anonymous
  • anonymous
Need verification: Apply newton's method with initial approximation x1=-4 to find the 2nd and 3rd approximations x2 and x3 to the real root of the equation
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[x^3 -9x +28 = 0\] my answer is -4 for both x2 and x3.. can anyone verify this?
ksaimouli
  • ksaimouli
can u explain ur steps
anonymous
  • anonymous
\[x2 = x1 - \frac{ f(x1) }{ f'(x1) }\] \[x3 = x2 - \frac{ f(x2) }{ f'(x2) }\]

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ksaimouli
  • ksaimouli
i am getting 0
ksaimouli
  • ksaimouli
x1
anonymous
  • anonymous
\[x2 = (-4) - (\frac{ (-4)^3 -9(-4) +28 }{ 3(-4) - 9 })\]
anonymous
  • anonymous
which is equal to -4, and repeats for x3
anonymous
  • anonymous
wait there is a power next to the -4 on the bottom
anonymous
  • anonymous
i just forgot that, but i still get -4 for x2
anonymous
  • anonymous
so its (-4) - (0/39)....right?
ksaimouli
  • ksaimouli
hmm but i got x1=0 so should'nt the x2=0-F(0)/(f'(0))
anonymous
  • anonymous
x1 is already given to be x1 = -4
anonymous
  • anonymous
or is initial approximation x0?
ksaimouli
  • ksaimouli
ohh yup then ur right
anonymous
  • anonymous
im confused though..so is initial approximation (-4) x0 or x1?
anonymous
  • anonymous
i know it says x1 in the question, but im not sure
anonymous
  • anonymous
i think it is probably x1
ksaimouli
  • ksaimouli
it is x1 clearly stated in question
anonymous
  • anonymous
do you get -4 for both x2 and x3 now when you try it?
ksaimouli
  • ksaimouli
yup
anonymous
  • anonymous
ok.. thanks

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