Linear approximation
Problem 3.116 · easy
Use the linear approximation of \( \displaystyle f(x) = \ln{\left(x \right)} \) at \( \displaystyle x = 1 \) to estimate \( \displaystyle f(\frac{19}{20}) \).
- The linearization at a is L(x) = f(a) + f'(a)(x − a).
- \[ \left. \ln{\left(x \right)} \right|_{\substack{ x=1 }} = 0 \]f(a).✓ Proved
- \[ \left. \frac{d}{d x} \ln{\left(x \right)} \right|_{\substack{ x=1 }} = 1 \]f'(a).✓ Proved
- \[ x - 1 \]The linearization.✓ Proved
- \[ - \frac{1}{20} \]Evaluate at x = \frac{19}{20}.✓ Proved
Answer \( L(\frac{19}{20}) = - \frac{1}{20} \approx -0.05 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the estimate is within Taylor's error bound of the true value |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/linear_approximation, checked 2026-09-26 with SymPy 1.14.0.