hcju/proofwikiqa huggingface dataset:
From Product of Generating Functions: :$\map G z = \paren {\map {G_1} z}^2$ where $\map {G_1} z$ is the generating function for $\displaystyle \sum_{k \mathop \ge 1} \dfrac 1 k$. From Generating Function for Sequence of Reciprocals of Natural Numbers: :$\map {G_1} z = \map \ln {\dfrac 1 {1 - z} }$ Hence: :$\map G z = \paren {\ln \dfrac 1 {1 - z} }^2$ Differentiating $\map G z$ {{WRT|Differentiation}} $z$ gives: {{begin-eqn}} {{eqn | l = \map {G'} z       | r = \dfrac \d {\d z} \paren {\paren {\ln \dfrac 1 {1 - z} }^2}       | c =  }} {{eqn | r = 2 \ln \dfrac 1 {1 - z} \dfrac \d {\d z} \paren {\ln \dfrac 1 {1 - z} }       | c = Chain Rule for Derivatives, Derivative of Power }} {{eqn | r = 2 \paren {\ln \dfrac 1 {1 - z} } \times \paren {\dfrac 1 {1 - z} }^{-1} \dfrac \d {\d z} \dfrac 1 {1 - z}       | c = Chain Rule for Derivatives, Derivative of Logarithm Function }} {{eqn | r = 2 \paren {\ln \dfrac 1 {1 - z} } \times \paren {\dfrac 1 {1 - z} }^{-1} \times \dfrac {-1} {\paren {1 - z}^2} \map {\dfrac \d {\d z} } {1 - z}       | c = Chain Rule for Derivatives, Derivative of Power }} {{eqn | r = 2 \paren {\ln \dfrac 1 {1 - z} } \times \paren {\dfrac 1 {1 - z} }^{-1} \times \dfrac {-1} {\paren {1 - z}^2} \times \paren {-1}       | c = Chain Rule for Derivatives, Derivative of Power }} {{eqn | r = 2 \paren {\dfrac 1 {1 - z} \ln \dfrac 1 {1 - z} }       | c = simplifying }} {{eqn | r = 2 \sum_{n \mathop \ge 0} H_n z^n       | c = Generating Function for Sequence of Harmonic Numbers }} {{end-eqn}} Integrating again {{WRT|Integration}} $z$ gives: {{begin-eqn}} {{eqn | l = \map G z       | r = \int_0^z \paren {2 \sum_{n \mathop \ge 0} H_n z^n}       | c =  }} {{eqn | r = 2 \sum_{n \mathop \ge 0} H_n \paren {\int_0^z z^n}       | c =  }} {{eqn | r = 2 \sum_{n \mathop \ge 0} H_n \dfrac {z^{n + 1} } {n + 1}       | c = Primitive of Power }} {{eqn | r = 2 \sum_{n \mathop \ge 1} H_{n - 1} \dfrac {z^n} n       | c = Translation of Index Variable of Summation }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = S_1       | o = :=       | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j       | c =  }} {{eqn | r = \sum_{j \mathop = a}^b \sum_{i \mathop = j}^b x_i x_j       | c = Summation of i from 1 to n of Summation of j from 1 to i }} {{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = i}^b x_i x_j       | c = Change of Index Variable of Summation }} {{eqn | o = =:       | r = S_2       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = 2 S_1       | r = S_1 + S_2       | c =  }} {{eqn | r = \sum_{i \mathop = a}^b \paren {\sum_{j \mathop = a}^i x_i x_j + \sum_{j \mathop = i}^b x_i x_j}       | c =  }} {{eqn | r = \sum_{i \mathop = a}^b \paren {\paren {\sum_{j \mathop = a}^b x_i x_j} + x_i x_i}       | c = Sum of Summations over Overlapping Domains: Example }} {{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^b x_i x_j + \sum_{i \mathop = a}^b x_i x_i       | c = Sum of Summations equals Summation of Sum }} {{eqn | r = \paren {\sum_{i \mathop = a}^b x_i} \paren {\sum_{j \mathop = a}^b x_j} + \paren {\sum_{i \mathop = a}^b {x_i}^2}       | c = Change of Index Variable of Summation }} {{end-eqn}} The result follows on multiplying by $\dfrac 1 2$. {{qed}}
