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henpen

  • 2 years ago

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?

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  1. henpen
    • 2 years ago
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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.

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

  3. henpen
    • 2 years ago
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    So\[\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}..... \le |\frac{x^5}{5}| \]?

  4. experimentX
    • 2 years ago
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    of course ... each preceding terms is greater than the succeeding term.

  5. henpen
    • 2 years ago
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    Minimum error?

  6. experimentX
    • 2 years ago
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    lol ... we don't calculate min error. we always bing error with max possible error.

  7. henpen
    • 2 years ago
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    So is there no way of getting the maximum of the minimum?

  8. henpen
    • 2 years ago
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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?

  9. experimentX
    • 2 years ago
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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 ...

  10. experimentX
    • 2 years ago
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    in this case ,, perhaps you maximum min error is given by magnitude of 7th term.

  11. experimentX
    • 2 years ago
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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

  12. henpen
    • 2 years ago
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    I doubt that it's given by the 7th term.

  13. experimentX
    • 2 years ago
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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,

  14. henpen
    • 2 years ago
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    I was referring to minimum calculations

  15. experimentX
    • 2 years ago
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    |dw:1350733606697:dw|

  16. henpen
    • 2 years ago
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    Yes, I understood that.

  17. henpen
    • 2 years ago
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    Thanks for the help- I'll read the POMN link

  18. experimentX
    • 2 years ago
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    sure!!

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