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I am trying to estimate the absolute error in approximating $\ln 1.09$ with the $3$rd-order Taylor polynomial centered at $0$. It's been a while since I've taken the Calculus and I'm afraid I need some refreshing, even for this textbook problem.

Here is my approach, but I am almost certain something is amiss: Taylor's Theorem states that the $n$th remainder polynomial for the nth Taylor polynomial is $$R_n(x)= \frac{f^{(n+1)}(c) \ (x-a)^{n+1} }{(n+1)!} ,$$ where $a$ is the center and the existence of $c \in [a,x]$ is guaranteed by the Mean Value Theorem. I found the $4^{th}$ derivative of $\ln (1+x)$ to be $-\frac{6}{(x+1)^4}.$ The absolute error is given by $M\frac{|x-a|^{n+1}}{(n+1)!}$, where $|f^{(n+1)}(x)|≤M, x\in[0, .09]$. Since the $4^{th}$ derivative is just a hyperbola shifted to the left $1$ unit and "scaled" by a factor of $-6$, the graph has two branches with no inflection points, therefore only the endpoints of the interval $[0,.09]$ need be checked. The value of $0$ gives the greatest magnitude, and so $$ |R_3(x)|= 6\frac{x^{3+1}}{(3+1)!}=\frac{x^4}{4} .$$ Evaluating this at $x=.09$ (because the original function was $\ln(x+1)$ and I am trying to estimate $\ln (1.09)$), we have $\frac{.09^4}{4} \approx 1.64 \times 10^{-5}$.

However, I am almost positive this is wrong. What did I do wrong? Any help would be greatly appreciated. Thank you.

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up vote 3 down vote accepted

I do not know why you say your reasoning is wrong. Anyway, the successive terms of the series expansions of $\log(1+x)$ have alternating signs when $x$ is positive. If, furthermore, their amplitudes are decreasing, which is the case for the whole sequence of terms if $x$ is in $(0,1)$, Leibniz criterion yields rigorous upper and lower bounds. For example, for every $x$ in $(0,1)$, $$ x-\frac12x^2+\frac13x^3-\frac14x^4\leqslant\log(1+x)\leqslant x-\frac12x^2+\frac13x^3, $$ hence the bound of your error term by $\frac14x^4$ is correct (and you get furthermore that it is one-sided).

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Insightful. It took you 1 line to derive what took me 10 lines. That is why, my good sir, you get that little green checkmark on the side. You win all my internetz for a week. Thanks. – Hautdesert Oct 31 '11 at 5:35

Technically, $y = \frac{-6}{(x+1)^4}$ is not "a hyperbola shifted to the left"; a hyperbola needs to be given by a quadratic polynomial in $x$ and $y$, and this is equivalent to a degree 5 polynomial in $x$ and $y$. It may look like a hyperbola, much like $y=x^4$ "looks like a parabola", but just like $y=x^4$ is not a parabola, neither is $y = -6/(x+1)^4$ a hyperbola.

Instead, note that the fifth derivative is $$\frac{24}{(x+1)^5}$$ which is positive on $[0,0.09]$. Since the fifth derivative is positive, the fourth derivative is strictly increasing, so it achieves its maximum at $0.09$ (not at $0$). But since both are negative, the largest absolute value occurs at $0$, exactly as you state. The rest is correct.

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But wouldn't we want the greatest magnitude since we want the absolute error? I'm sure that you are correct, but I just need this clarified for me. Thanks in advance. – Hautdesert Oct 31 '11 at 5:17
I just figured out what was wrong. You are right. Thanks. – Hautdesert Oct 31 '11 at 5:33
@Hautdesert: I agreed with you at the end: because both the value at $0$ and at $0.09$ are negative, the largest absolute value occurs at the smallest value, which is at $0$. – Arturo Magidin Oct 31 '11 at 13:02

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