hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \sum_{\map R {\map \pi j} } a_{\map \pi j}       | r = \sum_{j \mathop \in \Z} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }       | c = {{Defof|Summation by Iverson's Convention}} }} {{eqn | r = \sum_{j \mathop \in \Z} \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j}       | c =  }} {{eqn | r = \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sum_{j \mathop \in \Z} \sqbrk {i = \map \pi j}       | c =  }} {{eqn | r = \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i}       | c =  }} {{eqn | r = \sum_{\map R i} a_i       | c =  }} {{eqn | r = \sum_{\map R j} a_j       | c = Change of Index Variable of Summation }} {{end-eqn}} {{qed}}
Let $S = \left\{ {i \in \Z: R \left({i}\right)}\right\}$. Let $T = \left\{ {j \in \Z: R \left({j}\right)}\right\}$. Let $i \in S$. Then $R \left({i}\right)$. Let $j = i$. By Leibniz's Law, $R \left({j}\right)$. Thus $i \in T$. By definition of subset, $S \subseteq T$. Similarly, let $j \in T$. Then $R \left({j}\right)$. Let $i = j$. By Leibniz's Law, $R \left({i}\right)$. Thus $j \in S$. By definition of subset, $T \subseteq S$. Thus by definition of set equality: $S = T$ Thus: {{begin-eqn}} {{eqn | l = \prod_{R \left({i}\right)} a_i       | r = \prod a_i \left[{R \left({i}\right)}\right]       | c = {{Defof|Iverson's Convention}} }} {{eqn | r = \prod a_i \times \chi_S       | c = {{Defof|Characteristic Function of Set}} }} {{eqn | r = \prod a_j \times \chi_T       | c = as $S = T$ }} {{eqn | r = \prod a_j \left[{R \left({j}\right)}\right]       | c = {{Defof|Iverson's Convention}} }} {{eqn | r = \prod_{R \left({j}\right)} a_j       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \prod_{\map R {\map \pi j} } a_{\map \pi j}       | r = \prod_{j \mathop \in \Z} {a_{\map \pi j} }^{\sqbrk {\map R {\map \pi j} } }       | c = {{Defof|Product by Iverson's Convention}} }} {{eqn | r = \prod_{j \mathop \in \Z} \prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} \sqbrk {i = \map \pi j} }       | c =  }} {{eqn | r = \paren {\prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} } } \uparrow \paren {\prod_{j \mathop \in \Z} \sqbrk {i = \map \pi j} }       | c = using Knuth uparrow notation }} {{eqn | r = \prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} }       | c =  }} {{eqn | r = \prod_{\map R i} a_i       | c =  }} {{eqn | r = \prod_{\map R j} a_j       | c = Change of Index Variable of Product }} {{end-eqn}} {{qed}} {{Proofread}}
{{begin-eqn}} {{eqn | l = \prod_{R \left({j}\right)} a_j \times \prod_{S \left({j}\right)} a_j       | r = \prod_{j \mathop \in \Z} a_j^{\left[{R \left({j}\right)}\right]} \times \prod_{j \mathop \in \Z} a_j^{\left[{S \left({j}\right)}\right]}       | c = {{Defof|Product by Iverson's Convention}} }} {{eqn | r = \prod_{j \mathop \in \Z} a_j^{\left({\left[{R \left({j}\right)}\right] + \left[{S \left({j}\right)}\right]}\right)}       | c = Exponent Combination Laws: Product of Powers }} {{end-eqn}} Let: :$A := \left\{ {j \in \Z: R \left({j}\right)}\right\}$ :$B := \left\{ {j \in \Z: S \left({j}\right)}\right\}$ The result then follows from Cardinality of Set Union.
