Aristides Aristides - 6 months ago 8
Python Question

Newton's method: order of statements in loop

I'm trying to implement Newton's method for fun in Python but I'm having a problem conceptually understanding the placement of the check.

A refresher on Newton's method as a way to approximate roots via repeated linear approximation through differentiation:

Newton's Method

I have the following code:

# x_1 = x_0 - (f(x_0)/f'(x_0))
# x_n+1 - x_n = precision

def newton_method(f, f_p, prec=0.01):
x=1
x_p=1
tmp=0

while(True):
tmp = x
x = x_p - (f(x_p)/float(f_p(x_p)))
if (abs(x-x_p) < prec):
break;
x_p = tmp

return x


This works, however if I move the
if
statement in the loop to after the
x_p = tmp
line, the function ceases to work as expected. Like so:

# x_1 = x_0 - (f(x_0)/f'(x_0))
# x_n+1 - x_n = precision

def newton_method(f, f_p, prec=0.01):
x=1
x_p=1
tmp=0

while(True):
tmp = x
x = x_p - (f(x_p)/float(f_p(x_p)))
x_p = tmp
if (abs(x-x_p) < prec):
break;

return x


To clarify, function v1 (the first piece of code) works as expected, function v2 (the second) does not.

Why is this the case?

Isn't the original version essentially checking the current
x
versus the
x
from 2 assignments back, rather than the immediately previous
x
?

Here is the test code I am using:

def f(x):
return x*x - 5

def f_p(x):
return 2*x

newton_method(f,f_p)


EDIT

I ended up using this version of the code, which forgoes the
tmp
variable and is much clearer for me, conceptually:

# x_1 = x_0 - (f(x_0)/f'(x_0))
# x_n+1 - x_n = precision

def newton_method(f, f_p, prec=0.01):
x=1
x_p=1
tmp=0

while(True):
x = x_p - (f(x_p)/float(f_p(x_p)))
if (abs(x-x_p) < prec):
break;
x_p = x

return x

Answer

Let x[i] be the new value to be computed in an iteration.

What is happening in version 1:

The statement x = x_p - (f(x_p)/float(f_p(x_p))) translates to:

x[i] = x[i-2] - f(x[i-2])/f'(x[i-2]) - 1

But according to the actual mathematical formula, it should have been this:

x[i] = x[i-1] - f(x[i-1])/f'(x[i-1])

Similarly, x[i-1] = x[i-2] - f(x[i-2])/f'(x[i-2]) - 2

Comparing 1 and 2, we can see that the x[i] in 1 is actually x[i-1] according to the math formula.

The main point to note here is that x and x_p are always one iteration apart. That is, x is the actual successor to x_p, unlike what it might seem by just looking at the code.

Hence, it is working correctly as expected.

What is happening in version 2:

Just like the above case, the same thing happens at the statement x = x_p - (f(x_p)/float(f_p(x_p))).
But by the time we reach if (abs(x-x_p) < prec), x_p had changed its value to temp = x = x[i-1].

But as deduced in the case of version 1, x too is x[i-1] rather than x[i].

So, abs(x - x_p) translates to abs(x[i-1] - x[i-1]), which will turn out to be 0, hence terminating the iteration.

The main point to note here is that x and x_p are actually the same values numerically, which always results in the algorithm terminating after just 1 iteration itself.

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