hcju/proofwikiqa huggingface dataset:
Let: :$1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c$ where: :$0 < a \le b \le c$ and: {{AimForCont}} $a = 1$. Then: :$1 = \dfrac 1 1 + \dfrac 1 b + \dfrac 1 c$ and so: :$\dfrac 1 b + \dfrac 1 c = 0$ which contradicts the stipulation that $b, c > 0$. So there is no solution possible when $a = 1$. Therefore $a \ge 2$. === $a = 2$ === Let $a = 2$. ==== $b = 2$ ==== Let $b = 2$. Then: :$\dfrac 1 a + \dfrac 1 b = 1$ leaving no room for $c$. Hence there are no solutions where $a = 2$ and $b = 2$. {{qed|lemma}} ==== $b = 3$ ==== Let $b = 3$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 2 + \dfrac 1 3       | c =  }} {{eqn | r = \dfrac {3 + 2} 6       | c =  }} {{eqn | r = \dfrac 5 6       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | r = 1 - \dfrac 5 6       | c =  }} {{eqn | r = \dfrac 1 6       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | r = \dfrac 1 6       | c =  }} {{eqn | ll= \leadsto       | l = c       | r = 6       | c =  }} {{end-eqn}} Thus we have: :$(1): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6$ {{qed|lemma}} ==== $b = 4$ ==== Let $b = 4$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 2 + \dfrac 1 4       | c =  }} {{eqn | r = \dfrac {2 + 1} 4       | c =  }} {{eqn | r = \dfrac 3 4       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | r = 1 - \dfrac 3 4       | c =  }} {{eqn | r = \dfrac 1 4       | c =  }} {{eqn | ll= \leadsto       | l = c       | r = 4       | c =  }} {{end-eqn}} Thus we have: :$(2): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4$ ==== $b > 4$ ==== Let $b > 4$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c        | o = <       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Hence there are no solutions such that $a = 2, b > 4$. {{qed|lemma}} === $a = 3$ === Let $a = 3$. ==== $b = 3$ ==== Let $b = 3$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 3 + \dfrac 1 3       | c =  }} {{eqn | r = \dfrac 2 3       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | r = 1 - \dfrac 2 3       | c =  }} {{eqn | r = \dfrac 1 3       | c =  }} {{eqn | ll= \leadsto       | l = c       | r = 3       | c =  }} {{end-eqn}} Thus we have: :$(3): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 3$ {{qed|lemma}} === $a > 3$ === Let $a > 3$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | o = \le       | r = \dfrac 1 a + \dfrac 1 a + \dfrac 1 a       | c =  }} {{eqn | o = <       | r = 1       | c =  }} {{end-eqn}} Hence there are no solutions for $a>3$. {{qed|lemma}} === Summary === Hence our $3$ solutions: {{begin-eqn}} {{eqn | n = 1       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6       | o =  }} {{eqn | n = 2       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4       | o =  }} {{eqn | n = 3       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 3       | o =  }} {{end-eqn}} {{qed}}