This proof assumes the definition of $\exp$ as defined by an initial value problem. That is, suppose $\exp$ satisfies: :$(1): \quad D_x \exp x = \exp x$ :$(2): \quad \exp 0 = 1$ on $\R$. Consider the real function $f: \R \to \R$ defined by: :$\map f x := \dfrac {\map \exp {x + y} } {\map \exp y}$ From Exponential of Real Number is Strictly Positive, $f$ is well-defined. So: {{begin-eqn}} {{eqn | l = D_x \, \map f x       | r = D_x \frac {\map \exp {x + y} } {\map \exp y} }} {{eqn | r = \frac 1 {\map \exp y} D_x \, \map \exp {x + y}       | c = Derivative of Constant Multiple }} {{eqn | r = \frac {\map \exp {x + y} } {\map \exp y}       | c = Chain Rule for Derivatives }} {{eqn | r = \map f x }} {{end-eqn}} Thus $f$ satisfies $(1)$. Further: {{begin-eqn}} {{eqn | l = \map f 0       | r = \frac {\map \exp {0 + y} } {\map \exp y} }} {{eqn | r = \frac {\map \exp y} {\map \exp y} }} {{eqn | r = 1 }} {{end-eqn}} So $f$ satisfies $(2)$. From Exponential Function is Well-Defined: :$f = \exp$ That is: {{begin-eqn}} {{eqn | l = \frac {\map \exp {x + y} } {\map \exp y}       | r = \map \exp x }} {{eqn | ll= \leadsto       | l = \map \exp {x + y}       | r = \map \exp x \, \map \exp y }} {{end-eqn}} {{qed}}
This proof assumes the definition of $\exp$ as a series. Then: {{begin-eqn}} {{eqn | l = \map \exp {x + y}       | r = \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {x + y}^n }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac 1 {n!} \sum_{k \mathop = 0}^n \frac {n!} {k! \paren {n - k}!} x^k y^{n - k}       | c = Binomial Theorem }} {{eqn | r = \sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \paren {\frac 1 {k!} x^k} \paren {\frac 1 {\paren {n - k}!} y^{n - k} } }} {{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {y^n} {n!} }       | c = {{Defof|Cauchy Product}} }} {{eqn | r = \map \exp x \, \map \exp y }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^n j x^j       | r = x \sum_{1 \mathop \le j \mathop \le n}^n j x^{j - 1}       | c =  }} {{eqn | r = x \sum_{0 \mathop \le j \mathop \le n - 1} \paren {j + 1} x^j       | c =  }} {{eqn | r = \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} j x^j} + \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} x^j}       | c =  }} {{eqn | r = \paren {x \sum_{0 \mathop \le j \mathop \le n} j x^j} - n x^{n + 1} + \paren {\sum_{0 \mathop \le j \mathop \le n - 1} x^{j + 1} }       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x - 1} \sum_{0 \mathop \le j \mathop \le n} j x^j       | r = n x^{n + 1} - \paren {\sum_{1 \mathop \le j \mathop \le n} x^j}       | c =  }} {{eqn | r = n x^{n + 1} - \paren {\sum_{0 \mathop \le j \mathop \le n} x^j} + 1       | c =  }} {{eqn | r = n x^{n + 1} - \paren {\frac {x^{n - 1} - 1} {x - 1} } + 1       | c = Sum of Geometric Sequence }} {{eqn | ll= \leadsto       | l = \sum_{j \mathop = 0}^n j x^j       | r = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}       | c =  }} {{end-eqn}} {{qed}}
From Sum of Arithmetic-Geometric Sequence: :$\ds \sum_{j \mathop = 0}^n \paren {a + j d} x^j = \frac {a \paren {1 - x^{n + 1} } } {1 - x} + \frac {x d \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}$ Hence: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^n j x^j       | r = \frac {0 \paren {1 - x^{n + 1} } } {1 - x} + \frac {x \times 1 \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}       | c = putting $a = 0, d = 1$ }} {{eqn | r = \frac {x \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}       | c = initial simplification }} {{eqn | r = \frac {x - \paren {n + 1} x^{n + 1} + n x^{n + 2} } {\paren {x - 1}^2}       | c = further simplification }} {{end-eqn}} Hence the result. {{qed}}
