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Problem 33 Hard Difficulty

Explain why Newton's method fails when applied to the equation $ \sqrt[3]{x} = 0 $ with any initial approximation $ x_1 \not= 0 $. Illustrate your explanation with a sketch.

Answer

not converge

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Video Transcript

question 33. We're looking at our function as extra 1/3 but a key root of X and a prime of X and, of course, is 1/3 x to the negative 2/3. So x to the n plus one is an equal x minus x to the 1/3 over 1/3 X to the negative 2/3. So simplify that I would have, um, one there x to the to certain so back to the I'm just simply crying this part right here, x and we're when they're x 2/3 I'm going to invert and multiply, So I'm gonna have X to the 1/3 times three x to the 2/3 over one. So I'm just gonna multiply them when I must play and the exponents. So I have three x. So now I have X minus three X, which gives me and negative to X and negative X will not converge