Newton's method

**Makall** · March 10th, 2010, 12:24 AM

Use Newton's method to approximate to within 10^-4, the value of x that produces the point on the graph of y = 1/x that is closest to (2, 1).

I tried setting p0 = 2 and using the algorithm $p_{n} = p_{n-1} - f(p_{n-1})/f'(p_{n-1})$ where f(x) = 1/x and f'(x) = -1/x^2 but the number is actually increasing? It goes 4, 8, 16, 32, ...

I'm not sure what I'm doing wrong here.

**Archie Meade** · March 10th, 2010, 01:03 AM

Originally Posted by Makall

Use Newton's method to approximate to within 10^-4, the value of x that produces the point on the graph of y = 1/x that is closest to (2, 1).

I tried setting p0 = 2 and using the algorithm $p_{n} = p_{n-1} - f(p_{n-1})/f'(p_{n-1})$ where f(x) = 1/x and f'(x) = -1/x^2 but the number is actually increasing? It goes 4, 8, 16, 32, ...

I'm not sure what I'm doing wrong here.

Hi Makall,

You could develop a function for the distance from (2,1) to points on the graph.
Then differentiate that function, setting it equal to zero to find the value of x
producing that minimum distance.

A distance equation is

$d^2=(x-2)^2+(y-1)^2=x^2-4x+4+y^2-2y+1$

Now substitute $y=\frac{1}{x}$

to get

$d^2=x^2-4x+5+-2x^{-1}+x^{-2}$

Differentiate this, set it equal to zero and use Newton's method to find the root near x=2.

**Makall** · March 10th, 2010, 01:47 AM

So I got this $f = {\left(x - 2\right)}^2 + {\left(\frac{1}{x} - 1\right)}^2$ and differentiated it to get this $f' = 2\, x + \frac{2}{x^2} - \frac{2}{x^3} - 4$

Then I used this equation $x = x - \frac{{\left(x - 2\right)}^2 + {\left(\frac{1}{x} - 1\right)}^2}{2\, x + \frac{2}{x^2} - \frac{2}{x^3} - 4}$

Used initial value as 2 and got these numbers.

x = 2
>> for i=1:10, x = x - ((x - 2)^2 + (1/x - 1)^2)/(2*x + 2/x^2 - 2/x^3 - 4), end
x = 1
x = 1.500000000000000
x = 2.013157894736842
x = 1.090411751549197
x = 1.587061684245663
x = 2.164611788880735
x = 1.598178106645094
x = 2.188802634131807
x = 1.641701289644409
x = 2.300866462141473

The solution says p0 = 2, p2 = 1.866760 and the point is (1.866760, 0.535687). The number seems to be alternating so I'm not sure what's wrong with it.

**Archie Meade** · March 10th, 2010, 02:24 AM

Hi Makall,

to get the distance, you need to take the square root of the square of the distance,
differentiate that, equate to zero,
find the value of x using Newton's method that causes the derivative to be zero.

**Archie Meade** · March 10th, 2010, 08:58 AM

$d^2=x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}$

$d=\sqrt{x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}}$

To find the co-ordinate x at the minimum distance from (2,1)
differentiate d with respect to x and equate to zero
(finds x at the minimum value of the curve of d).

$d'=\frac{1}{2}\left(x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}\right)^{-\frac{1}{2}}\left(2x-4+\frac{2}{x^2}-\frac{2}{x^3}\right)$

$=\frac{x-2+\frac{1}{x^2}-\frac{1}{x^3}}{\sqrt{x^2-4x+5-\frac{2}{x}+\frac{1}{x^2}}}$

This =0 when the numerator =0

We need to find x for which

$x-2+\frac{1}{x^2}-\frac{1}{x^3}=0$

$g(x)=x^4-2x^3+x-1=0$

We use Newton's method to find the root of this near x=2.

$g'(x)=4x^3-6x^2+1$

$X_{n+1}=X_n-\frac{g(X_n)}{g'(X_n)}$

Taking x=2 as the initial estimate, $X_n$

$X_{n+1}=2-\frac{16-16+2-1}{32-24+1}=2-\frac{1}{9}$

Math Help Forum: Newton's method

LinkBack

Thread Tools

Display

Newton's method

Similar Math Help Forum Discussions

Newton's Method

Newton's method

Newton's Method

Newton's Method help!!

Newton's Method