Linear approximation in two variables
Problem 10.525 · hard
Find the linear approximation of \( \displaystyle f(x, y) = e^{x} \cos{\left(y \right)} \) at \( \displaystyle (0, 0) \), and use it to estimate \( \displaystyle f\left(\frac{1}{10}, \frac{1}{20}\right) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} e^{x} \cos{\left(y \right)}\\\frac{\partial}{\partial y} e^{x} \cos{\left(y \right)}\end{matrix}\right] = \left[\begin{matrix}e^{x} \cos{\left(y \right)}\\- e^{x} \sin{\left(y \right)}\end{matrix}\right] \]f_x and f_y.✓ Proved
- \[ \left[\begin{matrix}1\\1\\0\end{matrix}\right] \]f, f_x and f_y at the point.✓ Proved
- \[ x + 1 \]The linearization.✓ Proved
- \[ \frac{11}{10} \]The estimate; the true value is about 1.10379.✓ Proved
Answer \( L(x, y) = x + 1,\ f \approx \frac{11}{10} = 1.10000 \)
Every line of this solution was proved by the computer algebra system SymPy. 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | L agrees with f to within O(h²) in four directions at h = 1e−5 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The linear approximation formula is incorrect. The solution provides L(x, y) = x + 1, which corresponds to L(x, y) = f(0,0) + f_x(0,0)x, omitting the f_y(0,0)y term. While f_y(0,0) is indeed 0, the derivation steps do not explicitly show the evaluation of the gradient at (0,0) or the construction of the full linearization formula L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b). Step 2 lists values but does not connect them to the formula in Step 3, and Step 3 is a tautology that hides the missing logic. A student might incorrectly believe the y-term is always ignored or that the formula is simply f + f_x * x.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The linear approximation formula is incorrect. The solution provides L(x, y) = x + 1, which corresponds to L(x, y) = f(0,0) + f_x(0,0)x, omitting the f_y(0,0)y term. While f_y(0,0) is indeed 0, the derivation steps do not explicitly show the evaluation of the gradient at (0,0) or the construction of the full linearization formula L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b). Step 2 lists values but does not connect them to the formula in Step 3, and Step 3 is a tautology that hides the missing logic. A student might incorrectly believe the y-term is always ignored or that the formula is simply f + f_x * x.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12
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_approx_2var, checked 2026-10-12 with SymPy 1.14.0.