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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

Here and .

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]. and . . That makes 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. Tags conceptual, law, newton, problem ### conceptual problems on newtons law

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