id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-14400 | Exchange of Order of Summation/Example | Let the fiber of truth of both $R$ and $S$ be infinite.
Then it is not necessarily the case that:
:$\ds \sum_{\map R i} \sum_{\map S j} a_{i j} = \sum_{\map S j} \sum_{\map R i} a_{i j}$ | Proof by Counterexample:
{{proof wanted|No idea at the moment. Have to find a series of $a_{i j}$ which is conditionally convergent for a start.}} | Let the [[Definition:Fiber of Truth|fiber of truth]] of both $R$ and $S$ be [[Definition:Infinite Set|infinite]].
Then it is not necessarily the case that:
:$\ds \sum_{\map R i} \sum_{\map S j} a_{i j} = \sum_{\map S j} \sum_{\map R i} a_{i j}$ | [[Proof by Counterexample]]:
{{proof wanted|No idea at the moment. Have to find a series of $a_{i j}$ which is conditionally convergent for a start.}} | Exchange of Order of Summation/Example | https://proofwiki.org/wiki/Exchange_of_Order_of_Summation/Example | https://proofwiki.org/wiki/Exchange_of_Order_of_Summation/Example | [
"Summations"
] | [
"Definition:Fiber of Truth",
"Definition:Infinite Set"
] | [
"Proof by Counterexample"
] |
proofwiki-14401 | Sum of Geometric Sequence/Examples/Index to Minus 1 | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
still holds when $n = -1$:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = \frac {x^0 - 1} {x - 1}$ | The summation on the {{LHS}} is vacuous:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = 0$
while on the {{RHS}} we have:
{{begin-eqn}}
{{eqn | l = \frac {x^{\paren {-1} + 1} - 1} {x - 1}
| r = \frac {x^0 - 1} {x - 1}
| c =
}}
{{eqn | r = \frac 0 {x - 1}
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
as l... | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for [[Sum of Geometric Sequence]]:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
still holds when $n = -1$:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = \fr... | The [[Definition:Summation|summation]] on the {{LHS}} is [[Definition:Vacuous Summation|vacuous]]:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = 0$
while on the {{RHS}} we have:
{{begin-eqn}}
{{eqn | l = \frac {x^{\paren {-1} + 1} - 1} {x - 1}
| r = \frac {x^0 - 1} {x - 1}
| c =
}}
{{eqn | r = \frac 0 {x - 1}
... | Sum of Geometric Sequence/Examples/Index to Minus 1 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/Index_to_Minus_1 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/Index_to_Minus_1 | [
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field",
"Sum of Geometric Sequence"
] | [
"Definition:Summation",
"Definition:Summation/Vacuous Summation"
] |
proofwiki-14402 | Sum of Geometric Sequence/Examples/Index to Minus 2 | Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
breaks down when $n = -2$:
:$\ds \sum_{j \mathop = 0}^{-2} x^j \ne \frac {x^{-1} - 1} {x - 1}$ | The summation on the {{LHS}} is vacuous:
:$\ds \sum_{j \mathop = 0}^{-2} x^j = 0$
while on the {{RHS}} we have:
{{begin-eqn}}
{{eqn | l = \frac {x^{\paren {-2} + 1} - 1} {x - 1}
| r = \frac {x^{-1} - 1} {x - 1}
| c =
}}
{{eqn | r = \frac {1 / x - 1} {x - 1}
| c =
}}
{{eqn | r = \frac {\paren {1 - x}... | Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for [[Sum of Geometric Sequence]]:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
breaks down when $n = -2$:
:$\ds \sum_{j \mathop = 0}^{-2} x^j \ne \... | The [[Definition:Summation|summation]] on the {{LHS}} is [[Definition:Vacuous Summation|vacuous]]:
:$\ds \sum_{j \mathop = 0}^{-2} x^j = 0$
while on the {{RHS}} we have:
{{begin-eqn}}
{{eqn | l = \frac {x^{\paren {-2} + 1} - 1} {x - 1}
| r = \frac {x^{-1} - 1} {x - 1}
| c =
}}
{{eqn | r = \frac {1 / x - ... | Sum of Geometric Sequence/Examples/Index to Minus 2 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/Index_to_Minus_2 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/Index_to_Minus_2 | [
"Sum of Geometric Sequence"
] | [
"Definition:Standard Number Field",
"Sum of Geometric Sequence"
] | [
"Definition:Summation",
"Definition:Summation/Vacuous Summation"
] |
proofwiki-14403 | Sum of Geometric Sequence/Examples/Common Ratio 1 | Consider the Sum of Geometric Sequence defined on the standard number fields for all $x \ne 1$.
:$\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
When $x = 1$, the formula reduces to:
:$\ds \sum_{j \mathop = 0}^n a 1^j = a \paren {n + 1}$ | When $x = 1$, the {{RHS}} is undefined:
:$a \paren {\dfrac {1 - 1^{n + 1} } {1 - 1} } = a \dfrac 0 0$
However, the {{LHS}} degenerates to:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n a 1^j
| r = \sum_{j \mathop = 0}^n a
| c =
}}
{{eqn | r = a \paren {n + 1}
| c =
}}
{{end-eqn}}
{{qed}} | Consider the [[Sum of Geometric Sequence]] defined on the [[Definition:Standard Number Field|standard number fields]] for all $x \ne 1$.
:$\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
When $x = 1$, the formula reduces to:
:$\ds \sum_{j \mathop = 0}^n a 1^j = a \paren {n + 1}$ | When $x = 1$, the {{RHS}} is undefined:
:$a \paren {\dfrac {1 - 1^{n + 1} } {1 - 1} } = a \dfrac 0 0$
However, the {{LHS}} degenerates to:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n a 1^j
| r = \sum_{j \mathop = 0}^n a
| c =
}}
{{eqn | r = a \paren {n + 1}
| c =
}}
{{end-eqn}}
{{qed}} | Sum of Geometric Sequence/Examples/Common Ratio 1 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/Common_Ratio_1 | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/Common_Ratio_1 | [
"Sum of Geometric Sequence"
] | [
"Sum of Geometric Sequence",
"Definition:Standard Number Field"
] | [] |
proofwiki-14404 | Sum of Geometric Sequence/Examples/One Seventh from 1 to n | :$\ds \sum_{j \mathop = 0}^n \dfrac 1 {7^j} = \frac 7 6 \paren {1 - \frac 1 {7^{n + 1} } }$ | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n \dfrac 1 {7^j}
| r = \frac {1 - \frac 1 7^{n + 1} } {1 - \frac 1 7}
| c =
}}
{{eqn | r = \frac {7 - 7 \frac 1 7^{n + 1} } {7 - 1}
| c = multiplying top and bottom by $7$
}}
{{end-eqn}}
Hence the result.
{{qed}} | :$\ds \sum_{j \mathop = 0}^n \dfrac 1 {7^j} = \frac 7 6 \paren {1 - \frac 1 {7^{n + 1} } }$ | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n \dfrac 1 {7^j}
| r = \frac {1 - \frac 1 7^{n + 1} } {1 - \frac 1 7}
| c =
}}
{{eqn | r = \frac {7 - 7 \frac 1 7^{n + 1} } {7 - 1}
| c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $7$
}}
{{end-eqn}}
Hence the resu... | Sum of Geometric Sequence/Examples/One Seventh from 1 to n | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/One_Seventh_from_1_to_n | https://proofwiki.org/wiki/Sum_of_Geometric_Sequence/Examples/One_Seventh_from_1_to_n | [
"Sum of Geometric Sequence"
] | [] | [
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator"
] |
proofwiki-14405 | Sum of Sequence of n by 2 to the Power of n | :$\ds \sum_{j \mathop = 0}^n j \, 2^j = n \, 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2$ | From Sum of Arithmetic-Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n \paren {a + j d} r^j = \frac {a \paren {1 - r^{n + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {n + 1} r^n + n r^{n + 1} } } {\paren {1 - r}^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n j \, 2^j
| r = \frac {0 \paren {1 - 2^... | :$\ds \sum_{j \mathop = 0}^n j \, 2^j = n \, 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2$ | From [[Sum of Arithmetic-Geometric Sequence]]:
:$\ds \sum_{j \mathop = 0}^n \paren {a + j d} r^j = \frac {a \paren {1 - r^{n + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {n + 1} r^n + n r^{n + 1} } } {\paren {1 - r}^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n j \, 2^j
| r = \frac {0 \paren... | Sum of Sequence of n by 2 to the Power of n/Proof 1 | https://proofwiki.org/wiki/Sum_of_Sequence_of_n_by_2_to_the_Power_of_n | https://proofwiki.org/wiki/Sum_of_Sequence_of_n_by_2_to_the_Power_of_n/Proof_1 | [
"Arithmetic-Geometric Sequences",
"Sum of Sequence of n by 2 to the Power of n"
] | [] | [
"Sum of Arithmetic-Geometric Sequence"
] |
proofwiki-14406 | Sum of Sequence of n by 2 to the Power of n | :$\ds \sum_{j \mathop = 0}^n j \, 2^j = n \, 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2$ | From Sum of Sequence of Power by Index:
:$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \ds \sum_{j \mathop = 0}^n j \, 2^j
| r = \frac {n 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2} {\paren {2 - 1}^2}
| c = putting $... | :$\ds \sum_{j \mathop = 0}^n j \, 2^j = n \, 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2$ | From [[Sum of Sequence of Power by Index]]:
:$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \ds \sum_{j \mathop = 0}^n j \, 2^j
| r = \frac {n 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2} {\paren {2 - 1}^2}
| c = p... | Sum of Sequence of n by 2 to the Power of n/Proof 2 | https://proofwiki.org/wiki/Sum_of_Sequence_of_n_by_2_to_the_Power_of_n | https://proofwiki.org/wiki/Sum_of_Sequence_of_n_by_2_to_the_Power_of_n/Proof_2 | [
"Arithmetic-Geometric Sequences",
"Sum of Sequence of n by 2 to the Power of n"
] | [] | [
"Sum of Sequence of Power by Index"
] |
proofwiki-14407 | Summation of Unity over Elements | Let $S \subseteq \Z$ be a set of integers.
