Find interval $[-a,a]$ on which the Taylor Polynomial is accurate to within $0.0001$

Question

I am looking at a solution to a problem which asks us to find an interval $[-a,a]$ on which, if we approximate $\ln(1+x)$ near $x=0$ by using the third degree Taylor Polynomial $\ln(1+x) \approx x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$, we will have succeeded in doing so accurately to within $0.0001$.

In order to tackle this problem, we would need to look at the next higher order term, $f^{(4)}(x) = -\frac{6}{(1+x)^{4}}$ and then apply the Lagrange Error Bound formula: $E_{3}(x) \leq \frac{M}{4!}|x-a|^{4}$ where $M$ is some value satisfying $|f^{4}(x)|\leq M$ on the interval between $a$ and $x$.

The solution to this problem says that

"for $|a|<1$, the maximum of $|f^{(4)}(x)|$ is at $x = -a$, so $|f^{(4)}(x)|\leq\frac{6}{(1-a)^{4}}$".

Could someone please explain to me (1) where they got that $|a|$ must be $< 1$, and (2) how this tells us that the maximum of $|f^{(4)}(x)|$ occurs at $x = -a$ (I guess it tells us that $|f^{(4)}|$ is decreasing - but how?).?

I thank you in advance.

Answer 1

Both statements (1) and (2) follow from the properties of the $\ln x$ function.

(1) $\ln x$ is undefined at $x=0$. So $\ln(x+1)$ is undefined at $x=-1$. In order for $f(x)=\ln(x+1)$ to be defined at all $x\in[-a,a]$, we must take $|a|<1$.

(2) Since $|f^{(4)}|={6\over(x+1)^4}$, we see that $|f^{(4)}|$ indeed is decreasing because the numerator 6 is constant while the denominator $(x+1)^4$ is increasing for $x\in[-a,a]$ if $|a|<1$.

Answer 2

To calculate upper bound for the error of your Taylor-polynomial, you need to get an upper bound for $f^{(4)}(x)$ on your interval. However, if you look at $f^{(4)}(x)$ is has a singularity at -$1$. But you need to assure differentiability on the interval, for your Taylor-Series to have convergence. Therefore you can restrict the (symmetric) interval on $|a|<1$. The value still goes towards infinity, but it stays differentiable.

To the next Question: Why does the maximum for $M$ ocur for $x=-a$?

This is also quite easy to answer seeing, that the function will go towards infinity, the smaller $x$ is and it's absolute value will decrease, if you enlarge $x$. Keep in mind, that the denominator will grow very fast and keep the (absolute) value of $f^{(4)}$ small.

Combining this, will give you, that $x=-a$ with $|a|<1$ will give you the largest value of $M$.