By definition of increment of the functional: :$\ds J \sqbrk {u + h} - J \sqbrk u = \idotsint_R \paren {F \sqbrk {x, u + h, u_{\mathbf x} + h_{\mathbf x} } - F \sqbrk {x, u, u_{\mathbf x} } } \rd x_1 \cdots \rd x_n$ Use multivariate Taylor's Theorem on $F$ around the point $\tuple {\mathbf x, u, u_{\mathbf x} }$: :$F \sqbrk {\mathbf x, u + h, u_{\mathbf x} + h_{\mathbf x} } = F \sqbrk {\mathbf x, u, u_{\mathbf x} } + \dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} h + \dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } h_{\mathbf x} + \map \OO {h^2, h h_{\mathbf x}, h_{\mathbf x}^2}$ where $\OO$ stands for Big-O. Then: :$\ds \Delta J \sqbrk {u, h} = \idotsint_R \paren {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} h + \dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } h_{\mathbf x} + \map \OO {h^2, h h_{\mathbf x}, h_{\mathbf x}^2} } \rd x_1 \cdots \rd x_n$ By definition of variation of the functional: {{begin-eqn}} {{eqn | l = \delta J        | r = \idotsint_R \paren {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} + \dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } h_{\mathbf x} } \rd x_1 \cdots \rd x_n  }} {{eqn | r = \idotsint_R \paren {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} - \dfrac {\partial F_{u_{\mathbf x} } \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial \mathbf x} } \map h {\mathbf x} \rd x_1 \cdots \rd x_n + \idotsint_R \dfrac {\partial} {\partial \mathbf x} \sqbrk {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } \map h {\mathbf x} } \rd x_1 \cdots \rd x_n        | c = as $F_{u_{\mathbf x} } h_{\mathbf x} = \dfrac {\partial} {\partial \mathbf x} \sqbrk {F_{u_{\mathbf x} } h} - \dfrac {\partial F_{u_{\mathbf x} } } {\partial \mathbf x} h$ }} {{end-eqn}} By Green's theorem: :$\ds \idotsint_R \dfrac {\partial} {\partial \mathbf x} \sqbrk {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } \map h {\mathbf x} } \rd x_1 \cdots \rd x_n = \idotsint_\Gamma \map h {\mathbf x} F_{u_{\mathbf x} } \sqbrk {\mathbf x, u, u_{\mathbf x} } \boldsymbol \nu \rd \sigma$ where: :$\Gamma$ denotes the boundary of $R$ :$\boldsymbol \nu$ is an outward normal to $\Gamma$. Since the region is fixed, so are its boundary points.  Hence, the difference function $h$ has to vanish at the boundary. In other words: :$\forall \mathbf x \in \Gamma: \map h {\mathbf x} = 0$ This leaves only the first integral. :$\ds \delta J = \idotsint_R \paren {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} - \dfrac{\partial F_{u_{ \mathbf x } } \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial \mathbf x} } \map h {\mathbf x} \rd x_1 \cdots \rd x_n$ For arbitrary $h$ the first variation vanishes if the term in brackets vanishes: :$\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} - \dfrac {\partial F_{u_{\mathbf x} } \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial \mathbf x} = 0$ {{qed}}
We verify by brute force: :Starting on $n_0 \le 2916 = 4 \times 9^3$, we will end on $153$. {{qed|lemma}} First we prove that if $3 \divides n_0$, then $3 \divides \map f {n_0}$. From Divisibility by 3: The sum of digits of $n_0$ is divisible by $3$. By Fermat's Little Theorem: :$\forall x \in \Z: x^3 \equiv x \pmod 3$ Therefore sum of the cubes of the digits of $n_0$ is also divisible by $3$. That is the definition of $\map f {n_0}$. Thus $3 \divides \map f {n_0}$. {{qed|lemma}} It remains to be shown that for every $n_0 > 2916$, $\map f {n_0} < n_0$. For $n_0 \le 9999$: :$\map f {n_0} \le 9^3 + 9^3 + 9^3 + 9^3 = 2916 < n_0$ Now suppose $n_0$ is a $k$-digit number, where $k \ge 5$. Then: {{begin-eqn}} {{eqn | l = n_0       | o = >       | r = 10^{k - 1}       | c = the smallest $k$-digit number }} {{eqn | r = 10^4 \times 10^{k - 5} }} {{eqn | o = \ge       | r = 10^4 \paren {1 + 9 \paren {k - 