# Calculus using Functional Python

The aim this piece is to demonstrate some concepts of functional programming in Python, let’s dive-in!

*Part Derivative:*

The following function calculates the slope on a non-linear line.

Now let rewrite the code in Python.

`derivative = lambda f, x: ((f(x + 1e-3) - f(x)) / 1e-3)`

This function is equivalent to the prior:

`def derivative (f, x):`

return ((f(x + 1e-3) - f(x)) / 1e-3)

Functions in Python are **first-class functions**, which means that you can treat them as variables.

Note that the value *1e-3* (0.001) is the limit when *h* approaches zero (h → 0).

Now we use a **partial function:**

`derivative_of_sin = partial(derivative, math.sin)`

Partial functions allow us to pass fewer arguments to multiple arguments function.

In this example, we create a partial function from the function derivative, and its first argument (the sine function), leaving *x* for call time.

`x = 5`

derivative_of_sin(x) # --> 0.28

math.cos(x) # --> 0.28

Correct! since the derivative of sin is cos.

Let’s try x²

`derivative_of_x2 = partial(derivative, `**lambda** a: a * a)

As you can see, we passed a **lambda calculus **expression because our function derivative expects a function.

`derivative_of_x2(x) # --> 10`

Ten is the correct answer since the derivative of *x²* is *2 * x*.

Try to plug your function.

*Part Integral:*

We are going to use algebra to tackle the integral function.

This function has four important arguments; the range from a to b, the steps, and the input function.

In calculus, we find the integral of *x²* (*x³ / 3*) to find the solution.

With algebra, we stack multiple rectangles together to find the solution.

Using algebra:

| steps | equation | result |

|-----------+------------------------------+----------|

| 100 | (1e-1) * (10^2 + ... + 20^2) | 2318.349 |

| 1_000 | (1e-2) * (10^2 + ... + 20^2) | 2331.833 |

| 10_000 | (1e-3) * (10^2 + ... + 20^2) | 2333.183 |

| 100_000 | (1e-4) * (10^2 + ... + 20^2) | 2333.318 |

| 1_000_000 | (1e-5) * (10^2 + ... + 20^2) | 2333.331 |

| | | |*steps count = (20 - 10) / 1e-5 = 1_000_000

Using calculus:

`(1 / 3) * (20**3) - (1 / 3) * (10**3) # --> 2333.33`

The algebra method is approaching the correct solution, albeit it requires a lot of calculation than you transistors!

Let’s define the function:

defintegral (f, a, b):

delta = 1e-6

auc = 0 for i in np.arange(a, b, delta):

auc = auc + f(i) return auc * delta

Note that AUC stands for the area under the curve.

Integral is a **high-order function,** which means that it can take a function as an argument or return a function.

Now, let’s use partial again:

`integral_of_x2 = partial(integral, lambda x: x**2)`

if we pass our partial function with the limits a and b, we get:

`integral_of_x2(10, 20) # ==> 2333.331`

Hooray!

*Part Integral++:*

Another approach is using **closures:**

defintegral (f, s):

delta = sdefg (a, b):

auc = 0

for i in np.arange(a, b, delta):

nonlocal auc

auc += f(i)

return auc * delta

return g

The integral function will return the function g.

The function g holds the value of delta way after the execution of the outer function; this is closure.

Next, we pass a complex function and the number of steps delta.

`integral_of_complex = integral(lambda x: (5*x**2) - (8*x) + 5, 1e-5)`

Passing the limits:

`integral_of_complex2(10, 20) # --> 10515.66`

That’s correct!

*Final thoughts:*

Functional programming has many elegant concepts, some concepts covered:

- first-class functions
- partial functions
- lambda calculus
- high-order functions
- closures

You can find more examples on the GitHub link with additional functional techniques.

https://github.com/booletic/medium/blob/34b60709ce0bd0d657e1e2c88a58125a19b6ee91/calculus.ipynb

If you liked it, check out my other articles on functional programming: