Question

Need Help!

Use Newton's method to find all roots of the equation (x-2)^2=ln(x). Must be correct to 6 decimal places.

Answer 1

HI!!

Answer 2

is this an on line class or do you really have to use newton's method?

Answer 3

I really have to use newtons method...

Answer 4

ok then you have to start with something like
\[f(x)=\ln(x)-(x-2)^2\] and look for the zero of that one

Answer 5

better still \[f(x)=x^2-4x+4-\ln(x)\]

Answer 6

then you make a guess
you got a good guess in mind?

Answer 7

woah lot of algebra method
you can use that too, but it is sometimes easier not to

Answer 8

lets make a guess
i guess 3

Answer 9

\[f(3)=3^2-4\times 3+4-\ln(3)=-0.098612\]

Answer 10

pretty close to zero already

Answer 11

you gotta start somewhere right?

Answer 12

figure it is easier to start with an integer

Answer 13

for reference if needed... http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx

Answer 14

now for the second approxmition it is
\[3-\frac{f(3)}{f'(3)}\]

Answer 15

\[f'(x)=2x-4-\frac{1}{x}\] so \[f'(3)=2\times 3-4-\frac{1}{3}=\frac{5}{3}\]

Answer 16

so next guess is
\[3+\frac{-0.098612}{\frac{5}{3}}\] at this point you really need a calculator

Answer 17

this gets very very messy quick but you can do it with wolfram or a spreadsheet

Answer 18

i get \[x_2=2.9408328\]

Answer 19

now you have to repeat it so it is really going to be messy

Answer 20

Yea, didnt get it...

Answer 21

what did you not get?

Answer 22

why are you using 3? whats so special about it

Answer 23

oh , you have to make an educated guess first

you can guess, graph, use wolfram (what i did) you have to do something to guess a number close to the zero of the function

Answer 24

here is a picture of the function
http://www.wolframalpha.com/input/?i=x%5E2-4x%2B4-ln%28x%29
click on "real values plot" so as not to get too confused
you see it has two zeros

Answer 25

one is very near 3, so i picked 3 as \(x_1\)

Answer 26

ouch! I think it should be:
\(\large 3\color{red}{-}\frac{-0.098612}{\frac{5}{3}}\)

Answer 27

yeah you are right, it should be a plus

Answer 28

so \[x_2=3.0591672\]

Answer 29

you really want to make life easy on yourself
here is what you can do for newton's method
you are going to have to use a calculator anyways, so here is what we get if we make a first guess of 3
http://www.wolframalpha.com/input/?i=x-%28%28x-2%29%5E2-ln%28x%29%29%2F%282x-4-1%2Fx%29%2C+x%3D3

Answer 30

click on the numeric answer, then rewrite with it substituted where i wrote x = 3

Answer 31

i got this
http://www.wolframalpha.com/input/?i=x-%28%28x-2%29%5E2-ln%28x%29%29%2F%282x-4-1%2Fx%29%2C+x%3D3.05917

Answer 32

To resume how we could solve a problem using Newton's method.
1. define the equation in the form of f(x)=0, because Newton's method is for finding zeroes of an equation.
Here (x-2)^2=ln(x) becomes f(x)=(x-2)^2-ln(x)=0
2. Graph the function and visually search for zeroes, which will be the potential initial values of the solutions. Also, visually check the graph to be convinced that all possible roots are counted. for the above f(x), we see that the graph is basically a quadratic, slightly modulated by the function ln(x), and which also restricted the domain to x>0.
If we examine the graph between 1 and 4, we see that the roots are located near 1.5- and 3+. These will therefore be our initial approximations.
Note: if a root of higher multiplicity is suspected, a different procedure from the following is required.
3. set up the iterative equation, x(n+1)=x(n) - f(x(n))/f'(x(n))
Here we have
f(x)=(x-2)^2-ln(x)
f'(x)=2(x-2)-1/x
so the iterative equation is (x is used to stand for x(n) for simplicity)
x(n+1)=x(n)-f(x)/f'(x) =x(n)- ((x-2)^2-ln(x))/)2(x-2)-1/x)
=(xln(x)+x^3-5x)/(2x62-4x-1)
4. Apply the iteration solution a number of times to attain the required accuracy.
Keep at least 2 decimals beyond the required accuracy for all calculations.
Newton's method typically doubles the number of correct digits at each iteration, except for special circumstances such as multiple roots.
Use calculator, Excel, Wolfram or any other tool to help you calculate accurately.

Here the results are:
A. for initial approximation of 3:
3
3.059167373200866
3.0571060546916
3.057103549998436
3.057103549994738
B. for initial approximation of 1.5 :
1.5
1.406720935135101
1.412369957251193
1.412391171725005
1.412391172023884
Clearly, the roots are 3.057104 and 1.412391 each to 6 decimal places.