5} }       | c = Bernoulli's Inequality }} {{eqn | o = >       | r = 10 \times 9^3 \paren {9 k - 44}       | c = $10 > 9$ }} {{eqn | o = >       | r = 9^3 \paren {90 k - 440} }} {{eqn | o = \ge       | r = 9^3 \paren {k + 445 - 440}       | c = $k \ge 5$ }} {{eqn | o = >       | r = 9^3 \times k }} {{eqn | o = \ge       | r = \map f {n_0} }} {{end-eqn}} This shows that $\map f {n_0} < n_0$ for all $n_0 > 2916$. Thus for any $n_0 > 2916$, $s_n$ is eventually less than $2916$. Therefore we will eventually reach $153$. {{qed}}
We have: :$4^4 = 256 > 159$ :$3^4 = 81$ :$2^4 = 16$ :$1^4 = 1$ Let us attempt to construct an expression of $159$ as the sum of fewer than $19$ fourth powers: If no $3^4$ is used in our sum, the sum consists only of $2^4$ and $1^4$. Using $2^4$ is more efficient than using $1^4$, since $2^4$ can replace $16 \times 1^4$. So we have: :$159 = 9 \times 2^4 + 15 \times 1^4$ which uses $24$ fourth powers. If $3^4$ is used in our sum, the most efficient way is demonstrated above: :$159 = 14 \times 1^4 + 4 \times 2^4 + 3^4$ which uses $19$ fourth powers. We cannot use more than $2$ instances of $3^4$, as $2 \times 3^4 = 162 > 159$. {{qed}}
It will be proved that the largest integer which cannot be expressed as the sum of distinct pentagonal numbers is $159$. The remaining integers in the sequence can be identified by inspection. We prove this using a variant of Second Principle of Mathematical Induction. Let $\map P n$ be the proposition: :$n$ can be expressed as the sum of distinct pentagonal numbers. === Basis for the induction === We verify the result up to $P_{19} = 532$. Then we conclude all $n$ with $160 \le n \le 532$ can be expressed as the sum of distinct pentagonal numbers. So $\map P n$ is true for all $160 \le n \le 532$. This is the basis for the induction. === Induction Hypothesis === Suppose for some $k > 532$, $\map P j$ is true for all $160 \le j < k$. That is, every integer between $160$ and $k - 1$ can be expressed as the sum of distinct pentagonal numbers. This is the induction hypothesis. === Induction Step === This is the induction step: We find the largest integer $i$ such that $P_i < k \le P_{i + 1}$. We show that $k - P_{i - 4}$ can be expressed as the sum of distinct pentagonal numbers, and the sum does not involve $P_{i - 4}$. Then $k$ can be expressed as the sum of distinct pentagonal numbers. Since $k > 160 = P_{19}$, we must have $i \ge 19$. Hence $k > k - P_{i - 4} \ge P_i - P_{i - 4} = \paren {3 i - 2} + \paren {3 i - 5} + \paren {3 i - 8} + \paren {3 i - 11} \ge 55 + 52 + 49 + 47 = 202 > 160$. Thus $k - P_{i - 4}$ can be expressed as the sum of distinct pentagonal numbers by Induction Hypothesis. A sufficient condition such that the sum does not involve $P_{i - 4}$ is $P_{i - 4} > k - P_{i - 4}$. We have: {{begin-eqn}} {{eqn | l = P_{i - 4} - \paren {k - P_{i - 4} }       | o = \ge       | r = 2 P_{i - 4} - P_{i + 1}       | c = From $k \le P_{i + 1}$ }} {{eqn | r = 2 \paren {\frac {\paren {i - 4} \paren {3 i - 13} } 2} - \frac {\paren {i + 1} \paren {3 i + 2} } 2       | c = Closed Form for Pentagonal Numbers }} {{eqn | r = \frac 1 2 \paren {6 i^2 - 50 i + 104 - 3 i^2 - 5 i - 2} }} {{eqn | r = \frac 1 2 \paren {3 i^2 - 55 i + 102} }} {{eqn | r = \frac 3 2 \paren {i \paren {i - \frac {55} 3} + 34} }} {{eqn | o = \ge       | r = \frac 3 2 \paren {34}       | c = From $i \ge 19$ }} {{eqn | r = 51 }} {{eqn | o = >       | r = 0 }} {{end-eqn}} Therefore we have $P_{i - 4} > k - P_{i - 4}$. Hence $\map P k$ is true. By the Second Principle of Mathematical Induction, $\map P n$ is true for all $n \ge 160$. Thus every integer greater than $159$ can be expressed as the sum of distinct pentagonal numbers. {{Proofread}}
