Simple function with for loop

Question

I'm stuck in a very stupid point while reading Numeric Analysis.

So I have the following program in python. And I can't figure why I get these results.

Where do I use the i in heron(x,y) to get these results?

Because only the first one makes sense for me. Why are the numbers decreasing if the i isn't used at all at the function?

def heron(x,y):
    x=(x+y/x)*0.5
    return x

x=1
y=2
for i in range(5):
   x=heron(x,y)
   print('Approximation of square root : %.16f'%x)

And the results:

Approximation of square root :1.5000000000000000
Approximation of square root :1.4166666666666665
Approximation of square root :1.4142156862745097
Approximation of square root :1.4142135623746899
Approximation of square root :1.4142135623730949

Edit: The code was given by my professor in class and I guess the only use of it was to explain few basic things of Python?

Don't use global variable names for input arguments for heron — user378704
– user378704, Commented Nov 1, 2016 at 17:43
@kiran.koduru I'm pretty sure that's not true. I tested a similar function in interactive mode def foo(x): x = x**2 return x and it doesn't mutate the value of a global x. — Patrick Haugh
– Patrick Haugh, Commented Nov 1, 2016 at 17:51

score 3 · Accepted Answer · 2016-11-01 17:49:32Z

3

The line

for i in range(5):

only means:

Do the following five times.

The actual work is done in

x = heron(x,y)

which uses x as part of the arguments of heron and assigns the changed value back to it. So while y stays unchanged, x is changed with each call to heron. The changed x is then used as an argument to the next call.

Edit: I can't decide if this is a correct implementation because I don't know what algorithm you are trying to implement. But you only asked:

Why are the numbers decreasing if the i isn't used at all at the function?

edited Nov 1, 2016 at 17:49

answered Nov 1, 2016 at 17:46

user6999902

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3 Comments

Moses Koledoye Over a year ago

Better to use for _ in range(5) since i is never used

RedRose23 Over a year ago

Oh I see! So at first I have x=1 but after the first loop it the x equals 1.5, is that correct?

user6999902 Over a year ago

Yes, that's the result of the first call to heron. 1.5 is assigned to x and used in the 2nd call.

Guillaume Jacquenot · Accepted Answer · 2016-11-01 20:11:03Z

0

You are trying to implement Heron's algorithm to find the root square of a number.

It is an iterative algorithm where you increase precision of the result at each step.

In your implementation, x is an initial solution and y is the number you want to find the root square.

You are doing 5 iterations and the variable i is not required to do your iterations. You can use _ to declare a non needed variable.

You could define a wanted precision and iterate a various number of times to reach the wanted precision.

def heron(x,y):
    x=(x+y/x)*0.5
    return x

x=1
y=2
numberOfIterations = 5
for _ in range(numberOfIterations):
    x=heron(x,y)
    print('Approximation of square root : %.16f'%x)

edited Nov 1, 2016 at 20:11

answered Nov 1, 2016 at 17:56

Guillaume Jacquenot

11.8k6 gold badges45 silver badges50 bronze badges

Comments

Mangu Singh Rajpurohit · Accepted Answer · 2016-11-01 17:47:53Z

-1

I think, you must modify your code as follow :-

def heron(x,y):
    x=(x+y/x)*0.5
    return x

x=1
y=2
for i in range(5):
   z=heron(x,y)
   print 'Approximation of square root :%.16f'%z

answered Nov 1, 2016 at 17:47

Mangu Singh Rajpurohit

11.5k4 gold badges76 silver badges101 bronze badges

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