Ask your own question, for FREE!
Mathematics
OpenStudy (anonymous):

Using Newtons method on the square root of 16. The 2 relevant formulas are f(guess) = guess-squared - x and guess(new) = old-guess - (f(old-guess)/2 times old-guess). I will put my workings out below... but... at a certain point, when guess is 4.004, I get the answer 0. Does this indicate that 4.004 is the answer, where the x is 0? We all know that 4.004 is not exactly the square root of 16, is this because of my rounding? Or faulty calculation, or a faulty assumption that getting 0 is the end point? And how does he derive that the equation of the line for square root of 16 is guess-squared -

7 years ago
OpenStudy (anonymous):

f(3) = \[guess ^{2}\] - x f(3) = 9 - 16 = -7 guess(new) = 3 - (-7/6) = 4.1666 f(4.167) = 17.36 - 16 = 1.36 guess(new) = 4.167 - 1.36/8.334) = 4.004 f(4.004) = 16.032 - 16 = 0.032 guess(new) = 4.004 - (0.032/8.003) = 4.004 - 4.004 = 0

7 years ago
OpenStudy (cruffo):

Seems like round off error. The value 4.004 is within 10^-3 of the exact answer. If you keep more digits, you will usually get more accurate answers (but sometimes not :)

7 years ago
OpenStudy (anonymous):

Okay thanks, cool, so I am making the right assumption with regards, to the 0, I did it again with 8 decimal places and get to 4.00000139 before I get an answer on my calculator with an e which I can't understand, but which gets me to a much closer answer. Very happy with my progress today on this problem which has been holding me back for ages. So, what about the gap in knowledge, how does he get the equation guess-squared - x as the equation in the first place?

7 years ago
OpenStudy (cruffo):

Yep, that's what the 0 means. Usually you set a stopping condition to tell you when you have reached a numerically accurate answer, such as | previous - current | < 0.0001

7 years ago
OpenStudy (cruffo):

About the "e" on the calculator. Could be e for error, or it could represent scientific notation as in 200000 = 2*10^5 = 2 E5. Just depends on the calculator.

7 years ago
OpenStudy (anonymous):

okay, perfect, the stopping condition makes perfect sense.. I think the e is the scientific notation, I'm not too concerned about the understanding, for now that is close enough, I have too much knowledge gaps to fill... thankyou so much though... As to the equation, if I even knew what terms I could use to search for tutorials I could study it, but if someone was to explain it, it would be ace :) I have a lot of trouble with not having a clue what maths things are called, that I can't even search properly.

7 years ago
OpenStudy (cruffo):

They I looked at it was we are trying to find the root of \[f(x) = x^2-16\]

7 years ago
OpenStudy (cruffo):

Sorry - the way I looked at it was ...

7 years ago
OpenStudy (anonymous):

Yes, but why is the line graph of the problem of the square root of 16 go to that function? Or am I wrong in thinking that function is the square root of 16?

7 years ago
OpenStudy (anonymous):

Would the function of the square root of 9 look different?

7 years ago
OpenStudy (anonymous):

I mean, I'm now pretty satisfied that if I am given an equation, I can go off and use Newtons function, come out with a pretty good answer. But I don't know when I would use it myself, or if they told me to, say, use Newtons method to find the square root of 9.

7 years ago
OpenStudy (cruffo):

I think I get what you're saying about the line graph: A number line, and you put a solid dot on your initial guess, and at each successive answer.

7 years ago
OpenStudy (cruffo):

If you were asked to find the square root of any number, the process would be the same (no change in the equation, but obvious changes in the answers :)

7 years ago
OpenStudy (cruffo):

Newton's Method is a good numerical method for finding x-intercepts. But when you change the function, the equations you are working with will change.

7 years ago
OpenStudy (cruffo):

For instance, to find the real valued cube root of a number you would need the equations x = number \[f(\text{guess}) = \text{guess}^3 - x\] \[\text{guess}(\text{new}) = \text{guess} - \frac{f(\text{guess}) }{3\cdot \text{guess}^2}\]

7 years ago
OpenStudy (anonymous):

Oooooo! Okay, I am comprehending! the divide part, that is the derivative of f(guess), no? Ah, I see.. Thankyou SOOO much for the help you have given me today! I have been struggling with this for so long...

7 years ago
OpenStudy (cruffo):

Yep, the denominator is the derivative of the function you are working with. I'm glad you are so persistent.

7 years ago
OpenStudy (anonymous):

:) your Group Star title speaks accurately! I hope I can help you one day, though I doubt that.

7 years ago
OpenStudy (cruffo):

never know :)

7 years ago
Can't find your answer? Make a FREE account and ask your own question, OR you can help others and earn volunteer hours!