find the error bound using the Lagrange Remainder: $$\ln(x)\approx (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}, |x-1|<\dfrac{1}{64}$$

My attempt: \begin{align} R_3(x)&=\dfrac{f^{(4)}(c)}{4!}(x-1)^4\\ f^{4}(c)&=\dfrac{-6}{(x-1)^4}\\ \implies R_3(x)&=\dfrac{-6}{4!}\dfrac{(x-1)^4}{(c-1)^4}\\ \implies |R_3(x)|&\leq \dfrac{6}{4!}\dfrac{(\frac{1}{64})^4}{(\frac{-1}{64})^4}=.25 \end{align} Is this correct?


Note that,

\begin{equation} f(x) = log(x) \implies f^{(4)}(x) = \frac{-6}{x^4} \end{equation}

Therefore, \begin{equation} R_3(x) = -\frac{6}{4!c^4}(x-1)^4\quad; \quad \text{where } c \text{ is in between 1 and x} \end{equation}

Since $c$ depends on $x$, first find an upper bound as;

\begin{equation} |R_3(x)| \leq \begin{cases} \frac{6}{4!}(x-1)^4 \quad &if \quad x>1\\ \frac{6}{4!x^4}(x-1)^4 \quad &if \quad x<1 \end{cases} \end{equation}

Now we can bound for $|x-1|<1/64$,

\begin{equation} |R_3(x)| \leq \begin{cases} \frac{6}{4!64^4} \quad &if \quad x>1\\ \frac{6}{4!(1-1/64)^464^4} \quad &if \quad x<1 \end{cases} \end{equation}

Here is a plot done by wolframalpha.com. You can visaully see that we need two different bounds for $x<1$ and $x>1$. If you need a single bound, $x<1$ case will work.

  • $\begingroup$ If x was 0.3 in the above, isn't the Remainder bound then very large and gets worse as you add more terms? $\endgroup$ Apr 18 '18 at 4:03
  • $\begingroup$ @Cliff Stamp, I couldn't understand your question, can you explain what you mean by adding more terms? (There is a condition for last bound, $|x-1|<1/64$.) $\endgroup$
    – Atbey
    Apr 18 '18 at 5:34
  • $\begingroup$ Sorry, I should have been more clear, say the bound on x was between 0.3 and 1. $\endgroup$ Apr 18 '18 at 14:08
  • $\begingroup$ Then since $x<1$, $|R_3(x)| \leq \frac{6(0.7)^4}{4!(0.3)^4}$. $\endgroup$
    – Atbey
    Apr 19 '18 at 9:56
  • $\begingroup$ Thanks, that was what I thought. This was an old question on an AP Math exam, but it matches none of the given answers. $\endgroup$ Apr 19 '18 at 15:16

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