Let $\sequence {\epsilon_n}$ be the sequence defined by: :$\forall n \in \N : b_n - a_n$ From Linear Combination of Convergent Series, $\displaystyle \sum_{n \mathop = 0}^\infty \epsilon_n$ is convergent with sum $\epsilon > 0$. Then: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty b_n - \sum_{n \mathop = 0}^\infty a_n       | r = \sum_{n \mathop = 0}^\infty \paren {a_n + \epsilon_n} - \sum_{n \mathop = 0}^\infty a_n }} {{eqn | r = \sum_{n \mathop = 0}^\infty \epsilon_n       | c = Linear Combination of Convergent Series }} {{eqn | r = \epsilon }} {{eqn | o = >        | r = 0 }} {{end-eqn}} Hence the result, by definition of inequality. {{qed}} Category: Series l15hent94uzzn1e41bci3z6k85rp6oh
{{begin-eqn}} {{eqn | l = \map \exp 0       | r = 1       | c = Exponential of Zero }} {{eqn | ll= \leadsto       | l = \map \exp {x - x}       | r = 1       | c = as $\forall x \in \R : x - x = 0$ }} {{eqn | ll= \leadsto       | l = \map \exp {-x} \, \map \exp x       | r = 1       | c = Exponential of Sum: Real Numbers }} {{eqn | ll= \leadsto       | l = \map \exp {-x}       | r = \dfrac 1 {\exp x}       | c = dividing both sides by $\exp x$, which is non-zero by Exponential of Real Number is Strictly Positive }} {{end-eqn}} {{qed}} Category:Exponential Function Category:Reciprocals 9qdlcogsx46aqmeilw6mncowfgb3vsk
{{begin-eqn}} {{eqn | l = \displaystyle \sum_{k \mathop = m}^n a_k       | r = \sum_{k \mathop = m}^n \left({b_k - b_{k - 1} }\right)       | c =  }} {{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m}^n b_{k - 1}       | c =  }} {{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m - 1}^{n - 1} b_k       | c = Translation of Index Variable of Summation }} {{eqn | r = \left({\sum_{k \mathop = m}^{n - 1} b_k + b_n}\right) - \left({b_{m - 1} + \sum_{k \mathop = m}^{n - 1} b_k}\right)       | c =  }} {{eqn | r = b_n - b_{m - 1}       | c =  }} {{end-eqn}} {{Qed}}
A specific instance of the general result: {{:Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result}} where $b = 10$. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j}b^j + n + 1       | c =  }} {{eqn | r = n \paren {b - 1} \sum_{j \mathop = 0}^n b^j - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1       | c =  }} {{eqn | r = n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1       | c = Sum of Geometric Sequence }} {{eqn | r = n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1} \frac {n b^{n + 2} - \paren {n + 1} b^{n + 1} + b} {\paren {b - 1}^2} + n + 1       | c = Sum of Sequence of Power by Index }} {{eqn | r = \frac {n \paren {b - 1} \paren {b^{n + 1} - 1} - \paren {n b^{n + 2} - \paren {n + 1} b^{n + 1} + b} + \paren {b - 1} n + \paren {b - 1} } {b - 1}       | c = elementary simplification }} {{eqn | r = \frac {n b^{n + 2} - n b^{n - 1} - n b + n - n b^{n + 2} + n b^{n - 1} + b^{n + 1} - b + n b - n + b - 1} {b - 1}       | c = multiplying out }} {{eqn | r = \frac {b^{n + 1} - 1} {b - 1}       | c = simplification }} {{eqn | r = \sum_{j \mathop = 0}^n b^j       | c = Sum of Geometric Sequence }} {{end-eqn}} {{qed}}
This proof assumes the definition of $\exp$ as the solution to an initial value problem. That is, suppose $\exp$ satisfies: :$(1): \quad D_x \exp x = \exp x$ :$(2): \quad \map \exp 0 = 1$ on $\R$. {{AimForCont}} that $\exists \alpha \in \R: \exp \alpha = 0$. Suppose that $\alpha > 0$. Let $J = \closedint 0 \alpha$. From Exponential Function is Continuous, $\exp$ is continuous on $J$. From Max and Min of Function on Closed Real Interval: :$\exists K \in \R: \forall x \in J: \size {\exp x} < K$ Then, $\forall n \in \N : \exists c_n \in J$ such that: {{begin-eqn}} {{eqn | l = 1       | r = \exp 0       | c = by $(2)$ }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} \frac {\map {\exp^j} \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n       | c = Taylor's Theorem for Univariate Functions }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} \frac {\map \exp \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n       | c = by $(1)$ }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} \frac 0 {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n       | c = from our assumption aiming at a contradiction }} {{eqn | r = \frac {\exp c_n} {n!