g(x) = 1/2 x3 - 2x + 3x + 2 on the domain [-1, 1]
I have to perform iterations of Newton's method based on this theorem:
Theorem. Let \(\displaystyle f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}\) have the following properties:
. . . . .\(\displaystyle \mbox{1. }\, f(a)\, f(b)\, <\, 0,\)
. . . . .\(\displaystyle \mbox{2. }\, f\, \mbox{ has no critical points in }\, (a,\, b),\, \mbox{ and}\)
. . . . .\(\displaystyle \mbox{3. }\, f''\, \mbox{ exists, is continuous, and fulfills }\, f''\, \geq\, 0\, \mbox{ or }\, f''\, \leq\, 0\, \mbox{ on whole }\, (a,\, b).\)
Then \(\displaystyle f(x)\, =\, 0\) has exactly one solution \(\displaystyle x^*.\) The Newton sequence \(\displaystyle x_n\) converges always towards \(\displaystyle x^*\) as \(\displaystyle n\, \rightarrow\, \infty\) if the initial guess \(\displaystyle x_0\) is chosen according to
. . . . .\(\displaystyle \bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \leq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \geq\, 0\, \mbox{ then }\, x_0\, \in\, [a,\, x^*],\, \mbox{e.g.}\, x_0\, =\, a\)
. . . . .\(\displaystyle \bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \geq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \leq\, 0\, \mbox{ then }\, x_0\, \in\, [x^*,\, b],\, \mbox{e.g.}\, x_0\, =\, b \)
In both cases we have for all iterates \(\displaystyle x_n\) the estimate
. . . . .\(\displaystyle \displaystyle \big| x_n\, -\, x^{*} \big| \, \leq\, \dfrac{\big|f(x_n)\big|}{\min\limits_{a \leq x \leq b}\, \big|f'(x)\big|}\)
a) I have to show that the Conditions 1-3 hold, I proved that.
b) Compute the denominator of that expression in the end, min| g'(x) |, I completed this one also, the value is 1/2
c) choose an initial guess according to the theorem and perform iterations until the error |Xn - X*| lies below 10^-4
I do not understand this part how can I do the interations, if for example my Xo = 1, then how much is my X1, can someone do or explain just one or two interations, so then I am able to continue?
I have to perform iterations of Newton's method based on this theorem:
Theorem. Let \(\displaystyle f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}\) have the following properties:
. . . . .\(\displaystyle \mbox{1. }\, f(a)\, f(b)\, <\, 0,\)
. . . . .\(\displaystyle \mbox{2. }\, f\, \mbox{ has no critical points in }\, (a,\, b),\, \mbox{ and}\)
. . . . .\(\displaystyle \mbox{3. }\, f''\, \mbox{ exists, is continuous, and fulfills }\, f''\, \geq\, 0\, \mbox{ or }\, f''\, \leq\, 0\, \mbox{ on whole }\, (a,\, b).\)
Then \(\displaystyle f(x)\, =\, 0\) has exactly one solution \(\displaystyle x^*.\) The Newton sequence \(\displaystyle x_n\) converges always towards \(\displaystyle x^*\) as \(\displaystyle n\, \rightarrow\, \infty\) if the initial guess \(\displaystyle x_0\) is chosen according to
. . . . .\(\displaystyle \bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \leq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \geq\, 0\, \mbox{ then }\, x_0\, \in\, [a,\, x^*],\, \mbox{e.g.}\, x_0\, =\, a\)
. . . . .\(\displaystyle \bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \geq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \leq\, 0\, \mbox{ then }\, x_0\, \in\, [x^*,\, b],\, \mbox{e.g.}\, x_0\, =\, b \)
In both cases we have for all iterates \(\displaystyle x_n\) the estimate
. . . . .\(\displaystyle \displaystyle \big| x_n\, -\, x^{*} \big| \, \leq\, \dfrac{\big|f(x_n)\big|}{\min\limits_{a \leq x \leq b}\, \big|f'(x)\big|}\)
a) I have to show that the Conditions 1-3 hold, I proved that.
b) Compute the denominator of that expression in the end, min| g'(x) |, I completed this one also, the value is 1/2
c) choose an initial guess according to the theorem and perform iterations until the error |Xn - X*| lies below 10^-4
I do not understand this part how can I do the interations, if for example my Xo = 1, then how much is my X1, can someone do or explain just one or two interations, so then I am able to continue?
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