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Maclaurin series by substitution

Problem 7.456 · medium

Use a known Maclaurin series to find the first four nonzero terms of the Maclaurin series of \( \displaystyle f(x) = \ln{\left(3 x + 1 \right)} \).
  1. Start from a known series (ln(1 + u), u = cx) and substitute or multiply term by term.
  2. \[ \lim_{x \to 0^+}\left(\frac{\frac{81 x^{4}}{4} - 9 x^{3} + \frac{9 x^{2}}{2} - 3 x + \ln{\left(3 x + 1 \right)}}{x^{4}}\right) = 0 \]
    These terms match f through x^4.✓ Proved
Answer \( - \frac{81 x^{4}}{4} + 9 x^{3} - \frac{9 x^{2}}{2} + 3 x + \cdots \)

Lines: 1 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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0coefficients from derivatives at 0, and f − T is tiny at x = 0.01

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The stated answer has incorrect signs for the terms. The Maclaurin series for ln(1+u) is u - u^2/2 + u^3/3 - u^4/4 + ..., so substituting u=3x yields 3x - 9x^2/2 + 9x^3 - 81x^4/4 + ... The provided answer has positive coefficients for the odd powers and negative for the even powers (except the constant term which is 0), but specifically the signs of the x^3 and x^4 terms are flipped relative to the standard expansion (it lists +9x^3 and -81x^4/4, which is actually correct for the alternating series, wait. Let's re-evaluate. Standard: ln(1+u) = u - u^2/2 + u^3/3 - u^4/4 + ... Substitute u=3x: Term 1: 3x Term 2: -(3x)^2/2 = -9x^2/2 Term 3: +(3x)^3/3 = +27x^3/3 = +9x^3 Term 4: -(3x)^4/4 = -81x^4/4 So the correct series is 3x - 9/2 x^2 + 9x^3 - 81/4 x^4 + ... The stated answer is: - 81 x^{4}/4 + 9 x^{3} - 9 x^{2}/2 + 3 x + ... This matches the correct terms exactly. Let's look at the equation line: Limit((81*x**4/4 - 9*x**3 + 9*x**2/2 - 3*x + log(3*x + 1))/x**4, x, 0, dir='+') = 0 Let f(x) = log(3x+1). The series is S(x) = 3x - 9/2 x^2 + 9x^3 - 81/4 x^4. The equation checks if ( -S(x) + f(x) ) / x^4 -> 0? No, the equation has: 81*x**4/4 - 9*x**3 + 9*x**2/2 - 3*x + log(3*x + 1). This is equal to - (3x - 9/2 x^2 + 9x^3 - 81/4 x^4) + log(3x+1) = -S(x) + f(x). If f(x) ~ S(x), then f(x) - S(x) is O(x^5). So (f(x) - S(x))/x^4 -> 0. The equation calculates (f(x) - S(x))/x^4? Let's check the signs in the equation's numerator: Term: 81*x**4/4. This is +81/4 x^4. Term: -9*x**3. This is -9 x^3. Term: 9*x**2/2. This is +9/2 x^2. Term: -3*x. This is -3 x. Sum of polynomial part P(x) = -3x + 9/2 x^2 - 9x^3 + 81/4 x^4. This P(x) is exactly -S(x). So the numerator is P(x) + f(x) = f(x) - S(x). Since f(x) - S(x) = O(x^5), the limit is indeed 0. So the equation is correct and proves the stated answer is correct. Why did I think it was wrong initially? I misread the alternating signs or the order. Stated answer: - 81 x^{4}/4 + 9 x^{3} - 9 x^{2}/2 + 3 x Correct series: 3x - 9/2 x^2 + 9x^3 - 81/4 x^4 They are identical. The sentence says: "Start from a known series (ln(1 + u), u = cx) and substitute or multiply term by term." This is a valid method description. Is there any error? The problem asks for the first four nonzero terms. The solution provides them. The verification step confirms them. Wait, look closely at the equation line again. `Limit((81*x**4/4 - 9*x**3 + 9*x**2/2 - 3*x + log(3*x + 1))/x**4, x, 0, dir='+') = 0` If the limit is 0, it means `log(3x+1) + (81/4 x^4 - 9x^3 + 9/2 x^2 - 3x) = o(x^4)`. This implies `log(3x+1) = - (81/4 x^4 - 9x^3 + 9/2 x^2 - 3x) + o(x^4)` `log(3x+1) = 3x - 9/2 x^2 + 9x^3 - 81/4 x^4 + o(x^4)`. This matches the stated answer. So the solution is correct. Verdict: pass.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-09
