In calculus, the derivative of a function is "a measure of how a function changes as its input changes" (Wikipedia). For example, if an object is traveling at a constant velocity, that velocity is the same from moment to moment, so the derviative is zero. If an object is falling, its velocity changes a little bit as the object starts falling, and then falls faster and faster as time goes by.
You can calculate the rate of change of a function by calculating: (f(x + delta) - f(x)) / delta, where delta is the interval between measurements. As delta approaches zero, you get closer and closer to the true value of the derivative.
Write a module named Calculus with a function derivative/2. The first argument is the function whose derivative you wish to find, and the second argument is the point at which you are measuring the derivative.
What should you use for a value of delta? I used 1.0e-10, as that is a small number that approaches zero.
Here is some sample output.
iex(1)> c("calculus.ex")
[Calculus]
iex(2)> f1 = fn(x) -> x * x end
#Function<erl_eval.6.17052888>
iex(3)> f1.(7)
49
iex(4)> Calculus.derivative(f1, 3)
6.00000049644222599454
iex(5)> Calculus.derivative(fn(x) -> 3 * x * x + 2 * x + 1 end, 5)
32.00000264769187197089
iex(6)> Calculus.derivative(&:math.sin/1, 0)
1.0
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Line 3 is a test to see if the f1 function works.
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Line 5 shows that you don’t have to assign a function to a variable; you can define the function in line.
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Line 6 shows how to refer to a function in another module. You give its name, preceded by an ampersand, and use /1 to indicate its arity.
Is it possible to use pattern matching inside a list comprehension? Try it and find out.
Presume you have this list of people’s names, genders, and ages:
[{"Federico", "M", 22}, {"Kim", "F", 45}, {"Hansa", "F", 30},
{"Tran", "M", 47}, {"Cathy", "F", 32}, {"Elias", "M", 50}]
In iex (or in a module, if you prefer), write a list comprehension that creates a list consisting of the names of all the people who are male and over 40. Use pattern matching to separate the tuple into three variables, and use two guards to do the tests for age and gender. When you use multiple guards in a list comprehension, you get the moral equivalent of combining the conditions with and.
Then, write a list comprehension that selects the names of all the people who are male or over 40. You can’t use multiple guards here; you want a single guard that explicitly uses the or operator.
Add mean/1 and stdv/1 functions to the stats module which you created in Étude 6-1 to calculate the mean and standard deviation for a list of numbers.
iex(1)> c("stats.ex")
[Stats]
iex(2)> Stats.mean([7, 2, 9])
6.0
iex(3)> Stats.stdv([7, 2, 9])
3.60555127546398912486
The formula for the mean is simple; just add up all the numbers and divide by the number of items in the list (which you may find by using the Enum.count/1 function).Use List.foldl/3 to calculate the sum of the items in the list.
The following is the algorithm for calculating the standard deviation. Presume that n is the number of items in the list.
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Add up all the numbers in the list (call this the sum).
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Add the squares of the numbers in the list (call this the sum of squares).
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Multiply n times the sum of squares.
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Multiply the sum times itself.
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Subtract the result of step 4 from the result of step 3.
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Divide the result of step 5 by n * (n - 1).
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Take the square root of that result.
Thus, if your numbers are 7, 2, and 9, N would be three, and you would do these calculations:
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The sum is 7 + 2 + 9, or 18.
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The sum of squares is 49 + 4 + 81, or 134.
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N times the sum of squares is 134 * 3, or 402.
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The sum times itself is 18 * 18, or 324.
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402 - 324 is 78.
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78 divided by (3 * (3 - 1)) is 78 / 6, or 13.
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The standard deviation is the square root of 13, or 3.606.
In your code, you can do steps three through seven in one arithmetic expression. You’d have variables in your expression rather than constants, of course.
:math.sqrt((3 * 134 - 18 * 18)/(3 * (3 - 1))
Use List.foldl/3 to calculate the sum and the sum of squares. Bonus points if you can calculate both of them with one call to List.foldl/3. Hint: the argument for the accumulator doesn’t have to be a single number. It can be a list or a tuple.
Use h Enum.split to see how the Enum.split/2 function works, or try the following example and see if you can figure it out. Experiment to see what happens if the second argument is zero or greater than the length of the original list.
iex(1)> Enum.split([110, 220, 330, 440, 550, 600], 4)
{[110,220,330,440],[550,600]}
Use Enum.split/2 and List.foldl/3 to rewrite the Dates.julian/1 function from Étude 6-2. Hint: you’ll use those functions when calculating the total number of days up to (but not including) the month in question.
Back to list comprehensions. You can have more than one generator in a list comprehension. Try this in iex:
iex(1)> for x <- [3, 5, 7], y <- [2, 4, 6], do: x * y [6,12,18,10,20,30,14,28,42]
Using what you’ve learned from this example, write a module named Cards that contains a function make_deck/0. The function will use a list comprehension with two generators to create a deck of cards as a list 52 tuples in this form:
[{"A","Clubs"},
{"A","Diamonds"},
{"A","Hearts"},
{"A","Spades"},
{2,"Clubs"},
{2,"Diamonds"},
{2,"Hearts"},
{2,"Spades"},
...
{"K", "Clubs"},
{"K", "Diamonds"},
{"K", "Hearts"},
{"K", "Spades"}]
You need a way to shuffle the deck of cards. This is the code for doing a shuffle, adapted from the Erlang solution at the Literate Programs Wiki.
def shuffle(list) do
:random.seed(:erlang.now())
shuffle(list, [])
end
def shuffle([], acc) do
acc
end
def shuffle(list, acc) do
{leading, [h | t]} =
Enum.split(list, :random.uniform(Enum.count(list)) - 1)
shuffle(leading ++ t, [h | acc])
end
----
Wait a moment. If I've just given you the code, what's the purpose
of this étude? I want you to understand the code. The object of this
étude is to write the documentation for the algorithm.
If you aren't sure what the code does, try adding some
+IO.puts+ statements to see what is happening. If you're totally
stuck,
http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle#Fisher_and_Yates.27_original_method[see the original explanation of the method at Wikipedia].
<<SOLUTION08-ET06,See a suggested solution in Appendix A.>>