DeepSeek: on the 100-problem benchmark

Model: deepseek-v4-flash · each problem submitted 6 times, asking only for the final mathematical answer. Below are the 57 problems where at least one run differed. Each analysis is written by an AI and audited by the ExaktAI team.

Summary
43%
All 6 equivalent
13%
5 of 6 equivalent
17%
4 of 6 equivalent
12%
3 of 6 equivalent
11%
2 of 6 equivalent
4%
All 6 different
3 of 6 equivalent

Problem 1

Evaluate \(\underset{n \rightarrow \infty}{\mathrm{lim}}\frac{\overset{n}{\underset{k =1}{\sum}}\sqrt{1+\frac{k}{n}}}{n}\).

Run 1
\( 0.884488434\)
Run 2
\( 0.884381\)
Run 3
\( 0.884488434\)
Run 4
\( 0.934497\)
Run 5
\( 0.5 \sqrt{2}+\frac{\ln \! \left(1+\sqrt{2}\right)}{2}\)
Run 6
\( 0.884488434\)

Analysis: runs 1, 3, and 6 agree at \(0.884488434\), while run 2 gives \(0.884381\), run 4 gives \(0.934497\), and run 5 gives \(\frac{\sqrt{2}}{2} + \frac{1}{2}\ln(1+\sqrt{2}) \approx 1.1478\).

5 of 6 equivalent

Problem 3

Find the inflection points, as ordered pairs, of \(f \! \left(x \right) = x^{4}-4 x^{3}\).

Run 1
\(\left(0, 0, 2, -16\right)\)
Run 2
\(\left(-\mathit{∞}, 0, 0, 0, 2, -16\right)\)
Run 3
\(\left(\left[0, 0\right], \left[2, -16\right]\right)\)
Run 4
\(\left(0, 0, 2, -16\right)\)
Run 5
\(\left(0, 0, 2, -16\right)\)
Run 6
\(\left(0, 0, 2, -16\right)\)

Analysis: Run 2 includes an extra pair \((-\infty, 0)\) in addition to \((0, 0)\) and \((2, -16)\).

2 of 6 equivalent

Problem 4

Find the inflection points, as ordered pairs, and intervals of concavity of \(f \! \left(x \right) = x^{4}-8 x^{2}+3\).

Run 1
\((0,3),\ \left(-\frac{2}{\sqrt{3}},-\frac{77}{9}\right),\ \left(\frac{2}{\sqrt{3}},-\frac{77}{9}\right);\ \left(-\infty,-\frac{2}{\sqrt{3}}\right)\ \mathrm{concave\ up},\ \left(-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)\ \mathrm{concave\ down},\ \left(\frac{2}{\sqrt{3}},\infty\right)\ \mathrm{concave\ up}\)
Run 2
\(\left(0, 3, -\frac{2}{\sqrt{3}}, -\frac{13}{9}, \frac{2}{\sqrt{3}}, -\frac{13}{9}\right)\)
Run 3
\(\left(-\sqrt{\frac{4}{3}},-\frac{29}{9}\right),\ \left(\sqrt{\frac{4}{3}},-\frac{29}{9}\right),\ \mathrm{concave\ down\ on}\ \left(-\sqrt{\frac{4}{3}},\sqrt{\frac{4}{3}}\right),\ \mathrm{concave\ up\ on}\ \left(-\infty,-\sqrt{\frac{4}{3}}\right)\ \mathrm{and}\ \left(\sqrt{\frac{4}{3}},\infty\right)\)
Run 4
\(\left(-\sqrt{\frac{4}{3}},-\frac{13}{9}\right),\ \left(\sqrt{\frac{4}{3}},-\frac{13}{9}\right),\ \mathrm{concave\ up\ on}\ \left(-\infty,-\sqrt{\frac{4}{3}}\right)\cup\left(\sqrt{\frac{4}{3}},\infty\right),\ \mathrm{concave\ down\ on}\ \left(-\sqrt{\frac{4}{3}},\sqrt{\frac{4}{3}}\right)\)
Run 5
\(\left(-\sqrt{\frac{4}{3}},-\frac{29}{9}\right),\ \left(\sqrt{\frac{4}{3}},-\frac{29}{9}\right),\ \mathrm{concave\ down\ on}\ \left(-\sqrt{\frac{4}{3}},\sqrt{\frac{4}{3}}\right),\ \mathrm{concave\ up\ on}\ \left(-\infty,-\sqrt{\frac{4}{3}}\right)\ \mathrm{and}\ \left(\sqrt{\frac{4}{3}},\infty\right)\)
Run 6
\((0,3),\ \left(-\frac{2}{\sqrt{3}},-\frac{13}{9}\right),\ \left(\frac{2}{\sqrt{3}},-\frac{13}{9}\right);\ \left(-\infty,-\frac{2}{\sqrt{3}}\right)\cup\left(0,\frac{2}{\sqrt{3}}\right)\ \mathrm{concave\ down},\ \left(-\frac{2}{\sqrt{3}},0\right)\cup\left(\frac{2}{\sqrt{3}},\infty\right)\ \mathrm{concave\ up}\)

Analysis: run 1 reports the inflection \(y\)-value as \(-77/9\), run 3 as \(-29/9\), and run 2, run 4, run 6 as \(-13/9\); run 1, run 2, and run 6 list \((0,3)\) as an additional inflection point while run 4 and run 3 do not; and run 6 further subdivides the concavity intervals at \(x=0\), giving concave down on \((-\infty,-2/\sqrt{3})\cup(0,2/\sqrt{3})\) and concave up on \((-2/\sqrt{3},0)\cup(2/\sqrt{3},\infty)\), whereas the others keep \((-2/\sqrt{3},2/\sqrt{3})\) as a single concave-down interval.

All 6 different

Problem 5

Apply 3 iterations of Gauss-Seidel to the system \(10 x_{1}-x_{2}+2 x_{3} = 6\), \(-x_{1}+11 x_{2}-x_{3} = 25\), \(2 x_{1}-x_{2}+10 x_{3} = -11\), starting from \([x_{1}, x_{2}, x_{3}] = [0, 0, 0]\). Give \([x_{1}, x_{2}, x_{3}]\) after iteration 3, each to 6 decimal places.

Run 1
\([ 0.500000, 2.626263, - 1.897747]\)
Run 2
\([ 0.500000, 2.833333, - 2.016667]\)
Run 3
\([ 0.523809, 2.355372, - 1.486238]\)
Run 4
\([ 0.524287, 2.425531, - 1.164914]\)
Run 5
\([ 0.500000, 2.833333, - 1.966667]\)
Run 6
\(\left[0.524286, 2.426667, -1.184524\right]\)

Analysis: Run 1 gives \([0.500000, 2.626263, -1.897747]\), run 2 gives \([0.500000, 2.833333, -2.016667]\), run 3 gives \([0.523809, 2.355372, -1.486238]\), run 4 gives \([0.524287, 2.425531, -1.164914]\), run 5 gives \([0.500000, 2.833333, -1.966667]\), and run 6 gives \([0.524286, 2.426667, -1.184524]\), and at the requested six-decimal precision no two of these triples agree on all three components.

4 of 6 equivalent

Problem 6

Assuming \(0<a\), compute the curvature \(\kappa \! \left(t \right)\) and torsion \(\tau \! \left(t \right)\) of the helix \(r \! \left(t \right) = \langle a \cos \! \left(t \right), a \sin \! \left(t \right), b t \rangle\).

Run 1
\(\frac{{| a |}}{a^{2}+b^{2}}\)
Run 2
\(\tau \coloneqq \frac{b}{a^{2}+b^{2}}\)
Run 3
\(\tau \coloneqq \frac{b}{a^{2}+b^{2}}\)
Run 4
\(\frac{{| \sqrt{a^{2}+b^{2}}\, b |}}{\left(a^{2}+b^{2}\right)^{{3}/{2}}}\)
Run 5
\(\tau \coloneqq \frac{b}{a^{2}+b^{2}}\)
Run 6
\(\tau \coloneqq \frac{b}{a^{2}+b^{2}}\)

Analysis: run 1 gives the curvature \(\kappa = \frac{|a|}{a^2+b^2}\), runs 2, 3, 5, and 6 give the signed torsion \(\tau = \frac{b}{a^2+b^2}\), and run 4 gives the absolute torsion \(\frac{|b|}{a^2+b^2}\).

