Prove the following approximate formulas for sufficiently
Prove the following approximate formulas for sufficiently small $\alpha$: 1° $\bigl(1\pm\alpha\bigr)^{3}\approx 1\pm3\alpha$; 2° $\dfrac{1}{\sqrt{1+\alpha}}\approx 1-\tfrac12\alpha$.
Explanation of Method
Two more instances of the same binomial rule, with exact remainders.
Step 1: Part 1°, plus sign: $f(\alpha) = (1+\alpha)^{3}$
\[ f(0) = 1, \qquad f'(\alpha) = 3(1+\alpha)^{2}, \qquad f'(0) = 3, \]
\[ \bigl(1+\alpha\bigr)^{3} \approx 1+3\alpha. \]
Step 2: Part 1°, minus sign
No new work is needed: substituting $\beta = -\alpha$ into the result just proved, $\bigl(1-\alpha\bigr)^{3} = \bigl(1+\beta\bigr)^{3} \approx 1+3\beta = 1-3\alpha$.
The two cases are one statement, $\bigl(1\pm\alpha\bigr)^{3}\approx1\pm3\alpha$, with the signs read consistently.
Step 3: The remainder in Part 1° is exact
A cube expands in finitely many terms, so no estimate is required: \[\bigl(1+\alpha\bigr)^{3} = 1+3\alpha+3\alpha^{2}+\alpha^{3} \;\Longrightarrow\; \bigl(1+\alpha\bigr)^{3}-\bigl(1+3\alpha\bigr) = 3\alpha^{2}+\alpha^{3} = O\bigl(\alpha^{2}\bigr).\]
Step 4: Part 2°: $f(\alpha) = (1+\alpha)^{-1/2}$
\[ f(0) = 1, \qquad f'(\alpha) = -\tfrac12(1+\alpha)^{-3/2}, \qquad f'(0) = -\tfrac12, \]
\[ \frac{1}{\sqrt{1+\alpha}} \approx 1-\frac{\alpha}{2}. \]
Step 5: The remainder in Part 2°
Here the expansion is infinite, so the second derivative supplies the leading error term: $f''(\alpha) = \tfrac34(1+\alpha)^{-5/2}, \qquad f''(0) = \tfrac34$.
\[ \frac{1}{\sqrt{1+\alpha}} = 1-\frac{\alpha}{2}+\frac{3}{8}\alpha^{2}+O\bigl(\alpha^{3}\bigr), \]
so the error of the linear approximation is $\tfrac38\alpha^{2}+O(\alpha^{3})$, positive for small $\alpha$ of either sign.
Step 6: Verification at $\alpha = 0.01$
\[1.01^{3} = 1.030301 \quad\text{vs}\quad 1.03; \qquad \text{error } 0.000301 = 3\alpha^{2}+\alpha^{3} \;\;\checkmark\]
\[0.99^{3} = 0.970299 \quad\text{vs}\quad 0.97; \qquad \text{error } 0.000299 = 3\alpha^{2}-\alpha^{3} \;\;\checkmark\]
\[\frac{1}{\sqrt{1.01}} = 0.9950372 \quad\text{vs}\quad 0.995; \qquad \text{error } 0.0000372 \approx \frac38\alpha^{2} = 0.0000375 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\bigl(1\pm\alpha\bigr)^{3}\approx 1\pm3\alpha, \qquad \frac{1}{\sqrt{1+\alpha}}\approx 1-\frac{\alpha}{2}\,}\]
Remark
Remark. Both are again $\bigl(1+\alpha\bigr)^{n}\approx1+n\alpha$, now with $n = 3$ and $n = -\tfrac12$. Comparing with Problem 256 shows the pattern in the errors: the second-order coefficient is $\tfrac{n(n-1)}{2}$, which is $3$ for $n = 3$ and $\tfrac38$ for $n = -\tfrac12$ — exactly the numbers found above.