# Math Help - help with Newton

1. ## 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 Is it possible to explain it without using calculus?
dan

2. 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 $\approx{1.32471796}$

3. 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

4. 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.

5. ## 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)
Dan

6. 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)
Dan
I would be more strict and tell you to study "Pre-calculus" however they call the course.

7. 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)
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.

How will you explain the working of Newton's Method?I can't figure out why it works!!

Keep Smiling
Malay

9. Sorry, dan!

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: $No.$

10. 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.

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

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,......