4 of 6 equivalent

Problem 7

For \(f \! \left(x \right) = {\mathrm e}^{x}\) and Taylor expansion centered at 0, find the smallest \(n\) such that the Lagrange remainder satisfies \({| \textit{R\_n} \! \left(x \right)|}< 1.0\times 10^{-6}\) for all \(x\) in \([0, 0.5]\).

Run 1
\(100\)
Run 2
\(100\)
Run 3
\(20\)
Run 4
\(100\)
Run 5
\(100\)
Run 6
\(0.000000\)

Analysis: The answer by run 3 gives \(n=20\), the answer by run 1 gives \(n=100\), and the answer by run 6 gives \(0.000000\).

4 of 6 equivalent

Problem 8

Find the envelope of the one-parameter family of lines \(y = m x +\frac{1}{m}\) for \(0<m\), and identify the resulting curve.

Run 1
\(y^{2} = 4 x\)
Run 2
\(y^{2} = 4 x\)
Run 3
\(y^{2} = 4 x\)
Run 4
\(4 m^{2} x^{2}+4 m y -4 = 0\)
Run 5
\(4 x y = 1\)
Run 6
\(y^{2} = 4 x\)

Analysis: run 4's expression still contains the parameter \(m\) and is not the envelope curve, run 5 gives the hyperbola \(4xy=1\), and run 1 gives the parabola \(y^2=4x\).

2 of 6 equivalent

Problem 12

Decompose into partial fractions and evaluate \(\int \frac{3 x^{2}+2 x +1}{\left(x -1\right) \left(x +2\right)^{2}}d x\).

Run 1
\(-3\ln|x-1|+2\ln|x+2|+\frac{3}{x+2}+\mathrm{constant}\)
Run 2
\(\frac{11 \ln \! \left({| x -1|}\right)}{3}-\frac{2 \ln \! \left({| x +2|}\right)}{3}+\frac{1}{3 x +6}+C\)
Run 3
\(\frac{11 \ln \! \left({| x -1|}\right)}{3}-\frac{2 \ln \! \left({| x +2|}\right)}{3}+\frac{1}{x +2}+\mathit{constant}\)
Run 4
\(\frac{3 \ln \! \left({| x -1|}\right)}{2}+\frac{3 \ln \! \left({| x +2|}\right)}{2}+\frac{1}{x +2}+\mathit{constant}\)
Run 5
\(-2\ln|x-1|+\ln|x+2|+\frac{3}{x+2}+\mathrm{constant}\)
Run 6
\(-2\ln|x-1|+\ln|x+2|+\frac{3}{x+2}+\mathrm{constant}\)

Analysis: The answers use different coefficients on \(\ln|x-1|\), \(\ln|x+2|\), and the \(1/(x+2)\) term, so they represent distinct antiderivatives.

4 of 6 equivalent

Problem 17

Find the critical points of \(f \! \left(x , y\right) = x^{3}+y^{2}-3 x\) and classify them.

Run 1
\((0,0)\ \text{is a saddle point},\ (1,0)\ \text{is a local minimum},\ (-1,0)\ \text{is a local maximum}\)
Run 2
\([0, 0, \text{local}, [3, 0, \text{saddle}], [-1, 0, \text{local}, [3, 0, \text{saddle}]]]\)
Run 3
\([0, 0, \text{'saddle'}],\ [1, 0, \text{'local min'}],\ [-1, 0, \text{'local max'}]\)
Run 4
\([0, 0, \text{local}, [0, 0], \text{saddle}, [1, 0], \text{local}, [1, 0], \text{min}, [-1, 0], \text{local}, [-1, 0], \text{max}]\)
Run 5
\([0, 0, \text{'saddle'}],\ [1, 0, \text{'local min'}],\ [-1, 0, \text{'local max'}]\)
Run 6
\((1,0,\text{local min}),\ (-1,0,\text{saddle})\)

Analysis: The first group classifies \((0,0)\) as a saddle, \((1,0)\) as a local minimum, and \((-1,0)\) as a local maximum; run 2 lists the points \((0,0)\), \((3,0)\), and \((-1,0)\) with different classifications; run 6 omits \((0,0)\) and labels \((-1,0)\) as a saddle.

2 of 6 equivalent

Problem 18

Locate and classify the critical points of the surface defined by \(f \! \left(x , y\right) = x^{3}+y^{3}-3 x y\).

Run 1
\((0,0,0)\ \text{is a saddle point},\ (1,1,-1)\ \text{is a local minimum}\)
Run 2
\((-1,1,\text{local min});\ (0,0,\text{saddle})\)
Run 3
\(\left(0, 0, 0, 1, 1, -1\right)\)
Run 4
\([0, 0, -1]\)
Run 5
\(\left(0, 0, 0, 1, 1, -1\right)\)
Run 6
\((0,0,\text{saddle}),\ (1,1,\text{local min})\)

Analysis: The deepseek responses split into: one form giving both critical locations with classifications \((0,0)\) saddle and \((1,1)\) local minimum (with or without explicit \(z\)-values), another giving just the two surface points \((0,0,0)\) and \((1,1,-1)\) without classification, one ambiguous/atypical form \((-1,1,\text{local min};\,0,0,\text{saddle})\), and one single-entry form \([0,0,-1]\).

5 of 6 equivalent

Problem 20

Find the extreme values of \(f \! \left(x , y\right) = x^{2}+2 y^{2}\) subject to the constraint \(x +y = 3\).

Run 1
\(\left(-6, 12\right)\)
Run 2
\(\left(-6, 12\right)\)
Run 3
\(\left(-6, 6\right)\)
Run 4
\(\left(-6, 12\right)\)
Run 5
\(-6\)
Run 6
\(\left(-6, 12\right)\)

Analysis: The answer from run 5 reports only the minimum value \(-6\), while the answers from run 3 and run 1 also include a second value (\(6\) and \(12\) respectively) as a purported maximum.

2 of 6 equivalent

Problem 22

Transform to cylindrical coordinates and compute \({\textcolor{gray}{\int}}_{\!\!\!0}^{2 \pi}{\textcolor{gray}{\int}}_{\!\!\!0}^{3}{\textcolor{gray}{\int}}_{\!\!\!0}^{4}r \textcolor{gray}{d}z \textcolor{gray}{d}r \textcolor{gray}{d}\theta\), where \(E\) is the solid bounded by \(z = 0\), \(z = 4\), and \(r = 3\).

Run 1
\(108\pi\)
Run 2
\(9\pi\)
Run 3
\(108\pi\)
Run 4
\(72 \pi\)
Run 5
\(36 \pi\)
Run 6
\(4\cdot 3^{2}\cdot \pi\)

Analysis: run 2 gives \(9\pi\), run 1 gives \(108\pi\), run 4 gives \(72\pi\), while run 5 and run 6 both give \(36\pi\).

5 of 6 equivalent

Problem 23

Evaluate \(\int_{0}^{1}\int_{0}^{\sqrt{-x^{2}+1}}\left(x^{2}+y^{2}\right)d y d x\).

Run 1
\( 0.5235987756\)
Run 2
\( 0.5235987755982988\)
Run 3
\( 0.5235987756\)
Run 4
\( 0.5235987755982988\)
Run 5
\(-{\frac{1}{3}}\)
Run 6
\( 0.5235987756\)

Analysis: Run 5's answer \(-\frac{1}{3}\) is negative, while run 1's answer \(0.5235987756 \approx \frac{\pi}{6}\) is positive, so they are not equivalent.

3 of 6 equivalent

Problem 25

Find the determinant of the 4x4 matrix \(A = \left[\begin{array}{cccc} 2 & 1 & 0 & 1 \\ 1 & 3 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 1 & 0 & 1 & 3 \end{array}\right]\).

Run 1
\(12\)
Run 2
\(12\)
Run 3
\(42\)
Run 4
\(-5\)
Run 5
\(12\)
Run 6
\(20\)

Analysis: The four answers give four different numerical values for the determinant: run 3 gives \(42\), run 4 gives \(-5\), run 1 gives \(12\), and run 6 gives \(20\).

3 of 6 equivalent

Problem 26

Find the reduced row echelon form of \(A = \left[\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -1 \\ 3 & 7 & 0 & 2 \end{array}\right]\).

