Euler's method
Problem 6.256 · medium
Use Euler's method with step size \( \displaystyle h = \frac{1}{10} \) to approximate \( \displaystyle y(\frac{7}{5}) \) for \( \displaystyle y' = 1 - y \), \( \displaystyle y(1) = 1 \).
- Each step follows the tangent line: yₙ₊₁ = yₙ + h·f(tₙ, yₙ), tₙ₊₁ = tₙ + h.
- \[ 1 \cdot \frac{1}{10} \cdot 0 + 1 = 1 \]y1 = y0 + h·f(t0, y0) with t0 = 1, slope 0.✓ Proved
- \[ 1 \cdot \frac{1}{10} \cdot 0 + 1 = 1 \]y2 = y1 + h·f(t1, y1) with t1 = 11/10, slope 0.✓ Proved
- \[ 1 \cdot \frac{1}{10} \cdot 0 + 1 = 1 \]y3 = y2 + h·f(t2, y2) with t2 = 6/5, slope 0.✓ Proved
- \[ 1 \cdot \frac{1}{10} \cdot 0 + 1 = 1 \]y4 = y3 + h·f(t3, y3) with t3 = 13/10, slope 0.✓ Proved
Answer \( y(\frac{7}{5}) \approx 1 = 1 \)
Lines: 4 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 | Not checked | — | 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 | the steps rerun in floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution performs 4 steps, reaching t = 1.4, but the problem asks for the approximation at t = 7/5 = 1.4. While the final t-value is correct, the indexing is confusing (y4 corresponds to t4=1.4, but the label says t3=13/10 for the calculation of y4, implying y4 is at t4, which is correct, but the previous step y3 was at t3=1.3? No, t0=1, t1=1.1, t2=1.2, t3=1.3, t4=1.4. The calculation for y4 uses t3=1.3. The comment says 't3 = 13/10', which is 1.3. This is correct. However, the problem asks for y(7/5). 7/5 = 1.4. So we need y4. The solution provides y4. Wait, let's re-read carefully. Step 2 calculates y1. Step 3 calculates y2. Step 4 calculates y3. Step 5 calculates y4. The comment for step 5 says 't3 = 13/10'. 13/10 = 1.3. t0=1, t1=1.1, t2=1.2, t3=1.3. Correct. The result is y4. t4 = 1.4 = 7/5. So the value is correct. Why did I think it was wrong? Ah, the prompt says 'Stated answer: y(7/5) approx 1'. The solution arrives at 1. Is the logic sound? Yes. Is there an error? Let's check the slope. f(t,y) = 1-y. y0=1. f(1,1)=0. y1 = 1 + 0.1*0 = 1. y1=1. f(1.1, 1) = 0. y2=1. It stays 1. The solution is actually correct. Why would it fail? Let me re-read the 'adversarial' instruction. 'Find what is wrong'. Is there a subtle error? The comments say 'slope 0'. This is correct. The indices are correct. The final answer is correct. Maybe the issue is that it stops at y4 but the comment for line 5 says 't3 = 13/10'. 13/10 is 1.3. The next t is 1.4. The question asks for y(7/5) = y(1.4). y4 is the approximation at t4=1.4. So y4 is the answer. The solution provides y4. It seems correct. Let me look closer at the indices. Line 2: y1. Line 3: y2. Line 4: y3. Line 5: y4. The comment on line 5 says 't3 = 13/10'. This is the t used to calculate y4. This is correct. Is there a missing step? No. Is the answer format wrong? No. Wait, look at line 4 comment: 't2 = 6/5'. 6/5 = 1.2. Correct. Line 3 comment: 't1 = 11/10'. 1.1. Correct. Line 2 comment: 't0 = 1'. Correct. Everything seems correct. Is it possible the user considers the number of steps? h=1/10. From 1 to 1.4 is 4 steps. The solution does 4 steps. It seems perfect. Let me check if I missed a 'sentence' error. Sentence 1: 'Each step follows the tangent line...'. This is a correct description of Euler's method. Is there a style issue? 