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 -

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

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 :)

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?

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

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.

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.

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

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

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?

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

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.

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.

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 :)

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.

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

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...

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

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

never know :)

Join our real-time social learning platform and learn together with your friends!