Question

use newton's method with the specified initial approximation x1 to find x3, the thired approximation to the root of thr given equation. (give your answer to four places)
x^7+4=0, x1=-1

Answer 1

\[x_{n+1} = x_{n} -\frac{f(x)}{f'(x)}\]
\[f'(x)=7x^{6}\]
\[x_2 = (-1) - \frac{(-1)^{7}+4}{7(-1)^{6}}\]
...

Answer 2

\[y'=7x^6\]
for n=1
\[x2=\frac{ -1-((-1^7)+4) }{ 7(-1^6) }=\frac{ 4 }{ 7 }=0.571\]
\[x_{3}=\frac{ 4 }{ 7 }-\frac{ ((\frac{ 4 }{ 7 }) +4)}{ 7(\frac{ 4^6 }{ 7 } )}=0.03464\]

Answer 3

thats what i did but not sure if its correct or not

Answer 4

no you miscalculated....x_2 = -10/7

Answer 5

how did you got (-10)?

Answer 6

the formula, as dumb cow posted is
\[ x_{n+1} = x_{n} -\frac{f(x)}{f'(x)} \]
notice that you should replace xn with -1 the first time:
\[ x_{n+1} = -1 -\frac{f(x)}{f'(x)} \]
putting in f(x)=x^7+4 and f'(x)= 7x^6
\[ x_{n+1} = -1 -\frac{x^7+4}{7x^6} \]
we must replace x with -1. we get
\[ x_{n+1} = -1 -\frac{-1+4}{7} \\
x_{n+1} = -1 -\frac{3}{7} \\
x_{n+1} = -\frac{7}{7} -\frac{3}{7}\\
x_{n+1} = \frac{-7-3}{7} = -\frac{10}{7}
\]

Answer 7

oh okay got it. so the next step will be x= -10/7

Answer 8

\[x_3=\frac{ -10 }{ 7 }-\frac{ (\frac{ -10 }{ 7 })^7+4}{ 7+(\frac{ -10 }{ 7 })^6 }\]
\[=(\frac{ -10 }{ 7 })-(\frac{ -8.142656789 }{ 59.49901827 })\]
=-1.2917178
is that correct? Plus is this the final answer?

Answer 9

you are correct and yes since they want x_3, that is final answer

Answer 10

thanks!