hcju/proofwikiqa huggingface dataset:
Let $m, n$ be coprime integers. By definition of multiplicative functions, we have: :$(1): \quad \map f {m n} = \map f m \map f n$ :$(2): \quad \map g {m n} = \map g m \map g n$ {{begin-eqn}} {{eqn | l = \map {\paren {f * g} } {m n}       | r = \sum_{d \mathop \divides m n} \map f d \map g {\frac {m n} d}       | c = {{Defof|Dirichlet Convolution}} }} {{eqn | r = \sum_{\substack {r \mathop \divides m \\ s \mathop \divides n} } \map f {r s} \map g {\frac m r \frac n s}       | c = Divisors of Product of Coprime Integers }} {{eqn | r = \sum_{\substack {r \mathop \divides m \\ s \mathop \divides n} } \map f r \map f s \map g {\frac m r} \map g {\frac n s}       | c = $f, g$ are multiplicative }} {{eqn | r = \paren {\sum_{r \mathop \divides m} \map f r \map g {\frac m r} } \paren {\sum_{s \mathop \divides n} \map f s \map g {\frac n s} }       | c =  }} {{eqn | r = \map {\paren {f * g} } m \times \map {\paren {f * g} } n       | c = {{Defof|Dirichlet Convolution}} }} {{end-eqn}} {{qed}}
Since Sigma Function is Multiplicative, it follows easily that abundancy index is multiplicative as well. We have for any prime $p$ and positive integer $k$: {{begin-eqn}} {{eqn | l = \frac {\map \sigma {p^k} } {p^k}       | r = \frac {p^{k + 1} - 1} {p^k \paren {p - 1} }       | c = Sigma Function of Power of Prime }} {{eqn | r = \frac {p - p^{-k} } {p - 1} }} {{eqn | o = <       | r = \frac p {p - 1}       | c = as $p^{-k} > 0$ }} {{eqn | r = 1 + \frac 1 {p - 1} }} {{end-eqn}} In fact this is the limit of the abundancy index of a prime power. The greater the prime, the smaller this value is. Therefore finding the least number of prime factors a number needs to have its abundancy index exceed $9$ is to find the least $m$ such that: :$\displaystyle \prod_{i = 1}^m \frac {p_i} {p_i - 1} > 9$ where $p_i$ is the $i$th prime ordered by size. The result can be verified by direct computation. {{qed}}
{{begin-eqn}} {{eqn | l = P_n       | r = \sum_{k \mathop = 1}^n T_k       | c = {{Defof|Pentatope Number}} }} {{eqn | r = \sum_{k \mathop = 1}^n \frac {n \paren {n + 1} \paren {n + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \sum_{k \mathop = 1}^n \frac {\paren {n^3 + 3 n^2 + 2 n} } 6       | c =  }} {{eqn | r = \frac 1 6 \sum_{k \mathop = 1}^n n^3 + \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 3 \sum_{k \mathop = 1}^n n       | c = Summation is Linear }} {{eqn | r = \frac 1 6 \sum_{k \mathop = 1}^n n^3 + \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 3 \frac {n \paren {n + 1} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac 1 6 \sum_{k \mathop = 1}^n n^3 + \frac 1 2 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac 1 3 \frac {n \paren {n + 1} } 2       | c = Sum of Sequence of Squares }} {{eqn | r = \frac 1 6 \dfrac {n^2 \paren {n + 1}^2} 4 + \frac 1 2 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac 1 3 \frac {n \paren {n + 1} } 2       | c = Sum of Sequence of Cubes }} {{eqn | r = \dfrac {n^2 \paren {n + 1}^2 + 2 n \paren {n + 1} \paren {2 n + 1} + 4 n \paren {n + 1} } {24}       | c = putting over a common denominator }} {{eqn | r = \dfrac {n \paren {n + 1} \paren {n \paren {n + 1} + 2 \paren {2 n + 1} + 4} } {24}       | c = extracting $n \paren {n + 1}$ as a factor }} {{eqn | r = \dfrac {n \paren {n + 1} \paren {n^2 + 5 n + 6} } {24}       | c = simplifying }} {{eqn | r = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } {24}       | c = factorising }} {{end-eqn}} {{qed}}
