The binomial series
Problem 7.654 · medium
Find the first five terms of the binomial series for \( \displaystyle \frac{1}{\sqrt{\frac{x}{2} + 1}} \).
- (1 + u)ʳ = 1 + ru + r(r − 1)/2! u² + r(r − 1)(r − 2)/3! u³ + ⋯ with r = -1/2 and u = x/2.Reviewed
- \[ \frac{\left(1 \cdot \frac{1}{2}\right)^{2} \left(1 \cdot \frac{1}{2} + 1\right)}{2 \cdot 2} = \frac{3}{32} \]The x² coefficient, r(r − 1)/2 · c².✓ Proved
- \[ \lim_{x \to 0}\left(\frac{- \frac{35 x^{4}}{2048} + \frac{5 x^{3}}{128} - \frac{3 x^{2}}{32} + \frac{x}{4} - 1 + \frac{1}{\sqrt{\frac{x}{2} + 1}}}{x^{4}}\right) = 0 \]These terms agree with the function to order x⁴.✓ Proved
Answer \( \frac{35 x^{4}}{2048} - \frac{5 x^{3}}{128} + \frac{3 x^{2}}{32} - \frac{x}{4} + 1 + \cdots \)
Lines: 2 proved, 1 reviewed. 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 | 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f minus the polynomial at x = 0.01 and 0.001 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the binomial expansion parameters and verifies the coefficients algebraically and via limits. The stated answer matches the derived terms.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the binomial expansion parameters and verifies the coefficients algebraically and via limits. The stated answer matches the derived terms.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the binomial expansion parameters and verifies the coefficients. The stated answer matches the derived terms.gpt-oss:20b: fail (style) 2026-10-16 — The solution correctly sets up the binomial expansion and verifies the coefficient of x², but it never actually lists the first five terms of the series as requested. The missing explicit terms make the answer incomplete rather than incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The stated answer has the wrong signs for all non-constant terms. The binomial expansion of (1+u)^(-1/2) has alternating signs, so the x term should be negative, the x^2 term positive, etc. The stated answer lists -x/4 (correct sign) but +3x^2/32 (correct sign) but -5x^3/128 (correct sign) and +35x^4/2048 (correct sign). Wait, let me re-evaluate. Expansion of (1+u)^{-1/2} = 1 - (1/2)u + (3/8)u^2 - (5/16)u^3 + (35/128)u^4 - ... With u = x/2: Term 1: 1 Term 2: -(1/2)(x/2) = -x/4 Term 3: (3/8)(x/2)^2 = (3/8)(x^2/4) = 3x^2/32 Term 4: -(5/16)(x/2)^3 = -(5/16)(x^3/8) = -5x^3/128 Term 5: (35/128)(x/2)^4 = (35/128)(x^4/16) = 35x^4/2048 The stated answer is: 35 x^{4}/2048 - 5 x^{3}/128 + 3 x^{2}/32 - x/4 + 1. This matches my calculation. The signs are correct. Let's look at the solution steps. Step 1 sets up the formula correctly. Step 2 checks the x^2 coefficient. It calculates 3/32. This is correct. Step 3 checks the limit of the difference between the function and the polynomial divided by x^4. The polynomial in the limit is -35x^4/2048 + 5x^3/128 - 3x^2/32 + x/4 - 1. Wait. The stated answer is P(x) = 1 - x/4 + 3x^2/32 - 5x^3/128 + 35x^4/2048. The limit expression in Step 3 uses: -35x^4/2048 + 5x^3/128 - 3x^2/32 + x/4 - 1. This is -P(x). So the limit is ( -P(x) + f(x) ) / x^4. Since f(x) ~ P(x), f(x) - P(x) is O(x^5). So (f(x) - P(x))/x^4 -> 0. The expression in the limit is ( -P(x) + f(x) ) / x^4 = (f(x) - P(x))/x^4. This limit is indeed 0. So the solution steps are mathematically correct and support the stated answer. Is there any error? Step 1: Sentence. Correct setup. Step 2: Equation. Correct calculation for one coefficient. Step 3: Equation. Correct verification that the series matches the function to order 4. The verdict should be pass.gpt-oss:20b: pass 2026-10-15
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/binomial_series, checked 2026-10-18 with SymPy 1.14.0.