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ajprincess Group Title

Please help:) By applying Newton-Raphson method to \(f(x)=1-\large\frac{1}{ax}\) obtain the recurrence formula \(x_{i+1}=x_i(2-ax_i)\) for the iterative determination of the reciprocal of a. Show that if \(E_i\) denotes the error in the \(x_i\), there follows \(E_{i+1}=-a{E_i}^2\).

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    \[x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}\] \[x_{i+1}=x_i-\frac{1-\frac{1}{ax}}{-\frac{1}{ax^2}}\]and some algebra should work

    • 2 years ago
  2. satellite73 Group Title
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    i guess it should be \[x_{i+1}=x_i-\frac{1-\frac{1}{ax_i}}{-\frac{1}{ax_i^2}}\]

    • 2 years ago
  3. ajprincess Group Title
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    Ya I have proved that part. I need help with the error part

    • 2 years ago
  4. satellite73 Group Title
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    nope i am wrong again it should be \[x_{i+1}=x_i-\frac{1-\frac{1}{ax_i}}{\frac{1}{ax_i^2}}\]

    • 2 years ago
  5. satellite73 Group Title
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    damn i wish i could help, but i don't know an expression for the error. newton ralphson is always error squaring, i know that much but i am not sure how to prove it

    • 2 years ago
  6. satellite73 Group Title
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    i think it has something to do with taylor series, but i should really shut up

    • 2 years ago
  7. ajprincess Group Title
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    ohh that's k. Thankk u sooo much for helping me:)

    • 2 years ago
  8. satellite73 Group Title
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    on the other hand i am pretty good at googling. try looking at "error analysis" in the attached pdf, seems like exactly what you are looking for

    • 2 years ago
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  9. satellite73 Group Title
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    on second page for "division"

    • 2 years ago
  10. ajprincess Group Title
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    Thankkkk u sooooo much. It is really verryy useful:)

    • 2 years ago
  11. satellite73 Group Title
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    yw, and good luck

    • 2 years ago
  12. ajprincess Group Title
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    thank u:)

    • 2 years ago
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