} \paren {-\alpha}^n }} {{end-eqn}} So: :$\forall n \in \N: 1 \le K \dfrac {\alpha^n} {n!}$ That is, dividing both sides by $K$: :$\forall n \in \N: \dfrac 1 K \le \dfrac {\alpha^n} {n!}$ But from Power over Factorial, $\dfrac {\alpha^n} {n!} \to 0$. This contradicts our assumption. The same argument, mutatis mutandis proves the result for $\alpha < 0$. By hypothesis $(2)$: {{begin-eqn}} {{eqn | lll=\alpha = 0       | ll= \implies       | l = \exp \alpha        | r = 1 }} {{eqn | o = \ne       | r = 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Exponential of Real Number is Strictly Positive bwx8xkcv36pi3vwjjp1lnqhh642foz1
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$. Then: {{begin-eqn}} {{eqn | l = \prod_{R \paren i} \paren {b_i c_i}       | r = \prod_{R \paren i} \paren {a_{i 1} a_{i 2} }       | c = by definition }} {{eqn | r = \prod_{R \paren i} \paren {\prod_{1 \mathop \le j \mathop \le 2} a_{i j} }       | c = {{Defof|Product by Propositional Function}} }} {{eqn | r = \prod_{1 \mathop \le j \mathop \le 2} \paren {\prod_{R \paren i} a_{i j} }       | c = Exchange of Order of Product }} {{eqn | r = \paren {\prod_{R \paren i} a_{i 1} } \paren {\prod_{R \paren i} a_{i 2} }       | c = {{Defof|Product by Propositional Function}} }} {{eqn | r = \paren {\prod_{R \paren i} b_i} \paren {\prod_{R \paren i} c_i}       | c = by definition }} {{end-eqn}} {{qed}}
The proof proceeds by induction. First let: :$S := \left\{ {a_i: R \left({i}\right)}\right\}$ We have that $S$ is finite. Hence the contents of $S$ can be well-ordered, by Finite Totally Ordered Set is Well-Ordered. Let $S$ have $m$ elements, identified as: :$S = \left\{ {s_1, s_2, \ldots, s_m}\right\}$ For all $n \in \Z_{\ge 0}$ such that $n \le m$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \log_b \left({\prod_{i \mathop = 1}^n s_i}\right) = \sum_{i \mathop = 1}^n \log_b s_i$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \log_b \left({\prod_{i \mathop = 1}^0 s_i}\right)       | r = \log_b 1       | c = {{Defof|Vacuous Product}} }} {{eqn | r = 0       | c = Logarithm of 1 is 0 }} {{eqn | r = \sum_{i \mathop = 1}^n \log_b s_i       | c = {{Defof|Vacuous Summation}} }} {{end-eqn}} === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \log_b \left({\prod_{i \mathop = 1}^1 s_i}\right)       | r = \log_b s_1       | c = {{Defof|Product Notation (Algebra)|Product}} }} {{eqn | r = \sum_{i \mathop = 1}^1 \log_b s_i       | c = {{Defof|Summation}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \log_b \left({\prod_{i \mathop = 1}^k s_i}\right) = \sum_{i \mathop = 1}^k \log_b s_i$ from which it is to be shown that: :$\displaystyle \log_b \left({\prod_{i \mathop = 1}^{k + 1} s_i}\right) = \sum_{i \mathop = 1}^{k + 1} \log_b s_i$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \log_b \left({\prod_{i \mathop = 1}^{k + 1} s_i}\right)       | r = \log_b \left({\left({\prod_{i \mathop = 1}^k s_i}\right) \times s_{k + 1} }\right)       | c = {{Defof|Product Notation (Algebra)|Product}} }} {{eqn | r = \log_b \left({\prod_{i \mathop = 1}^k s_i}\right) + \log_b s_{k + 1}       | c = Sum of General Logarithms }} {{eqn | r = \sum_{i \mathop = 1}^k \log_b s_i + \log_b s_{k + 1}       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = 1}^{k + 1} \log_b s_i       | c = {{Defof|Summation}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \log_b \left({\prod_{i \mathop = 1}^n s_i}\right) = \sum_{i \mathop = 1}^n \log_b s_i$ Hence, by definition of $S$: :$\displaystyle \log_b \left({\prod_{R \left({i}\right)} a_i}\right) = \sum_{R \left({i}\right)} \log_b a_i$ {{qed}}
