So I have this question here.
http://25.media.tumblr.com/tumblr_mc0i7e…
I understand the answer when X0 = 5, but I don't understand X0 = 1. The answer key says that the results would be 1,2,1,2,1,...
http://25.media.tumblr.com/tumblr_mc0i7e…
I understand the answer when X0 = 5, but I don't understand X0 = 1. The answer key says that the results would be 1,2,1,2,1,...
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Newton's method doesn't always converge. The actual formula for the graph in the picture isn't given, but it appears that with a starting value of 1, the tangent to the curve where x=1 will intersect the x axis at around x=2, so 2 would be the next value to try. Then the tangent from the point where x=2 will point back to x=1, and the results will oscillate between those two points.
The moral of the story is that Newton's method requires a good starting value.
The moral of the story is that Newton's method requires a good starting value.
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The function is positive, but the derivative is negative. Therefore, Newton's method will go over to the right.
At this new point, both the derivative and function are positive, Newton's method will go to the left.
The exact point each will hit depends on the slope and point-value. Since you don't know any of these, you can't accurately determine where Newton's method will go. Depending on the slope, the approximation could possibly go over to the root. I wouldn't expect to see this question on a test.
EDIT: I see something I missed. The point-value at 1 and 2 are both equal, and the slope looks like it's supposed to be opposite. If Newton's method went from 1 to 2, it would certainly go back to 1. However, you can't be sure Newton's method will go from 1 to 2. I suppose you can assume it will, when faced with a graph like this.
At this new point, both the derivative and function are positive, Newton's method will go to the left.
The exact point each will hit depends on the slope and point-value. Since you don't know any of these, you can't accurately determine where Newton's method will go. Depending on the slope, the approximation could possibly go over to the root. I wouldn't expect to see this question on a test.
EDIT: I see something I missed. The point-value at 1 and 2 are both equal, and the slope looks like it's supposed to be opposite. If Newton's method went from 1 to 2, it would certainly go back to 1. However, you can't be sure Newton's method will go from 1 to 2. I suppose you can assume it will, when faced with a graph like this.