=== Necessary condition === Differentiate the first-order system {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = \mathbf y''       | r = \frac {\d \boldsymbol \psi} {\d x} }} {{eqn | r = \frac {\partial \boldsymbol \psi} {\partial x} + \sum_{i \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_i} \frac {\d y_i} {\d x} }} {{eqn | r = \frac {\partial \boldsymbol \psi} {\partial x} + \sum_{i \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_i} \psi_i }} {{end-eqn}} This can be rewritten as the following system of equations: :$\mathbf y'' = \map {\mathbf f} {x, \mathbf y, \mathbf y'}$ :$\displaystyle \frac {\partial \boldsymbol \psi} {\partial x} + \sum_{i \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_i} \psi_i = \map {\mathbf f} {x, \mathbf y, \mathbf y'}$ By assumption, the first-order system is valid in $\closedint a b$. For the second-order system to be valid in the same interval, the corresponding Hamilton-Jacobi equation has to hold as well. {{qed|lemma}} === Sufficient condition === {{ProofWanted}}
Let $n$ be even such that $n = 2 m$. We have: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2       | r = \sum_{k \mathop = 0}^m \paren {2 m - 2 k}^2       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^m \paren {2 k}^2       | c = Permutation of Indices of Summation }} {{eqn | r = \frac {2 m \paren {m + 1} \paren {2 m + 1} } 3       | c = Sum of Sequence of Even Squares }} {{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6       | c =  }} {{eqn | r = H_{2 m}       | c = Closed Form for Tetrahedral Numbers }} {{end-eqn}} Let $n$ be odd such that $n = 2 m + 1$. We have: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2       | r = \sum_{k \mathop = 0}^m \paren {2 m + 1 - 2 k}^2       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^m \paren {2 k + 1}^2       | c = Permutation of Indices of Summation }} {{eqn | r = \frac {\paren {m + 1} \paren {2 m + 1} \paren {2 m + 3} } 3       | c = Sum of Sequence of Odd Squares: Formulation 1 }} {{eqn | r = \frac {\paren {2 m + 1} \paren {2 m + 2} \paren {2 m + 3} } 6       | c =  }} {{eqn | r = H_{2 m + 1}       | c = Closed Form for Tetrahedral Numbers }} {{end-eqn}} {{qed}} Category:Tetrahedral Numbers Category:Square Numbers Category:Tetrahedral Number as Sum of Squares 7s51qp2538g2fo2vmilix4kes88mu89
{{begin-eqn}} {{eqn | l = \sum_{i \mathop = 0}^n \paren {2 i + 1}^2       | r = \sum_{i \mathop = 0}^n \paren {2 i}^2 + \sum_{i \mathop = 0}^n 4 i + \sum_{i \mathop = 0}^n 1       | c =  }} {{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \sum_{i \mathop = 0}^n i + \sum_{i \mathop = 0}^n 1       | c = Sum of Sequence of Even Squares }} {{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 0}^n 1       | c = adjustment of indices: $4 i = 0$ when $i = 0$ }} {{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \frac {n \paren {n + 1} } 2 + \sum_{i \mathop = 0}^n 1       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 2 n \paren {n + 1} + \paren {n + 1}       | c = further simplification }} {{eqn | r = \frac {\paren {n + 1} \paren {2 n \paren {2 n + 1} + 6 n + 3} } 3       | c = factorising }} {{eqn | r = \frac {\paren {n + 1} \paren {4 n^2 + 8 n + 3} } 3       | c = multiplying out }} {{eqn | r = \frac {\paren {n + 1} \paren {2 n + 1} \paren {2 n + 3} } 3       | c = factorising }} {{end-eqn}} {{qed}} Category:Sum of Sequence of Odd Squares e1c38kqvljyn1fsv8ucyex9ei9l4jrg
