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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
 one year ago
 one year 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
 one year ago
 one year ago

This Question is Closed

TuringTestBest ResponseYou've already chosen the best response.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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
btw it should return the root x_0, not eval_poly(poly, x_0)
 one year ago

claraluciaBest ResponseYou've already chosen the best response.0
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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...
 one year ago

claraluciaBest ResponseYou've already chosen the best response.0
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I will try to draw what is happening something likedw:1351214431286:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
dw:1351214499069:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
dw:1351214573121:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
dw:1351214610832:dwand so we are seesawing back and forth around a minimum, which keeps us from selling on a value
 one year ago

claraluciaBest ResponseYou've already chosen the best response.0
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!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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 ;)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
for the future, if you are gonna test a function you should already have an expected answer
 one year ago

claraluciaBest ResponseYou've already chosen the best response.0
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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 :)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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"
 one year ago

claraluciaBest ResponseYou've already chosen the best response.0
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?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

claraluciaBest ResponseYou've already chosen the best response.0
Oh, I suppose you could add something like while numguess <= 40 to the conditions of the loop.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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 :)
 one year ago

claraluciaBest ResponseYou've already chosen the best response.0
Well, that's encouraging. My math's a little rusty but it's coming by quickly by necessity. Gotta run but thanks again!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
see, I didn't even need to ad the iters vairable, I forgot we already had the equivalent variable numguesses welcome!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

bwCABest ResponseYou've already chosen the best response.0
@Turing Test http://www.youtube.com/watch?v=x2KbdoxrQ6o
 one year ago
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