$$f(x) = \ln (1-x) $$ I should find third power series solution in point $0$. First, I find the derivatives: $$ f'(x) = -\frac1{1-x} $$ $$ f''(x) = -\frac1{(1-x)^2} $$ $$ f'''(x) = -\frac2{(1-x)^3}$$ Now put them in the formula: $$ f(x) \approx \ln 1 - \frac11(x-0) - \frac{1}{2!}(x-0)^2 - \frac{2}{3!}(x-0)^3 = \ln 1 - x - \frac{x^2}{2} - \frac{2x^3}{6}$$ I should now estimate it's error if $|x| \leq 0.5$

What exactly should I do after the formula part.

  • $\begingroup$ Note: some of your derivatives have sign errors. $\endgroup$ – David H Oct 27 '13 at 21:20
  • $\begingroup$ You have a sign error, $f'(x) = -\dfrac{1}{1-x}$, the sign goes through to the higher derivatives, and $f'''(x) = - \dfrac{2}{(1-x)^3}$, not $(1-x)^2$. $\endgroup$ – Daniel Fischer Oct 27 '13 at 21:21
  • $\begingroup$ I edited the 2 derivatives and changed a sign error in the formula. $\endgroup$ – user101077 Oct 27 '13 at 21:23
  • $\begingroup$ Now it should be correct. $\endgroup$ – user101077 Oct 27 '13 at 21:30
  • $\begingroup$ I guess I shoudn't use wolfram alpha to find derivatives. Lesson learned... $\endgroup$ – user101077 Oct 27 '13 at 21:39

Some observations:

Calculate the next few terms, and reduce the fractions to lowest terms. Do you see a pattern? Can you see why it's there?

If $x$ is negative, how do the signs behave? How can you use that to get an error bound really easily (when $x$ is negative)?

What do you know about the value of the Taylor remainder?

  • $\begingroup$ I guess I should find fourth derivative and use it Taylor remainder formula. I should get: $$f^{(4)}=-\frac{6}{(1-x)^6} $$ $$\alpha_4 = \frac{-\frac{6}{(1-z)^6}}{4!}x^4$$ $\endgroup$ – user101077 Oct 27 '13 at 21:59
  • $\begingroup$ @user101077, can you see how to maximize the error for relevant $x$ and $z$? $\endgroup$ – dfeuer Oct 27 '13 at 22:43

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