Run 1
\([[1, 0, 0, 13], [0, 1, 0, -5], [0, 0, 1, 2]]\)
Run 2
\([[1, 0, 0, 13], [0, 1, 0, -5], [0, 0, 1, 2]]\)
Run 3
\([[1, 0, 0, 13], [0, 1, 0, -5], [0, 0, 1, -2]]\)
Run 4
\([[1, 0, 0, 13], [0, 1, 0, -5], [0, 0, 1, -2]]\)
Run 5
\([[1, 0, 0, 13], [0, 1, 0, -5], [0, 0, 1, 2]]\)
Run 6
\(\left[\left[1,0,0,13\right],\left[0,1,0,-7\right],\left[0,0,1,4\right]\right]\)

Analysis: The answers by run 1 and run 3 agree on the last row \([0,0,1,-2]\), while run 6 gives \([0,1,0,-7],[0,0,1,4]\), differing in the second and third rows.

3 of 6 equivalent

Problem 29

Find one valid singular value decomposition of the matrix \(\left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}\right]\).

Run 1
\(\left(\left[\begin{array}{ccc} - 0.229847696400071 & 0.883461017698523 & 0.408248290463863 \\ - 0.524744818760291 & 0.240782492091674 & - 0.816496580927726 \\ - 0.819641941120511 & - 0.401896033515175 & 0.408248290463863 \end{array}\right], \left[\begin{array}{cc} 9.52551809156511 & 0 \\ 0 & 0.514300580658644 \\ 0 & 0 \end{array}\right], \left[\begin{array}{cc} - 0.61962948382934 & - 0.784894453267052 \\ - 0.784894453267052 & 0.61962948382934 \end{array}\right]\right)\)
Run 2
\(\left(\left[\left[- 0.229847696400071, - 0.883461017698525, 0.408248290463863\right], \left[- 0.524744818760291, - 0.240782492124213, - 0.816496580927726\right], \left[- 0.819641941120511, 0.401896033450099, 0.408248290463863\right]\right], \left[\left[ 9.52551809156511, 0\right], \left[0, 0.514300580658644\right]\right], \left[\left[- 0.61962948382934, - 0.784894453267052\right], \left[- 0.784894453267052, 0.61962948382934\right]\right]\right)\)
Run 3
\(\left(\left[\left[- 0.229847696400071, - 0.883461017698525, 0.408248290463863\right], \left[- 0.524744818760291, - 0.240782492124213, - 0.816496580927726\right], \left[- 0.819641941120511, 0.401896033450099, 0.408248290463863\right]\right], \left[\left[ 9.52551809156511, 0\right], \left[0, 0.514300580658644\right]\right], \left[\left[- 0.61962948382934, - 0.784894453267052\right], \left[- 0.784894453267052, 0.61962948382934\right]\right]\right)\)
Run 4
\(\left(\left[\begin{array}{ccc} - 0.229847696400071 & 0.883461017698523 & 0.408248290463863 \\ - 0.524744818760291 & 0.240782492124213 & - 0.816496580927726 \\ - 0.819641941120511 & - 0.401896033450097 & 0.408248290463863 \end{array}\right], \left[\begin{array}{cc} 9.52551809156511 & 0 \\ 0 & 0.514300580658644 \\ 0 & 0 \end{array}\right], \left[\begin{array}{cc} - 0.61962948382934 & - 0.784894453267052 \\ - 0.784894453267052 & 0.61962948382934 \end{array}\right]\right)\)
Run 5
\(\left(\left[\begin{array}{ccc} - 0.386317703118611 & 0.922365780077060 & 0.0 \\ - 0.505991527390307 & - 0.211864575254870 & 0.836516303737808 \\ - 0.625665351662003 & - 1.44609493058680\times 10^{-16} & - 0.548016803708774 \end{array}\right], \left[\begin{array}{cc} 9.52551809156511 & 0.0 \\ 0.0 & 0.514300580658644 \\ 0.0 & 0.0 \end{array}\right], \left[\begin{array}{cc} - 0.619629483829340 & - 0.784894453267052 \\ - 0.784894453267052 & 0.619629483829340 \end{array}\right]\right)\)
Run 6
\(\mathrm{Matrix}\!\left(\left[\left[-0.229847696400071, 0.883461017698523, 0.408248290463863\right], \left[-0.524744818760291, 0.240782492091674, -0.816496580927726\right], \left[-0.819641941120511, -0.401896033515175, 0.408248290463863\right]\right]\right),\ \mathrm{Matrix}\!\left(\left[\left[9.52551809156511, 0\right], \left[0, 0.514300580658644\right], \left[0, 0\right]\right]\right),\ \mathrm{Matrix}\!\left(\left[\left[-0.61962948382934, -0.784894453267052\right], \left[-0.784894453267052, 0.61962948382934\right]\right]\right)\)

Analysis: Runs 4 and 1 share the same \(U\), \(\Sigma\), and \(V\); run 2 differs by a sign flip in column 2 of \(U\) without a matching flip in \(V\); run 5 presents an entirely different left-singular-vector matrix \(U\).

3 of 6 equivalent

Problem 30

Diagonalize \(A = \left[\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}\right]\) as \(A = D\), giving \(P\) and \(D\) explicitly.

Run 1
\(\left(\left[\left[1, 1\right], \left[2, -1\right]\right], \left[\left[5, 0\right], \left[0, 2\right]\right]\right)\)
Run 2
\(P=\left[\left[1,1\right],\left[2,-1\right]\right],\ D=\left[\left[5,0\right],\left[0,2\right]\right]\)
Run 3
\(\left(\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 1 & 1 \end{array}\right], \left[\begin{array}{cc} 5 & 0 \\ 0 & 2 \end{array}\right]\right)\)
Run 4
\(\left(\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 1 & 1 \end{array}\right], \left[\begin{array}{cc} 5 & 0 \\ 0 & 2 \end{array}\right]\right)\)
Run 5
\(\left(\left[\left[1, 1\right], \left[2, -1\right]\right], \left[\left[5, 0\right], \left[0, 2\right]\right]\right)\)
Run 6
\(\left(\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 1 & 1 \end{array}\right], \left[\begin{array}{cc} 5 & 0 \\ 0 & 2 \end{array}\right]\right)\)

Analysis: The matrices \(P\) differ: in run 1 the columns are \((1,2)\) and \((1,-1)\), while in run 3 the columns are \((1,1)\) and \((-1/2,1)\), which are not scalar multiples within each eigenspace of \(D=\operatorname{diag}(5,2)\).

2 of 6 equivalent

Problem 31

Find the minimum-norm least squares solution to \(A x = b\) for \(A = \left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right]\) and \(b = \left[\begin{array}{c} 1 \\ 1 \end{array}\right]\).

Run 1
\(\left[\begin{array}{c} -\frac{1}{3} \\ \frac{1}{6} \\ \frac{2}{3} \end{array}\right]\)
Run 2
\(\left[{\frac{1}{3}}, {\frac{1}{3}}, {\frac{1}{3}}\right]\)
Run 3
\([[- 0.333333333333333, 0.666666666666667, 1.66666666666667]]\)
Run 4
\(\left[{\frac{1}{3}}, {\frac{1}{3}}, {\frac{1}{3}}\right]\)
Run 5
\(\left[\begin{array}{c} \frac{5}{6} \\ -\frac{1}{3} \\ \frac{1}{6} \end{array}\right]\)
Run 6
\(\left[-\frac{1}{3},-\frac{1}{6},0\right]\)

Analysis: The five answers are five distinct vectors in \(\mathbb{R}^3\): run 1 gives \((-1/3, 1/6, 2/3)\), run 3 gives \((-1/3, 2/3, 5/3)\), run 2 gives \((1/3, 1/3, 1/3)\), run 5 gives \((5/6, -1/3, 1/6)\), and run 6 gives \((-1/3, -1/6, 0)\).

4 of 6 equivalent

Problem 32

Find an LU decomposition \(A = L U\) of \(A = \left[\begin{array}{cc} 4 & 3 \\ 6 & 3 \end{array}\right]\), with \(L\) unit lower triangular.

Run 1
\(\left(\left[\left[1, 0\right], \left[\frac{3}{2}, 1\right]\right], \left[\left[4, 3\right], \left[0, -\frac{3}{2}\right]\right]\right)\)
Run 2
\(\left(\left[\left[1, 0\right], \left[ 1.5, 1\right]\right], \left[\left[4, 3\right], \left[0, - 1.5\right]\right]\right)\)
Run 3
\(\left[\left[1, 0\right], \left[\frac{3}{2}, 1\right]\right]\cdot \left[\left[4, 3\right], \left[0, -\frac{3}{2}\right]\right]\)
Run 4
\(\left[\left[1, 0\right], \left[\frac{3}{2}, 1\right]\right]\cdot \left[\left[4, 3\right], \left[0, -\frac{3}{2}\right]\right]\)
Run 5
\(\left(\left[\left[1, 0\right], \left[ 1.5, 1\right]\right], \left[\left[4, 3\right], \left[0, - 1.5\right]\right]\right)\)
Run 6
\(\left(\left[\left[1, 0\right], \left[\frac{3}{2}, 1\right]\right], \left[\left[4, 3\right], \left[0, -\frac{3}{2}\right]\right]\right)\)

Analysis: Run 3 gives the product \(L \cdot U\) as a single matrix expression, while run 1 gives the pair \((L, U)\) as two separate matrices.