{{begin-eqn}} {{eqn | l = 18 \times \tau \left({18}\right)       | r = 18 \times 6       | c = Tau Function of 18 }} {{eqn | r = \left({2 \times 3^2}\right) \times \left({2 \times 3}\right)       | c =  }} {{eqn | r = 2^2 \times 3^3       | c =  }} {{eqn | r = 108       | c =  }} {{eqn | l = 27 \times \tau \left({27}\right)       | r = 27 \times 4       | c = Tau Function of 27 }} {{eqn | r = 2^2 \times 3^3       | c =  }} {{eqn | r = 108       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 24 \times \tau \left({24}\right)       | r = 24 \times 8       | c = Tau Function of 24 }} {{eqn | r = \left({2^3 \times 3}\right) \times 2^3       | c =  }} {{eqn | r = 2^6 \times 3       | c =  }} {{eqn | r = 192       | c =  }} {{eqn | l = 32 \times \tau \left({32}\right)       | r = 32 \times 6       | c = Tau Function of 32 }} {{eqn | r = 2^5 \times \left({2 \times 3}\right)       | c =  }} {{eqn | r = 192       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 56 \times \tau \left({56}\right)       | r = 56 \times 8       | c = Tau Function of 56 }} {{eqn | r = \left({2^3 \times 7}\right) \times 2^3       | c =  }} {{eqn | r = 2^6 \times 7       | c =  }} {{eqn | r = 448       | c =  }} {{eqn | l = 64 \times \tau \left({64}\right)       | r = 64 \times 7       | c = Tau Function of 64 }} {{eqn | r = 2^6 \times 7       | c =  }} {{eqn | r = 448       | c =  }} {{end-eqn}} {{qed}}
Let $t \in S_2$, $s \in S_1$. We will prove that :$t \preceq_2 f\left({s}\right) \implies f^{-1}\left({t}\right) \preceq_1 s$ Assume that :$t \preceq_2 f\left({s}\right)$ By Inverse of Order Isomorphism is Order Isomorphism: :$f^{-1}$ is an order isomorphism. By definition of order isomorphism: :$f^{-1}$ is an order embedding. By definition of order embedding: :$f^{-1}\left({t}\right) \preceq_1 f^{-1}\left({f\left({s}\right)}\right)$ Thus by definition of bijection: :$f^{-1}\left({t}\right) \preceq_1 s$ {{qed|lemma}} We will prove that :$f^{-1}\left({t}\right) \preceq_1 s \implies t \preceq_2 f\left({s}\right)$ Assume that :$f^{-1}\left({t}\right) \preceq_1 s$ By definition of order embedding: :$f\left({f^{-1}\left({t}\right)}\right) \preceq_2 f\left({s}\right)$ Thus by definition of bijection: :$t \preceq_2 f\left({s}\right)$ {{qed}}
{{:Definition:Centered Hexagonal Number/Sequence}} We have that: {{begin-eqn}} {{eqn | l = 1       | r = 1^2       | c = {{Defof|Square Number}} }} {{eqn | r = 3 \times 1 \times \paren {1 - 1} + 1       | c = Closed Form for Centered Hexagonal Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 169       | r = 13^2       | c = {{Defof|Square Number}} }} {{eqn | r = 3 \times 8 \times \paren {8 - 1} + 1       | c = Closed Form for Centered Hexagonal Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 32 \, 761       | r = 181^2       | c = {{Defof|Square Number}} }} {{eqn | r = 3 \times 105 \times \paren {105 - 1} + 1       | c = Closed Form for Centered Hexagonal Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 6 \, 355 \, 441       | r = 2521^2       | c = {{Defof|Square Number}} }} {{eqn | r = 3 \times 1456 \times \paren {1456 - 1} + 1       | c = Closed Form for Centered Hexagonal Numbers }} {{end-eqn}} {{qed}}