'Stated answer: y(7/5) approx 1 = 1'. The solution just lists equations. It doesn't explicitly state 'Therefore y(7/5) approx 1'. But the last line is y4=1. This is implied. Let's look for a 'misleading' aspect. The solution assumes the student knows y4 corresponds to t=1.4. It doesn't explicitly state t4=7/5. This might be considered 'style' or 'misleading' if a student doesn't connect y4 to the target time. However, usually, explicit connection is preferred. Let's look for an 'error'. Did it calculate y(7/5) or y(1.4)? They are the same. Is 7/5 exactly 1.4? Yes. Is there a floating point issue? No, fractions are used. I will mark it as 'pass' unless I find a real error. Wait, look at the last line comment: 't3 = 13/10'. 13/10 = 1.3. The step calculates y4. y4 is at t4 = 1.4. The comment identifies the *previous* t. This is standard. I see no error. I will pass it.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution performs 4 steps, reaching t = 1.4, but the problem asks for the approximation at t = 7/5 = 1.4. While the final t-value is correct, the indexing is confusing (y4 corresponds to t4=1.4, but the label says t3=13/10 for the calculation of y4, implying y4 is at t4, which is correct, but the previous step y3 was at t3=1.3? No, t0=1, t1=1.1, t2=1.2, t3=1.3, t4=1.4. The calculation for y4 uses t3=1.3. The comment says 't3 = 13/10', which is 1.3. This is correct. However, the problem asks for y(7/5). 7/5 = 1.4. So we need y4. The solution provides y4. Wait, let's re-read carefully. Step 2 calculates y1. Step 3 calculates y2. Step 4 calculates y3. Step 5 calculates y4. The comment for step 5 says 't3 = 13/10'. 13/10 = 1.3. t0=1, t1=1.1, t2=1.2, t3=1.3. Correct. The result is y4. t4 = 1.4 = 7/5. So the value is correct. Why did I think it was wrong? Ah, the prompt says 'Stated answer: y(7/5) approx 1'. The solution arrives at 1. Is the logic sound? Yes. Is there an error? Let's check the slope. f(t,y) = 1-y. y0=1. f(1,1)=0. y1 = 1 + 0.1*0 = 1. y1=1. f(1.1, 1) = 0. y2=1. It stays 1. The solution is actually correct. Why would it fail? Let me re-read the 'adversarial' instruction. 'Find what is wrong'. Is there a subtle error? The comments say 'slope 0'. This is correct. The indices are correct. The final answer is correct. Maybe the issue is that it stops at y4 but the comment for line 5 says 't3 = 13/10'. 13/10 is 1.3. The next t is 1.4. The question asks for y(7/5) = y(1.4). y4 is the approximation at t4=1.4. So y4 is the answer. The solution provides y4. It seems correct. Let me look closer at the indices. Line 2: y1. Line 3: y2. Line 4: y3. Line 5: y4. The comment on line 5 says 't3 = 13/10'. This is the t used to calculate y4. This is correct. Is there a missing step? No. Is the answer format wrong? No. Wait, look at line 4 comment: 't2 = 6/5'. 6/5 = 1.2. Correct. Line 3 comment: 't1 = 11/10'. 1.1. Correct. Line 2 comment: 't0 = 1'. Correct. Everything seems correct. Is it possible the user considers the number of steps? h=1/10. From 1 to 1.4 is 4 steps. The solution does 4 steps. It seems perfect. Let me check if I missed a 'sentence' error. Sentence 1: 'Each step follows the tangent line...'. This is a correct description of Euler's method. Is there a style issue? 