Mathematical expression as argument to function?

Question

I'm very much a beginner, so please be gentle.

I am tinkering with some Python exercises, and I have code looking like this.

def newton(x0, Tol):
    def F(x):
        return (x**3)+898
    def dF(x):
        return 3*x**2
    x=[x0]
    for n in range(400):
        x.append(x[n]-(F(x[n]))/dF(x[n]))
        if abs((x[n+1])-(x[n]))<Tol:
            conv='Converge'
            print n
            break
    if abs((x[n+1])-(x[n]))>=Tol:
        conv='No converge'
    return x[n+1], conv

I define a function F(x) and its derivative dF(x) and add values to a list x.

The task is to check if the series converge or not, which I think I have succeeded with.

But the question I have is regarding having the functions (x**3)+898 and 3*x**2 as arguments to the function Newton.

I imagined it would be something like this

def newton(f, df, x0, Tol):
    def F(x):
        return f
    def dF(x):
        return df
    *calculations*
    return x[n+1], conv

And you would call it with

newton((x**3)+898, 3*x**2, x0=something, Tol=something)

So that the functions F(x) and dF(x) are defined in the process.

However, x is not defined so it does not work.

Note that having f and df as parameters of 'newton' is required in the excercise.

How would you go about solving this?

Thanks.

Answer 1

You can use lambdas, which are basically simple functions.

You would call it like this:

newton(lambda x: (x**3)+898, lambda x: 3*x**2, x0=something, Tol=something)

This would be the same as doing something like this:

def f1(x):
    return (x**3)+898
def f2(x):
    return 3*x**2
newton(f1, f2, x0=something, Tol=something)

The only difference is that you don't "give the lambda a name", ie assign it to a variable. This is handy for functions you only need to use once, especially as key arguments to functions like sorted and max.

Answer 2

Use lambda:

newton(lambda x: (x**3)+898, lambda x: 3*x**2, x0=something, Tol=something)