We note the following: {{begin-eqn}} {{eqn | l = 169       | r = 13^2       | c =  }} {{eqn | r = 12^2 + 5^2       | c =  }} {{eqn | r = 12^2 + 4^2 + 3^2       | c =  }} {{eqn | r = 8^2 + 8^2 + 5^2 + 4^2       | c =  }} {{eqn | r = 8^2 + 8^2 + 4^2 + 4^2 + 3^2       | c =  }} {{end-eqn}} Let $41 \le n \le 155$. Let $n \equiv 0 \pmod 3$. Then $169$ can be represented as: :$169 = 2 \times 3^2 + \left({37 - \dfrac {n - 42} 3}\right) \times 2^2 + \left({\dfrac {4 \left({n - 42}\right)} 3 + 3}\right) \times 1^2$ Let $n \equiv 1 \pmod 3$. Then $169$ can be represented as: :$169 = \left({42 - \dfrac {n - 43} 3}\right) \times 2^2 + \left({\dfrac {4 \left({n - 43}\right)} 3 + 1}\right) \times 1^2$ Let $n \equiv 2 \pmod 3$. Then $169$ can be represented as: :$169 = 3^2 + \left({40 - \dfrac {n - 41} 3}\right) \times 2^2 + \left({\dfrac {4 \left({n - 41}\right)} 3}\right) \times 1^2$ Thus in each case $169$ has been expressed in terms of $n$ squares: a combination of varying numbers of $1^2$, $2^2$ and $3^2$. For $n > 155$ this construction does not work, because it results in a negative number of instances of $2^2$. Similarly, for $n < 41$ this construction does not work, because it results in a negative number of instances of $1^2$. For $n < 41$, we have: :$169 = 25 + k$ where $k$ equals one of: :$1 \times 12^2$ :$9 \times 4^2$ :$16 \times 3^2$ :$14 \times 3^2 + 3 \times 2^2 + 6 \times 1^2$ :$10 \times 3^2 + 11 \times 2^2 + 10 \times 1^2$ That is, $169$ equals $25$ plus $1$, $9$, $16$, $23$ or $31$ other squares. But from 25 as Sum of 4 to 11 Squares, $25$ can also be written as a sum of $n$ squares for $4 \le n \le 11$. Thus, by using judicious combinations of: : sums of squares totalling $25$ : sums of squares totalling $169 - 25$ we can obtain a representation of $169$ as the sum of $n$ squares where $5 \le n \le 40$. In summary, therefore, there exists a representation of $169$ as the sum of $n$ squares for $1 \le n \le 155$. {{qed}}
Define $C = \left({K\left({P}\right), \preceq}\right)$ as an ordered subset of $P$ where $K\left({P}\right)$ denotes the compact subset of $P$. Thus by Lattice of Power Set is Algebraic: :$P$ is algebraic. It remains to prove that :$K\left({P}\right)$ is meet closed. Let $x, y \in K\left({P}\right)$. By definition of compact subset: :$x$ is compact. By Element is Finite iff Element is Compact in Lattice of Power Set :$x$ is finite. By Intersection is Subset: :$x \cap y \subseteq x$ By Subset of Finite Set is Finite: :$x \cap y$ is finite. By Element is Finite iff Element is Compact in Lattice of Power Set :$x \cap y$ is compact. Thus by definition of compact subset: :$x \cap y \in K\left({P}\right)$ {{qed}}
