hcju/proofwikiqa huggingface dataset:
As $p$ and $q$ are distinct prime numbers, it follows that $p$ and $q$ are coprime. Thus by Euler Phi Function is Multiplicative: :$\map \phi n = \map \phi p \, \map \phi q$ From Euler Phi Function of Prime: :$\map \phi p = p - 1$ :$\map \phi q = q - 1$ Hence the result. {{qed}}
A semiprime with distinct prime factors is a square-free integer. :$\map \phi n = \displaystyle \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$ where $p \divides n$ denotes the primes which divide $n$. As there are $2$ prime factors: $p$ and $q$, this devolves to: :$\map \phi n = \paren {p - 1} \paren {q - 1}$ except when $p = 2$, in which case: :$\map \phi n = q - 1$ But when $p = 2$, we have that $p - 1 = 1$ and so: :$\paren {p - 1} \paren {q - 1} = q - 1$ Hence the result. {{qed}}
We have that the Euler Phi Function is Multiplicative. Let the prime decomposition of $n$ be: :$\displaystyle n = \prod_{1 \mathop \le i \mathop \le r} p_i =  p_1 p_2 \cdots p_r$ From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$. From Euler Phi Function of Prime, we have: :$\map \phi {p_i} = \paren {p_i - 1}$ Thus: :$\map \phi n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} \paren {p_i - 1}$ or: :$\map \phi n = \displaystyle \prod_{p \mathop \divides n} \paren {p - 1}$ where $p$ ranges over all prime numbers When $p = 2$ we have that: :$p - 1 = 1$ and so: :$\displaystyle \prod_{p \mathop \divides n} \paren {p - 1} = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$ Hence the result. {{qed}}
From Euler Phi Function of Integer: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ As $n$ is square-free: :$\displaystyle n = \prod_{p \mathop \divides n} p$ Hence: :$\displaystyle \map \phi n = \prod_{p \mathop \divides n} p \paren {1 - \frac 1 p}$ and so: :$\displaystyle \map \phi n = \prod_{p \mathop \divides n} \paren {p - 1}$ When $p = 2$ we have that: :$p - 1 = 1$ and so: :$\displaystyle \prod_{p \mathop \divides n} \paren {p - 1} = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$ Hence the result. {{qed}}
By definition $n$ is a semiprime. As $p$ is an odd prime, $n$ is not square. Thus from Euler Phi Function of Non-Square Semiprime: :$\map \phi n = \paren {2 - 1} \paren {p - 1}$ Hence the result. {{qed}}
{{AimForCont}} $a$ is such a set. Then $a \in a$ and $a \in \set a$. $a \ne \O$ because the empty set has no elements by definition. It is also seen that: {{begin-eqn}} {{eqn | l = a \cap \set a       | r = \set {x: x \in a \land x \in \set a}       | c = {{Defof|Set Intersection}} }} {{eqn | r = \set {x: x \in a \land x = a}       | c = {{Defof|Singleton}} }} {{eqn | r = a }} {{end-eqn}} By the Axiom of Foundation: :$a \cap \set a = \O$ Thus $a = \O$. But it was previously established that $a \ne \O$. From this contradiction it follows that there cannot exist such a set. Hence the result. {{qed}} {{AoF}} Category:Axiom of Foundation sn2kwjz7ehyjfm6xgawfy59j4ksudqp
A tetrahedron has: :$6$ edges and: :$4$ faces. Each face is a triangle, and so has $3$ edges. Hence: :$6 \times 4 \times 3 = 72$ {{qed}}
A cube has: :$12$ edges and: :$6$ faces. Each face is a square, and so has $4$ edges. Hence: :$12 \times 6 \times 4 = 288$ {{qed}}
A regular octahedron has: :$12$ edges and: :$8$ faces. Each face is a triangle, and so has $3$ edges. Hence: :$12 \times 8 \times 3 = 288$ {{qed}}
Let $A$ be a set of ordinals. Let $\Epsilon {\restriction_A}$ denote the epsilon restriction on $A$. It is to be shown that $\Epsilon {\restriction_A}$ is antireflexive. Suppose there is a $a \in A$ such that $a \in a$. Then by ordinal is element of ordinal iff proper subset of ordinal, we have $a \subsetneq a$. And so: :$\exists x \in a: x \notin a$ But this is a contradiction, so $\Epsilon {\restriction_A}$ must be antireflexive. By Strict Subset Relation is Transitive, it follows that $\Epsilon {\restriction_A}$ is transitive. By Ordinals are Well-Ordered, it follows that for any non-empty subset $B \subseteq A$ there is a smallest element $x \in B$ such that: :$\forall y \in B: x \in y \lor x = y$ It has been shown that $\Epsilon {\restriction_A}$ is a strict ordering such that every non-empty subset of $A$ has a smallest element. Hence the result. {{qed}} Category:Ordinals Category:Ordinals are Well-Ordered 806muh777tqq80xmyb86iuar3fnux8f
A regular dodecahedron has: :$30$ edges and: :$12$ faces. Each face is a pentagon, and so has $5$ edges. Hence: :$30 \times 12 \times 5 = 1800$ {{qed}}
A regular icosahedron has: :$30$ edges and: :$20$ faces. Each face is a triangle, and so has $3$ edges. Hence: :$30 \times 20 \times 3 = 1800$ {{qed}}
From the sequence of nontotients: {{:Definition:Nontotient/Sequence}} Hence, by inspection, it can be seen that $74$ and $76$ are the smallest such pair. {{qed}}
We have $1^4 = 1, 2^4 = 16, 3^4 = 81 > 79$. Hence for each $n < 79$, we can only use $1^4$ and $2^4$ in our sum. Write $n = 2^4 a + 1^4 b$. We can use the greedy algorithm to generate these expressions, since replacing $2^4$ with $16 \times 1^4$ increases the number of fourth powers required. Suppose $n < 64$. By Division Theorem, there is a unique way to write $n = 16 q + r$, with $q \in \Z$, $0 \le r < 16$. {{begin-eqn}} {{eqn | l = 16 q + r       | r = n }} {{eqn | ll = \leadsto       | l = 16 q + r       | o = <       | r = 64 }} {{eqn | ll = \leadsto       | l = 16 q       | o = <       | r = 64 }} {{eqn | ll = \leadsto       | l = q       | o = <       | r = 4 }} {{end-eqn}} Thus $q + r \le 3 + 15 = 18$. It follows that each positive integer less than $64$ can be expressed in not more than $18$ fourth powers. Suppose $64 \le n \le 78$. We cannot use more than $4$ $2^4$, since $5 \times 2^4 = 80 > n$. Thus we write $n = 4 \times 2^4 + \paren {n - 64} \times 1^4$. Since $n - 64 \le 78 - 64 = 14$, we can use not more than $18$ fourth powers to express $n$. For $n = 79$, we still cannot use more than $4$ $2^4$, since $5 \times 2^4 = 80 > n$. Thus $n = 4 \times 2^4 + 15 \times 1^4$ uses the least number of fourth powers. Hence $79$ is the smallest positive integer which cannot be expressed as the sum of fewer than $19$ fourth powers. {{qed}}
Performing the calculation using long division: <pre>    0.0123456790...   ---------------- 81)1.0000000000000      81      --      190      162      ---       280       243       ---        370        324        ---         460         405         ---          550          486          ---           640           567            --            730            729            ---              100               81              ---              ... </pre> {{qed}}
{{begin-eqn}} {{eqn | l = n^2       | r = \left({b - 1}\right)^2       | c =  }} {{eqn | r = b^2 - 2 b + 1       | c =  }} {{eqn | r = b \left({b - 2}\right) + 1       | c =  }} {{end-eqn}} The result follows by definition of number base. {{qed}}
{{begin-eqn}} {{eqn | l = \sqrt 0       | r = 0       | c =  }} {{eqn | l = \sqrt 1       | r = 1       | c =  }} {{eqn | l = \sqrt {81}       | r = 9       | c =  }} {{eqn | r = 8 + 1       | c =  }} {{end-eqn}} By considering the square roots, we are looking for positive integers with a square for which its sum of digits is equal to the original number. It is easy to verify the result up to $36$. We prove that for $n > 36$, the sum of digits of $n^2$ cannot equal $n$. Let $n$ be a $d$-digit number. Then $n < 10^d$. Thus $n^2 < 10^{2 d}$. Therefore $n^2$ has at most $2 d$ digits. Hence the sum of digits of $n^2$ is at most $18 d$. If $n$ is a $2$-digit number, we have $n > 36 = 18 d$. Now suppose $d \ge 3$. Then: {{begin-eqn}} {{eqn | l = n       | o = \ge       | r = 10^{d - 1} }} {{eqn | r = 10 \times 10^{d - 2} }} {{eqn | o = \ge       | r = 10 \paren {1 + 9 \paren {d - 2} }       | c = Bernoulli's Inequality }} {{eqn | r = 90 d - 170 }} {{eqn | o = \ge       | r = 18 d + 216 - 170       | c = as $d \ge 3$ }} {{eqn | o = >       | r = 18 d }} {{end-eqn}} Therefore for $n > 36$, the sum of digits of $n^2$ cannot equal $n$. {{qed}}
