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henpen Group Title

Estimate the error, as a function of x, in approximating \[ ln(x+1) \] as \[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} \]. That is, what is \[ ln(x+1)- (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}) \], or \[\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}..... \] expressed in a more compact form?

  • one year ago
  • one year ago

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  1. henpen Group Title
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    Or, if impossible to express directly as a function of x, find information about it, like the maximum and minimum values it could take for a certain x, for example.

    • one year ago
  2. experimentX Group Title
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    this is alternating series ... the maximum error of nth order is given by |n+1 term|

    • one year ago
  3. henpen Group Title
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    So\[\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}..... \le |\frac{x^5}{5}| \]?

    • one year ago
  4. experimentX Group Title
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    of course ... each preceding terms is greater than the succeeding term.

    • one year ago
  5. henpen Group Title
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    Minimum error?

    • one year ago
  6. experimentX Group Title
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    lol ... we don't calculate min error. we always bing error with max possible error.

    • one year ago
  7. henpen Group Title
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    So is there no way of getting the maximum of the minimum?

    • one year ago
  8. henpen Group Title
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    And isn't \[\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}-\frac{x^8}{8}+\frac{x^9}{9}..... \le |\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}|\] a better approximation of the error?

    • one year ago
  9. experimentX Group Title
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    no ... you don't need min error. you are trying to estimate ln(1+x) ... you evaluate it up to 5 terms ... that will give you approximation. but there will always be error. the max error is given by magnitude of 6th term. we don't calculate min error. It might be possible to set lower bound for error. but it's rarely used ...

    • one year ago
  10. experimentX Group Title
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    in this case ,, perhaps you maximum min error is given by magnitude of 7th term.

    • one year ago
  11. experimentX Group Title
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    this is a good place to learn about error approximation ... although i've forgotten most of it http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx

    • one year ago
  12. henpen Group Title
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    I doubt that it's given by the 7th term.

    • one year ago
  13. experimentX Group Title
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    7 term is greater than sum of rest of term ... 6 term is greater than sum of rest of term ... error is less than 6th term,

    • one year ago
  14. henpen Group Title
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    I was referring to minimum calculations

    • one year ago
  15. experimentX Group Title
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    |dw:1350733606697:dw|

    • one year ago
  16. henpen Group Title
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    Yes, I understood that.

    • one year ago
  17. henpen Group Title
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    Thanks for the help- I'll read the POMN link

    • one year ago
  18. experimentX Group Title
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    sure!!

    • one year ago
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