{{begin-eqn}} {{eqn | l = \sum_{i \mathop = 0}^n \left({2 i}\right)^2       | r = 4 \sum_{i \mathop = 1}^n i^2       | c = adjustment of indices: $4 i^2 = 0$ when $i = 0$ }} {{eqn | r = 4 \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6       | c = Sum of Sequence of Squares }} {{eqn | r = \frac {2 n \left({n + 1}\right) \left({2 n + 1}\right)} 3       | c = simplifying }} {{end-eqn}} {{qed}} Category:Square Numbers Category:Sums of Sequences dbt2ajh62bj39euk59wtzk7x1nbzprz
=== Lemma === Let $R$ be a ring with unity. Let $n$ be a positive integer. Let $E_{ij}$ denote the matrix with only zeroes except a $1$ at the $(i,j)$th position. Then for all $X, Y \in R^{n \times n}$: :$\displaystyle \sum_{1 \mathop \le i \mathop < j \le n} \operatorname{tr} \left({\left({X \left({E_{ij} - E_{ji} }\right) Y}\right)^t \left({E_{ij} - E_{ji} }\right)}\right) = \operatorname{tr} \left({Y}\right) \operatorname {tr} \left({X}\right) - \operatorname {tr} \left({Y^t X}\right)$ === Proof === Use Trace of Alternating Product of Matrices and Almost Zero Matrices. ---- Use Definition:Frobenius Inner Product and Trace in Terms of Orthonormal Basis and the fact that the $\left({E_{ij} - E_{ji} }\right) / \sqrt 2$ are an orthonormal basis of $\mathfrak {so}_n$. {{ProofWanted}}
=== Lemma === Let $R$ be a ring with unity. Let $n$ be a positive integer. Let $E_{ij}$ denote the matrix with only zeroes except a $1$ at the $\left({i, j}\right)$th position. Let $X, Y \in R^{2 n \times 2 n}$. Let $X = \begin {pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix}$ and $Y = \begin{pmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22}\end{pmatrix}$ Then: :$\displaystyle \sum_{i, j \mathop = 1}^{2 n} \operatorname {tr} \left({\left({X E_{i j} Y}\right)^t E_{i j} }\right) = \operatorname{tr} \left({Y}\right) \operatorname{tr} \left({X}\right)$ :$\displaystyle \sum_{i, j \mathop = 1}^n \operatorname{tr} \left({\left({X E_{ij} Y}\right)^t E_{j + n, i + n} }\right) = \operatorname{tr} \left({Y_{1 2}^t X_{2 1} }\right)$ === Proof === Use Trace of Alternating Product of Matrices and Almost Zero Matrices. ---- Use Definition:Frobenius Inner Product and Trace in Terms of Orthonormal Basis and the fact that the $\left({E_{i j} }\right)_{i \le n, j \ge n + 1}, \left({E_{i j} }\right)_{i \ge n + 1, j \le n}, \left({E_{i j} - E_{j + n, i + n} }\right) / \sqrt2$ are an orthonormal basis of $\mathfrak{sp}_{2 n}$. {{ProofWanted}}
Let $\displaystyle \map f {e_i} = \sum_{j \mathop = 1}^n c_{ij} e_j$ Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$. Then by the above assumption, $A_{ij} = c_{ij}$. Then: {{begin-eqn}} {{eqn | l = \map \tr f       | r = \map \tr A       | c = {{Defof|Trace of Linear Operator}} }} {{eqn | r = \sum_{i \mathop = 1}^n A_{ii}       | c = {{Defof|Trace of Matrix}} }} {{eqn | r = \sum_{i \mathop = 1}^n c_{ii}       | c = From above }} {{end-eqn}} Now it remains to show that $c_{ii} = \innerprod {\map f {e_i} } {e_i}$: {{begin-eqn}} {{eqn | l = \innerprod {\map f {e_i} } {e_i}       | r = \innerprod {\sum_{j \mathop = 1}^n c_{ij} e_j} {e_i}       | c = From above assumption }} {{eqn | r = \sum_{j \mathop = 1}^n c_{ij} \innerprod {e_j} {e_i}       | c = Axiom $(2)$ of Inner Product: Bilinearity }} {{eqn | r = c_{ii} (1)       | c = {{Defof|Orthonormal Subset}}: other terms vanish }} {{eqn | r = c_{ii}       | c =  }} {{end-eqn}} {{qed}}