We have that: :$\displaystyle \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}$ From Summation of Products of n Numbers taken m at a time with Repetitions: :$\displaystyle \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$ where: :$S_r = \displaystyle \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$. Setting $m = 2$: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}       | r = \sum_{\substack {k_1, k_2 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop = 2} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !}       | c =  }} {{eqn | r = \dfrac { {S_1}^2 {S_2}^0} {\left({1^2 \times 2!}\right) \left({2^0 \times 0!}\right)} + \dfrac { {S_1}^0 {S_2}^1} {\left({1^0 \times 0!}\right) \left({2^1 \times 1!}\right)}       | c = as $k_1 = 2, k_2 = 0$ and $k_1 = 0, k_2 = 1$ are the only $k_1$ and $k_2$ to fulfil the criteria }} {{eqn | r = \frac 1 2 \left({ {S_1}^2 + S_2}\right)       | c = simplifying }} {{end-eqn}} {{qed}}
We have that: :$\displaystyle \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}$ From Summation of Products of n Numbers taken m at a time with Repetitions: :$\displaystyle \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$ where: :$S_r = \displaystyle \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$. Setting $m = 3$: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}       | r = \sum_{\substack {k_1, k_2, k_3 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + 3 k_3 \mathop = 3} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !}       | c =  }} {{end-eqn}} We need to find all sets of $k_1, k_2, k_3 \in \Z_{\ge 0}$ such that: :$k_1 + 2 k_2 + 3 k_3 = 3$ Thus $\left({k_1, k_2, k_3}\right)$ can be: :$\left({3, 0, 0}\right)$ :$\left({1, 1, 0}\right)$ :$\left({0, 0, 1}\right)$ Hence: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}       | r = \dfrac { {S_1}^3 {S_2}^0 {S_3}^0} {\left({1^3 \times 3!}\right) \left({2^0 \times 0!}\right) \left({3^0 \times 0!}\right)} + \dfrac { {S_1}^1 {S_2}^1 {S_3}^0} {\left({1^1 \times 1!}\right) \left({2^1 \times 1!}\right) \left({3^0 \times 0!}\right)} + \dfrac { {S_1}^0 {S_2}^0 {S_3}^1} {\left({1^0 \times 0!}\right) \left({2^0 \times 0!}\right) \left({3^1 \times 1!}\right)}       | c =  }} {{eqn | r = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3       | c =  }} {{end-eqn}} {{qed}}
From Summation of Products of n Numbers taken m at a time with Repetitions: :$\displaystyle \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$ where: :$S_j = \displaystyle \sum_{k \mathop = a}^b {x_k}^j$ for $j \in \Z_{\ge 0}$. Setting $m = 4$: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4}       | r = \sum_{\substack {k_1, k_2, k_3, k_4 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + 3 k_3 \mathop + 4 k_4 \mathop = 4} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \dfrac { {S_4}^{k_4} } {4^{k_4} k_4 !}       | c =  }} {{end-eqn}} We need to find all sets of $k_1, k_2, k_3, k_4 \in \Z_{\ge 0}$ such that: :$k_1 + 2 k_2 + 3 k_3 + 4 k_4 = 4$ Thus $\tuple {k_1, k_2, k_3, k_4}$ can be: :$\tuple {4, 0, 0, 0}$ :$\tuple {2, 1, 0, 0}$ :$\tuple {0, 2, 0, 0}$ :$\tuple {1, 0, 1, 0}$ :$\tuple {0, 0, 0, 1}$ Hence: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}       | r = \dfrac { {S_1}^4 {S_2}^0 {S_3}^0 {S_4}^0} {\paren {1^4 \times 4!} \paren {2^0 \times 0!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^2 {S_2}^1 {S_3}^0 {S_4}^0} {\paren {1^2 \times 2!} \paren {2^1 \times 1!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^0 {S_2}^2 {S_3}^0 {S_4}^0} {\paren {1^0 \times 0!} \paren {2^2 \times 2!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^1 {S_2}^0 {S_3}^1 {S_4}^0} {\paren {1^1 \times 1!