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
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
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 $\displaystyle \approx{1.32471796}$
The TeX system seems to not like long (not that long) strings of TeX.Quote:
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
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.
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.Quote:
Originally Posted by 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.Quote:
Originally Posted by dan
Good luck.
Someone Please help.
How will you explain the working of Newton's Method?I can't figure out why it works!!
Keep Smiling
Malay
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: $\displaystyle No.$
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,......