=== Definition 1 implies Definition 2 === Let $H_n$ be a heptagonal number by definition 1. Let $n = 0$. By definition: :$H_0 = 0$ By vacuous summation: :$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4} = 0$ By definition of summation: {{begin-eqn}} {{eqn | l = H_{n - 1}       | r = \sum_{i \mathop = 1}^{n - 1} \paren {5 i - 4}       | c =  }} {{eqn | r = 1 + 6 + \cdots + 5 \paren {n - 1} + 4       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = H_n       | r = H_{n - 1} + 5 n - 4       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {5 n - 4}       | c =  }} {{eqn | r = 1 + 6 + \cdots + 5 n - 4       | c =  }} {{end-eqn}} Thus $H_n$ is a heptagonal number by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $H_n$ be a heptagonal number by definition 2. Then: {{begin-eqn}} {{eqn | l = H_n       | r = \sum_{i \mathop = 1}^n \paren {5 i - 4}       | c =  }} {{eqn | r = 1 + 6 + \cdots + \paren {5 \paren {n - 1} - 4} + \paren {5 n - 4}       | c =  }} {{eqn | r = H_{n - 1} + 5 n - 4       | c =  }} {{end-eqn}} Then: :$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4}$ is a vacuous summation and so: :$H_0 = 0$ Thus $H_n$ is a heptagonal number by definition 1. {{qed|lemma}} === Definition 1 equivalent to Definition 3 === We have by definition that $H_0 = 0 = \map P {7, 0}$. Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = \map P {7, n}       | r = \map P {7, n - 1} + \paren {7 - 2} \paren {n - 1} + 1       | c =  }} {{eqn | r = \map P {7, n - 1} + 5 \paren {n - 1} + 1       | c =  }} {{eqn | r = \map P {7, n - 1} + 5 n - 4       | c =  }} {{end-eqn}} Thus $\map P {7, n}$ and $H_n$ are generated by the same recurrence relation. {{qed}} Category:Heptagonal Numbers j1nufkda0hxuuykaaoridcuh4s2ftl6
Heptagonal numbers are $k$-gonal numbers where $k = 7$. From Closed Form for Polygonal Numbers we have that: :$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$ Hence: {{begin-eqn}} {{eqn | l = H_n       | r = \frac n 2 \paren {\paren {7 - 2} n - 7 + 4}       | c = Closed Form for Polygonal Numbers }} {{eqn | r = \dfrac {n \paren {5 n - 3} } 2       | c =  }} {{end-eqn}} Hence the result. {{qed}}
By definition, a Cunningham chain of the first kind is a sequence of prime numbers $\tuple {p_1, p_2, \ldots, p_n}$ such that: : $p_{k + 1} = 2 p_k + 1$ : $\dfrac {p_1 - 1} 2$ is not prime : $2 p_n + 1$ is not prime. Thus each term except the last is a Sophie Germain prime. {{:Definition:Sophie Germain Prime/Sequence}} Let $P: \Z \to \Z$ be the mapping defined as: :$\map P n = 2 n + 1$ Applying $P$ iteratively to each of the smallest Sophie Germain primes in turn: {{begin-eqn}} {{eqn | l = \map P 2       | r = 5       | c =  }} {{eqn | l = \map P 5       | r = 11       | c =  }} {{eqn | l = \map P {11}       | r = 23       | c =  }} {{eqn | l = \map P {23}       | r = 47       | c =  }} {{eqn | l = \map P {47}       | r = 95       | c = which is not prime }} {{end-eqn}} Thus $\tuple {2, 5, 11, 23, 47}$ is a Cunningham chain of the first kind of length $5$. {{begin-eqn}} {{eqn | l = \map P 3       | r = 7       | c =  }} {{eqn | l = \map P 7       | r = 15       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map P {29}       | r = 59       | c =  }} {{eqn | l = \map P {59}       | r = 119       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map P {41}       | r = 83       | c =  }} {{eqn | l = \map P {83}       | r = 167       | c =  }} {{eqn | l = \map P {167}       | r = 335       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map P {53}       | r = 107       | c =  }} {{eqn | l = \map P {107}       | r = 215       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map P {83}       | r = 167       | c =  }} {{eqn | l = \map P {167}       | r = 335       | c = which is not prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map P {89}       | r = 179       | c =  }} {{eqn | l = \map P {179}       | r = 359       | c =  }} {{eqn | l = \map P {359}       | r = 719       | c =  }} {{eqn | l = \map P {719}       | r = 1439       | c =  }} {{eqn | l = \map P {1439}       | r = 2879       | c =  }} {{eqn | l = \map P {2879}       | r = 5759       | c = which is not prime. }} {{end-eqn}} It is noted that $\dfrac {89 - 1} 2 = 44$ which is not prime. Hence the sequence of $6$: :$\tuple {89, 179, 359, 719, 1439, 2879}$ {{qed}}