} \paren {2^0 \times 0!} \paren {3^1 \times 1!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^0 {S_2}^0 {S_3}^0 {S_4}^1} {\paren {1^0 \times 0!} \paren {2^0 \times 0!} \paren {3^0 \times 0!} \paren {4^1 \times 1!} }       | c =  }} {{eqn | r = \dfrac { {S_1}^4} {24} + \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 + \dfrac {S_4} 4       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \ln {\map G z}       | r = \sum_{k \mathop \ge 1} \dfrac {S_k z^k} k       | c = Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2 }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d z} } {\map \ln {\map G z} }       | r = \map {\dfrac \d {\d z} } {\sum_{k \mathop \ge 1} \dfrac {S_k z^k} k}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 {\map G z} \map {\dfrac \d {\d z} } {\map G z}       | r = \sum_{k \mathop \ge 1} \map {\dfrac \d {\d z} } {\dfrac {S_k z^k} k}       | c = Derivative of Logarithm Function, Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \dfrac 1 {\sum_{m \mathop \ge 0} h_m z^m} \sum_{m \mathop \ge 0} \paren {m + 1} h_{m + 1} z^m       | r = \sum_{k \mathop \ge 1} S_k z^{k - 1}       | c = Derivative of Generating Function, Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \sum_{m \mathop \ge 0} \paren {m + 1} h_{m + 1} z^m       | r = \sum_{k \mathop \ge 1} S_k z^{k - 1} \sum_{m \mathop \ge 0} h_m z^m       | c =  }} {{end-eqn}} {{finish|Follows from Product of Generating Functions}}
For each $k \in \set {a, a + 1, \ldots, b}$, the product of $x_k$ taken $m$ at a time is simply ${x_k}^m$. Thus for $n = 1$ we have: :$h_m = {x_k}^m$ Let the generating function for such a $\sequence {h_m}$ be $\map {G_k} z$. From Generating Function for Sequence of Powers of Constant: :$\map {G_k} z = \dfrac 1 {1 - x_k z}$ By Product of Summations, we have: :$\displaystyle \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \prod_{k \mathop = a}^b \sum_{j \mathop = 1}^m x_j$ Hence: {{begin-eqn}} {{eqn | l = \map G z       | r = \sum_{k \mathop \ge 0} h_k z^k       | c = {{Defof|Generating Function}} }} {{eqn | r = \prod_{k \mathop = a}^b \dfrac 1 {1 - x_k z}       | c = Product of Generating Functions: General Rule }} {{eqn | r = \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \dotsm \paren {1 - x_b z} }       | c =  }} {{end-eqn}} {{qed}} {{Proofread}}
{{begin-eqn}} {{eqn | l = G \left({z}\right)       | r = \dfrac 1 {\left({1 - x_a z}\right) \left({1 - x_{a + 1} z}\right) \cdots \left({1 - x_b z}\right)}       | c = Lemma 1 }} {{eqn | ll= \leadsto       | l = \ln \left({G \left({z}\right)}\right)       | r = \ln \dfrac 1 {1 - x_a z} + \ln \dfrac 1 {1 - x_{a + 1} z} + \cdots + \ln \dfrac 1 {1 - x_b z}       | c = Sum of Logarithms }} {{eqn | r = \left({\sum_{k \mathop \ge 1} \dfrac { {x_a}^k z^k} k}\right) + \left({\sum_{k \mathop \ge 1} \dfrac { {x_{a + 1} }^k z^k} k}\right) + \cdots + \left({\sum_{k \mathop \ge 1} \dfrac { {x_b}^k z^k} k}\right)       | c = Generating Function for Sequence of Reciprocals of Natural Numbers }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {z^k} k \left({ {x_a}^k + {x_{a + 1} }^k + \cdots + {x_b}^k}\right)       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {z^k} k \sum_{j \mathop = a}^b {x_j}^k       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {S_k z^k} k       | c = by definition of $S_k$ }} {{end-eqn}} {{qed}}