  • qwen3.6:27b-mlx: fail (error) 2026-10-09 — The stated answer has incorrect signs for the terms. The Maclaurin series for ln(1+u) is u - u^2/2 + u^3/3 - u^4/4 + ..., so substituting u=3x yields 3x - 9x^2/2 + 9x^3 - 81x^4/4 + ... The provided answer has positive coefficients for the odd powers and negative for the even powers (except the constant term which is 0), but specifically the signs of the x^3 and x^4 terms are flipped relative to the standard expansion (it lists +9x^3 and -81x^4/4, which is actually correct for the alternating series, wait. Let's re-evaluate. Standard: ln(1+u) = u - u^2/2 + u^3/3 - u^4/4 + ... Substitute u=3x: Term 1: 3x Term 2: -(3x)^2/2 = -9x^2/2 Term 3: +(3x)^3/3 = +27x^3/3 = +9x^3 Term 4: -(3x)^4/4 = -81x^4/4 So the correct series is 3x - 9/2 x^2 + 9x^3 - 81/4 x^4 + ... The stated answer is: - 81 x^{4}/4 + 9 x^{3} - 9 x^{2}/2 + 3 x + ... This matches the correct terms exactly. Let's look at the equation line: Limit((81*x**4/4 - 9*x**3 + 9*x**2/2 - 3*x + log(3*x + 1))/x**4, x, 0, dir='+') = 0 Let f(x) = log(3x+1). The series is S(x) = 3x - 9/2 x^2 + 9x^3 - 81/4 x^4. The equation checks if ( -S(x) + f(x) ) / x^4 -> 0? No, the equation has: 81*x**4/4 - 9*x**3 + 9*x**2/2 - 3*x + log(3*x + 1). This is equal to - (3x - 9/2 x^2 + 9x^3 - 81/4 x^4) + log(3x+1) = -S(x) + f(x). If f(x) ~ S(x), then f(x) - S(x) is O(x^5). So (f(x) - S(x))/x^4 -> 0. The equation calculates (f(x) - S(x))/x^4? Let's check the signs in the equation's numerator: Term: 81*x**4/4. This is +81/4 x^4. Term: -9*x**3. This is -9 x^3. Term: 9*x**2/2. This is +9/2 x^2. Term: -3*x. This is -3 x. Sum of polynomial part P(x) = -3x + 9/2 x^2 - 9x^3 + 81/4 x^4. This P(x) is exactly -S(x). So the numerator is P(x) + f(x) = f(x) - S(x). Since f(x) - S(x) = O(x^5), the limit is indeed 0. So the equation is correct and proves the stated answer is correct. Why did I think it was wrong initially? I misread the alternating signs or the order. Stated answer: - 81 x^{4}/4 + 9 x^{3} - 9 x^{2}/2 + 3 x Correct series: 3x - 9/2 x^2 + 9x^3 - 81/4 x^4 They are identical. The sentence says: "Start from a known series (ln(1 + u), u = cx) and substitute or multiply term by term." This is a valid method description. Is there any error? The problem asks for the first four nonzero terms. The solution provides them. The verification step confirms them. Wait, look closely at the equation line again. `Limit((81*x**4/4 - 9*x**3 + 9*x**2/2 - 3*x + log(3*x + 1))/x**4, x, 0, dir='+') = 0` If the limit is 0, it means `log(3x+1) + (81/4 x^4 - 9x^3 + 9/2 x^2 - 3x) = o(x^4)`. This implies `log(3x+1) = - (81/4 x^4 - 9x^3 + 9/2 x^2 - 3x) + o(x^4)` `log(3x+1) = 3x - 9/2 x^2 + 9x^3 - 81/4 x^4 + o(x^4)`. This matches the stated answer. So the solution is correct. Verdict: pass.
  • gpt-oss:20b: pass 2026-10-09
  • qwen3.6:27b-mlx: fail (error) 2026-10-09 — The stated answer has incorrect signs for all terms. The Maclaurin series for ln(1+u) is u - u^2/2 + u^3/3 - ..., so substituting u=3x yields 3x - 9x^2/2 + 9x^3 - 81x^4/4. The provided answer has the opposite signs for every term.

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/maclaurin_substitution, checked 2026-10-09 with SymPy 1.14.0.

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