By Galois Connection implies Upper Adjoint is Surjection iff Lower Adjoint is Injection: :$d$ is an injection. Define $d' = d:T \to g\left[{T}\right]$ By definition: :$d'$ is an injection. By definition of Galois connection: :$d$ is an increasing mapping. We will prove that :$d$ is order embedding. Let $x, y \in T$. By definition of increasing mapping: :$x \precsim y \implies d\left({x}\right) \preceq d\left({y}\right)$ Thus by definition of $d'$ and ordered subset: :$x \precsim y \implies d'\left({x}\right) \preceq' d'\left({y}\right)$ Assume that :$d'\left({x}\right) \preceq' d'\left({y}\right)$ By Upper Adjoint of Galois Connection is Surjection implies Lower Adjoint at Element is Minimum of Preimage of Singleton of Element :$\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{\left\{ {t}\right\} }\right]}\right)$ By Lower Adjoint at Element is Minimum of Preimage of Singleton of Element implies Composition is Identity: :$g \circ d = I_T$ By definition of Galois connection: :$g$ is an increasing mapping. Thus by definition of $d'$ and ordered subset: :$d\left({x}\right) \preceq d\left({y}\right)$ By definition of increasing mapping: :$g\left({d\left({x}\right)}\right) \precsim g\left({d\left({y}\right)}\right)$ Thus by definitions of composition of mappings and identity mapping: :$x \precsim y$ {{qed|lemma}} By definition: :$d'$ is a surjection. Hence $d'$ is order isomorphism between $R$ and $N$. {{qed}}
Let $\displaystyle \map f {e_i} = \sum_{j \mathop = 1}^n c_{ij} e_j$ Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$. Then by the above assumption, $A_{ij} = c_{ij}$. Then: {{begin-eqn}} {{eqn | l = \map \tr f       | r = \map \tr A       | c = {{Defof|Trace of Linear Operator}} }} {{eqn | r = \sum_{i \mathop = 1}^n A_{ii}       | c = {{Defof|Trace of Matrix}} }} {{eqn | r = \sum_{i \mathop = 1}^n c_{ii}       | c = from above }} {{end-eqn}} Now it remains to show that $c_{ii} = e_i^* \paren {\map f {e_i} }$: {{begin-eqn}} {{eqn | l = e_i^* \paren {\map f {e_i} }       | r = e_i^* \paren {\sum_{j \mathop = 1}^n c_{ij} e_j}       | c = From above assumption }} {{eqn | r = \sum_{j \mathop = 1}^n c_{ij} e_i^* \paren {e_j}       | c = $e_i$ is a linear form }} {{eqn | r = \map {c_{ii} } 1       | c = {{Defof|Ordered Dual Basis}}: other terms vanish }} {{eqn | r = c_{ii}       | c =  }} {{end-eqn}} {{qed}}
By definition, an equilateral triangle is a regular polygon with $3$ sides. From Internal Angles of Regular Polygon, the size $A$ of each internal angle of a regular $n$-gon is given by: :$A = \dfrac {\paren {n - 2} 180^\circ} n$ Thus: :$A = \dfrac {180^\circ} n = 60^\circ$ From Value of Degree in Radians: :$1^\circ = \dfrac {\pi} {180^\circ} \mathrm {rad}$ and so: :$A = 60^\circ \times \dfrac {\pi} {180^\circ} = \dfrac \pi 3 \mathrm {rad}$ {{qed}}