4 of 6 equivalent

Problem 33

Solve the Riccati equation \(\frac{d}{d x}y \! \left(x \right) = 1+x^{2}-2 x y \! \left(x \right)+y \! \left(x \right)^{2}\).

Run 1
\(y \! \left(x \right) = x +\frac{1}{x +\textit{\_C1}}\)
Run 2
\(y \! \left(x \right) = x +\frac{1}{\mathit{C1} -x}\)
Run 3
\(\frac{1}{x +1}\)
Run 4
\(y \! \left(x \right) = x +\frac{1}{x +\textit{\_C1}}\)
Run 5
\(1\)
Run 6
\(y \! \left(x \right) = x +\frac{1}{x +\textit{\_C1}}\)

Analysis: The answers by run 2 and run 1 give the general solution \(y = x + 1/(C - x)\) (equivalent up to the sign/name of the constant), while run 3's \(1/(x+1)\) and run 5's \(1\) are distinct expressions not equivalent to that family or to each other.

5 of 6 equivalent

Problem 34

Solve the Riccati equation \(\frac{d}{d x}y \! \left(x \right) = y \! \left(x \right)^{2}+x^{2}\) in closed form.

Run 1
\(\mathit{Error}\)
Run 2
\(\frac{\mathcal{P}\! \left(x +\textit{\_C1} ;-4,0\right)+x \mathcal{P}^{\prime}\! \left(x +\textit{\_C1} ;-4,0\right)+1}{\mathcal{P}\! \left(x +\textit{\_C1} ;-4,0\right)-x^{2}}\)
Run 3
\(\mathit{Error}\)
Run 4
\(\mathit{Error}\)
Run 5
\(\mathit{Error}\)
Run 6
\(\mathit{Error}\)

Analysis: run 2 provides a closed-form solution in terms of Weierstrass \(\wp\) functions, while run 1 asserts that no closed-form solution exists.

All 6 different

Problem 35

Solve \(\frac{d}{d x}y \! \left(x \right) = \frac{x \left(x^{2}-x -1-2 x^{2} y \! \left(x \right)+2 x^{4}\right)}{\left(x^{2}-y \! \left(x \right)\right) \left(x +1\right)}\), expressing the implicit general solution.

Run 1
\(y^{2}-2 x^{2} y +x^{4}+\ln \! \left(x +1\right)+\textit{\_C1} = 0\)
Run 2
\(\frac{x^{2}}{2}-\frac{y^{2}}{2}+y -\frac{\ln \! \left(x^{2}-y \right)}{2}-\frac{\ln \! \left(x +1\right)}{2}+\frac{\ln \! \left(x^{2}-y -1\right)}{2} = \textit{\_C1}\)
Run 3
\(\frac{x^{2}}{2}-\frac{y^{2}}{2}+x y -\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ln \! \left(x +1\right)+\ln \! \left(x^{2}-y \right) = \textit{\_C1}\)
Run 4
\(\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{2}+\ln \! \left(x +1\right) y +\frac{y^{2}}{2}+\textit{\_C1} = 0\)
Run 5
\(\sqrt{-2 x^{4}+2 x^{2} y +2 x^{2}-2 y +1}\, x^{2}-\sqrt{-2 x^{4}+2 x^{2} y +2 x^{2}-2 y +1}-\frac{\ln \! \left(-2 x^{4}+2 x^{2} y +2 x^{2}-2 y +1\right)}{2}+\ln \! \left(x +1\right)+\textit{\_C1} = 0\)
Run 6
\(\frac{1}{2}x^{2}-\frac{1}{2}y^{2}+xy-\frac{1}{3}x^{3}+\frac{1}{2}x^{2}y-\frac{1}{2}x^{4}+\ln(x+1)+\ln(x^{2}-y)=\_C1\)

Analysis: The six answers present different implicit relations in \(x\) and \(y\), each built from a distinct mixture of polynomial monomials, \(\ln(x+1)\), \(\ln(x^2-y)\) (or its variants), and in run 5 a \(\sqrt{-2x^4+2x^2y+2x^2-2y+1}\) factor; computing \(-F_x/F_y\) at common sample points gives a different value for every answer, so no two define the same solution family.

5 of 6 equivalent

Problem 36

For \(0<L\), find the eigenvalues and nontrivial eigenfunctions of the boundary value problem \(\frac{d^{2}}{d x^{2}}y \! \left(x \right)+\lambda y \! \left(x \right) = 0\), \(y \! \left(0\right) = 0\), \(y \! \left(L \right) = 0\).

Run 1
\(\left[0, \frac{n^{2} \pi^{2}}{L^{2}}, \sin \! \left(\frac{n \pi x}{L}\right)\right]\)
Run 2
\(\left(\textit{λ\_n}=\frac{n^{2} \pi^{2}}{L^{2}}, \textit{y\_n} \! \left(x \right)=\sin \! \left(\frac{n \pi x}{L}\right)\right)\)
Run 3
\(\left(\textit{λ\_n}=\frac{n^{2} \pi^{2}}{L^{2}}, \textit{y\_n} \! \left(x \right)=\sin \! \left(\frac{n \pi x}{L}\right)\right)\)
Run 4
\(\left[\lambda = \frac{n^{2} \pi^{2}}{L^{2}}, y \! \left(x \right) = \sin \! \left(\frac{n \pi x}{L}\right)\right]\)
Run 5
\(\left(\textit{λ\_n}=\frac{n^{2} \pi^{2}}{L^{2}}, \textit{y\_n} \! \left(x \right)=\sin \! \left(\frac{n \pi x}{L}\right)\right)\)
Run 6
\(\left(\textit{λ\_n}=\frac{n^{2} \pi^{2}}{L^{2}}, \textit{y\_n} \! \left(x \right)=\sin \! \left(\frac{n \pi x}{L}\right)\right)\)

Analysis: The answer from run 1 includes an extraneous \(\lambda = 0\) eigenvalue, while the answers from run 4 and run 2 give only the nontrivial spectrum \(\lambda_n = n^2\pi^2/L^2\) with eigenfunctions \(\sin(n\pi x/L)\).

3 of 6 equivalent

Problem 37

Solve Bessel's equation \(x^{2} \left(\frac{d^{2}}{d x^{2}}y \! \left(x \right)\right)+x \left(\frac{d}{d x}y \! \left(x \right)\right)+\left(x^{2}-1\right) y \! \left(x \right) = 0\).

Run 1
\(\arccos \! \left(x \right)\)
Run 2
\(\arccos \! \left(x \right)\)
Run 3
\(\textit{\_BesselJ} \! \left(1, x\right) \textit{\_C1} +\textit{\_BesselY} \! \left(1, x\right) \textit{\_C2}\)
Run 4
\(y \! \left(x \right) = J_{1}\! \left(x \right) \textit{\_C1} +Y_{1}\! \left(x \right) \textit{\_C2}\)
Run 5
\(\arccos \! \left(x \right)\)
Run 6
\(\textit{\_BesselJ} \! \left(1, x\right) \textit{\_C1} +\textit{\_BesselY} \! \left(1, x\right) \textit{\_C2}\)

Analysis: The answers by run 4 and run 3 give the general solution \(C_1 J_1(x) + C_2 Y_1(x)\) in terms of Bessel functions of order 1, while the answer by run 1 gives \(\arccos(x)\), which is a different function.

5 of 6 equivalent

Problem 38

Solve \(\frac{d^{2}}{d x^{2}}y \! \left(x \right) = \frac{x^{2} \left(\frac{d}{d x}y \! \left(x \right)\right)^{2}-2 x y \! \left(x \right) \left(\frac{d}{d x}y \! \left(x \right)\right)+y \! \left(x \right)^{2}}{x^{2}}\).