The summation for $\map G z$ is a finite sum $m = 0, 1, \ldots, n$, which settles convergence issues. Begin with Viete's Formulas: :$\displaystyle \prod_{k \mathop = a}^b \paren {x - x_k} = x^n + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } U \, x^m$ Change variables $x = -1 / z$: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^n \paren {-\frac 1 z  - x_k}       | r = \paren {-z}^{-n} + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } U \, \paren {-z}^{-m} }} {{eqn | ll= \leadsto       | l = \prod_{k \mathop = 1}^n \paren {1 + x_k z}       | r = z^n + \sum_{m \mathop = 0}^{n - 1} \map {e_{n - m} } U \, \paren z^{n - m}       | c = all signs cancel }} {{eqn | ll= \leadsto       | l = \prod_{k \mathop = 1}^n \paren {1 + x_k z}       | r = \sum_{m \mathop = 0}^n \map {e_m} U \, z^m }} {{end-eqn}} {{qed}}
Apply mathematical induction on $n$. Let $\map P n$ be the statement that  {{begin-eqn}} {{eqn | l = \map G z       | o = \equiv        | r = \displaystyle \sum_{m \mathop = 0}^{n+1} \map {e_m} { \set {x_1,\ldots,x_n} } z^{m} }} {{eqn | r = \displaystyle \prod_{k \mathop = 1}^n \paren {1+x_kz} }} {{end-eqn}} Basis for the induction: Set $U = \set {x_1}$ for $n=1$. Expand the formal series: {{begin-eqn}} {{eqn | l = \map G z        | r = \map {e_0} U + \map {e_1} {U} z + \displaystyle\sum_{m \mathop = 2}^\infty \map {e_m} {U} z^m }} {{eqn | ll = \leadsto       | r = \map {e_0} U + \map {e_1} {U} z        | c = The summation has all zero terms. }} {{eqn | ll = \leadsto       | r = 1 + x_1 z }} {{end-eqn}} Then $\map P 1$ holds. Induction step: Assume $\map P n$ holds. Let's prove $\map P {n+1}$ holds. The induction step uses a recursion relation: {{begin-eqn}} {{eqn | n = 1       | l = e_m \paren { \set {x_1,\ldots,x_n,x_{n+1} } }        | r = x_{n+1}  e_{m-1} \paren { \set {x_1,\ldots,x_n} } + e_m \paren { \set {x_1,\ldots,x_n} }       | c = Elementary Symmetric Function/Examples/Recursion }} {{end-eqn}} Let $\map G z$ be defined by statement $\map P n$. Let $\map {G^*} z$ be defined by statement $\map P {n+1}$. Then: {{begin-eqn}} {{eqn | l = \map {G^*} z        | r = \displaystyle \prod_{k \mathop = 1}^{n+1} \paren {1+x_kz} }} {{eqn | ll = \leadsto       | r = \map G z \paren { 1 + x_{n+1} z } }} {{eqn | ll = \leadsto       | r = \map G z + x_{n+1} z \map G z }} {{eqn | ll = \leadsto       | r = \sum_{m \mathop = 0}^n \map {e_m} {U} z^m +  \sum_{m \mathop = 1}^{n+1} x_{n+1} \map {e_{m-1} } {U} z^{m}       | c = use hypothesis $\map P n$ }} {{eqn | ll = \leadsto       | r = \map {e_0} {U} + \sum_{m \mathop = 1}^{n+1} \paren { \map {e_m} {U} + x_{n+1} \map {e_{m-1} } {U} } z^{m}       | c = because $\map {e_{n+1} } {U} = 0$ }} {{eqn | ll = \leadsto       | r = \map {e_0} {U} + \sum_{m \mathop = 1}^{n+1} \map {e_m} {\set {x_1,\ldots,x_n,x_{n+1} } } z^{m}       | c = by recursion relation (1) }} {{eqn | ll = \leadsto       | r = \sum_{m \mathop = 0}^{n+1} \map {e_m} {\set {x_1,\ldots,x_n,x_{n+1} } } z^{m}       | c = because $\map {e_0} {X} = 1$ for all sets $X$ }} {{end-eqn}} Then $\map P {n+1}$ holds, completing the induction. {{qed}}