=== Necessary condition === Take the partial derivative of $(3)$ {{WRT}} $x$: :$(4): \quad \frac {\partial^2 \map g {x, \mathbf y} } {\partial \mathbf y \partial x} = -\dfrac \partial {\partial \mathbf y} \map H {x, \mathbf y, \dfrac {\partial g} {\partial \mathbf y} }$ By the Schwarz-Clairaut Theorem, the order of partial derivatives of $g$ can be exchanged: :$(5): \quad \dfrac {\partial^2 \map g {x, \mathbf y} } {\partial \mathbf y \partial x} = \dfrac {\partial \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x}$ Since $\map g {x, \mathbf y}$ depends on $\mathbf y$ only in a direct way, it follows that: :$\dfrac {\partial g} {\partial \mathbf y} = g_{\mathbf y}$ Hence, from $(2)$, $(4)$ and $(5)$ it follows that: :$\dfrac {\partial \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x} = -\dfrac \partial {\partial \mathbf y} \map H {x, \mathbf y, \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } }$ These are the consistency equations. {{qed|lemma}} === Sufficient condition === {{ProofWanted}}
The sequence obtained by iterating $r$ on $89$ is: :$89, 187, 968, 1837, 9218, 17347, 91718, 173437, 907808, 1716517, 8872688,$ :$17735476, 85189247, 159487405, 664272356, 1317544822, 3602001953, 7193004016, 13297007933,$ :$47267087164, 93445163438, 176881317877, 955594506548, 170120002107, 8713200023178$ {{OEIS|A033670}} Note that $\map r {89} = \map r {98} = 187$, so the sequence obtained by iterating $r$ on $98$ is the same. To show that no other $2$-digit integers have longer sequences, we define an equivalence relation on the $2$-digit integers as follows: :$\sqbrk {a_1 a_2} \sim \sqbrk {b_1 b_2} \iff a_1 + a_2 = b_1 + b_2$ For each $2$-digit integer $\sqbrk {a_1 a_2}$: :$\map r {\sqbrk {a_1 a_2} } = \sqbrk {a_1 a_2} + \sqbrk {a_2 a_1} = 11 \paren {a_1 + a_2}$ Notice that if $\sqbrk {a_1 a_2} \sim \sqbrk {b_1 b_2}$: :$\map r {\sqbrk {a_1 a_2} } = 11 \paren {a_1 + a_2} = 11 \paren {b_1 + b_2} = \map r {\sqbrk {b_1 b_2} }$ So the value of $r$ is equal in each equivalence class. For each $2$-digit integer $\sqbrk {a_1 a_2}$, we have: :$18 = 9 + 9 \ge a_1 + a_2 \ge 1 + 0 = 1$ For $2$-digit integers where $a_1 + a_2 \le 9$: :$\map r {\sqbrk {a_1 a_2} } = 11 \paren {a_1 + a_2}$ is a $2$-digit multiple of $11$, which is a palindrome. So we consider each equivalence class of $2$-digit integers with $a_1 + a_2 \ge 10$. We have the sequences: :$19, 110, 121$ :$29, 121$ :$39, 132, 363$ :$49, 143, 484$ :$59, 154, 605, 1111$ :$69, 165, 726, 1353, 4884$ :$79, 176, 847, 1595, 7546, 14003, 44044$ :$89, \cdots, 8713200023178$, with $24$ iterations :$99$ So: :every $2$-digit integer not in the equivalence class of $89$ has a shorter iteration and: :$\eqclass {89} \sim = \set {89, 98}$ {{qed}}
Performing the calculation using long division: <pre>    0.0112359550561797752808988764044943820224719101...   ------------------------------------------------------ 89)1.0000000000000000000000000000000000000000000000000      89       534       712       801        89      --       ---       ---       ---       ---        110       160       880       390       810       89        89       801       356       801      ---       ---       ---       ---       ---       210       710       790       340        90       178       623       712       267        89       ---       ---       ---       ---        --        320       870       780       730        100        267       801       712       712         89        ---       ---       ---       ---        ---          530       690       680       180       ...         