For y = x² at x = 1, compute the increment Δ y
For $y = x^{2}$ at $x = 1$, compute the increment $\Delta y$ and the differential $dy$ for 1° $\Delta x = 0.1$; 2° $\Delta x = 0.01$; 3° $\Delta x = 0.001$. Determine the absolute and relative errors committed when $\Delta y$ is replaced by $dy$.
Explanation of Method
Watching the error of the linear approximation shrink by a factor of 100 each time.
Step 1: General formulas
For $y = x^{2}$ the exact increment expands with no remainder: \[\Delta y = \bigl(x+\Delta x\bigr)^{2}-x^{2} = 2x\,\Delta x + \bigl(\Delta x\bigr)^{2}, \qquad dy = y'\,\Delta x = 2x\,\Delta x.\]
At $x = 1$ these become $\Delta y = 2\Delta x+\bigl(\Delta x\bigr)^{2}$ and $dy = 2\Delta x$.
Step 2: The two errors, in closed form
The differential drops exactly one term, so
\[ \bigl|\Delta y - dy\bigr| = \bigl(\Delta x\bigr)^{2}, \]
\[ \frac{\bigl|\Delta y-dy\bigr|}{\bigl|\Delta y\bigr|} = \frac{\bigl(\Delta x\bigr)^{2}}{2\Delta x+\bigl(\Delta x\bigr)^{2}} = \frac{\Delta x}{2+\Delta x}. \]
The absolute error is quadratic in $\Delta x$; the relative error is essentially linear, roughly $\Delta x/2$.
Step 3: Case 1°: $\Delta x = 0.1$
\[ \Delta y = 0.2+0.01 = 0.21, \qquad dy = 0.20, \]
\[ \bigl|\Delta y-dy\bigr| = 0.01, \qquad \frac{0.01}{0.21} \approx 4.76\%. \]
Step 4: Case 2°: $\Delta x = 0.01$
\[ \Delta y = 0.02+0.0001 = 0.0201, \qquad dy = 0.02, \]
\[ \bigl|\Delta y-dy\bigr| = 0.0001, \qquad \frac{0.0001}{0.0201} \approx 0.50\%. \]
Step 5: Case 3°: $\Delta x = 0.001$
\[ \Delta y = 0.002+0.000001 = 0.002001, \qquad dy = 0.002, \]
\[ \bigl|\Delta y-dy\bigr| = 10^{-6}, \qquad \frac{10^{-6}}{0.002001} \approx 0.050\%. \]
Step 6: Summary
\[\begin{array}{c|c|c|c|c} \Delta x & \Delta y & dy & \bigl|\Delta y-dy\bigr| & \text{relative} \\ \hline 0.1 & 0.210000 & 0.200000 & 10^{-2} & 4.76\% \\ 0.01 & 0.020100 & 0.020000 & 10^{-4} & 0.50\% \\ 0.001 & 0.002001 & 0.002000 & 10^{-6} & 0.050\% \end{array}\]
\[\text{Each tenfold reduction of } \Delta x \text{ divides the absolute error by } 100 \text{ and the relative error by } 10 \;\;\checkmark\]
\[\text{Formula check: } \frac{\Delta x}{2+\Delta x} = \frac{0.1}{2.1} = 4.7619\%,\ \frac{0.01}{2.01} = 0.4975\%,\ \frac{0.001}{2.001} = 0.0500\% \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\bigl|\Delta y-dy\bigr| = \bigl(\Delta x\bigr)^{2}, \qquad \frac{\bigl|\Delta y-dy\bigr|}{\bigl|\Delta y\bigr|} = \frac{\Delta x}{2+\Delta x}\,}\]
Remark
Remark. This is the definition of the differential made concrete. Saying that $dy$ is the principal linear part of $\Delta y$ means precisely that $\Delta y - dy = o(\Delta x)$; here the discarded piece is $(\Delta x)^{2}$, which is not merely smaller than $\Delta x$ but smaller by a factor of $\Delta x$. That is why halving the step quarters the absolute error.