Newton's method (intersection points)

pari786
Messages
29
Reaction score
0

Homework Statement



Given two functions; y= ln(x) and y=(x^2)/8 - 2

Use Newton's Method to approximate all intersection points of the given functions, each with 3 decimal places.

Homework Equations



Xn+1 = Xn - f(x) / f`(x)

The Attempt at a Solution



Step 1: I equated the two equations, and I got

f(x) = ln(x) - (x^2)/8 +2

Step 2: I found the derivative as f`(x) = 1/x - x/4

Step 3: then I drew the graphs, and I found that they only intersect at a point near 5.4 (because ln(x) doesn't have any graph in the -x direction)

so I started off x1 = 5.4 and then using the equation given above I found the following values:

X2 = 5.435
X3 = 5.436
X4 = 5.435
X5 = 5.436


Can you guys please tell me if I did the solution right and at which point do I stop ... they keep going ... I'm not going same 3 decimal places for any of them ?
I hope someone would answer soon because I have it due tomorrow early morning. Thanks and can you also tell that whatever I did was right and nothing else was to be done!

:confused:
Thanks
 
Physics news on Phys.org
Yes, you are doing fine with one proviso- if you want the answer correct to 3 decimal places, use more than 3 decimal places, 4 should do, during your calculation. Continue until the 4 decimal places are the same or at least give the same answer rounded to 3 decimal places.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top