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henpen
 3 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?
henpen
 3 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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henpen
 3 years ago
Best ResponseYou've already chosen the best response.0Or, 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2this is alternating series ... the maximum error of nth order is given by n+1 term

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0So\[\frac{x^5}{5}\frac{x^6}{6}+\frac{x^7}{7}..... \le \frac{x^5}{5} \]?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2of course ... each preceding terms is greater than the succeeding term.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2lol ... we don't calculate min error. we always bing error with max possible error.

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0So is there no way of getting the maximum of the minimum?

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0And 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?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2no ... 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 ...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2in this case ,, perhaps you maximum min error is given by magnitude of 7th term.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2this 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

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0I doubt that it's given by the 7th term.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.27 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,

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0I was referring to minimum calculations

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1350733606697:dw

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for the help I'll read the POMN link
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