{{begin-eqn}} {{eqn | l = 60       | r = 7 + 53       | c = $6$ ways }} {{eqn | r = 13 + 47       | c =  }} {{eqn | r = 17 + 43       | c =  }} {{eqn | r = 19 + 41       | c =  }} {{eqn | r = 23 + 37       | c =  }} {{eqn | r = 29 + 31       | c =  }} {{end-eqn}} It is determined that there are no smaller numbers with this property by inspection: {{begin-eqn}} {{eqn | l = 58       | r = 53 + 5       | c = $3$ ways }} {{eqn | r = 47 + 11       | c =  }} {{eqn | r = 41 + 17       | c =  }} {{eqn | r = 29 + 29       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 56       | r = 53 + 3       | c = $3$ ways }} {{eqn | r = 43 + 13       | c =  }} {{eqn | r = 37 + 19       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 54       | r = 47 + 7       | c = $5$ ways }} {{eqn | r = 43 + 11       | c =  }} {{eqn | r = 41 + 13       | c =  }} {{eqn | r = 37 + 17       | c =  }} {{eqn | r = 31 + 23       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 52       | r = 47 + 5       | c = $3$ ways }} {{eqn | r = 41 + 11       | c =  }} {{eqn | r = 29 + 23       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 50       | r = 47 + 3       | c = $4$ ways }} {{eqn | r = 43 + 7       | c =  }} {{eqn | r = 37 + 13       | c =  }} {{eqn | r = 31 + 19       | c =  }} {{end-eqn}} From the workings of Smallest Positive Integer which is Sum of 2 Odd Primes in n Ways, $48$ is the smallest positive integer which is the sum of $2$ odd primes in $5$ different ways. Thus there is no need to explore smaller numbers. {{qed}}
Follows directly from the definition of change of basis matrix. {{qed}} Category:Change of Basis 3ob4isdrjumnvvaa4nbh17c8srbaahz
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^0 j^7       | r = 0       | c =  }} {{eqn | r = \frac {0^2 \paren {0 + 1}^2 \paren {3 \times 0^4 + 6 \times 0^3 - 0^2 - 4 \times 0 + 2} } {24}       | c =  }} {{end-eqn}} Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | r = \frac {1^2 \paren {1 + 1}^2 \paren {3 \times 1^4 + 6 \times 1^3 - 1^2 - 4 \times 1 + 2} } {24}       | o =        | c =  }} {{eqn | r = \frac {1 \times 2^2 \paren {3 + 6 - 1 - 4 + 2} } {24}       | c =  }} {{eqn | r = \frac {4 \times 6} {24}       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = 1^7       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^1 j^7       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^k j^7 = \dfrac {k^2 \paren {k + 1}^2 \paren {3 k^4 + 6 k^3 - k^2 - 4 k + 2} } {24}$ from which it is to be shown that: :$\displaystyle \sum_{j \mathop = 0}^{k + 1} j^7 = \dfrac {\paren {k + 1}^2 \paren{k + 2}^2 \paren {3 \paren {k + 1}^4 + 6 \paren {k + 1}^3 - \paren {k + 1}^2 - 4 \paren {k + 1} + 2} } {24}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{k + 1} j^7       | r = \sum_{j \mathop = 0}^k j^7 + \paren {k + 1}^7       | c =  }} {{eqn | r = \frac {k^2 \paren {k + 1}^2 \paren {3 k^4 + 6 k^3 - k^2 - 4 k + 2} } {24}       | c = Induction Hypothesis }} {{eqn | o =        | ro= +       | r = \paren {k + 1}^2 \paren {k^5 + 5 k^4 + 10 k^3 + 10 k^2 + 5 k + 1}       | c = Binomial Theorem }} {{eqn | r = \frac {\paren {k + 1}^2} {24} \paren {3 k^6 + 6 k^5 - k^4 - 4 k^3 + 2 k^2 + 24 k^5 + 120 k^4 + 240 k^3 + 240 k^2 + 120 k + 24} }} {{eqn | r = \frac {\paren {k + 1}^2} {24} \paren {3 k^6 + 30 k^5 + 119 k^4 + 236 k^3 + 242 k^2 + 120 k + 