Newtons Method

[janeann]

a)Use Newton's Method to determine sqrt(5)+sqrt(7) to six decimal places of accuracey.
b) Find a point close to the orgin as an initial guess which is in the domain of convergence of sqrt(5)+sqrt(7). How do you know the initial guess is in the domain of convergence?

 

[galactus]

We know that $\displaystyle 2^{2}=4$ and $\displaystyle 3^{2}=9$

So, $\displaystyle \sqrt{5}$ has to be between 2 and 3.

Make an initial guess of 2. It will eventually converge.

Here we have something known as the Mechanic's Rule/Babylonian Method of finding a square root by hand.

It can be derived from Newton's method, but goes back further.

One of the ways they done it before calculators.

To find $\displaystyle \sqrt{a}$ with an intial guess of $\displaystyle x_{n}$

$\displaystyle x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)$

Let's take $\displaystyle x_{n}=2$, since we know it is going to be 2 and some change.

$\displaystyle \frac{1}{2}\left(2+\frac{5}{2}\right)=\frac{9}{4}=2.25$

$\displaystyle \frac{1}{2}\left(2.25+\frac{5}{2.25}\right)=\frac{161}{72}=2.236\overline{1}$

Keep going. This converges very fast, so not many iterations are needed.

Keep going until you get within the desired accuracy.

Do it for $\displaystyle \sqrt{7}$ and add them.


To use Newton's Method: $\displaystyle x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$

To approximate $\displaystyle \sqrt{5}$, apply Newton to $\displaystyle x^{2}-5$

To approximate $\displaystyle \sqrt{7}$, apply Newton to $\displaystyle x^{2}-7$

$\displaystyle x_{n+1}=x_{n}-\frac{(x_{n})^{2}-5}{2x_{n}}$

Make an initial guess and keep iterating. As before, it converges fast.

Notice the similarity to the first method. It is the same thing. A little algebraic fanaggling and we have it.

 

[kjbohn]

thank you!

 

[DrMike]

Alternatively, if x=sqrt(5)+sqrt(7), then x^2 = 12+2sqrt(35), so (x^2-12)^2=140. You could use Newton's method once on x^4-24x^2+4, instead of twice on x^2-5 and x^2-7. make sure you check that you've converged to sqrt(5)+sqrt(7) and not to sqrt(5)-sqrt(7), sqrt(7)-sqrt(5) or -sqrt(5)-sqrt(7)! It's enough to check that your root is bigger 1, since sqrt(7)-sqrt(5) < sqrt(9)-sqrt(4)=1, and the other two roots are negative. Hint : choose a big enough x0 and you'll hit the correct root first try!

 

[DrMike]

For part (b), what theorems have you been given about the domain of convergence of newton's method?

 

[kjbohn]

I don't think we have had anything about convergence theroms:/

 

[galactus]

Here is an animated graph of what Newton is doing using Dr. Mikes suggestion.

See the lines converging on the point?.