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November 11th, 2012, 05:04 PM   #1
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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.
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November 11th, 2012, 08:16 PM   #2
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Re: Newton's Law (conceptual problem)...

please.
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November 11th, 2012, 08:27 PM   #3
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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?
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November 11th, 2012, 09:03 PM   #4
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Re: Newton's Law (conceptual problem)...

i forgot to add. for equation six, a=2 and b=4.

it is copied correctly otherwise.
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November 11th, 2012, 09:04 PM   #5
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Re: Newton's Law (conceptual problem)...

that represents that there is a root between 2 and 4.
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November 12th, 2012, 07:35 AM   #6
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Re: Newton's Law (conceptual problem)...

nothing?...
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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.
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November 12th, 2012, 11:01 AM   #8
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Re: Newton's Law (conceptual problem)...

Well, that's the problem. I don't know what else to tell you.
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