anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
is this an on line class or do you really have to use newton's method?
anonymous
  • anonymous
I really have to use newtons method...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

misty1212
  • misty1212
ok then you have to start with something like \[f(x)=\ln(x)-(x-2)^2\] and look for the zero of that one
misty1212
  • misty1212
better still \[f(x)=x^2-4x+4-\ln(x)\]
misty1212
  • misty1212
then you make a guess you got a good guess in mind?
misty1212
  • misty1212
woah lot of algebra method you can use that too, but it is sometimes easier not to
misty1212
  • misty1212
lets make a guess i guess 3
misty1212
  • misty1212
\[f(3)=3^2-4\times 3+4-\ln(3)=-0.098612\]
misty1212
  • misty1212
pretty close to zero already
misty1212
  • misty1212
you gotta start somewhere right?
misty1212
  • misty1212
figure it is easier to start with an integer
DanJS
  • DanJS
for reference if needed... http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx
misty1212
  • misty1212
now for the second approxmition it is \[3-\frac{f(3)}{f'(3)}\]
misty1212
  • misty1212
\[f'(x)=2x-4-\frac{1}{x}\] so \[f'(3)=2\times 3-4-\frac{1}{3}=\frac{5}{3}\]
misty1212
  • misty1212
so next guess is \[3+\frac{-0.098612}{\frac{5}{3}}\] at this point you really need a calculator
misty1212
  • misty1212
this gets very very messy quick but you can do it with wolfram or a spreadsheet
misty1212
  • misty1212
i get \[x_2=2.9408328\]
misty1212
  • misty1212
now you have to repeat it so it is really going to be messy
anonymous
  • anonymous
Yea, didnt get it...
misty1212
  • misty1212
what did you not get?
anonymous
  • anonymous
why are you using 3? whats so special about it
misty1212
  • misty1212
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
misty1212
  • misty1212
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
misty1212
  • misty1212
one is very near 3, so i picked 3 as \(x_1\)
mathmate
  • mathmate
ouch! I think it should be: \(\large 3\color{red}{-}\frac{-0.098612}{\frac{5}{3}}\)
misty1212
  • misty1212
yeah you are right, it should be a plus
misty1212
  • misty1212
so \[x_2=3.0591672\]
misty1212
  • misty1212
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
misty1212
  • misty1212
click on the numeric answer, then rewrite with it substituted where i wrote x = 3
misty1212
  • misty1212
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
mathmate
  • mathmate
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.