Question

I need help on the Newton question (#3) on the problem set 2!

I use the eval_poly and compute_deriv function from the first few questions. Those check out, but I just can't get the while loop to work. It runs forever, for reasons I don't understand. Here's the code:

deriv = compute_deriv(poly)
numguess = 0

while abs(eval_poly(poly, x_0)) >= epsilon:
numguess += 1
x_0 = x_0 - (eval_poly(poly, x_0))/(eval_poly(deriv, x_0))
else: print (eval_poly(poly, x_0), numguess)
return

Answer 1

not sure, looks okay to me
try inserting a print statement or two:

deriv = compute_deriv(poly)
numguess = 0

while abs(eval_poly(poly, x_0)) >= epsilon:
numguess += 1
x_0 = x_0 - (eval_poly(poly, x_0))/(eval_poly(deriv, x_0))
print x_0, (eval_poly(poly, x_0)), (eval_poly(deriv, x_0))
else: print (eval_poly(poly, x_0), numguess)
return
that ought to be enlightening

Answer 2

btw it should return the root x_0, not eval_poly(poly, x_0)

Answer 3

Thanks for your help! I put in the print command and I found that the values don't seem to be converging.

0.373076923077 0.671405325444 -0.238461538462
2.44249379653 23.7823154313 16.6549627792
1.01455212253 6.11705227303 8.08731273516
0.25817574964 1.71631565239 3.54905449784
-0.225422249047 0.701601073002 0.647466505718
-1.30903208339 3.52263081924 -5.85419250033
-0.707304207333 1.08622931047 -2.243825244
-0.223207099783 0.703050028615 0.660757401303
-1.28721338595 3.39632813098 -5.72328031567
-0.693790044146 1.05645378778 -2.16274026488

This is using the test values:
poly = (1, 2, 3)
x_0 = .1
epsilon = .0001

Weirdly enough, though, when I put in the original test values from the problem:
poly = (-13.39, 0.0, 17.5, 3.0, 1.0)
x_0 = 0.1
epsilon = .0001

the program works.

I'm not sure what to make of this, because I don't quite understand Newtons formula (x_1 = f(x_0)/f'(x_0) itself.

Answer 4

if it works with the test values then there is a mathematical explanation to why the other one is not working: newton's method fails for certain initial values of x_0
let me try it in my program and see what happens...

Answer 5

Thanks. I suspected it was an issue with restricted application of the formula itself. I need to read up on it, since the problem set provides no info.

Answer 6

yes, the same exact thing happens to mine, I even get the same print values for x_0
it is a result of the fallibility of newtons method, and has to do with having a bad initial choice for x_0
If you want to understand what is happening you could look at the section on Newton's method in the single variable calc section of OCW

Answer 7

I will try to draw what is happening
something like