## claralucia 3 years ago 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

1. TuringTest

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

2. TuringTest

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

3. claralucia

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.

4. TuringTest

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...

5. claralucia

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.

6. TuringTest

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

7. TuringTest

I will try to draw what is happening something like|dw:1351214431286:dw|

8. TuringTest

|dw:1351214499069:dw|

9. TuringTest

|dw:1351214573121:dw|

10. TuringTest

|dw:1351214610832:dw|and so we are see-sawing back and forth around a minimum, which keeps us from selling on a value

11. TuringTest

settling*

12. claralucia

Ah, I think I see - so that's why the x guesses kept cycling up and down and not converging. In any case, this has been really informative, thanks so much!

13. TuringTest

actually, your problem in this case is much simpler: the function you chose has no real zeroes :P http://www.wolframalpha.com/input/?i=plot%20y%3D1%2B2x%2B3x%5E2&t=crmtb01 so of course you will never get a final answer it also could have been what I said though, just for the record ;)

14. TuringTest

for the future, if you are gonna test a function you should already have an expected answer

15. claralucia

Ha, why didn't I think of that... I wonder if there would be any way to built a "no root" answer into the program.

16. TuringTest

That sounds like a darn good wuestion, but I suppose you could use some logic and say that if x_0 oscillates around its initial value some limit of n times, the loop stops and claims there is no zero to be found it wouldn't be fool proof though, maybe there's a better way :)

17. TuringTest

newtons method usually only requires 5 iterations to get within 0.00001 or so of a root, so at 40 iterations it would be logical to stop the program and say "cannot find root"

18. claralucia

Sounds right. I'm a total beginner by the way, so this is very helpful. Is there a way to tell the loop to terminate after a certain number of iterations?

19. TuringTest

sure, just do something like deriv = compute_deriv(poly) numguess = 0 iters=0 while abs(eval_poly(poly, x_0)) >= epsilon and iters<n: numguess += 1 iters+1 x_0 = x_0 - (eval_poly(poly, x_0))/(eval_poly(deriv, x_0)) if abs(eval_poly(poly, x_0)) >= epsilon: print 'method failed' else: print (eval_poly(poly, x_0), numguess) return when iters gets over some value n the loop stops

20. claralucia

Oh, I suppose you could add something like while numguess <= 40 to the conditions of the loop.

21. TuringTest

exactly btw I'm a total beginner too. I'm doing this course for edX right now, that's why I could check your answer against mine so quickly. I have a good math background though, so I may have the edge on a few things like this :)

22. claralucia

Well, that's encouraging. My math's a little rusty but it's coming by quickly by necessity. Gotta run but thanks again!

23. TuringTest

see, I didn't even need to ad the iters vairable, I forgot we already had the equivalent variable numguesses welcome!

24. TuringTest

deriv = compute_deriv(poly) numguess = 0 while abs(eval_poly(poly, x_0)) >= epsilon and numguess<n: numguess += 1 x_0 = x_0 - (eval_poly(poly, x_0))/(eval_poly(deriv, x_0)) if abs(eval_poly(poly, x_0)) >= epsilon: print 'method failed' else: print (eval_poly(poly, x_0), numguess) return

25. bwCA