This question (from Hildebrand's Introduction to Analysis) states:
Show that the number $2.46^{\frac{1}{64}}$ is known within less than one unit in the place of its fifth significant digit if $2.46$ is known only to be corrected to three digits
What does the part
one unit in the place of its fifth significant digit
mean? My understand was that this means that $\left|E(2.46^{\frac{1}{64}})\right| \leq 1 \times 10^{-5}$. My final answer was $\left|E(2.46^{\frac{1}{64}})\right| \leq 1.5878087... \times 10^{-5}$. So either my understanding of the question is off, or my answer is off.
My work:
Assume that $2.46$ is only known to be rounded correctly to three digits. Then $E(2.45) \leq 5 \times 10 ^{-3}$ by (1.4.5). Define $f(N) = N^{\frac{1}{64}}$. Then $f'(N) = \frac{1}{64} N^{-\frac{63}{64}}$. By (1.5.2), \begin{align*} E(f(\overline{N})) &\leq \left| f'(\eta) \right|_{\text{max}} \left| E(\overline{N})\right| && \text{Where } \eta \text{ between } N - E(\overline{N}) \text{ and } N + E(\overline{N})\\ E(f(2.46)) &\leq \left| \frac{1}{64} \eta ^{-\frac{63}{64}} \right|_{\text{max}} \cdot 2.46 \times 10^{-3}\\ &\leq \left| \frac{1}{64} (2.46 - 5 \times 10^{-3}) \right| \cdot 2.46 \times 10^{-3}\\ &\leq 1.5878087... \times 10^{-5} \end{align*}
Where
$E$ = Error
(1.4.5) $\left| E \right| \leq 5 \times 10^{r - n}$ where $N = N^* \times 10^r$ (N is expressed in scientific notation)
(1.5.2) $\left| E(f(\overline{N}) \right| \leq \left| f'(\eta) \right|_{\text{max}} \left| E(\overline{N}) \right|$ where $\overline{N}$ approximates $N$ and $\eta$ is between $N$ and $\overline{N}$