By Bottom is Way Below Any Element: :$\bot \ll x$ By definition of way below closure: :$\bot \in x^\ll$ Thus by definition: :$x^\ll$ is a non-empty set. Let $y, z \in x^\ll$ By definition of way below closure: :$y \ll x$ and $z \ll x$ By Join is Way Below if Operands are Way Below :$y \vee z \ll x$ By definition of way below closure: :$y \vee z \in x^\ll$ By Join Succeeds Operands: :$y \preceq y \vee z$ and $z \preceq y \vee z$ Thus by definition :$x^\ll$ is directed. {{qed}}
Let $D$ be directed subset of $S$ such that :$D$ admits a supremum and $x \preceq \sup D$ Because $D$ is non-empty, therefore :$\exists a: a \in D$ By definition of smallest element: :$\bot \preceq a$ Thus by definition of way below relation; :$\bot \ll x$ {{qed}}
Let: {{begin-eqn}} {{eqn | l = P_1       | o = :=       | r = \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j }} {{eqn | l = P_2       | o = :=       | r = \prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = P_1 P_2       | r = \left({\prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j}\right) \left({\prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j}\right)       | c =  }} {{eqn | r = \prod_{i \mathop = 0}^n \left({\left({\prod_{j \mathop = 0}^i a_i a_j}\right) \left({\prod_{j \mathop = i}^n a_i a_j}\right)}\right)       | c =  }} {{eqn | r = \prod_{i \mathop = 0}^n \left({\left({\prod_{j \mathop = 0}^n a_j a_j}\right) {a_i}^2}\right)       | c = Product of Products over Overlapping Domains }} {{eqn | r = \prod_{i \mathop = 0}^n \left({ {a_i}^{n + 1} \left({\prod_{j \mathop = 0}^n a_j}\right) {a_i}^2}\right)       | c =  }} {{eqn | r = \left({\prod_{i \mathop = 0}^n {a_i}^{n + 1} }\right) \left({\prod_{j \mathop = 0}^n \left({\prod_{j \mathop = 0}^n a_j}\right) {a_i}^2}\right)       | c =  }} {{eqn | r = \left({\prod_{i \mathop = 0}^n {a_i}^{n + 1} }\right) \left({\prod_{j \mathop = 0}^n {a_i}^{n + 1} {a_i}^2}\right)       | c =  }} {{eqn | r = \prod_{i \mathop = 0}^n {a_i}^{2 n + 4}       | c =  }} {{eqn | ll= \leadsto       | l = P_1       | r = \prod_{i \mathop = 0}^n {a_i}^{n + 2}       | c =  }} {{end-eqn}} {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^0 \left({1 - a_j}\right)       | r = 1       | c = Definition of Vacuous Product }} {{eqn | o = \ge       | r = 1 - 0       | c =  }} {{eqn | r = 1 - \sum_{j \mathop = 1}^0 a^j       | c = Definition of Vacuous Summation }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^1 \left({1 - a_j}\right)       | r = 1 - a_j       | c =  }} {{eqn | r = 1 - \sum_{j \mathop = 1}^1 a^j       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \prod_{j \mathop = 1}^k \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^k a^j$ from which it is to be shown that: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^{k + 1} a^j$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^{k + 1} \left({1 - a_j}\right)       | r = \left({\prod_{j \mathop = 1}^k \left({1 - a_j}\right)}\right) \left({1 - a_{k + 1} }\right)       | c = Definition of Product }} {{eqn | o = \ge       | r = \left({1 - \sum_{j \mathop = 1}^k a^j}\right) \left({1 - a_{k + 1} }\right)       | c = Induction Hypothesis }} {{eqn | r = 1 - \sum_{j \mathop = 1}^k a^j - a_{k + 1} + a_{k + 1} \sum_{j \mathop = 1}^k a^j       | c =  }} {{eqn | r = 1 - \sum_{j \mathop = 1}^{k + 1} a^j + a_{k + 1} \sum_{j \mathop = 1}^k a^j       | c =  }} {{eqn | o = \ge       | r = 1 - \sum_{j \mathop = 1}^{k + 1} a^j       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$ {{qed}}