24} }} {{eqn | r = \frac {\paren {k + 1}^2 \paren {k + 2}^2} {24} \paren {3 k^4 + 18 k^3 + 35 k^2 + 24 k + 6} }} {{eqn | r = \frac {\paren {k + 1}^2 \paren {k + 2}^2} {24} \paren {3 k^4 + 12 k^3 + 18 k^2 + 12 k + 3 + 6 k^3 + 18 k^2 + 18 k + 6 - k^2 - 2 k - 1 - 4 k - 4 + 2} }} {{eqn | r = \frac {\paren {k + 1}^2 \paren {k + 2}^2 \paren {3 \paren {k + 1}^4 + 6 \paren {k + 1}^3 - \paren {k + 1}^2 - 4 \paren {k + 1} + 2} } {24}       | c = Binomial Theorem }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \displaystyle \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$ {{qed}}
Intuitively, when we consider $\BB$ and $\CC$ as row vectors, this is because: :$\CC = \BB \cdot \mathbf M_{\BB, \CC}$ and: :$\BB \cdot \sqbrk m_\BB =  \CC \cdot \sqbrk m_\CC$ imply: :$\BB \cdot \sqbrk m_\BB = \BB \cdot \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$. Because $\BB$ is a basis, this implies $\sqbrk m_\BB = \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$. This can be formalized by giving $R \times M$ the structure of a ring. Alternatively, this can be verified directly, which boils down to re-proving that that matrix multiplication is associative. {{ProofWanted}}
=== Necessary Condition === By assumption the boundary conditions are self-adjoint. Then there exists $\map g {x, \mathbf y}$ such that: :$\map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } = \dfrac {\partial \map g {x \mathbf y} } {\partial y_i}$ Because $\mathbf p \in C^1$: :$g \in C^2$ {{handwaving|Why?}} Differentiate both sides {{WRT|Differentiation}} $y_k$: :$\dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} = \dfrac {\partial^2 \map g {x, \mathbf y} } {\partial y_k \partial y_i}$ By the Schwarz-Clairaut Theorem, partial derivatives commute, hence indices can be mutually replaced: :$\dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} = \dfrac {\partial \map {p_k} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_i}$ Fixing $x = a$ provides the result. {{qed|lemma}} === Sufficient condition === By assumption: :$\valueat {\dfrac {\partial p_i} {\partial y_j} } {x \mathop = a} = \valueat {\dfrac {\partial p_j} {\partial y_i} } {x \mathop = a}$ Then: :$\exists \map g {x, \mathbf y} \in C^2: \valueat {\dfrac {\partial p_i} {\partial y_j} } {x \mathop = a} = \valueat {\dfrac {\partial p_j} {\partial y_i} } {x \mathop = a} = \valueat {\dfrac {\partial^2 g} {\partial y_i \partial y_j} } {x \mathop = a}$ In other words: :$\bigvalueat {p_i} {x \mathop = a} = \valueat {\dfrac {\partial g} {\partial y_i} } {x \mathop = a}$ {{handwaving|why?}} Hence, the boundary conditions are self-adjoint. {{qed}}
Performing the calculation using long division: <pre>    0.01639344262295081967213114754098360655737704918032786885245901...   -------------------------------------------------------------------- 61)1.00000000000000000000000000000000000000000000000000000000000000000      61    122      61     61     488     183      183    305              --    ---     ---     --     ---     ---      ---    ---      390    380     590     90     220     470      170    150      366    366     549     61     183     427      122    122      ---    ---     ---     --     ---     ---      ---    ---       240    140     410    290     370     430      480    280       183    