Linear approximation
Problem 3.111 · easy
Use the linear approximation of \( \displaystyle f(x) = \ln{\left(x \right)} \) at \( \displaystyle x = 1 \) to estimate \( \displaystyle f(\frac{51}{50}) \).
- 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.
- \[ 1 \cdot \frac{1}{50} = \frac{1}{50} \]Evaluate at x = \frac{51}{50}.
Answer \( L(\frac{51}{50}) = \frac{1}{50} \approx 0.02 \)
Lines: 2 proved, 3 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
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 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 5 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 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 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — Line 4 is labeled 'The linearization' but contains only the term (x - 1), omitting the f(a) term which is 0. While the final numerical result is correct because f(1)=0, the statement that the linearization is just (x-1) is technically incomplete and misleading regarding the structure of L(x).
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — Line 4 is labeled 'The linearization' but contains only the term (x - 1), omitting the f(a) term which is 0. While the final numerical result is correct because f(1)=0, the statement that the linearization is just (x-1) is technically incomplete and misleading regarding the structure of L(x).gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the linear approximation formula, computes the derivative and function value at the point of tangency, and evaluates the resulting linear function at the target point.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the linearization formula, computes the derivative and function value at the point of tangency, constructs the linear approximation, and evaluates it at the specified point. The steps are logically sound and algebraically correct.gpt-oss:20b: pass 2026-09-26
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 2026-10-13 with SymPy 1.14.0.