Newton's Method

[Violagirl]

Hi I was wondering if someone could check my work for this problem:

Use Newton's Method to find a zero of the function f(x)=3-x ln x correct to three decimal places. X[sub:esya7svb]1[/sub:esya7svb]=3.

I've completed three iterations and on the third, have gotten an answer of -22.13. And plugging it into the function, it does not exist. What would you have to do at this point? :?

 

[Violagirl]

Nevermind I see now what I did wrong in the first problem, I derived the equation the wrong way. However I'm confused with one other problem:

Use Newton's Method to approximate -3[sup:17og97vn]1/3[/sup:17og97vn] correct to five decimal places, X[sub:17og97vn]1[/sub:17og97vn]=1.2.

So for the eqauation, I'm confused how you would find the derivative of -3^{1/3[/sup without any x's in the equation. :?}

 

[Violagirl]

I have no idea how to start this problem:

Apply Newton's method to the equation x[sup:1hloxyc6]2[/sup:1hloxyc6]-a=0 to derive the following square-root algorithm (used by the ancient Babylonians to compute a[sup:1hloxyc6]1/2[/sup:1hloxyc6]:

X[sub:1hloxyc6]n+1[/sub:1hloxyc6]= 1/2(X[sub:1hloxyc6]n[/sub:1hloxyc6]+a/X[sub:1hloxyc6]n[/sub:1hloxyc6])

 

[BigGlenntheHeavy]

\(\displaystyle Hint: \ -3^{1/3} \ \implies \ x^{3}+3 \ = \ 0 \ = \ f(x).\)

 

[Violagirl]

Thanks bigglenntheheavy, that helped me a lot! Now if I could get help with the other problem, that would be great too! It's the one listed below:

I have no idea how to start this problem:

Apply Newton's method to the equation x[sup:2qu0sdy6]2[/sup:2qu0sdy6]-a=0 to derive the following square-root algorithm (used by the ancient Babylonians to compute a[sup:2qu0sdy6]1/2[/sup:2qu0sdy6]:

X[sub:2qu0sdy6]n+1[/sub:2qu0sdy6]= 1/2(X[sub:2qu0sdy6]n[/sub:2qu0sdy6]+a/X[sub:2qu0sdy6]n[/sub:2qu0sdy6])

 

[BigGlenntheHeavy]

\(\displaystyle Newton's \ Method:\)

\(\displaystyle x_{n+1} \ = \ x_n \ - \ \frac{f(x_n)}{f \ ' \ (x_n)}\)

\(\displaystyle f(x_n) \ = \ x^{2}_n-a, \ f \ ' \ (x_n) \ = \ 2x_n\)

\(\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.\)

 

[Violagirl]

Thanks BigGlennTheHeavy, that did help a little bit on the last problem. I am still confused by how you're supposed to derive the square root-algorithm or how to start that. I don't see where to start for it. :?

 

[Violagirl]

I looked at the algorithem problem again and am very unclear about what they want me to do for it to find the answer or how to derive the algorithem part of the problem.

 

[galactus]

Just use your initial guess of 3 and plug it in the formula.

\(\displaystyle f(x)=3-xln(x), \;\ f'(x)=-ln(x)-1\)

\(\displaystyle 3-\frac{3-3ln(3)}{\underbrace{-ln(3)-1}_{\text{derivative}}}=\frac{6}{ln(3)+1}\approx 2.85903\)

Now, plug that value in and keep going until there is an accetable amount of accuracy.

The actual solution to \(\displaystyle 3-xln(x)=0\) is 2.85739.

Just 2 or 3 interations and it's done.