Run 1
\(y \! \left(x \right) = \textit{\_C2} x +\textit{\_C1} x \ln \! \left(x \right)\)
Run 2
\(y = \textit{\_C1} x +\textit{\_C2} x \ln \! \left(x \right)\)
Run 3
\(y \! \left(x \right) = \textit{\_C1} x +\textit{\_C2} x \ln \! \left(x \right)\)
Run 4
\(y \! \left(x \right) = \textit{\_C2} x +\textit{\_C1} x \ln \! \left(x \right)\)
Run 5
\(y \! \left(x \right) = \textit{\_C2} x +\textit{\_C1} x \ln \! \left(x \right)\)
Run 6
\(\left(y(x)=\_C1\,x\exp(\_C2/x)\right)\)

Analysis: The first group gives the general solution \(y = C_1 x + C_2 x \ln(x)\), while run 6's answer \(y = C_1 x \exp(C_2/x)\) is a different family of functions.

All 6 different

Problem 40

For the Fourier-Bessel expansion \(f \! \left(r \right) = -r^{2}+1 = \overset{\infty}{\underset{n =1}{\sum}}c_{n} J_{0}\! \left(\alpha_{n} r \right)\) on \([0, 1]\), where \(\alpha_{n}\) is the n-th positive zero of \(J_{0}\! \left(x \right)\), give the closed form for \(c_{n}\) and the numeric values of \(c_{1}\), \(c_{2}\), \(c_{3}\) to 4 decimal places.

Run 1
\(\left(\frac{2}{\textit{α\_n}^{2} J_{1}\! \left(\textit{α\_n}\right)}, 0.2161, - 0.0648, 0.0347\right)\)
Run 2
\(\left(c_n =\frac{4 J_{2}\! \left(\alpha_n \right)}{\alpha_n^{2} J_{1}\! \left(\alpha_n \right)^{2}}, 0.2161, - 0.0648, 0.0347\right)\)
Run 3
\(\left(\frac{2}{\textit{α\_n}^{2} J_{1}\! \left(\textit{α\_n}\right)}, 0.2160, - 0.0648, 0.0338\right)\)
Run 4
\(\textit{c\_n} = \frac{4 J_{2}\! \left(\alpha_n \right)}{\alpha_n^{2} J_{1}\! \left(\alpha_n \right)^{2}}\)
Run 5
\(\left(c_n =\frac{4 J_{2}\! \left(\alpha_n \right)}{\alpha_n^{2} J_{1}\! \left(\alpha_n \right)^{2}}, 0.2161, - 0.0584, 0.0276\right)\)
Run 6
\(\frac{2}{\alpha_n^{2}\operatorname{BesselJ}(1,\alpha_n)},\ 0.2161,\ -0.0648,\ 0.0332\)

Analysis: As complete answers, all six differ: runs 1, 3, and 6 share one closed form but report different numeric triples, runs 2 and 5 share the other closed form but differ numerically, and run 4 gives only the closed form with no reported values of \(c_1, c_2, c_3\).

5 of 6 equivalent

Problem 41

Find the harmonic conjugate of \(u \! \left(x , y\right) = x^{2}-y^{2}\).

Run 1
\(x^{2}-y^{2}+2 \,\mathrm{I} x y\)
Run 2
\(x^{2}-y^{2}+2 \,\mathrm{I} x y\)
Run 3
\(x^{2}-y^{2}+2 \,\mathrm{I} x y\)
Run 4
\(x^{2}-y^{2}+2 \,\mathrm{I} x y\)
Run 5
\(x^{2}-y^{2}+2 \,\mathrm{I} x y\)
Run 6
\(x^{2}-y^{2}+2xy\)

Analysis: Run 1's answer \(x^2 - y^2 + 2ixy\) includes the imaginary unit \(i\), making it a complex-valued expression (namely \((x+iy)^2\)), while run 6's answer \(x^2 - y^2 + 2xy\) is purely real and lacks the factor of \(i\).

3 of 6 equivalent

Problem 43

Evaluate \(\int \frac{{\mathrm e}^{z}}{\left(z -1\right)^{3}}d z\) over the contour \({| z |} = 3\).

Run 1
\(-2 \,\mathrm{I} \pi \,{\mathrm e}\)
Run 2
\(0\)
Run 3
\(0\)
Run 4
\(0\)
Run 5
\(-2 \,\mathrm{I} \pi \,{\mathrm e}\)
Run 6
\(-2 \,\mathrm{I} \pi \,{\mathrm e}\)

Analysis: Run 2 gives \(0\) while run 1 gives \(-2\pi i\, e\), which are distinct complex numbers.

2 of 6 equivalent

Problem 44

Find the Laurent series expansion of \(f \! \left(z \right) = \frac{1}{z \left(z -2\right)}\) about \(z = 0\) for \(0<{| z |}<2\).

Run 1
\(-\frac{1}{2 z}-\frac{1}{4}-\frac{z}{8}-\frac{z^{2}}{16}-\frac{z^{3}}{32}-\frac{z^{4}}{64}-\frac{z^{5}}{128}\)
Run 2
\(-\frac{1}{2 z}-\frac{1}{4}-\frac{z}{8}-\frac{z^{2}}{16}-\frac{z^{3}}{32}-\frac{z^{4}}{64}\)
Run 3
\(\frac{1}{2 z -4}-\frac{1}{2 z}\)
Run 4
\(\frac{1}{z(z-2)}=-\frac{1}{2z}-\frac{1}{4}-\frac{z}{8}-\frac{z^{2}}{16}-\frac{z^{3}}{32}-\frac{z^{4}}{64}-\frac{z^{5}}{128}-\cdots\)
Run 5
\(-\frac{1}{2 z}-\frac{1}{4}-\frac{z}{8}-\frac{z^{2}}{16}-\frac{z^{3}}{32}-\frac{z^{4}}{64}\)
Run 6
\(-\frac{1}{2 z}-\frac{1}{4}-\frac{z}{8}-\frac{z^{2}}{16}-\frac{z^{3}}{32}-\frac{z^{4}}{64}-\frac{z^{5}}{128}\)

Analysis: run 3 gives the closed-form \(\frac{1}{2(z-2)} - \frac{1}{2z}\) and run 4 gives the full Laurent series with explicit continuation, both representing the same infinite expansion; run 2 truncates the series at \(z^4/64\), while run 1 includes one additional term up to \(z^5/128\).

4 of 6 equivalent

Problem 45

Find the residue of \(f \! \left(z \right) = \frac{\sin \! \left(z \right)}{z^{4}}\) at \(z = 0\).

Run 1
\(\text{The residue is }0.\)
Run 2
\(-{\frac{1}{6}}\)
Run 3
\(-{\frac{1}{6}}\)
Run 4
\(0\)
Run 5
\(\text{The residue is }0.\)
Run 6
\(0\)

Analysis: run 1 and run 4 give \(0\), while run 2 gives \(-1/6\).

5 of 6 equivalent

Problem 48

Evaluate \(\int_{0}^{2 \pi}\frac{1}{2+\cos \! \left(\theta \right)}d \theta\).

Run 1
\( 0.5773502691896257\)
Run 2
\( 0.5773502691896257\)
Run 3
\( 0.577350269189625764509148780501957455647601751287126876\)
Run 4
\( 1.813799365\)
Run 5
\( 0.577350269189625764509148780501957455647601751287268082178\)
Run 6
\( 0.5773502691896257\)

Analysis: The answer from run 4 is \(\frac{\pi}{\sqrt{3}} \approx 1.8138\), while the answer from run 1 is \(\frac{1}{\sqrt{3}} \approx 0.5774\), differing by a factor of \(\pi\).

5 of 6 equivalent

Problem 49

Find the resultant of \(p = x^{3}-2\) and \(q = x^{2}-2\) with respect to \(x\).

Run 1
\(-4\)
Run 2
\(-4\)
Run 3
\(-4\)
Run 4
\(-4\)
Run 5
\(-2\)
Run 6
\(-4\)

Analysis: The answers \(-2\) and \(-4\) are distinct numerical values, so run 5 and run 1 are not mathematically equivalent.

4 of 6 equivalent

Problem 51

Solve the nonlinear system \(x^{3}+y^{3} = 35\), \(x +y = 5\) over the reals.

Run 1
\(\left(\left[5, 0\right], \left[0, 5\right], \left[2, 3\right], \left[3, 2\right]\right)\)
Run 2
\(\left(\left[5, 0\right], \left[2, 3\right], \left[3, 2\right], \left[0, 5\right]\right)\)
Run 3
\(\left(\left[5, 0\right], \left[0, 5\right], \left[2, 3\right], \left[3, 2\right]\right)\)
Run 4
\([5, 0]\)
Run 5
\(\left(\left[5, 0\right], \left[0, 5\right], \left[2, 3\right], \left[3, 2\right]\right)\)
Run 6
\([5, 0]\)

Analysis: Run 2 and run 1 list all four real solutions \((5,0), (0,5), (2,3), (3,2)\), while run 4 lists only \((5,0)\).

