Data commit

Task/Numerical-integration/Liberty-BASIC/numerical-integration.basic

@@ -0,0 +1,117 @@
+while 1
+    read x$
+    if x$ ="end" then print "**Over**": end
+
+    read a, b, N, knownValue
+
+    print " Function y ="; x$; " from "; a; " to "; b; " in "; N; " steps"
+    print " Known exact value ="; knownValue
+
+    areaLR = IntegralByLeftRectangle(   x$, a, b, N)
+    areaRR = IntegralByRightRectangle(  x$, a, b, N)
+    areaMR = IntegralByMiddleRectangle( x$, a, b, N)
+    areaTr = IntegralByTrapezium(       x$, a, b, N)
+    areaSi = IntegralBySimpsonRule(     x$, a, b, N)
+
+    print "Left rectangle method   "; using( "##########.##########", areaLR); " diff "; knownValue-areaLR; tab(70); (knownValue-areaLR)/knownValue*100;" %"
+    print "Right rectangle method  "; using( "##########.##########", areaRR); " diff "; knownValue-areaRR; tab(70); (knownValue-areaRR)/knownValue*100;" %"
+    print "Middle rectangle method "; using( "##########.##########", areaMR); " diff "; knownValue-areaMR; tab(70); (knownValue-areaMR)/knownValue*100;" %"
+    print "Trapezium  method       "; using( "##########.##########", areaTr); " diff "; knownValue-areaTr; tab(70); (knownValue-areaTr)/knownValue*100;" %"
+    print "Simpson's Rule          "; using( "##########.##########", areaSi); " diff "; knownValue-areaSi; tab(70); (knownValue-areaSi)/knownValue*100;" %"
+
+    print
+
+wend
+
+end
+
+'------------------------------------------------------
+    'we have N sizes, that gives us N+1 points
+    'point 0 is a
+    'point N is b
+    'point i is xi =a +i *h
+    'Often, precision is (sharper?) then single step area
+    'So there should be EXACT number of steps, hence loop by integer i.
+
+function IntegralByLeftRectangle( x$, a, b, N)
+    h = ( b -a) /N
+    s = 0
+    for i = 0 to N -1
+        x = a +i *h
+        s = s + h *eval( x$)
+    next
+    IntegralByLeftRectangle = s
+end function
+
+function IntegralByRightRectangle( x$, a, b, N)
+    h =( b -a) /N
+    s = 0
+    for i =1 to N
+        x = a +i *h
+        s = s + h *eval( x$)
+    next
+    IntegralByRightRectangle = s
+end function
+
+function IntegralByMiddleRectangle( x$, a, b, N)
+    h =( b -a) /N
+    s = 0
+    for i =0 to N -1
+        x = a +i *h +h /2
+        s = s + h *eval( x$)
+    next
+    IntegralByMiddleRectangle = s
+end function
+
+function IntegralByTrapezium( x$, a, b, N)
+'Formula is h*((f(a)+f(b))/2 + sum_{i=1}^{N-1} (f(x_i)))
+    h  =( b -a) /N
+    x  = a
+    fa =eval( x$)
+    x  =b
+    fb =eval( x$)
+    s = h *( fa +fb) /2
+    for i =1 to N -1
+        x = a +i *h
+        s = s + h *eval( x$)
+    next
+    IntegralByTrapezium = s
+end function
+
+function IntegralBySimpsonRule( x$, a, b, N)
+    'Simpson
+    'N should be even.
+    if N mod 2 then N =N +1
+    'It really doesn't look right to double number of points from N to 2N -
+    ' - this method is most accurate of all presented!
+    'So we use NN as N/2, and N will be 2NN
+    'Formula is h/6*( f(a)+f(b) + 4*(f(x_1)+f(x_3)+...+f(x_{2NN-1})+ 2*(f(x_2)+f(x_4)+...+f(x_{2NN-2})) )
+    'Somehow I messed up h/6, h/3 and what is h, regarding "n=number of double intervals of size 2h"
+    NN =N /2
+
+    h  =( b -a) /N
+    x  =a
+    fa =eval (x$)
+    x  =b
+    fb =eval( x$)
+    s = h /3 *( fa +fb)
+    for i =1 to 2 *NN -1 step 2
+        x = a +i *h
+        s = s + h /3 *4 *eval( x$)  'odd points
+    next
+    for i =2 to 2 *NN -2 step 2
+        x = a +i *h
+        s = s + h /3 *2 *eval( x$)  'even points
+    next
+
+    IntegralBySimpsonRule = s
+end function
+
+'=======================================================
+data "x^3",  0,    1,     100,          0.25
+data "x^-1", 1,  100,    1000,          4.605170
+data "x",    0, 5000,    1000,   12500000.0   '   should use 5 000 000 steps
+data "x",    0, 6000,    1000,   18000000.0   '   should use 6 000 000 steps
+data "end"
+
+end