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

Printable View

- Jul 15th 2006, 11:22 AMdanhelp 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 - Jul 15th 2006, 12:08 PMgalactus
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 $\displaystyle x^{3}-x-1=0$

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

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

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

Make something in that region your initial guess.

Try 1.5

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

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.**.

$\displaystyle x_{3}=1.34782609-$$\displaystyle \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:

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

See, the last two are the same.

No need to continue. . The solution is $\displaystyle \approx{1.32471796}$ - Jul 15th 2006, 01:11 PMCaptainBlackQuote:

Originally Posted by**galactus**

Placing pairs of [ /math] and [ math] (without extra spaces) s seems to

fix the problems usually (as I have done in your post).

RonL - Jul 15th 2006, 01:49 PMgalactus
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.

- Jul 15th 2006, 06:17 PMdanhelp 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 - Jul 15th 2006, 06:22 PMThePerfectHackerQuote:

Originally Posted by**dan**

- Jul 16th 2006, 03:25 AMgalactusQuote:

Originally Posted by**dan**

Good luck. - Jul 16th 2006, 04:26 AMmalaygoel
Someone Please help.

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

Keep Smiling

Malay - Jul 16th 2006, 04:35 AMSoroban
Sorry, dan!

Quote:

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

You're saying, "Can someone explain a $\displaystyle B\flat\text{ minor}$ chord*without*using Music Theory?

. . You see, I don't read music."

Answer: $\displaystyle No.$

- Jul 16th 2006, 04:45 AMgalactus
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, $\displaystyle x_{1}$ be our initial approximation, then we can improve by moving along the tsngent line to y=f(x) at $\displaystyle x_{1}$ until we meet it at a point $\displaystyle x_{2}$.

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:

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

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

Sub this in our point-slope form:

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

Solve for $\displaystyle x_{2}$

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

We keep going until we see that the approximation is

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