We have by definition of generating function that: :$G \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$ We have that: :$a_0 = 1$ {{explain|We need $a_0 {{=}} 1$ but I can't get my head around as to why at the moment}} Suppose $n=1$. Let $G \left({z}\right)$ be the generating function for $\left\langle{a_m}\right\rangle$ under this condition. Then: :$1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le 1$ can be fulfilled by only one set $\left\{ {j_1, j_2, \ldots, j_m}\right\}$, that is: :$j_1 = 1$ Thus in this case: {{begin-eqn}} {{eqn | l = a_m       | r = x_1 \delta_{m 1}       | c = where $\delta_{m 1}$ is the Kronecker delta. }} {{eqn | ll= \leadsto       | l = G \left({z}\right)       | r = \sum_{n \mathop \ge 0} x_1 \delta_{n 1} z^n       | c =  }} {{eqn | ll= \leadsto       | l = G \left({z}\right)       | r = 1 + x_1 z       | c =  }} {{end-eqn}} Then by Product of Generating Functions, it follows that: :$G \left({z}\right) = \left({1 + x_1 z}\right) \left({1 + x_{2} z}\right) \cdots \left({1 + x_n z}\right)$ {{finish|Explain exactly how. I'm having difficulty getting my head around this. Help would be appreciated here.}}
This is an instance of Generating Function for Elementary Symmetric Function. {{qed}} Category:Newton-Girard Formulas dfvh7pgmd8wzkiufbotpjv2dbef24qm
{{begin-eqn}} {{eqn | l = G \left({z}\right)       | r = \left({1 + x_a z}\right) \left({1 + x_{a + 1} z}\right) \cdots \left({1 + x_b z}\right)       | c = Lemma 1 }} {{eqn | ll= \leadsto       | l = \ln \left({G \left({z}\right)}\right)       | r = \ln \left({1 + x_a z}\right) + \ln \left({1 + x_{a + 1} z}\right) + \cdots + \ln \left({1 + x_b z}\right)       | c = Sum of Logarithms }} {{eqn | r = \left({\sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k + 1} {x_a}^k z^k} k}\right) + \left({\sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k + 1} {x_{a + 1} }^k z^k} k}\right) + \cdots + \left({\sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k + 1} {x_b}^k z^k} k}\right)       | c = Power Series Expansion for Logarithm of 1 + x }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k + 1} z^k} k \left({ {x_a}^k + {x_{a + 1} }^k + \cdots + {x_b}^k}\right)       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k + 1} z^k} k \sum_{j \mathop = a}^b {x_j}^k       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k + 1} S_k z^k} k       | c = Definition of $S_k$ }} {{end-eqn}} {{qed}} Category:Newton-Girard Formulas tclji054x3r8itwdjm7cvyjafx6m035
We have that: From Newton-Girard Formulas: :$\displaystyle \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\left({-S_2}\right)^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac {\left({\left({-1}\right)^{m + 1} S_m}\right)^{k_m} } {m^{k_m} k_m !}$ where: :$S_r = \displaystyle \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$. Setting $m = 2$, and setting $x_i := x_{j_1}, x_j := x_{j_2}$: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le i \mathop < j \mathop \le b} x_{j_1} x_{j_2}       | r = \sum_{\substack {k_1, k_2 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop = 2} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\left({-S_2}\right)^{k_2} } {2^{k_2} k_2 !}       | c =  }} {{eqn | r = \dfrac { {S_1}^2 \left({-S_2}\right)^0} {\left({1^2 \times 2!}\right) \left({2^0 \times 0!}\right)} + \dfrac { {S_1}^0 \left({-S_2}\right)^1} {\left({1^0 \times 0!}\right) \left({2^1 \times 1!}\right)}       | c = as $k_1 = 2, k_2 = 0$ and $k_1 = 0, k_2 = 1$ are the only $k_1$ and $k_2$ to fulfil the criteria }} {{eqn | r = \frac 1 2 \left({ {S_1}^2 - S_2}\right)       | c = simplifying }} {{end-eqn}} {{qed}}