122     366    244     366     427      427    244       ---    ---     ---    ---     ---     ---      ---    ---        570    180     440    460      400     300     530    360        549    122     427    427      366     244     488    305        ---    ---     ---    ---      ---     ---     ---    ---         210    580     130    330      340     560     420    550         183    549     122    305      305     549     366    549         ---    ---     ---    ---      ---     ---     ---    ---          270    310      80    250      350     110     540     100          244    305      61    244      305      61     488      61          ---    ---      --    ---      ---     ---     ---     ---           260     500    190     600     450     490     520    ...           244     488    183     549     427     488     488            --     ---    ---     ---     ---     ---     ---            160     120     70     510     230      200    320            122      61     61     488     183      183    305            ---     ---     --     ---     ---      ---    --- </pre> {{qed}}
By definition of Galois connection: :$g_1$, $g_2$, $d_2$, and $d_1$ are increasing mappings. Thus by Composition of Increasing Mappings is Increasing: :$g_2 \circ g_1$ and $d_1 \circ d_2$ are increasing mappings. Let $s \in S_3, t \in S_1$. We will prove that :$s \preceq_3 \left({g_2 \circ g_1}\right)\left({t}\right) \implies \left({d_1 \circ d_2}\right)\left({s}\right) \preceq_1 t$ Assume that :$s \preceq_3 \left({g_2 \circ g_1}\right)\left({t}\right)$ By definition of composition of mappings: :$s \preceq_3 g_2\left({g_1\left({t}\right)}\right)$ By definition of Galois connection: :$d_2\left({s}\right) \preceq_2 g_1\left({t}\right)$ By definition of increasing mapping: :$d_1\left({d_2\left({s}\right)}\right) \preceq_1 d_1\left({g_1\left({t}\right)}\right)$ By Galois Connection Implies Order on Mappings :$d_1 \circ g_1 \preceq_1 I_{S_1}$ By definitions of ordering on mappings and composition of mappings: :$g_1\left({d_1\left({t}\right)}\right) \preceq_1 I_{S_1}\left({t}\right)$ By definition of identity mapping: :$g_1\left({d_1\left({t}\right)}\right) \preceq_1 t$ By definition of transitivity: :$d_1\left({d_2\left({s}\right)}\right) \preceq_1 t$ Thus by definition of composition of mappings: :$\left({d_1 \circ d_2}\right)\left({s}\right) \preceq_1 t$ {{qed|lemma}} Assume that :$\left({d_1 \circ d_2}\right)\left({s}\right) \preceq_1 t$ By definition of composition of mappings: :$d_1\left({d_2\left({s}\right)}\right) \preceq_1 t$ By definition of Galois connection: :$d_2\left({s}\right) \preceq_2 g_1\left({t}\right)$ By definition of increasing mapping: :$g_2\left({d_2\left({s}\right)}\right) \preceq_1 g_2\left({g_1\left({t}\right)}\right)$ By Galois Connection Implies Order on Mappings :$I_{S_3} \preceq_3 g_2 \circ d_2$ By definitions of ordering on mappings and composition of mappings: :$I_{S_3}\left({s}\right) \preceq_3 g_2\left({d_2\left({s}\right)}\right)$ By definition of identity mapping: :$s \preceq_3 g_2\left({d_2\left({s}\right)}\right)$ By definition:Transitivity|transitivity]]: :$s \preceq_3 g_2\left({g_1\left({t}\right)}\right)$ Thus by definition of composition of mappings: :$s \preceq_3 \left({g_2 \circ g_1}\right)\left({t}\right)$ {{qed}}