# Application of Newton's Method

• Oct 16th 2012, 11:32 AM
alane1994
Application of Newton's Method
Hello!
I have a homework problem that is involving Newton's Method.
You have two equations that make up a graph. I had to graph them.

$y=\frac{1}{x}$ and $y=64-25x^2$

From there I had to approximate where the functions intercept(x value). On the online homework program that my class uses, I approximated the values of -1.6,0.01, and 1.6 for where they intercept.

Then it asks the graphs intersect when x is approximately_________.
There are three answers, I got them all wrong.
The answers were -1.607756 , 0.015626 , 1.592130.
What I want to know is how do you find those values? Please help, this confuses me greatly.
• Oct 16th 2012, 12:35 PM
TheEmptySet
Re: Application of Newton's Method
Quote:

Originally Posted by alane1994
Hello!
I have a homework problem that is involving Newton's Method.
You have two equations that make up a graph. I had to graph them.

$y=\frac{1}{x}$ and $y=64-25x^2$

From there I had to approximate where the functions intercept(x value). On the online homework program that my class uses, I approximated the values of -1.6,0.01, and 1.6 for where they intercept.

Then it asks the graphs intersect when x is approximately_________.
There are three answers, I got them all wrong.
The answers were -1.607756 , 0.015626 , 1.592130.
What I want to know is how do you find those values? Please help, this confuses me greatly.

Newtons method is used to approximate real zeros of a function.

We can define a new function by subtracting the two equations gives to get

$f(x)=25x^2+\frac{1}{x}-64=\frac{25x^3-64x+1}{x}$

We want to find the zero's of this function.

Newton's method is given by

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

So $f'(x)=50x-\frac{1}{x^2}=\frac{50x^3-1}{x^2}$

This gives

$x_{n+1}=x_n-\frac{\frac{25x_{n}^3-64x_{n}+1}{x_{n}}}{\frac{50x_{n}^3-1}{x_{n}^2}}=\frac{25x_{n}^4-64x^2_{n}+x_{n}}{50x_{n}^3-1}$

This will zero in on each root. Each time you do the process again you will get closer to exact answer.

If you use your seed of 1.6 you will get the accuracy you want after 2 interations.

Let $x_1=1.6$ then plug this into the above to get $x_2$ and repeat...

Here is a spread sheet that will allow you to play around with different seed values.

Attachment 25251
• Oct 16th 2012, 12:45 PM
MarkFL
Re: Application of Newton's Method
I would set:

$\frac{1}{x}=64-25x^2$ and then define:

$f(x)=25x^2-64+\frac{1}{x}=0$ and we find:

$f'(x)=50x-\frac{1}{x^2}$

and so we will have:

$x_{n+1}=x_n-\frac{25x_n^2-64+\frac{1}{x_n}}{50x_n-\frac{1}{x_n^2}}$

Looking at a graph of $f(x)$ we see the smallest root is near $x=-1.6$ so we set:

$x_0=-1.6$

What I do on my TI-89 Titanium is the following:

-1.6 [ENTER]

result: -1.6

ans(1)-(25ans(1)^2-64+1/ans(1))/(50ans(1)-1/ans(1)^2) [ENTER]

result: -1.60777453839

Now, just keep hitting [ENTER] to get successive approximations:

-1.60775601298

-1.60775601287

-1.60775601287

The limit of accuracy of the calculator has been reached.

Now, the second root appears to be just above zero, so we'll use $x_0=0.01$

0.01 [ENTER]

result: 0.01

ans(1)-(25ans(1)^2-64+1/ans(1))/(50ans(1)-1/ans(1)^2) [ENTER]

result: 0.013600430022

0.015363748612

0.015622071623

0.015626489293

0.015626490543

0.015626490543

Now, the third root seem to be near $x_0=1.6$

1.6 [ENTER]

result: 1.6

ans(1)-(25ans(1)^2-64+1/ans(1))/(50ans(1)-1/ans(1)^2) [ENTER]

result: 1.59214916585

1.59212952245

1.59212952233

1.59212952233

So, we know the 3 roots are approximately:

$x=-1.60775601287,\,0.015626490543,\,1.59212952233$

Round as required.
• Oct 16th 2012, 01:20 PM
alane1994
Re: Application of Newton's Method
Thanks alot to both of you. So basically, you just do really monotonous work over and over until you get two like values. Then you have your root? Also, another question? When does calculus become second nature? You guys seem to just know how to do it right off! I love math and always will, but calculus is just really difficult for me to pick up for some reason...
• Oct 16th 2012, 01:40 PM
MarkFL
Re: Application of Newton's Method
It just takes time and practice. You'll get it...very few people just pick it right up!
• Oct 17th 2012, 04:06 PM
uperkurk
Re: Application of Newton's Method
Quote:

Originally Posted by alane1994
Thanks alot to both of you. So basically, you just do really monotonous work over and over until you get two like values. Then you have your root? Also, another question? When does calculus become second nature? You guys seem to just know how to do it right off! I love math and always will, but calculus is just really difficult for me to pick up for some reason...

Exactly how I feel about algebra hahaha. Those guys to you, is like you to me haha