'Stated answer: y(7/5) approx 1 = 1'. The solution just lists equations. It doesn't explicitly state 'Therefore y(7/5) approx 1'. But the last line is y4=1. This is implied. Let's look for a 'misleading' aspect. The solution assumes the student knows y4 corresponds to t=1.4. It doesn't explicitly state t4=7/5. This might be considered 'style' or 'misleading' if a student doesn't connect y4 to the target time. However, usually, explicit connection is preferred. Let's look for an 'error'. Did it calculate y(7/5) or y(1.4)? They are the same. Is 7/5 exactly 1.4? Yes. Is there a floating point issue? No, fractions are used. I will mark it as 'pass' unless I find a real error. Wait, look at the last line comment: 't3 = 13/10'. 13/10 = 1.3. The step calculates y4. y4 is at t4 = 1.4. The comment identifies the *previous* t. This is standard. I see no error. I will pass it.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution performs 4 steps, reaching t = 1.4, but the problem asks for y(7/5) = y(1.4). While the final t-value is correct, the indexing is confusing (t3 is listed as 13/10=1.3, so y4 corresponds to t=1.4). More critically, the problem asks for y(7/5). 7/5 = 1.4. The steps are: t0=1, t1=1.1, t2=1.2, t3=1.3, t4=1.4. So 4 steps are needed. The solution calculates y1, y2, y3, y4. y4 is the approximation at t4=1.4. The calculation is correct (y stays 1 because y'=0 when y=1). However, the label 't2 = 6/5' in step 3 is wrong. 6/5 = 1.2. t2 should be 1.2. Wait, t0=1, t1=1.1, t2=1.2. 1.2 = 6/5. So t2=6/5 is correct. t3=1.3=13/10. Correct. t4=1.4=7/5. The solution stops at y4. The question asks for y(7/5). 7/5 = 1.4. So y4 is the answer. The arithmetic is correct. The setup is correct. Why fail? Let's re-read carefully. Step 3 says 't2 = 6/5'. 6/5 = 1.2. Correct. Step 4 says 't3 = 13/10'. 13/10 = 1.3. Correct. Step 5 calculates y4. It doesn't explicitly state t4, but implies it. The result is 1. The exact solution is y(t) = 1 + (y0-1)e^{-(t-1)} = 1. So the approximation is exact. The solution is technically correct. Is there a subtle error? 't2 = 6/5' in step 3. Step 3 calculates y3. y3 is at t3? No, y_n is at t_n. y1 is at t1. y2 is at t2. y3 is at t3. y4 is at t4. Step 3 calculates y3 using t2. It says 't2 = 6/5'. This is correct. Step 4 calculates y4 using t3. It says 't3 = 13/10'. This is correct. The final answer is y4. The problem asks for y(7/5). 7/5 = 1.4. t4 = 1 + 4*(0.1) = 1.4. So y4 is the value at t=1.4. The solution is correct. Wait, look at step 3 again. 'y3 = y2 + h*f(t2, y2) with t2 = 6/5'. 6/5 = 1.2. Correct. Look at step 4. 'y4 = y3 + h*f(t3, y3) with t3 = 13/10'. 13/10 = 1.3. Correct. The solution seems fine. Is there a style issue? The prompt asks to find what is wrong. If it's correct, pass. But wait, step 2 says 't0 = 1'. Step 3 says 't1 = 11/10'. Step 4 says 't2 = 6/5'. Step 5 says 't3 = 13/10'. The indices for t in the comments are t0, t1, t2, t3. The y indices are y1, y2, y3, y4. This is consistent. y_{n+1} uses t_n. So y1 uses t0. y2 uses t1. y3 uses t2. y4 uses t3. The comments are correct. The math is correct. The answer is correct. I will pass it.gpt-oss:20b: fail (error) 2026-10-06 — The solution stops one step early: after t₃=13/10 it should compute y₄ at t₄=7/5. Also the sentence claiming the slope is 0 at each step is correct, but the missing final step means the approximation is not fully carried out.
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/eulers_method, checked 2026-10-06 with SymPy 1.14.0.