445       623       623       178         ---       ---       ---       ---          850       670       570        200          801       623       534        178          ---       ---       ---        ---           490       470       360        220           445       445       356        178            --       ---       ---        ---            450       250        400       420            445       178        356       356            ---       ---        ---       ---              500      720        440       640              445      712        356       623              ---      ---        ---       ---               550       800       840       170               534       712       801        89 </pre> {{qed}}
We have that: :$89 = 10^2 - 10 - 1$ So: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \dfrac {F_k} {10^{k + 1} }       | r = \dfrac 1 {10} \sum_{k \mathop \ge 0} F_k \paren {\dfrac 1 {10} }^k }} {{eqn | r = \dfrac 1 {10} \cdot \dfrac {\dfrac 1 {10} } {1 - \dfrac 1 {10} - \paren {\dfrac 1 {10} }^2 }       | c = Generating Function for Fibonacci Numbers }} {{eqn | r = \dfrac 1 {10^2 - 10 - 1}       | c =  }} {{eqn | r = \dfrac 1 {89}       | c =  }} {{end-eqn}} {{qed}}
By definition of a Fermat pseudoprime, we need to check for $a < 91$: :$a^{90} \equiv 1 \pmod {91}$ is satisfied or not. Note that: {{begin-eqn}} {{eqn | l = \paren {91 - a}^{90}       | r = {-a}^{90}       | c = Congruence of Powers }} {{eqn | r = a^{90} }} {{end-eqn}} So we essentially need to check $2 \le a \le 45$ only. Since $91 = 7 \times 13$, we can also exclude all multiples of $7$ and $13$. {{ProofWanted}}
Performing the calculation using long division: <pre>    0.01030927835051546391752577319587628865979381443298969072164948453608247422680412371134020618556701...   ------------------------------------------------------------------------------------------------------ 97)1.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000      97      485     873    679    582    679    291     679    776     388     388    291      485      --      ---     ---    ---    ---    ---    ---     ---    ---     ---     ---    ---      ---       300      500    170    310    280    910    290     210    440     720     120    390      650       291      485     97    291    194    873    194     194    388     679      97    388      582       ---      ---    ---    ---    ---    ---    ---     ---    ---     ---     ---    ---      ---         900     150    730    190    860    370    960     160    520     410     230     200     680         873      97    679     97    776    291    983      97    485     388     194     194     679         ---     ---    ---    ---    ---    ---    ---     ---    ---     ---     ---     ---     ---          270     530    510    930    840    790    870     630    350     220     360      600     100          194     485    485    873    776    776    776     582    291     194     291      582      97          ---     ---    ---    ---    ---    ---    ---     ---    ---     ---     ---      ---     ---           760     450    250    570    640    140    940     480    590     260     690      180    ...           679     388    194    485    582     97    873     388    582     194     679       97           ---     ---    ---    ---    ---    ---    ---     ---    ---     ---     ---      ---            810     620    560    850    580    430    670     920     800    660     110      830            776     582    485    776    485    388    582     873     776    582      97      776            ---     ---    ---    ---    ---    ---    ---     ---     ---    ---     ---      ---             340     380    750    740    950    420    880     470     240    780     130      540             291     291    679    679    873    388    873     388     194    776      97      485             ---     ---    ---    ---    ---    ---    ---     ---     ---    ---     ---      ---              490     890    710    610    770    320     700    820     460     400    330      550              485     873    679    582    679    291     679    776     388     388    291      485 </pre> {{qed}}
