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Suppose you attempt to use Newton's method to approximate a solution of the equation z* + a° – 32? – 2x + 2 = 0. Let zo = 1 be the initial approximation, and then calculate T1, F2, T3, and z4. If a value is undefined, enter DNE. I3 = 24 = What do you notice? O Newton's method fails because the approximations are alternating between two numbers. O Newton's method is successful at approaching a number so far. O Newton's method fails because the approximations are getting larger in absolute value. O Newton's method fails because f'(zo) = 0 but f(zo) + 0. %3D

Suppose you attempt to use Newton's method to approximate a solution of the equation z* + a° – 32? – 2x + 2 = 0. Let zo = 1 be the initial approximation, and then calculate T1, F2, T3, and z4. If a value is undefined, enter DNE. I3 = 24 = What do you notice? O Newton's method fails because the approximations are alternating between two numbers. O Newton's method is successful at approaching a number so far. O Newton's method fails because the approximations are getting larger in absolute value. O Newton's method fails because f'(zo) = 0 but f(zo) + 0. %3D

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Suppose you attempt to use Newton's method to approximate a solution of the equation a* + a* – 3z? – 2x + 2 = 0. Let ro = 1 be the initial approximation, and then calculate a1, x2, 13, and a4. If a value is undefined, enter DNE. 22 = 23 = 24 = What do you notice? O Newton's method fails because the approximations are alternating between two numbers. ONewton's method is successful at approaching a number so far. O Newton's method fails because the approximations are getting larger in absolute value. O Newton's method fails because f'(x0) = 0 but f(xo) # 0.
Transcribed Image Text:Suppose you attempt to use Newton's method to approximate a solution of the equation a* + a* – 3z? – 2x + 2 = 0. Let ro = 1 be the initial approximation, and then calculate a1, x2, 13, and a4. If a value is undefined, enter DNE. 22 = 23 = 24 = What do you notice? O Newton's method fails because the approximations are alternating between two numbers. ONewton's method is successful at approaching a number so far. O Newton's method fails because the approximations are getting larger in absolute value. O Newton's method fails because f'(x0) = 0 but f(xo) # 0.
Use Newton's method to approximate a solution of the equation a + x + 6 = 0. Let ro = - 1 be the initial approximation, and then calculate z1 and r2. 21 = 22 =
Transcribed Image Text:Use Newton's method to approximate a solution of the equation a + x + 6 = 0. Let ro = - 1 be the initial approximation, and then calculate z1 and r2. 21 = 22 =
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