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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
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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henpenBest ResponseYou've already chosen the best response.0
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

experimentXBest ResponseYou've already chosen the best response.2
this is alternating series ... the maximum error of nth order is given by n+1 term
 one year ago

henpenBest ResponseYou've already chosen the best response.0
So\[\frac{x^5}{5}\frac{x^6}{6}+\frac{x^7}{7}..... \le \frac{x^5}{5} \]?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
of course ... each preceding terms is greater than the succeeding term.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
lol ... we don't calculate min error. we always bing error with max possible error.
 one year ago

henpenBest ResponseYou've already chosen the best response.0
So is there no way of getting the maximum of the minimum?
 one year ago

henpenBest ResponseYou've already chosen the best response.0
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

experimentXBest ResponseYou've already chosen the best response.2
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

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

experimentXBest ResponseYou've already chosen the best response.2
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

henpenBest ResponseYou've already chosen the best response.0
I doubt that it's given by the 7th term.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
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

henpenBest ResponseYou've already chosen the best response.0
I was referring to minimum calculations
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1350733606697:dw
 one year ago

henpenBest ResponseYou've already chosen the best response.0
Yes, I understood that.
 one year ago

henpenBest ResponseYou've already chosen the best response.0
Thanks for the help I'll read the POMN link
 one year ago
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