Numerical integration: Difference between revisions
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Write functions to calculate the definite integral of a function (f(x)) using rectangular (left, right, and midpoint), trapezium, and Simpson's methods. Your functions should take in the upper and lower bounds (<tt>a</tt> and <tt>b</tt>) and the number of approximations to make in that range (<tt>n</tt>). Assume that your example already has a function that gives values for f(x). |
Write functions to calculate the definite integral of a function (<tt>f(x)</tt>) using rectangular (left, right, and midpoint), trapezium, and Simpson's methods. Your functions should take in the upper and lower bounds (<tt>a</tt> and <tt>b</tt>) and the number of approximations to make in that range (<tt>n</tt>). Assume that your example already has a function that gives values for <tt>f(x)</tt>. |
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Simpson's method is defined by the following pseudocode: |
Simpson's method is defined by the following pseudocode: |
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answer = (h / 6) * (f(a) + f(b) + 4 * sum1 + 2 * sum2) |
answer = (h / 6) * (f(a) + f(b) + 4 * sum1 + 2 * sum2) |
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What it will be doing is computing integrals for multiple quadratics (like the one shown in [http://en.wikipedia.org/wiki/Simpsons_rule the wiki]) and summing them. |
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=={{header|Ada}}== |
=={{header|Ada}}== |
Revision as of 14:52, 3 January 2008
You are encouraged to solve this task according to the task description, using any language you may know.
Write functions to calculate the definite integral of a function (f(x)) using rectangular (left, right, and midpoint), trapezium, and Simpson's methods. Your functions should take in the upper and lower bounds (a and b) and the number of approximations to make in that range (n). Assume that your example already has a function that gives values for f(x).
Simpson's method is defined by the following pseudocode:
h = (b - a) / n sum1 = sum2 = 0 loop on i from 0 to (n - 1) sum1 = sum1 + f(a + h * i + h / 2) loop on i from 1 to (n - 1) sum2 = sum2 + f(a + h * i) answer = (h / 6) * (f(a) + f(b) + 4 * sum1 + 2 * sum2)
What it will be doing is computing integrals for multiple quadratics (like the one shown in the wiki) and summing them.
Ada
This solution creates a generic package into which the function F(X) is passed during generic instantiation. The first part is the package specification. The second part is the package body.
generic with function F(X : Long_Float) return Long_Float; package Integrate is function Left_Rect(A, B, N : Long_Float) return Long_Float; function Right_Rect(A, B, N : Long_Float) return Long_Float; function Mid_Rect(A, B, N : Long_Float) return Long_Float; function Trapezium(A, B, N : Long_Float) return Long_Float; function Simpson(A, B, N : Long_Float) return Long_Float; end Integrate;
package body Integrate is --------------- -- Left_Rect -- --------------- function Left_Rect (A, B, N : Long_Float) return Long_Float is H : constant Long_Float := (B - A) / N; Sum : Long_Float := 0.0; X : Long_Float := A; begin while X <= B - H loop Sum := Sum + (H * F(X)); X := X + H; end loop; return Sum; end Left_Rect; ---------------- -- Right_Rect -- ---------------- function Right_Rect (A, B, N : Long_Float) return Long_Float is H : constant Long_Float := (B - A) / N; Sum : Long_Float := 0.0; X : Long_Float := A + H; begin while X <= B - H loop Sum := Sum + (H * F(X)); X := X + H; end loop; return Sum; end Right_Rect; -------------- -- Mid_Rect -- -------------- function Mid_Rect (A, B, N : Long_Float) return Long_Float is H : constant Long_Float := (B - A) / N; Sum : Long_Float := 0.0; X : Long_Float := A; begin while X <= B - H loop Sum := Sum + (H / 2.0) * (F(X) + F(X + H)); X := X + H; end loop; return Sum; end Mid_Rect; --------------- -- Trapezium -- --------------- function Trapezium (A, B, N : Long_Float) return Long_Float is H : constant Long_Float := (B - A) / N; Sum : Long_Float := F(A) + F(B); X : Long_Float := 1.0; begin while X <= N - 1.0 loop Sum := Sum + 2.0 * F(A + X * (B - A) / N); X := X + 1.0; end loop; return (B - A) / (2.0 * N) * Sum; end Trapezium; ------------- -- Simpson -- ------------- function Simpson (A, B, N : Long_Float) return Long_Float is H : constant Long_Float := (B - A) / N; Sum1 : Long_Float := 0.0; Sum2 : Long_Float := 0.0; Limit : Integer := Integer(N) - 1; begin for I in 0..Limit loop Sum1 := Sum1 + F(A + H * Long_Float(I) + H / 2.0); end loop; for I in 1..Limit loop Sum2 := Sum2 + F(A + H * Long_Float(I)); end loop; return H / 6.0 * (F(A) + F(B) + 4.0 * Sum1 + 2.0 * Sum2); end Simpson; end Integrate;
Java
The function in this example is assumed to be f(double x).
public class Integrate{ public static double leftRect(double a, double b, double n){ double h = (b - a) / n; double sum = 0; for(double x = a;x <= b - h;x += h) sum += h * (f(x)); return sum; } public static double rightRect(double a, double b, double n){ double h = (b - a) / n; double sum = 0; for(double x = a + h;x <= b - h;x += h) sum += h * (f(x)); return sum; } public static double midRect(double a, double b, double n){ double h = (b - a) / n; double sum = 0; for(double x = a;x <= b - h;x += h) sum += (h / 2) * (f(x) + f(x + h)); return sum; } public static double trap(double a, double b, double n){ double h = (b - a) / n; double sum = f(a) + f(b); for(int i = 1;i <= n-1;i++) sum += 2 * f((a + i * (b - a) / n)); return (b - a) / (2 * n) * sum; } public static double simpson(double a, double b, double n){ double h = (b - a) / n; double sum1 = 0; double sum2 = 0; for(int i = 0;i < n;i++) sum1 += f(a + h * i + h / 2); for(int i = 1;i <= n - 1;i++) sum2 += f(a + h * i); return h / 6 * (f(a) + f(b) + 4 * sum1 + 2 * sum2); } //assume f(double x) elsewhere in the class }