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 2 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
 2 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

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TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1not 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1btw it should return the root x_0, not eval_poly(poly, x_0)

claralucia
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks 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.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1if 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...

claralucia
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks. 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.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1yes, 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1I will try to draw what is happening something likedw:1351214431286:dw

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1351214499069:dw

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1351214573121:dw

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

claralucia
 2 years ago
Best ResponseYou've already chosen the best response.0Ah, 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!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1actually, 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 ;)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1for the future, if you are gonna test a function you should already have an expected answer

claralucia
 2 years ago
Best ResponseYou've already chosen the best response.0Ha, why didn't I think of that... I wonder if there would be any way to built a "no root" answer into the program.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1That 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 :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1newtons 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"

claralucia
 2 years ago
Best ResponseYou've already chosen the best response.0Sounds 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?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1sure, 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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1exactly 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 :)

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1deriv = 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

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.0@Turing Test http://www.youtube.com/watch?v=x2KbdoxrQ6o
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