We have:  {{begin-eqn}} {{eqn	| l = \prod_{j \mathop = 2}^n \paren {1 - \frac 1 {j^2} } 	| r = \prod_{j \mathop = 2}^n \paren {\frac {\paren {j - 1} \paren {j + 1} } {j^2} } 	| c = Difference of Two Squares }} {{eqn	| r = \paren {\prod_{j \mathop = 2}^n \paren {j - 1} } \paren {\prod_{j \mathop = 2}^n \paren {j + 1} } \paren {\prod_{j \mathop = 2}^n \frac 1 j}^2 	| c = Product of Products }} {{eqn	| r = \paren {\prod_{j \mathop = 1}^{n - 1} j} \paren {\prod_{j \mathop = 3}^{n + 1} j} \frac 1 {\paren {\prod_{j \mathop = 2}^n j}^2} }} {{eqn	| r = \paren {n - 1}! \paren {\frac 1 2 \prod_{j \mathop = 2}^{n + 1} j} \frac 1 {\paren {n!}^2} 	| c = {{Defof|Factorial}} }} {{eqn	| r = \frac {\paren {n - 1}! \paren {n + 1}!} {2 \times n! \times n!} 	| c = {{Defof|Factorial}} }} {{eqn	| r = \frac {\paren {n - 1}! n! \paren {n + 1} } {2 \paren {n - 1}! \times n \times n!} 	| c = Gamma Difference Equation }} {{eqn	| r = \frac {n + 1} {2 n} }} {{end-eqn}} {{qed}}
Take the Binet-Cauchy Identity: :$\displaystyle \left({\sum_{i \mathop = 1}^n a_i c_i}\right) \left({\sum_{j \mathop = 1}^n b_j d_j}\right) = \left({\sum_{i \mathop = 1}^n a_i d_i}\right) \left({\sum_{j \mathop = 1}^n b_j c_j}\right) + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)$ Make the following assignments: {{begin-eqn}} {{eqn | ll= 1 \le i \le n:       | l = a_i       | o = :=       | r = u_i }} {{eqn | ll= 1 \le i \le n:       | l = c_i       | o = :=       | r = v_i }} {{eqn | ll= 1 \le j \le n:       | l = b_j       | o = :=       | r = 1 }} {{eqn | ll= 1 \le j \le n:       | l = d_j       | o = :=       | r = 1 }} {{end-eqn}} Then we have: :$\displaystyle \left({\sum_{i \mathop = 1}^n u_i v_i}\right) \left({\sum_{j \mathop = 1}^n 1 \times 1}\right) = \left({\sum_{i \mathop = 1}^n u_i \times 1}\right) \left({\sum_{j \mathop = 1}^n 1 \times v_j}\right) + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({u_i \times 1 - u_j \times 1}\right) \left({v_i \times 1 - v_j \times 1}\right)$ and the result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j}       | r = \sum_{i \mathop = 1}^m a_{i 1} \sum_{i \mathop = 1}^m a_{i 2} \cdots  \sum_{i \mathop = 1}^m a_{i n}       | c = Definition of Product }} {{eqn | r = \sum_{i_1 \mathop = 1}^m a_{i_1 1} \sum_{i_2 \mathop = 1}^m a_{i_2 2} \cdots  \sum_{i_n \mathop = 1}^m a_{i_n n}       | c =  }} {{eqn | r = \sum_{i_1 \mathop = 1}^m \sum_{i_2 \mathop = 1}^m \cdots  \sum_{i_n \mathop = 1}^m a_{i_1 1} a_{i_2 2} \cdots a_{i_n n}       | c = General Distributivity Theorem }} {{eqn | r = \sum_{1 \mathop \le i_1, \mathop \ldots \mathop , i_n \le m} a_{i_1 1} \cdots a_{i_n n}       | c =  }} {{end-eqn}} {{qed}}
By the Residue Theorem: :$\displaystyle \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} } \end{pmatrix} = \dfrac 1 {2 \pi i} \int \limits_{\left\lvert{z}\right\rvert \mathop = R} \dfrac {z^r \rd z} {\left({z - z_1}\right) \cdots \left({z - z_n}\right)}$ where $R > \left\lvert{z_1}\right\rvert, \ldots, \left\lvert{z_n}\right\rvert$. The Laurent series of the integrand converges uniformly on $\left\lvert{z}\right\rvert = R$: {{begin-eqn}} {{eqn | o =        | r = z^{r - n} \left({\dfrac 1 {1 - x_1 / z} }\right) \cdots \left({\dfrac 1 {1 - x_n / z} }\right)       | c =  }} {{eqn | r = z^{r - n} + \left({x_1 + \cdots x_n}\right) z^{r - n - 1} + \left({x_1^2 + x_1 x_2 + \cdots}\right) z^{r - n - 2} + \cdots       | c =  }} {{end-eqn}} On integrating term my term, everything vanishes except the coefficient of $z^{-1}$. Thus: :$\displaystyle \sum_{\substack {j_1 \mathop + \mathop \cdots \mathop + j_n \mathop = r \mathop - n \mathop + 1 \\ j_1, \mathop \ldots j_n \mathop \ge 0} } {x_1}^{j_1} {x_2}^{j_2} \cdots {x_n}^{j_n}$ {{qed}}
Instance of Summation of Powers over Product of Differences: :$\displaystyle \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases}0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ where: :$n = 3$ :$x_1 = a$ :$x_2 = b$ :$x_3 = c$ {{qed}}