# help with Newton

• July 15th 2006, 12:22 PM
dan
help with Newton
Is there some one who would get a kick out of explaining Newton’s method to me? I've read some different things about it but I didn’t understand the calculus :confused: Is it possible to explain it without using calculus?
dan
• July 15th 2006, 01:08 PM
galactus
What seems to be the trouble?. Newton's Method is one of the easiest things in calculus.

It's just a matter of iterations.

Let's give an example:

Use Newton to find the 'real' solutions of $x^{3}-x-1=0$

Let $f(x)=x^{3}-x-1\;\ and\;\ f'(x)=3x^{2}-1$

Graph the function, $x^{3}-x-1$

You can see that y=0 when x is between 1 and 2.

Make something in that region your initial guess.

Try 1.5

$x_{2}=1.5{-}\frac{(1.5)^{3}-1.5-1}{3(1.5)^{2}-1}=1.34782609$

Now use the result you get from this and sub back into the equation:

What is wrong with this website that I keep getting these errors saying the image is too big?. I've never seen this on another site. The code is fine as far as I can tell..

$x_{3}=1.34782609-$ $\frac{(1.34782609)^{3}-(1.34782609)-1}{3(1.34782609)^{2}-1}=1.32520040$

Continue in this fashion until you arrive at approximations which are so close they are virtually unchanged.

We end up with:

$x_{1}=1.5\;\ x^{2}=1.34782609\;\ x^{3}$ $=1.32520040\;\ x^{4}=$ $1.32471817\;\ x^{5}=1.32471796\;\ x^{6}=1.32471796$

See, the last two are the same.

No need to continue. . The solution is $\approx{1.32471796}$
• July 15th 2006, 02:11 PM
CaptainBlack
Quote:

Originally Posted by galactus

What is wrong with this website that I keep getting these errors saying the image is too big?. I've never seen this on another site. The code is fine as far as I can tell..

The TeX system seems to not like long (not that long) strings of TeX.
Placing pairs of [ /math] and [ math] (without extra spaces) s seems to
fix the problems usually (as I have done in your post).

RonL
• July 15th 2006, 02:49 PM
galactus
Thank you Cap'N. I will keep that in mind. I have not run into that problem before, regardless of the length of the string. Thanks for the fix up.
• July 15th 2006, 07:17 PM
dan
help with newton
Ok,
What seems to be the problem is...I know practically no calculus and my dad wont let me take a course until I finish my college algebra book :( so... I don't even know what f'(x) means and how you get there from f(x) :confused: :eek:
Dan
• July 15th 2006, 07:22 PM
ThePerfectHacker
Quote:

Originally Posted by dan
Ok,
What seems to be the problem is...I know practically no calculus and my dad wont let me take a course until I finish my college algebra book :( so... I don't even know what f'(x) means and how you get there from f(x) :confused: :eek:
Dan

I would be more strict and tell you to study "Pre-calculus" however they call the course.
• July 16th 2006, 04:25 AM
galactus
Quote:

Originally Posted by dan
Ok,
What seems to be the problem is...I know practically no calculus and my dad wont let me take a course until I finish my college algebra book :( so... I don't even know what f'(x) means and how you get there from f(x) :confused: :eek:
Dan

Poor Dan. I didn't realize that. Math is done in steps; You need to finish your algebra or pre-caculus and then go to Calc I. BTW, f'(x) is the derivative of a function f(x). What brought up Newton anyway?. You need to learn differentiation before you can use Newton's Method. Do a Google search.
Good luck.
• July 16th 2006, 05:26 AM
malaygoel
How will you explain the working of Newton's Method?I can't figure out why it works!!

Keep Smiling
Malay
• July 16th 2006, 05:35 AM
Soroban
Sorry, dan!

Quote:

I don't even know what $f'(x)$ means . . .

You're saying, "Can someone explain a $B\flat\text{ minor}$ chord without using Music Theory?
. . You see, I don't read music."

Answer: $No.$
• July 16th 2006, 05:45 AM
galactus
Most any calc book will explain how it works. It's not that complicated.

Off the top of my head.

The solutions of f(x)=0 are the values of x where the graph crosses the x-axis.

Suppose that x=c is some solution we are looking for. Even if we can't find c exactly, it is usually possible to approximate it by graphing and using the Intermediate Value Theorem.

If we let, say, $x_{1}$ be our initial approximation, then we can improve by moving along the tsngent line to y=f(x) at $x_{1}$ until we meet it at a point $x_{2}$.

Repeat.

One thing we have to do is derive some sort of formula so we can use ol' Newton.

Start with point-slope form of a line:

$y-f(x_{1})=f'(x_{1})(x-x_{1})$

If $f'(x_{1})\neq{0}$, then this line is not parallel to the x-axis and crosses it at some point $(x_{2},0)$.

Sub this in our point-slope form:

$-f(x)=f'(x_{1})(x_{2}-x_{1})$

Solve for $x_{2}$

$x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{2})}$

We keep going until we see that the approximation is

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$, n=1,2,3,......