Prove the following approximate formulas for sufficiently

Prove the following approximate formulas for sufficiently small $\alpha$: 1° $\dfrac{1}{1+\alpha}\approx 1-\alpha$; 2° $\sqrt{1+\alpha}\approx 1+\tfrac12\alpha$; 3° $\dfrac{1}{(1+\alpha)^{2}}\approx 1-2\alpha$.

Step 1: The method

Each formula is the linearisation of a function $f$ at $\alpha = 0$: $f(\alpha) \approx f(0)+f'(0)\,\alpha$.

Proving a formula therefore means computing two numbers, $f(0)$ and $f'(0)$, and then checking that the discarded remainder really is of order $\alpha^{2}$.

Step 2: Part 1°: $f(\alpha) = (1+\alpha)^{-1}$

\[ f(0) = 1, \qquad f'(\alpha) = -(1+\alpha)^{-2}, \qquad f'(0) = -1, \]

\[ \frac{1}{1+\alpha} \approx 1-\alpha. \]

The remainder can be computed exactly, with no appeal to Taylor's theorem: \[\frac{1}{1+\alpha}-\bigl(1-\alpha\bigr) = \frac{1-\bigl(1-\alpha\bigr)\bigl(1+\alpha\bigr)}{1+\alpha} = \frac{1-\bigl(1-\alpha^{2}\bigr)}{1+\alpha} = \frac{\alpha^{2}}{1+\alpha} = O\bigl(\alpha^{2}\bigr).\]

It is positive, so the estimate always falls short.

Step 3: Part 2°: $f(\alpha) = (1+\alpha)^{1/2}$

\[ f(0) = 1, \qquad f'(\alpha) = \tfrac12(1+\alpha)^{-1/2}, \qquad f'(0) = \tfrac12, \]

\[ \sqrt{1+\alpha} \approx 1+\tfrac12\alpha. \]

Multiplying and dividing by the conjugate sum gives the remainder exactly: \[\sqrt{1+\alpha}-\left(1+\frac{\alpha}{2}\right) = \frac{\bigl(1+\alpha\bigr)-\left(1+\dfrac{\alpha}{2}\right)^{2}}{\sqrt{1+\alpha}+1+\dfrac{\alpha}{2}} = \frac{-\dfrac{\alpha^{2}}{4}}{\sqrt{1+\alpha}+1+\dfrac{\alpha}{2}} = O\bigl(\alpha^{2}\bigr).\]

Here the sign is negative: the estimate always overshoots, which is the concavity of the square root.

Step 4: Part 3°: $f(\alpha) = (1+\alpha)^{-2}$

\[ f(0) = 1, \qquad f'(\alpha) = -2(1+\alpha)^{-3}, \qquad f'(0) = -2, \]

\[ \frac{1}{(1+\alpha)^{2}} \approx 1-2\alpha. \]

For the remainder, expand the product $\bigl(1-2\alpha\bigr)\bigl(1+\alpha\bigr)^{2}$: \[\bigl(1-2\alpha\bigr)\bigl(1+2\alpha+\alpha^{2}\bigr) = 1+2\alpha+\alpha^{2}-2\alpha-4\alpha^{2}-2\alpha^{3} = 1-3\alpha^{2}-2\alpha^{3},\]

\[ \frac{1}{(1+\alpha)^{2}}-\bigl(1-2\alpha\bigr) = \frac{3\alpha^{2}+2\alpha^{3}}{\bigl(1+\alpha\bigr)^{2}} = O\bigl(\alpha^{2}\bigr). \]

Step 5: Verification at $\alpha = 0.02$

\[\frac{1}{1.02} = 0.9803922 \quad\text{vs}\quad 0.98; \qquad \text{error } +0.00039 = \frac{0.0004}{1.02} \;\;\checkmark\]

\[\sqrt{1.02} = 1.0099505 \quad\text{vs}\quad 1.01; \qquad \text{error } -0.0000495 \approx -\frac{0.0004}{4\cdot2.01} \;\;\checkmark\]

\[\frac{1}{1.02^{2}} = 0.9611688 \quad\text{vs}\quad 0.96; \qquad \text{error } +0.0011688 \approx \frac{3\cdot0.0004}{1.0404} \;\;\checkmark\]