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Use Newton’s method with the specified initial approximation x1 to find x3. |
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| Jul14-10, 10:15 PM | #1 |
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Use Newton’s method with the specified initial approximation x1 to find x3.
Please verify my answer.
1. The problem statement, all variables and given/known data Use Newton’s method with the specified initial approximation to find , the third approximation to the root of the given equation. (Give your answer to four decimal places.) x^5+2=0, x_1=-1 2. Relevant equations 3. The attempt at a solution x5+2=0, x1=-1 y'=5x4 x(n+1)=xn-(x5+2)/(5x4 ) For n=1 x2=-1-((-1)5+2)/(5(-1)4 )=-6/5=-1.2 For n=2 x3=-1.2-((-1.2)5+2)/(5(-1.2)4 )= -1.2-0.4883/10.368=-1.2+0.047=-1.1530 |
| Jul15-10, 06:52 AM | #2 |
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x(n+1)=xn-(xn5+2)/(5xn4 ) |
| Jul15-10, 11:34 AM | #3 |
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Hi! What do you mean by
The task says to do the third approximation, which I found already.... Do you mean I should do the fourth...and the fifth?... |
| Jul15-10, 12:34 PM | #4 |
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Mentor
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Use Newton’s method with the specified initial approximation x1 to find x3.
I think that HallsOfIvy missed the part about the third approximation, so you're done.
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