From Newton-Girard Formulas: :$\displaystyle \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\left({-S_2}\right)^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \cdots \dfrac {\left({\left({-1}\right)^{m + 1} S_m}\right)^{k_m} } {m^{k_m} k_m !}$ where: :$S_r = \displaystyle \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$. Setting $m = 3$, and setting $x_i := x_{j_1}, x_j := x_{j_2}, x_k := x_{j_3}$: {{begin-eqn}} {{eqn | l = \sum_{a \mathop < j_1 \mathop < j_2 \mathop < j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}       | r = \sum_{\substack {k_1, k_2, k_3 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + 3 k_3 \mathop = 3} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\left({-S_2}\right)^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !}       | c =  }} {{end-eqn}} We need to find all sets of $k_1, k_2, k_3 \in \Z_{\ge 0}$ such that: :$k_1 + 2 k_2 + 3 k_3 = 3$ Thus $\left({k_1, k_2, k_3}\right)$ can be: :$\left({3, 0, 0}\right)$ :$\left({1, 1, 0}\right)$ :$\left({0, 0, 1}\right)$ Hence: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}       | r = \dfrac { {S_1}^3 \left({-S_2}\right)^0 {S_3}^0} {\left({1^3 \times 3!}\right) \left({2^0 \times 0!}\right) \left({3^0 \times 0!}\right)} + \dfrac { {S_1}^1 \left({-S_2}\right)^1 {S_3}^0} {\left({1^1 \times 1!}\right) \left({2^1 \times 1!}\right) \left({3^0 \times 0!}\right)} + \dfrac { {S_1}^0 \left({-S_2}\right)^0 {S_3}^1} {\left({1^0 \times 0!}\right) \left({2^0 \times 0!}\right) \left({3^1 \times 1!}\right)}       | c =  }} {{eqn | r = \dfrac { {S_1}^3} 6 - \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3       | c =  }} {{end-eqn}} {{qed}}
From Newton-Girard Formulas: :$\displaystyle \sum_{a \mathop \le j_1 \mathop < \mathop \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \cdots \dfrac {\paren {\paren {-1}^{m + 1} S_m}^{k_m} } {m^{k_m} k_m !}$ where: :$S_j = \displaystyle \sum_{k \mathop = a}^b {x_k}^j$ for $j \in \Z_{\ge 0}$. Setting $m = 4$: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop < j_2 \mathop < j_3 \mathop < j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4}       | r = \sum_{\substack {k_1, k_2, k_3, k_4 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + 3 k_3 \mathop + 4 k_4 \mathop = 4} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-S_2}^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \dfrac {\paren {-S_4}^{k_4} } {4^{k_4} k_4 !}       | c =  }} {{end-eqn}} We need to find all ordered quadruples of $k_1, k_2, k_3, k_4 \in \Z_{\ge 0}$ such that: :$k_1 + 2 k_2 + 3 k_3 + 4 k_4 = 4$ Thus $\tuple {k_1, k_2, k_3, k_4}$ can be: :$\tuple {4, 0, 0, 0}$ :$\tuple {2, 1, 0, 0}$ :$\tuple {0, 2, 0, 0}$ :$\tuple {1, 0, 1, 0}$ :$\tuple {0, 0, 0, 1}$ Hence: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}       | r = \dfrac { {S_1}^4 \paren {-S_2}^0 {S_3}^0 \paren {-S_4}^0} {\paren {1^4 \times 4!} \paren {2^0 \times 0!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^2 \paren {-S_2}^1 {S_3}^0 \paren {-S_4}^0} {\paren {1^2 \times 2!} \paren {2^1 \times 1!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^0 \paren {-S_2}^2 {S_3}^0 \paren {-S_4}^0} {\paren {1^0 \times 0!} \paren {2^2 \times 2!} \paren {3^0 \times 0!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^1 \paren {-S_2}^0 {S_3}^1 \paren {-S_4}^0} {\paren {1^1 \times 1!}\ \paren {2^0 \times 0!} \paren {3^1 \times 1!} \paren {4^0 \times 0!} }       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac { {S_1}^0 \paren {-S_2}^0 {S_3}^0 \paren {-S_4}^1} {\paren {1^0 \times 0!} \paren {2^0 \times 0!} \paren {3^0 \times 0!} \paren {4^1 \times 1!} }       | c =  }} {{eqn | r = \dfrac { {S_1}^4} {24} - \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 - \dfrac {S_4} 4       | c =  }} {{end-eqn}} {{qed}}