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 November 11th, 2012, 05:04 PM #1 Newbie   Joined: Nov 2012 Posts: 29 Thanks: 0 Newton's Law (conceptual problem)... 17. Show that Newton's method applied to f(x) in Eq. 6 leads to xsub1 = r + h if xnought = r - h and to xsub1 = r-h if xnought = r+h. Interpret the result geometrically. Eq 6 is f(x) = (2x+1)^1/2 - (x + 4)^1/2 I know how to do Newton's method. I just don't understand where this talk about the height and radius comes into play. I don't know how to go about solving this problem. Thank you.
 November 11th, 2012, 08:16 PM #2 Newbie   Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... please.
 November 11th, 2012, 08:27 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Newton's Law (conceptual problem)... I just took a look at it, spent some time with it, and do not get the intended result. Are you sure the function is copied correctly?
 November 11th, 2012, 09:03 PM #4 Newbie   Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... i forgot to add. for equation six, a=2 and b=4. it is copied correctly otherwise.
 November 11th, 2012, 09:04 PM #5 Newbie   Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... that represents that there is a root between 2 and 4.
 November 12th, 2012, 07:35 AM #6 Newbie   Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... nothing?...
November 12th, 2012, 08:47 AM   #7
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Re: Newton's Law (conceptual problem)...

Quote:
 Originally Posted by nicoleb 17. Show that Newton's method applied to f(x) in Eq. 6 leads to xsub1 = r + h if xnought = r - h and to xsub1 = r-h if xnought = r+h. Interpret the result geometrically. Eq 6 is f(x) = (2x+1)^1/2 - (x + 4)^1/2 I know how to do Newton's method. I just don't understand where this talk about the height and radius comes into play. I don't know how to go about solving this problem. Thank you.
There is nothing said about "height and radius". If you are refering to r and h, they are just arbitrary numbers.

Quote:
 i forgot to add. for equation six, a=2 and b=4.
? There are no "a" and "b" in equation six.

Newtons method for solve f(x)= 0 is to construct a sequence of numbers recursively defined by $x_{n+1}= x_n- \frac{f(x_n)}{f#39;(x_n)}$

Here $f(x)= (2x+ 1)^{1/2}- (x+ 4)^{1/2}$ and $f'(x)= (2x+ 1)^{-1/2}- \frac{1}{2}(x+ 4)^{-1/2}$.

You have already been told, by MarkFL, that what you are trying to prove is not true.. To see that, let's use specific numbers: r= 2, h= 1. Then [latex]x_0= r- h= 1[latex]. $f(1)= \sqrt{3}- \sqrt{5}=-0.5040$ and $f'(1)= \frac{1}{\sqrt{3}}- \frac{1}{2\sqrt{5}}= 0.3537$. $\frac{f(1)}{f'(1)}= \frac{-.5040}{.3537}= -0.3537$. That makes $x_1= 1- 0.3537= 0.6463$ which is NOT equal to r+ h= 3.

 November 12th, 2012, 11:01 AM #8 Newbie   Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... Well, that's the problem. I don't know what else to tell you.

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