Home›Calculus 2›Taylor series about a point›Problem 7.311

Taylor series about a point

Problem 7.311 · medium

Find the Taylor polynomial of degree 3 for \( \displaystyle f(x) = \ln{\left(x \right)} \) centered at \( \displaystyle a = 1 \).

\[ \left[\begin{matrix}\left. \ln{\left(x \right)} \right|_{\substack{ x=1 }}\\\left. \frac{d}{d x} \ln{\left(x \right)} \right|_{\substack{ x=1 }}\\\left. \frac{d^{2}}{d x^{2}} \ln{\left(x \right)} \right|_{\substack{ x=1 }}\\\left. \frac{d^{3}}{d x^{3}} \ln{\left(x \right)} \right|_{\substack{ x=1 }}\end{matrix}\right] = \left[\begin{matrix}0\\1\\-1\\2\end{matrix}\right] \]
f and its first three derivatives at a.✓ Proved
T(x) = Σ f⁽ʲ⁾(a)/j! · (x − a)ʲ.
\[ \lim_{x \to 1^+}\left(\frac{- x - \frac{\left(x - 1\right)^{3}}{3} + \frac{\left(x - 1\right)^{2}}{2} + \ln{\left(x \right)} + 1}{\left(x - 1\right)^{3}}\right) = 0 \]
T agrees with f to third order at a.✓ Proved

Answer \( T_3(x) = x + \frac{\left(x - 1\right)^{3}}{3} - \frac{\left(x - 1\right)^{2}}{2} - 1 \)

Lines: 2 proved, 1 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.

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Not checked	—	a sentence; read, not computed
3	✓ 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's series() expansion about a

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The stated answer is incorrect; the linear term should be (x-1), not (x-1). The solution claims the limit of the difference is 0, but the provided T_3(x) has a coefficient of 1 for the linear term instead of 1, and the constant term is -1 instead of 0. Specifically, f(1)=0, f'(1)=1, so the first two terms are 0 + 1*(x-1). The stated answer has x-1 which is correct for the linear part, but the constant term in the stated answer is -1, whereas f(1)=0. Wait, let's re-evaluate. f(1)=0. f'(1)=1. f''(1)=-1. f'''(1)=2. T_3(x) = 0 + 1(x-1) - 1/2(x-1)^2 + 2/6(x-1)^3 = (x-1) - 1/2(x-1)^2 + 1/3(x-1)^3. The stated answer is x + (x-1)^3/3 - (x-1)^2/2 - 1. Let's expand the stated answer: x - 1 - 1/2(x-1)^2 + 1/3(x-1)^3. Since x-1 is the linear term, this matches. However, the constant term in the Taylor polynomial centered at 1 is f(1)=0. The expression x-1-1 is -1 + (x-1). At x=1, this is -1. But f(1)=0. So the stated answer evaluates to -1 at x=1, which is wrong. The stated answer is actually T_3(x) = (x-1) - 1/2(x-1)^2 + 1/3(x-1)^3. The provided text says `x + ... - 1`. `x-1` is `(x-1)`. So `x-1 - 1/2...` is correct. Wait. `x - 1` is the linear term. The constant term of the polynomial `x - 1 - 1/2(x-1)^2 + 1/3(x-1)^3` is `1 - 1 - 0 + 0 = 0`. The stated answer is `x + (x-1)^3/3 - (x-1)^2/2 - 1`. This simplifies to `(x-1) - 1/2(x-1)^2 + 1/3(x-1)^3`. This is the correct polynomial. Why did I think it was wrong? Let's check the limit equation. The limit checks if `f(x) - T(x)` is `o((x-1)^3)`. If T(x) is correct, the limit is 0. The equation is marked proved. So the algebra is correct. Is there a sentence error? Sentence 2 states the formula. Sentence 1 computes derivatives. The setup is correct. The answer is correct. Why would it fail? Let me re-read carefully. `T_3(x) = x + (x-1)^3/3 - (x-1)^2/2 - 1`. This is equal to `(x-1) - 1/2(x-1)^2 + 1/3(x-1)^3`. This is the correct Taylor polynomial. The solution seems sound. Is there a subtle issue? The problem asks for the Taylor polynomial. The solution provides it. The limit confirms it. I see no error.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-05 — The stated answer is incorrect; the linear term should be (x-1), not (x-1). The solution claims the limit of the difference is 0, but the provided T_3(x) has a coefficient of 1 for the linear term instead of 1, and the constant term is -1 instead of 0. Specifically, f(1)=0, f'(1)=1, so the first two terms are 0 + 1*(x-1). The stated answer has x-1 which is correct for the linear part, but the constant term in the stated answer is -1, whereas f(1)=0. Wait, let's re-evaluate. f(1)=0. f'(1)=1. f''(1)=-1. f'''(1)=2. T_3(x) = 0 + 1(x-1) - 1/2(x-1)^2 + 2/6(x-1)^3 = (x-1) - 1/2(x-1)^2 + 1/3(x-1)^3. The stated answer is x + (x-1)^3/3 - (x-1)^2/2 - 1. Let's expand the stated answer: x - 1 - 1/2(x-1)^2 + 1/3(x-1)^3. Since x-1 is the linear term, this matches. However, the constant term in the Taylor polynomial centered at 1 is f(1)=0. The expression x-1-1 is -1 + (x-1). At x=1, this is -1. But f(1)=0. So the stated answer evaluates to -1 at x=1, which is wrong. The stated answer is actually T_3(x) = (x-1) - 1/2(x-1)^2 + 1/3(x-1)^3. The provided text says `x + ... - 1`. `x-1` is `(x-1)`. So `x-1 - 1/2...` is correct. Wait. `x - 1` is the linear term. The constant term of the polynomial `x - 1 - 1/2(x-1)^2 + 1/3(x-1)^3` is `1 - 1 - 0 + 0 = 0`. The stated answer is `x + (x-1)^3/3 - (x-1)^2/2 - 1`. This simplifies to `(x-1) - 1/2(x-1)^2 + 1/3(x-1)^3`. This is the correct polynomial. Why did I think it was wrong? Let's check the limit equation. The limit checks if `f(x) - T(x)` is `o((x-1)^3)`. If T(x) is correct, the limit is 0. The equation is marked proved. So the algebra is correct. Is there a sentence error? Sentence 2 states the formula. Sentence 1 computes derivatives. The setup is correct. The answer is correct. Why would it fail? Let me re-read carefully. `T_3(x) = x + (x-1)^3/3 - (x-1)^2/2 - 1`. This is equal to `(x-1) - 1/2(x-1)^2 + 1/3(x-1)^3`. This is the correct Taylor polynomial. The solution seems sound. Is there a subtle issue? The problem asks for the Taylor polynomial. The solution provides it. The limit confirms it. I see no error.
gpt-oss:20b: pass 2026-10-05
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The stated answer is incorrect; the linear term should be (x-1), not x-1. The solution fails to explicitly construct the polynomial from the computed derivatives, and the verification limit in line 3 uses an expression that does not match the stated answer (signs are flipped and the linear term is inconsistent).
gpt-oss:20b: pass 2026-10-05

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_series_center, checked 2026-10-05 with SymPy 1.14.0.