Here's the question you clicked on:
henpen
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?
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.
this is alternating series ... the maximum error of nth order is given by |n+1 term|
So\[\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}..... \le |\frac{x^5}{5}| \]?
of course ... each preceding terms is greater than the succeeding term.
lol ... we don't calculate min error. we always bing error with max possible error.
So is there no way of getting the maximum of the minimum?
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?
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 ...
in this case ,, perhaps you maximum min error is given by magnitude of 7th term.
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
I doubt that it's given by the 7th term.
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,
I was referring to minimum calculations
|dw:1350733606697:dw|
Thanks for the help- I'll read the POMN link