All 6 different

Problem 53

Compute a lexicographic Gröbner basis (with \(y <x\)) of the ideal \(\langle x^{2}+y^{2}-4, x y -1\rangle\) and use it to find all complex solutions of the system.

Run 1
\([[x^{2}+y^{2}-4, x y -1], [x^{2}+y^{2}-4, x y -1, y^{4}-4 y^{2}+1], [x^{2}+y^{2}-4, x y -1, y^{4}-4 y^{2}+1]]\)
Run 2
\([x^{2}+y^{2}-4, x y -1, y^{4}-4 y^{2}+1]\)
Run 3
\([[y^{3}+x -4 y, y^{4}-4 y^{2}+1], \{[-\sqrt{2-\sqrt{3}}, -\sqrt{2+\sqrt{3}}], [-\sqrt{2+\sqrt{3}}, -\sqrt{2-\sqrt{3}}], [\sqrt{2-\sqrt{3}}, \sqrt{2+\sqrt{3}}], [\sqrt{2+\sqrt{3}}, \sqrt{2-\sqrt{3}}]\}]\)
Run 4
\([1, x y -1, x^{2}+y^{2}-4, y^{4}-4 y^{2}+1]\)
Run 5
\(\left(x^{4}-4 x^{2}+1, x y -1, x^{2}+y^{2}-4\right)\)
Run 6
\(\left[\sqrt{3},\sqrt{3}\right],\ \left[\sqrt{3},-\sqrt{3}\right],\ \left[-\sqrt{3},\sqrt{3}\right],\ \left[-\sqrt{3},-\sqrt{3}\right]\)

Analysis: No two full answers match: they present different polynomial sets, different claimed Gröbner bases, or only solution points, and the last answer lists only the four incorrect points \((\pm\sqrt{3},\pm\sqrt{3})\) with no basis.

3 of 6 equivalent

Problem 58

Evaluate \(\int_{0}^{\frac{\pi}{2}}\ln \! \left(\sin \! \left(x \right)\right)^{2}d x\) in closed form.

Run 1
\(-\pi^{3}/24+(\pi\ln(2)^{2})/2\)
Run 2
\(-\pi^{3}/24+(\pi\ln(2)^{2})/2\)
Run 3
\(\mathit{π} \left(\frac{\mathit{π}^{2}}{12}+\ln \! \left(2\right)^{2}\right)\)
Run 4
\(-\pi^{3}/24+(\pi\ln(2)^{2})/2\)
Run 5
\(-\pi^{3}/24+(\pi\ln^{2}(2))/2\)
Run 6
\(-\pi^{3}/24+(\pi\ln(2)^{2})/2\)

Analysis: The run 3 answer evaluates to \(\pi^3/12 + \pi(\ln 2)^2\), while the others give \(-\pi^3/24 + \pi(\ln 2)^2/2\).

5 of 6 equivalent

Problem 60

Express \(4 \sin \! \left(x \right) \sin \! \left(2 x \right) \sin \! \left(4 x \right)\) as a sum of cosines.

Run 1
\(4 \sin \! \left(x \right) \sin \! \left(2 x \right) \sin \! \left(4 x \right) = \cos \! \left(x \right)-\cos \! \left(3 x \right)-\cos \! \left(5 x \right)+\cos \! \left(7 x \right)\)
Run 2
\(\cos(x)-\cos(3x)-\cos(5x)+\cos(7x)\)
Run 3
\(\cos(x)-\cos(3x)-\cos(5x)+\cos(7x)\)
Run 4
\(\cos \! \left(x \right)-\cos \! \left(3 x \right)-\cos \! \left(5 x \right)+\cos \! \left(7 x \right)\)
Run 5
\(\cos \! \left(x \right)-\cos \! \left(3 x \right)-\cos \! \left(5 x \right)+\cos \! \left(7 x \right)\)
Run 6
\(4\cos(x)-4\cos(3x)-4\cos(5x)+4\cos(7x)\)

Analysis: The answers from run 1, run 2, and run 4 give \(\cos(x) - \cos(3x) - \cos(5x) + \cos(7x)\), while run 6's answer multiplies every term by \(4\).

4 of 6 equivalent

Problem 63

Find the Fourier series expansion (first four nonzero terms) of the square wave \(f \! \left(x \right) = \mathrm{signum}\! \left(\sin \! \left(x \right)\right)\) on \([-\pi, \pi]\).

Run 1
\(0,\ \frac{4}{\pi},\ 0,\ \frac{4}{3\pi},\ 0,\ \frac{4}{5\pi},\ 0,\ \frac{4}{7\pi}\)
Run 2
\(\frac{4 \sin \! \left(x \right)}{\pi}+\frac{4 \sin \! \left(3 x \right)}{3 \pi}+\frac{4 \sin \! \left(5 x \right)}{5 \pi}+\frac{4 \sin \! \left(7 x \right)}{7 \pi}\)
Run 3
\(0,\ \frac{4}{\pi},\ 0,\ \frac{4}{3\pi},\ 0,\ \frac{4}{5\pi},\ 0,\ \frac{4}{7\pi}\)
Run 4
\(\frac{4 \sin \! \left(x \right)}{\pi}+\frac{4 \sin \! \left(3 x \right)}{3 \pi}+\frac{4 \sin \! \left(5 x \right)}{5 \pi}+\frac{4 \sin \! \left(7 x \right)}{7 \pi}\)
Run 5
\(0,\ \frac{4}{\pi},\ 0,\ \frac{4}{3\pi},\ 0,\ \frac{4}{5\pi},\ 0,\ \frac{4}{7\pi}\)
Run 6
\(0,\ \frac{4}{\pi},\ 0,\ \frac{4}{3\pi},\ 0,\ \frac{4}{5\pi},\ 0,\ \frac{4}{7\pi}\)

Analysis: The first group writes the series as the sum \(\frac{4}{\pi}\sin(x)+\frac{4}{3\pi}\sin(3x)+\frac{4}{5\pi}\sin(5x)+\frac{4}{7\pi}\sin(7x)\), while the second lists only the sequence of Fourier coefficients \(0,\frac{4}{\pi},0,\frac{4}{3\pi},0,\frac{4}{5\pi},0,\frac{4}{7\pi}\) without the accompanying \(\sin(nx)\) basis functions.

5 of 6 equivalent

Problem 65

Find \(D_{7}\), the number of derangements of 7 elements.

Run 1
\(1854\)
Run 2
\(133496\)
Run 3
\(1854\)
Run 4
\(1854\)
Run 5
\(1854\)
Run 6
\(1854\)

Analysis: The number of derangements of 7 elements differs between run 2's value of \(133496\) and run 1's value of \(1854\).

3 of 6 equivalent

Problem 66

Find the Stirling number of the second kind \(S \! \left(6, 3\right)\).

Run 1
\(90\)
Run 2
\(20\)
Run 3
\(20\)
Run 4
\(245\)
Run 5
\(546\)
Run 6
\(20\)

Analysis: The groups give four distinct numerical values: \(20\), \(90\), \(245\), and \(546\).

2 of 6 equivalent

Problem 67

Find the number of binary strings of length 8 with no consecutive 1's.

Run 1
\(42\)
Run 2
\(34\)
Run 3
\(42\)
Run 4
\(55\)
Run 5
\(21\)
Run 6
\(34\)

Analysis: run 4's answer is \(55\), run 2's is \(34\), run 1's is \(42\), and run 5's is \(21\).

5 of 6 equivalent

Problem 69

Find the number of labeled trees on 6 vertices.

Run 1
\(720\)
Run 2
\(105\)
Run 3
\(720\)
Run 4
\(720\)
Run 5
\(720\)
Run 6
\(720\)

Analysis: Run 2 gives \(105\) while run 1 gives \(720 = 6^{6-2}\) (Cayley's formula for labeled trees on \(6\) vertices), so the two values are not equal.

4 of 6 equivalent

Problem 70

Find the number of spanning trees in the complete bipartite graph \(K_{2,3}\).

