For y = x³-x at x = 2, compute Δ y and dy for 1° Δ x = 1

For $y = x^{3}-x$ at $x = 2$, compute $\Delta y$ and $dy$ for 1° $\Delta x = 1$; 2° $\Delta x = 0.1$; 3° $\Delta x = -0.01$. Find the absolute and relative errors committed in replacing $\Delta y$ by $dy$.

Step 1: The exact increment at $x = 2$

Expanding $\bigl(2+\Delta x\bigr)^{3} = 8+12\Delta x+6\bigl(\Delta x\bigr)^{2}+\bigl(\Delta x\bigr)^{3}$ and subtracting $y(2) = 6$: \[\Delta y = \bigl(2+\Delta x\bigr)^{3}-\bigl(2+\Delta x\bigr)-6 = 11\,\Delta x + 6\bigl(\Delta x\bigr)^{2}+\bigl(\Delta x\bigr)^{3}.\]

Step 2: The differential

\[ y' = 3x^{2}-1, \qquad y'(2) = 11, \qquad dy = 11\,\Delta x. \]

So the differential keeps exactly the linear term, and the two higher-order terms are what get discarded.

Step 3: The errors in closed form

\[ \bigl|\Delta y-dy\bigr| = \bigl|6(\Delta x)^{2}+(\Delta x)^{3}\bigr| = (\Delta x)^{2}\bigl|6+\Delta x\bigr|, \qquad \frac{\bigl|\Delta y-dy\bigr|}{\bigl|\Delta y\bigr|} = \frac{(\Delta x)^{2}\bigl|6+\Delta x\bigr|}{\bigl|\Delta x\bigr|\,\bigl|11+6\Delta x+(\Delta x)^{2}\bigr|}. \]

Step 4: Case 1°: $\Delta x = 1$

\[ \Delta y = 11+6+1 = 18, \qquad dy = 11, \qquad \bigl|\Delta y-dy\bigr| = 7, \qquad \frac{7}{18} \approx 38.9\%. \]

Step 5: Case 2°: $\Delta x = 0.1$

\[ \Delta y = 1.1+0.06+0.001 = 1.161, \qquad dy = 1.1, \]

\[ \bigl|\Delta y-dy\bigr| = 0.061, \qquad \frac{0.061}{1.161} \approx 5.25\%. \]

Step 6: Case 3°: $\Delta x = -0.01$

Now the quadratic term is positive and the cubic term negative, so they partly offset: \[\Delta y = -0.11+0.0006-0.000001 = -0.109401, \qquad dy = -0.11,\]

\[ \bigl|\Delta y-dy\bigr| = 0.000599, \qquad \frac{0.000599}{0.109401} \approx 0.55\%. \]

Step 7: Summary

\[\begin{array}{c|c|c|c|c} \Delta x & \Delta y & dy & \bigl|\Delta y-dy\bigr| & \text{relative} \\ \hline 1 & 18 & 11 & 7 & 38.9\% \\ 0.1 & 1.161 & 1.100 & 0.061 & 5.25\% \\ -0.01 & -0.109401 & -0.110000 & 0.000599 & 0.55\% \end{array}\]

\[\text{Check of the closed form: } (\Delta x)^{2}\bigl|6+\Delta x\bigr| = 1\cdot 7 = 7;\ 0.01\cdot 6.1 = 0.061;\ 0.0001\cdot 5.99 = 0.000599 \;\;\checkmark\]

\[\text{Note } dy \text{ overshoots for } \Delta x\lt 0 \text{ and undershoots for } \Delta x>0, \text{ since } y \text{ is convex at } x=2 \;\;\checkmark\]