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Taylor and Maclaurin polynomials

Problem 7.99 · medium

Find the Maclaurin polynomial of degree 2 for \( \displaystyle f(x) = \cos{\left(x \right)} \).

p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 1 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = -1 \]
f⁽2⁾(0).✓ Proved
\[ \left(-1\right) x^{2} \cdot 1 \cdot \frac{1}{2} + 1 = 1 - \frac{x^{2}}{2} \]
Assemble the polynomial.✓ Proved

Answer \( p_{2}(x) = 1 - \frac{x^{2}}{2} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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	sympy.series expands f on its own and matches

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-27
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26
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/taylor_polynomial, checked 2026-09-28 with SymPy 1.14.0.