Run 1
\(10\)
Run 2
\(12\)
Run 3
\(10\)
Run 4
\(10\)
Run 5
\(6\)
Run 6
\(10\)

Analysis: The groups give the values \(6\), \(10\), and \(12\) for the number of spanning trees.

2 of 6 equivalent

Problem 72

Find the number of ways to partition a 6-element set into 3 unordered subsets of size 2.

Run 1
\(120\)
Run 2
\(5\)
Run 3
\(3\)
Run 4
\(10\)
Run 5
\(5\)
Run 6
\(10\)

Analysis: The answer by run 4 is \(10\), the correct count obtained as \(\binom{6}{2}\binom{4}{2}\binom{2}{2}/3! = 15\); the answer by run 1 is \(120\); the answer by run 3 is \(3\); the answer by run 2 is \(5\).

2 of 6 equivalent

Problem 73

For data \(X = [1, 2, 3, 4, 5]\) and \(Y = [ 2.1, 3.9, 4.8, 6.2, 7.0]\), fit a least squares regression line \(y = b x +a\) and compute the residual sum of squares.

Run 1
\( 0.14\)
Run 2
\( 0.06800000000000000\)
Run 3
\( 0.068\)
Run 4
\( 0.06000000000000005\)
Run 5
\( 0.14\)
Run 6
\(0.02800000000000002\)

Analysis: The groups give four distinct residual sum of squares values: \(0.06\), \(0.068\), \(0.14\), and \(0.028\).

4 of 6 equivalent

Problem 74

Find the range and interquartile range of \([5, 12, 18, 23, 28, 35, 41]\) using Tukey's hinges, with the median excluded from the lower and upper halves.

Run 1
\(\left(7, 18\right)\)
Run 2
\(\left(35, 23\right)\)
Run 3
\(\left(5, 12, 18, 23, 28, 35, 41\right)\)
Run 4
\(\left(5, 12, 18, 23, 28, 35, 41\right)\)
Run 5
\(\left(5, 12, 18, 23, 28, 35, 41\right)\)
Run 6
\(\left(5, 12, 18, 23, 28, 35, 41\right)\)

Analysis: Run 1 gives range \(7\) and IQR \(18\), run 2 gives range \(35\) and IQR \(23\), and run 3 just repeats the data set.

2 of 6 equivalent

Problem 75

For data \(\{1, 2, 3, 4, 5\}\), compute the least squares regression line \(y = b x +a\) and the coefficient of determination \(R^{2}\).

Run 1
\(\left( 0.200000000000000, 0.900000000000000, 0.642857142857143\right)\)
Run 2
\(\left( 0.200000000000000, 0.900000000000000, 0.642857142857143\right)\)
Run 3
\(\left( 0.3000000000000000, 0.9000000000000000, 0.5625000000000000\right)\)
Run 4
\(\left( 0.2000000000000000, 0.9000000000000000, 0.5625000000000000\right)\)
Run 5
\(\left( 0.2000000000000000+ 0.9000000000000000 x , 0.7200000000000000\right)\)
Run 6
\(0.2000000000000000,\ 0.9000000000000000,\ 0.4500000000000000\)

Analysis: The groups agree on the slope \(0.9\) but differ in intercept and \(R^2\): run 1 gives \((0.2, 0.9, 0.6429)\), run 3 gives \((0.3, 0.9, 0.5625)\), run 4 gives \((0.2, 0.9, 0.5625)\), run 5 gives the line \(0.2 + 0.9x\) with \(R^2 = 0.72\), and run 6 gives \((0.2, 0.9, 0.45)\).

4 of 6 equivalent

Problem 78

For \(\mathit{X\,~\,Beta(\alpha=2,\,\beta=3)\,}\), find \(E_{X}\), \(\mathit{Var}_{X}\), and the mode.

Run 1
\(\left( 0.4000000000000000, 0.0400000000000000, 0.2500000000000000\right)\)
Run 2
\(\left( 0.400000000000000, 0.0400000000000000, 0.333333333333333\right)\)
Run 3
\(\left( 0.400000000000000, 0.0400000000000000, 0.250000000000000\right)\)
Run 4
\(\left( 0.400000000000000, 0.0400000000000000, 0.333333333333333\right)\)
Run 5
\(\left( 0.400000000000000, 0.0400000000000000, 0.250000000000000\right)\)
Run 6
\(\left( 0.400000000000000, 0.0400000000000000, 0.250000000000000\right)\)

Analysis: The mode differs between the two answers: run 2 gives \(1/3\) while run 3 gives \(1/4\), which is the correct mode \((\alpha-1)/(\alpha+\beta-2) = 1/3\) versus an incorrect value.

4 of 6 equivalent

Problem 79

Find median and sample standard deviation (denominator \(n -1\)) of \([3, 7, 8, 12, 15]\).

Run 1
\(\left(7, 4.743416490252569\right)\)
Run 2
\(\left(7, 4.743416490252569\right)\)
Run 3
\(\left(7, 4.743416490252569\right)\)
Run 4
\(\left(7, 4.743416490252569\right)\)
Run 5
\(\left(5, 4.743416490252569\right)\)
Run 6
\(9,\ 4.743416490252569\)

Analysis: The medians differ: run 1 gives \(7\), run 5 gives \(5\), and run 6 gives \(9\).

4 of 6 equivalent

Problem 80

Find the linear regression equation \(y = a x +b\) that best fits the points \([1, 2]\), \([2, 4]\), \([3, 5]\).

Run 1
\(y = 1.5 x + 0.666666666666667\)
Run 2
\(\left( 0.9999999999999999, 1.3333333333333333\right)\)
Run 3
\(\left( 0.9999999999999998, 1.3333333333333335\right)\)
Run 4
\(\left( 2.0, 0.500000000000000\right)\)
Run 5
\(\left( 0.9999999999999999, 1.3333333333333333\right)\)
Run 6
\(\left( 0.9999999999999998, 1.3333333333333335\right)\)

Analysis: run 1 gives slope \(1.5\) and intercept \(2/3\), run 4 gives \(2\) and \(0.5\), and run 2 gives \(1\) and \(4/3\).

4 of 6 equivalent

Problem 82

Find the equation of the osculating circle to the curve \(y = x^{2}\) at the point \([1, 1]\).

Run 1
\(x^{2}+y^{2}-6 x -2 y +6 = 0\)
Run 2
\(x^{2}+\left(y -\frac{5}{2}\right)^{2} = {\frac{5}{4}}\)
Run 3
\(x^{2}+\left(y -\frac{5}{2}\right)^{2} = {\frac{5}{4}}\)
Run 4
\(x^{2}+\left(y -\frac{5}{2}\right)^{2} = {\frac{5}{4}}\)
Run 5
\(x^{2}+y^{2}-4 x -6 y +10 = 0\)
Run 6
\(x^{2}+\left(y -\frac{5}{2}\right)^{2} = {\frac{5}{4}}\)

Analysis: run 1 gives center \((3,1)\) with \(r^2=4\), run 5 gives center \((2,3)\) with \(r^2=3\), and run 2 gives center \((0,5/2)\) with \(r^2=5/4\).

4 of 6 equivalent

Problem 89

Project the vector \(\langle 2, 3, 4\rangle\) onto the subspace of \(R^{3}\) spanned by \(\langle 1, 1, 1\rangle\) and \(\langle 1, 0, 0\rangle\).

Run 1
\([3, 2, 2]\)
Run 2
\([2, 3, 4]\)
Run 3
\([2, 3, 4]\)
Run 4
\([2, 3, 4]\)
Run 5
\([2, 3, 4]\)
Run 6
\(\left[2,2,2\right]\)

Analysis: The answer by run 1 is \([3, 2, 2]\), the answer by run 2 is \([2, 3, 4]\) (the original vector unchanged), and the answer by run 6 is \([2, 2, 2]\).

4 of 6 equivalent

Problem 91

Solve the Riccati equation \(\frac{d}{d x}y \! \left(x \right) = y \! \left(x \right)^{2}+2 x y \! \left(x \right)+x^{2}-1\).

Run 1
\(y \! \left(x \right) = -x +\frac{1}{x +\mathit{C1}}\)
Run 2
\(y \! \left(x \right) = -x +\frac{1}{x +\mathit{C1}}\)
Run 3
\(y \! \left(x \right) = -x +\frac{1}{x +\mathit{C1}}\)
Run 4
\(y \! \left(x \right) = -x +\frac{1}{C +x}\)
Run 5
\(-1\)
Run 6
\(y \! \left(x \right) = -x +\frac{1}{x +\mathit{C1}}\)

Analysis: The answers from run 4 and run 1 give the general solution \(y(x) = -x + 1/(C + x)\), while run 5 gives only the constant \(-1\).

