------------------(i)
+//    exp(1) - [a0+a1] = p1 -----------------------(ii)
+//    exp(x3) - [a0+a1*x3] = -sigma1 --------------(iii)
+//    exp(x3) - a1 = 0 ----------------------------(iv)
+
+//    These have the solution
+
+a1 = (exp(1) - exp(-1))/2
+x3 = log(a1)
+sigma1 = 0.5*exp(-1) + x3*(exp(1) - exp(-1))/4
+a0 = sigma1 + (1-x3)*a1
+
+x = poly(0,"x");
+//    Thus,
+q1 = a0 + a1*x
+
+deff('[y1]=f(x)','y1=1.2643+1.1752*x')
+
+xset('window',0);
+x=-1:.01:1;                // defining the range of x.
+y=feval(x,f);
+ 
+a=gca(); 
+ 
+a.y_location = "origin";
+ 
+a.x_location = "origin"; 
+plot(x,y)                // instruction to plot the graph