During this activity, students should be able to:
This activity helps the student develop the following skills, values and attitudes: ability to analyze and synthesize, capacity for identifying and solving problems, and efficient use of computer systems.
Individually, solve the following set of programming
exercises using Erlang. Place all your functions in a module
called highorder.
multiple_apply(ProcList, Val), in which ProcList is a list that contains zero or more one-argument functions, and Val is any value. It returns a list that results from applying every function in ProcList to Val. Examples:
> P = [fun (X) -> X + X end,
fun (X) -> X * X end,
fun erlang:is_integer/1,
fun (X) -> [X] end,
fun (X) -> X end].
[#Fun<erl_eval.6.13229925>,#Fun<erl_eval.6.13229925>,
#Fun<erlang.is_number.1>,#Fun<erl_eval.6.13229925>,
#Fun<erl_eval.6.13229925>]
> highorder:multiple_apply(P, 20).
[40,400,true,[20],20]
> highorder:multiple_apply(P, 1.5).
[3.0,2.25,false,[1.5],1.5]
> highorder:multiple_apply([fun erlang:is_number/1,
fun erlang:is_atom/1,
fun erlang:is_tuple/1], what).
[false,true,false]
The bisection method is a root-finding algorithm which works by repeatedly dividing an interval in half and then selecting the subinterval in which the root exists.
Suppose we want to solve the equation f(x) = 0. Given two points a and b such that f(a) and f(b) have opposite signs, f must have at least one root in the interval [a, b] as long as f is continuous on this interval. The bisection method divides the interval in two by computing c = (a+b) / 2. There are now two possibilities: either f(a) and f(c) have opposite signs, or f(c) and f(b) have opposite signs. The bisection algorithm is then applied to the sub-interval where the sign change occurs.
Write the function bisection(A,
B, F) that finds the corresponding root using
the bisection method. The algorithm must stop when a value of
X is found such that:
abs(F(X)) < +1.0e-10.
Examples:
> highorder:bisection(1, 4, fun (X) -> (X - 3) * (X + 4) end). 3.00000 > highorder:bisection(4, 1, fun math:sin/1). 3.14159 > highorder:bisection(-1, 1, fun (X) -> math:pow(X, 5) + 5 * X + 1 end). -0.199936
The derivate of a function f(x) with respect to variable x is defined as:
Where f must be a continuous function. Write the
function derive(F, H) that
returns a new function that takes X as input, and which
represents the derivate of F given a certain value for
H. For example:
> F = fun (X) -> X * X * X end. #Fun<erl_eval.6.49591080> > DF = highorder:derive(F, 0.001). #Fun<highorder.0.442056> > DF(5). 75.0150 > DDF = highorder:derive(DF, 0.001). #Fun<highorder.0.442056> > DDF(5). 30.0060
Simpson's rule is a method for numeric integration:
Where h = (b - a) ÷
n, for a given even positive integer n (if you
increment the value of n you get a better
approximation), and yk =
f(a + k × h). Write the
function integral(A, B,
N, F) that returns the value of the integral,
using Simpson's rule.
For example:
would be written in Erlang as follows (with n = 10):
> highorder:integral(0, 1, 10, fun (X) -> X * X * X end). 0.250000
The double integral:
would be written in Erlang as follows (with n = 10):
> highorder:integral(
1, 2, 10,
fun (X) -> highorder:integral(
3, 4, 10,
fun (Y) -> X * Y end)
end).
5.25000
Using the Online
Assignment Delivery System (SETA), deliver the file called
highorder.erl. No assignments will be accepted through
e-mail or any other means.
IMPORTANT: The program source file must include at the top the author's personal information (name and student id) within comments. For example:
%% ITESM CEM, February 6, 2009.
%% Erlang Source File
%% Activity: High-Order Functions
%% Author: Steve Rogers, 449999
.
. (The rest of the program goes here)
.
Due date: Friday, February 6.
This activity will be evaluated using the following criteria:
| -10 | The program doesn't contain within comments the author's personal information. |
|---|---|
| 10 | The program contains syntax errors. |
| DA | The program was plagiarized. |
| 10-100 | Depending on the amount of exercises that were solved correctly. |