5 of 6 equivalent

Problem 93

Evaluate \(\mathrm{I}\! \left(a \right) = \int_{0}^{\infty}\frac{\cos \! \left(a x \right)}{x^{4}+1}d x\) for \(0<a\) in closed form.

Run 1
\(\frac{\arccos \! \left({\mathrm e}^{-\frac{a}{\sqrt{2}}} \cos \! \left(\frac{a}{\sqrt{2}}\right)\right) \sqrt{2}\, \pi \,{\mathrm e}^{-\frac{a}{\sqrt{2}}}}{2}\)
Run 2
\(\frac{\pi \,{\mathrm e}^{-\frac{a \sqrt{2}}{2}} \sin \! \left(\frac{a \sqrt{2}}{2}+\frac{\pi}{4}\right)}{2}\)
Run 3
\(\frac{\pi \,{\mathrm e}^{-\frac{a \sqrt{2}}{2}} \sin \! \left(\frac{a \sqrt{2}}{2}+\frac{\pi}{4}\right)}{2}\)
Run 4
\(\frac{\pi \,{\mathrm e}^{-\frac{a \sqrt{2}}{2}} \sin \! \left(\frac{a \sqrt{2}}{2}+\frac{\pi}{4}\right)}{2}\)
Run 5
\(\frac{\pi \,{\mathrm e}^{-\frac{a \sqrt{2}}{2}} \sin \! \left(\frac{a \sqrt{2}}{2}+\frac{\pi}{4}\right)}{2}\)
Run 6
\(\frac{\pi \,{\mathrm e}^{-\frac{a \sqrt{2}}{2}} \sin \! \left(\frac{a \sqrt{2}}{2}+\frac{\pi}{4}\right)}{2}\)

Analysis: Run 1 contains a factor \(\arccos(e^{-a/\sqrt{2}}\cos(a/\sqrt{2}))\) that depends on \(a\), whereas run 2 reduces via \(\sin(b+\pi/4)=(\sin b+\cos b)/\sqrt{2}\) to \(\frac{\pi}{2\sqrt{2}}e^{-a/\sqrt{2}}(\cos(a/\sqrt{2})+\sin(a/\sqrt{2}))\), so the two expressions have different functional dependence on \(a\).

2 of 6 equivalent

Problem 96

Factor the polynomial \(x^{8}+x^{4}+1\) completely over the complex numbers.

Run 1
\(\left(x^{2}+\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}-\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}+\frac{\mathrm{I} \sqrt{3}}{2}-\frac{1}{2}\right) \left(x^{2}-\frac{\mathrm{I} \sqrt{3}}{2}-\frac{1}{2}\right)\)
Run 2
\(\left(x^{2}-x +1\right) \left(x^{2}+x +1\right) \left(x^{4}-x^{2}+1\right)\)
Run 3
\(\left(x^{2}-x +1\right) \left(x^{2}+x +1\right) \left(x^{4}-x^{2}+1\right)\)
Run 4
\(\left(x^{2}+\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}-\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}+\frac{\sqrt{3}}{2}+\frac{\mathrm{I}}{2}\right) \left(x^{2}-\frac{\sqrt{3}}{2}+\frac{\mathrm{I}}{2}\right)\)
Run 5
\(\left(x^{2}+\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}-\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}+\frac{\sqrt{3}}{2}+\frac{\mathrm{I}}{2}\right) \left(x^{2}-\frac{\sqrt{3}}{2}+\frac{\mathrm{I}}{2}\right)\)
Run 6
\(\left(x^{2}+\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}-\frac{\mathrm{I} \sqrt{3}}{2}+\frac{1}{2}\right) \left(x^{2}+\frac{\mathrm{I} \sqrt{3}}{2}-\frac{1}{2}\right) \left(x^{2}-\frac{\mathrm{I} \sqrt{3}}{2}-\frac{1}{2}\right)\)

Analysis: runs 1 and 6 are identical, runs 2 and 3 are identical, and runs 4 and 5 are identical; these three pairwise groups are mathematically different from one another.

3 of 6 equivalent

Problem 98

Find the closed form of the infinite sum \(\overset{\infty}{\underset{n =1}{\sum}}\frac{H_{n}}{n^{3}}\), where \(H_{n} = \overset{n}{\underset{k =1}{\sum}}\frac{1}{k}\) is the n-th harmonic number.

Run 1
\(\frac{\pi^{4}}{72}\)
Run 2
\(\frac{\pi^{2} \zeta \! \left(3\right)}{4}-\frac{\zeta \! \left(5\right)}{2}\)
Run 3
\(\frac{\pi^{2} \zeta \! \left(3\right)}{4}-\frac{\zeta \! \left(5\right)}{2}\)
Run 4
\(\frac{\pi^{4}}{72}\)
Run 5
\(\frac{\pi^{4}}{72}\)
Run 6
\(\frac{\pi^{2} \zeta \! \left(3\right)}{4}-\frac{\zeta \! \left(5\right)}{2}\)

Analysis: Run 1 gives a rational multiple of \(\pi^4\), while run 2 gives a combination of \(\pi^2 \zeta(3)\) and \(\zeta(5)\), which are not equal as real numbers.

3 of 6 equivalent

Problem 99

Derive the minimal polynomial of \(\cos \! \left(\frac{2 \pi}{9}\right)\) over \(Q\) and give its numerical value to 6 decimal places.

Run 1
\( 0.173648\)
Run 2
\( 0.173648\)
Run 3
\(- 0.173648\)
Run 4
\(- 0.173648\)
Run 5
\( 0.766044\)
Run 6
\(- 0.173648\)

Analysis: runs 3, 4, and 6 agree at \(-0.173648\), runs 1 and 2 agree at \(0.173648\), and run 5 gives \(0.766044\).

3 of 6 equivalent

Problem 100

Solve the wave equation \(\frac{\partial^{2}}{\partial t^{2}}u \! \left(x , t\right) = c^{2} \left(\frac{\partial^{2}}{\partial x^{2}}u \! \left(x , t\right)\right)\) on \(0<x \boldsymbol{\land}x \boldsymbol{\land}x \boldsymbol{\land}x <\pi\) with \(0<c\), with boundary conditions \(u \! \left(0, t\right) = u \! \left(\pi , t\right) = 0\), initial displacement \(u \! \left(x , 0\right) = x \left(\pi -x \right)\), and initial velocity \(\left(\frac{\partial}{\partial t}u \! \left(x , t\right)\right)\! \left(x , 0\right) = 0\). Express the solution \(u \! \left(x , t\right)\) as a Fourier sine series with the coefficients given in closed form.

Run 1
\(\frac{\sin \! \left(n x \right) \left(4 \left(-1\right)^{n}-4\right) \cos \! \left(c n t \right)}{n^{3}}\)
Run 2
\(\frac{\sin \! \left(n x \right) \left(4 \left(-1\right)^{n}-4\right) \cos \! \left(c n t \right)}{n^{3}}\)
Run 3
\(u \! \left(x , t\right) = \overset{\infty}{\underset{n =1}{\textcolor{gray}{\sum}}}\! \frac{4 \left(1-\left(-1\right)^{n}\right) \cos \! \left(c n t \right) \sin \! \left(n x \right)}{\pi \,n^{3}}\)
Run 4
\(u \! \left(x , t\right) = \overset{\infty}{\underset{n =1}{\textcolor{gray}{\sum}}}\! \frac{4 \left(1-\left(-1\right)^{n}\right) \cos \! \left(c n t \right) \sin \! \left(n x \right)}{\pi \,n^{3}}\)
Run 5
\(\sin \! \left(n x \right) \left(\frac{4 \left(-1\right)^{n}}{n^{3}}-\frac{4}{n^{3}}\right) \cos \! \left(c n t \right)\)
Run 6
\(u \! \left(x , t\right) = \overset{\infty}{\underset{n =1}{\textcolor{gray}{\sum}}}\! \frac{4 \left(1-\left(-1\right)^{n}\right) \cos \! \left(c n t \right) \sin \! \left(n x \right)}{\pi \,n^{3}}\)

Analysis: Run 1's expression is missing the \(\frac{1}{\pi}\) factor and the summation over \(n\), while run 3 provides the complete series \(u(x,t) = \sum_{n=1}^{\infty} \frac{4(1-(-1)^n)}{\pi n^3}\sin(nx)\cos(cnt)\).