Proof by induction: For all $n \in \N_{>1}$, let $\map P n$ be the proposition: :$\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$ === Basis for the Induction === $\map P 1$ is true, as this just says: :$\dfrac \d {\d x} \ln x = \dfrac 1 x$ This follows by Derivative of Natural Logarithm Function This is our basis for the induction. === Induction Hypothesis === Now we need to show 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 our induction hypothesis: :$\dfrac {\d^k} {\d x^k} \ln x = \dfrac {\paren {k - 1}! \paren {-1}^{k - 1} } {x^k}$ Then we need to show: :$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \ln x = \dfrac {k! \paren {-1}^k} {x^{k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \frac {\d^{k + 1} } {\d x^{k + 1} } \ln x       | r = \map {\frac \d {\d x} } {\frac {\d^k} {\d x^k} \ln x}       | c =  }} {{eqn | r = \map {\frac \d {\d x} } {\paren {k - 1}! \paren {-1}^{k - 1} x^{-k} }       | c = Induction Hypothesis }} {{eqn | r = \paren {k - 1}! \paren {-1}^{k - 1} \map {\frac \d {\d x} } {x^{-k} }       | c = Derivative of Constant Multiple }} {{eqn | r = \paren {k - 1}! \paren {-1}^{k - 1} \paren {k x^{-k - 1} }       | c = Power Rule for Derivatives }} {{eqn | r = k! \paren {-1}^k x^{-\paren {k + 1} }       | c = {{Defof|Factorial}} }} {{eqn | r = \frac {k! \paren {-1}^k} {x^{k + 1} }       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$ {{Qed}} Category:Derivatives Category:Natural Logarithms Category:Proofs by Induction 4kn1vgarampqz7xu9rh0ek5g1vjhroe
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^{69} k^3       | r = \paren {\dfrac {69 \paren {69 + 1} } 2}^2       | c = Sum of Sequence of Cubes }} {{eqn | r = 5 \, 832 \, 225       | c =  }} {{eqn | l = \sum_{k \mathop = 1}^5 k^3       | r = \paren {\dfrac {5 \paren {5 + 1} } 2}^2       | c = Sum of Sequence of Cubes }} {{eqn | r = 225       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 6}^{69} k^3       | r = \sum_{k \mathop = 1}^{69} k^3 - \sum_{k \mathop = 1}^5 k^3       | c =  }} {{eqn | r = 5 \, 832 \, 225 - 225       | c =  }} {{eqn | r = 5 \, 832 \, 000       | c =  }} {{eqn | r = 180^3       | c =  }} {{end-eqn}} {{qed}}
The sequence obtained by iterating $r$ on $187$ is: :$187, 968, 1837, 9218, 17347, 91718, 173437, 907808, 1716517, 8872688,$ :$17735476, 85189247, 159487405, 664272356, 1317544822, 3602001953, 7193004016, 13297007933,$ :$47267087164, 93445163438, 176881317877, 955594506548, 170120002107, 8713200023178$ {{OEIS|A033670}} Note that $r \left({187}\right) = r \left({781}\right) = 968$, so the sequence obtained by iterating $r$ on $781$ is the same. The rest of the numbers converge on the same sequence immediately: :$286, 968, \ldots$ :$385, 968, \ldots$ :$880, 88, \ldots$ :$869, 1837, \ldots$ as do their reversals. {{qed}}
From Numbers not Sum of Distinct Squares, the following positive integers cannot be expressed as the sum of distinct squares at all: :$2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128$ {{OEIS|A001422}} {{ProofWanted|It remains to be shown that all other positive integers less than $188$ can be expressed as the sum of no more than 5 distinct squares, and all numbers over $188$ can also be so expressed.}}
{{begin-eqn}} {{eqn | l = Q_n       | r = \sum_{k \mathop = 1}^n H_n       | c = {{Defof|Heptagonal Pyramidal Number}} }} {{eqn | r = \sum_{k \mathop = 1}^n \dfrac {k \paren {5 k - 3} } 2       | c = Closed Form for Heptagonal Numbers }} {{eqn | r = \dfrac 1 2 \paren {5 \sum_{k \mathop = 1}^n k^2 - 3 \sum_{k \mathop = 1}^n k}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {5 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \sum_{k \mathop = 1}^n k}       | c = Sum of Sequence of Squares }} {{eqn | r = \dfrac 1 2 \paren {5 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \dfrac {n \paren {n + 1} } 2}       | c = Closed Form for Triangular Numbers }} {{eqn | r = \dfrac 1 2 \paren {\frac {5 n \paren {n + 1} \paren {2 n + 1} - 9 n \paren {n + 1} } 6}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {5 \paren {2 n + 1} - 9} } 6}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {10 n + 5 - 9} } 6}       | c =  }} {{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {10 n - 4} } 6}       | c =  }} {{eqn | r = \frac {n \paren {n + 1} \paren {5 n - 2} } 6       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \dfrac {0 \left({0 + 1}\right) \left({5 \times 0 - 2}\right)} 6       | o =        | c = Closed Form for Heptagonal Pyramidal Numbers }} {{eqn | r = \dfrac {0 \times 1 \times \left({- 2}\right)} 6       | c =  }} {{eqn | r = 0       | c =  }} {{eqn | r = 0^2       | c = {{Defof|Square Number}} }} {{end-eqn}} {{begin-eqn}} {{eqn | r = \dfrac {1 \left({1 + 1}\right) \left({5 \times 1 - 2}\right)} 6       | o =        | c = Closed Form for Heptagonal Pyramidal Numbers }} {{eqn | r = \dfrac {1 \times 2 \times 3} 6       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = 1^2       | c = {{Defof|Square Number}} }} {{end-eqn}} {{begin-eqn}} {{eqn | r = \dfrac {6 \left({6 + 1}\right) \left({5 \times 6 - 2}\right)} 6       | o =        | c = Closed Form for Heptagonal Pyramidal Numbers }} {{eqn | r = \dfrac {6 \times 7 \times 28} 6       | c =  }} {{eqn | r = 7 \times \left({2^2 \times 7}\right)       | c =  }} {{eqn | r = \left({2 \times 7}\right)^2       | c = {{Defof|Square Number}} }} {{eqn | r = 196       | c =  }} {{end-eqn}} {{qed}}
By definition of minimizing sequence: :$\ds \inf_y J \sqbrk y = \lim_{n \mathop \to \infty} J \sqbrk {y_n}$ Any mapping from this sequence either minimises the functional or not. This is true for the limit mapping as well: :$\ds J \sqbrk {\hat y} \ge \inf_y J \sqbrk y$ By assumption, $J$ is lower semicontinuous at $\hat y$: :$\ds \forall \epsilon_1 > 0: \forall n \in \N: \exists N \in \N: \paren {n > N} \implies \paren {J \sqbrk {\hat y} - J \sqbrk {y_n} < \epsilon_1}$ Hence, for sufficiently large $n$: :$\ds \inf_y J \sqbrk y \le J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon_1$ By the definition of minimizing sequence, where the label $n$ has been replaced with $m$: :$\ds \forall \epsilon_2: \forall m \in \N: \exists M \in \N: \paren {m > M} \implies \paren {\size {J \sqbrk {y_m} - \inf_y J \sqbrk y} < \epsilon_2}$ From this, for sufficiently large $m$: :$\ds J \sqbrk {y_m} - \epsilon_2 < \inf_y J \sqbrk y$   Then: :$J \sqbrk {y_m} - \epsilon_2 < J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon_1$ Here $\epsilon_1$, $n$ have similar properties like $\epsilon_2$, $m$, but are otherwise arbitrary and independent. Let $\epsilon_1 = \epsilon_2 = \epsilon$, $n = m$, $M = N$. Arbitrariness is still not affected.  Then: :$J \sqbrk {y_n} - \epsilon < J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon$ Subtract $J \sqbrk {y_n}$ from all the terms. Hence: :$-\epsilon < J \sqbrk {\hat y} - J \sqbrk {y_n} < \epsilon$ or: :$\size {J \sqbrk {y_n} - J \sqbrk {\hat y} } < \epsilon$ Therefore this relation inherits all the constraints on its values $n$, $N$, $\epsilon$, and by definition is a limit: :$\ds \lim_{n \mathop \to \infty} J \sqbrk {y_n} = J \sqbrk {\hat y}$ {{qed}} {{Proofread|may contain mistakes, since original was not satisfactory enough}}