derivative and newton raphson

[DDCane]

i have made a code for finding a derivative and now im trying to use it to help me with a code for the newton raphson method:

def derivative (f,x,h):
    import math
    return float(1/(2*h)) * (f(x+h) - f(x-h))
 
def solve (f,x0,h):
    delta= f(x(n))/fp(x(n)
        for x(n+1) in solve():
            x(n)-delta

the first def works fine but i cant get the second def to work can anyone see what im doing wrong?

[bvdet]

i have made a code for finding a derivative and now im trying to use it to help me with a code for the newton raphson method:

def derivative (f,x,h):
    import math
    return float(1/(2*h)) * (f(x+h) - f(x-h))
 
def solve (f,x0,h):
    delta= f(x(n))/fp(x(n)
        for x(n+1) in solve():
            x(n)-delta

the first def works fine but i cant get the second def to work can anyone see what im doing wrong?

In function solve(), you have passed arguments 'f', 'x0' and 'h'. Where are 'x', 'fp', and 'n' defined? You are calling function 'solve()' recursively, but you are not passing any arguments to it. 'solve()' requires three arguments. The format of a 'for' loop is:

for item in iterable:
    ......code......

'item' cannot be an expression.

>>> for x+1 in range(10):
...     print x
Traceback (SyntaxError: can't assign to operator
>>> 

[DDCane]

how am im supposed to fix it? fp in solve is the fisrt function. u use the first function to get the derivative of f in the first function. do u know how i fix this?

In function solve(), you have passed arguments 'f', 'x0' and 'h'. Where are 'x', 'fp', and 'n' defined? You are calling function 'solve()' recursively, but you are not passing any arguments to it. 'solve()' requires three arguments. The format of a 'for' loop is:

for item in iterable:
    ......code......

'item' cannot be an expression.

>>> for x+1 in range(10):
...     print x
Traceback (SyntaxError: can't assign to operator
>>> 

[bvdet]

how am im supposed to fix it? fp in solve is the fisrt function. u use the first function to get the derivative of f in the first function. do u know how i fix this?

I am not familiar with the calculation you are attempting. I know that it will not work if 'fp()' is not defined inside of 'solve()' or globally to your module.

[DDCane]

globally in the module?

I am not familiar with the calculation you are attempting. I know that it will not work if 'fp()' is not defined inside of 'solve()' or globally to your module.

[bvdet]

globally in the module?

Let's say your code is in a file called function.py.

# code.py
def fp():
    ....code....
 
def derivative (f,x,h):
    import math
    return float(1/(2*h)) * (f(x+h) - f(x-h))
 
def solve (f,x0,h):
    delta= f(x(n))/fp(x(n)
        for x(n+1) in solve():
            x(n)-delta
 
if __name__ == __main__:
    ....call your functions...

Function 'fp()' is global to module 'function'.

[DDCane]

ok. is there a simpler way to write the newton raphsons method in python? ive looked at the discussions in the forums here and havent found anything that could help m efurther than what i already know. do u have any clue on how to write the method?

Let's say your code is in a file called function.py.

# code.py
def fp():
    ....code....
 
def derivative (f,x,h):
    import math
    return float(1/(2*h)) * (f(x+h) - f(x-h))
 
def solve (f,x0,h):
    delta= f(x(n))/fp(x(n)
        for x(n+1) in solve():
            x(n)-delta
 
if __name__ == __main__:
    ....call your functions...

Function 'fp()' is global to module 'function'.

[bvdet]

ok. is there a simpler way to write the newton raphsons method in python? ive looked at the discussions in the forums here and havent found anything that could help m efurther than what i already know. do u have any clue on how to write the method?

The Newton Raphson method uses an iterative process to approximate the root of a function. That's all I know about the subject, and I had to look that up.

[KaezarRex]

ok. is there a simpler way to write the newton raphsons method in python? ive looked at the discussions in the forums here and havent found anything that could help m efurther than what i already know. do u have any clue on how to write the method?

Here's what I have come up with:
import math

def derivative (f, x, h):
    return float((f(x + h) - f(x))) / h
 
def solve(f, x0, h, depth):
    if depth > 0:
        delta = f(x0) / derivative(f, x0, h)
        return solve(f, x0 - delta, h, depth - 1)
    else:
        return x0

I changed the formulas in your function to ones I'm more familiar with, and I had to add the parameter "depth" to tell the solve function when to stop. That's necessary because when working with the Newton-Raphson Method, you have to choose how many times you use it.
To use the solve function, you can either do something like this:

def function(x):
    return math.cos(x) - x**3
 
solve(function, 0.5, 0.001, 6)

or:

function = lambda x: math.cos(x) - x**3
solve(function, 0.5, 0.001, 6)

Either way in that example your approximating the root of "cos(3) - x**3" by applying the Newton-Raphson Method 6 times with a starting guess of "0.5". The "0.001" is the "h" value used to approximate your derivative. The lower it is the more accurate the derivative will be.

[bvdet]

Here's what I have come up with:
import math

def derivative (f, x, h):
    return float((f(x + h) - f(x))) / h
 
def solve(f, x0, h, depth):
    if depth > 0:
        delta = f(x0) / derivative(f, x0, h)
        return solve(f, x0 - delta, h, depth - 1)
    else:
        return x0

I changed the formulas in your function to ones I'm more familiar with, and I had to add the parameter "depth" to tell the solve function when to stop. That's necessary because when working with the Newton-Raphson Method, you have to choose how many times you use it.
To use the solve function, you can either do something like this:

def function(x):
    return math.cos(x) - x**3
 
solve(function, 0.5, 0.001, 6)

or:

function = lambda x: math.cos(x) - x**3
solve(function, 0.5, 0.001, 6)

Either way in that example your approximating the root of "cos(3) - x**3" by applying the Newton-Raphson Method 6 times with a starting guess of "0.5". The "0.001" is the "h" value used to approximate your derivative. The lower it is the more accurate the derivative will be.

That's great KaezarRex - nice solution to an interesting problem.
BV