Suppose: :$\ds \map g {x, \mathbf y} = \int_c^{\paren {x, \mathbf y} } \map F {x, \hat {\mathbf y}, \hat {\mathbf y}'} \rd x$ where $\hat {\mathbf y}$ is an extremal of $J$ connecting points $c$ and $\tuple {x, \mathbf y}$. {{explain|Extremum or extremal? Can this be made consistent?}} Define a field of directions in $R$ by the following: :$\map {\mathbf p} {x, \mathbf y, \mathbf y'} = \map {g_{\mathbf y} } {x, \mathbf y}$ where $\mathbf p$ stands for momentum. Note, that it does not depend on the path defined by $\hat {\mathbf y}$, only on its endpoint at $\tuple {x, \mathbf y}$. By definition, $ \map g {x, \mathbf y}$ is a geodetic distance $S$. Hence, $g_\mathbf y$ does not explicitly depend on $\mathbf y'$. Then: :$\map {g_{\mathbf y} } {x, \mathbf y} = \map {\mathbf p} {x, \mathbf y, \mathbf z}$ where $\mathbf z = \map {\mathbf z} {x, \mathbf y}$ denotes slope of the curve joining $c$ and $\tuple {x, \mathbf y}$ at the point $\tuple {x, \mathbf y}$. Furthermore, since $g$ is a geodetic distance, it satisfies Hamilton-Jacobi equation: :$\ds \frac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {g_{\mathbf y} } {x, \mathbf y} } = 0$ which together with the previous relation results in a system of equations: :$\begin {cases} \dfrac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {\mathbf p} {x, \mathbf y, \mathbf z} } = 0 \\ \map {\mathbf z} {x, \mathbf y} = \map{\mathbf y'} x  \end {cases}$ Differentiation {{WRT|Differentiation}} $\mathbf y$ yields: :$\begin {cases} \dfrac {\partial \mathbf p} {\partial x} = -\dfrac {\partial H} {\partial \mathbf y} \\ \map {\mathbf z} {x,\mathbf y} = \map{\mathbf y'} x  \end {cases}$ These are Euler's equations in canonical coordinates with a constraint on derivative. {{explain|the last part is not exactly clear; where is the second Euler equation?}} Hence, the previously defined field of directions coincides with the field of functional. {{qed}}
Performing the calculation using long division: <pre>    0.01020408163265306122448979591836734693877551...   -------------------------------------------------- 98)1.00000000000000000000000000000000000000000000000      98       196     196    490    294    686      --       ---     ---    ---    ---    ---       200      640     440    900    460    540       196      588     392    882    392    490       ---      ---     ---    ---    ---    ---         400     520     480    180    680    500         392     490     392     98    588    490         ---     ---     ---    ---    ---    ---           800    300     880    820    920    100           784    294     784    784    882     98           ---    ---     ---    ---    ---    ---            160     600    960    360    380   ...             98     588    882    294    294            ---     ---    ---    ---    ---             620     120    780    660    860             588      98    686    588    784             ---     ---    ---    ---    ---              320     220    940    720    760              294     196    882    686    686              ---     ---    ---    ---    ---               260     240    580    340    740               196     196    490    294    686 </pre> We have that: :$0 \cdotp 01 + 0 \cdotp 0002 + 0 \cdotp 000004 + 0 \cdotp 00000008 + \cdots$ is nothing else but: :$\dfrac 1 2 \paren {\dfrac 1 {50} + \paren {\dfrac 1 {50} }^2 + \paren {\dfrac 1 {50} }^3 + \paren {\dfrac 1 {50} }^4 + \cdots} = \displaystyle \dfrac 1 2 \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k$ Hence: {{begin-eqn}} {{eqn | l = \dfrac 1 2 \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k       | r = \dfrac 1 2 \dfrac {1 / 50} {1 - 1 / 50}       | c = Sum of Infinite Geometric Sequence: Corollary 1 }} {{eqn | r = \dfrac 1 2 \dfrac 1 {50 - 1}       | c =  }} {{eqn | r = \dfrac 1 2 \dfrac 1 {49}       | c =  }} {{eqn | r = \dfrac 1 {98}       | c =  }} {{end-eqn}} {{qed}}