Let:
:$n := \ds \sum_{j \mathop \in S} 1$
Then $n$ is equal to the cardinality of $S$. | {{ProofWanted|It's hard to prove obvious things.}} | Let $S \subseteq \Z$ be a [[Definition:Set|set]] of [[Definition:Integer|integers]].
Let:
:$n := \ds \sum_{j \mathop \in S} 1$
Then $n$ is equal to the [[Definition:Cardinality|cardinality]] of $S$. | {{ProofWanted|It's hard to prove obvious things.}} | Summation of Unity over Elements | https://proofwiki.org/wiki/Summation_of_Unity_over_Elements | https://proofwiki.org/wiki/Summation_of_Unity_over_Elements | [
"Summations"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Cardinality"
] | [] |
proofwiki-14408 | Exchange of Order of Summation with Dependency on Both Indices/Example | Let $n \in \Z$ be an integer.
Let $R: \Z \to \set {\T, \F}$ be the propositional function on the set of integers defining:
:$\forall i \in \Z: \map R 1 := \paren {n = k i \text { for some } k \in \Z}$
Let $S: \Z \times \Z \to \set {\T, \F}$ be a propositional function on the Cartesian product of the set of integers wit... | From Exchange of Order of Summation with Dependency on Both Indices:
:$\map {S'} j$ denotes the propositional function:
::there exists an $i$ such that both $\map R i$ and $\map S {i, j}$ hold
:$\map {R'} {i, j}$ denotes the propositional function:
::both $\map R i$ and $\map S {i, j}$ hold.
The definition of $\map {R'... | Let $n \in \Z$ be an [[Definition:Integer|integer]].
Let $R: \Z \to \set {\T, \F}$ be the [[Definition:Propositional Function|propositional function]] on the [[Definition:Integer|set of integers]] defining:
:$\forall i \in \Z: \map R 1 := \paren {n = k i \text { for some } k \in \Z}$
Let $S: \Z \times \Z \to \set {\... | From [[Exchange of Order of Summation with Dependency on Both Indices]]:
:$\map {S'} j$ denotes the [[Definition:Propositional Function|propositional function]]:
::there exists an $i$ such that both $\map R i$ and $\map S {i, j}$ hold
:$\map {R'} {i, j}$ denotes the [[Definition:Propositional Function|propositional fu... | Exchange of Order of Summation with Dependency on Both Indices/Example | https://proofwiki.org/wiki/Exchange_of_Order_of_Summation_with_Dependency_on_Both_Indices/Example | https://proofwiki.org/wiki/Exchange_of_Order_of_Summation_with_Dependency_on_Both_Indices/Example | [
"Summations"
] | [
"Definition:Integer",
"Definition:Propositional Function",
"Definition:Integer",
"Definition:Propositional Function",
"Definition:Cartesian Product",
"Definition:Integer",
"Definition:Summation",
"Definition:Propositional Function",
"Definition:Propositional Function"
] | [
"Exchange of Order of Summation with Dependency on Both Indices",
"Definition:Propositional Function",
"Definition:Propositional Function",
"Absolute Value of Integer is not less than Divisors"
] |
proofwiki-14409 | Hausdorff's Maximal Principle implies Zorn's Lemma | Hausdorff's Maximal Principle implies Zorn's Lemma. | Let $\struct {\PP, \preceq}$ be a partially ordered set.
Hausdorff's Maximal Principle states that there is a maximal chain in $\struct {\PP, \preceq}$.
If this maximal chain has an upper bound in $\PP$, then such an upper bound is a maximal element of $\PP$ itself, by the maximality of the chain.
{{qed}} | [[Hausdorff's Maximal Principle]] implies [[Zorn's Lemma]]. | Let $\struct {\PP, \preceq}$ be a [[Definition:Partially Ordered Set|partially ordered set]].
[[Hausdorff's Maximal Principle]] states that there is a [[Definition:Maximal Chain|maximal chain]] in $\struct {\PP, \preceq}$.
If this [[Definition:Maximal Chain|maximal chain]] has an upper bound in $\PP$, then such an up... | Hausdorff's Maximal Principle implies Zorn's Lemma | https://proofwiki.org/wiki/Hausdorff's_Maximal_Principle_implies_Zorn's_Lemma | https://proofwiki.org/wiki/Hausdorff's_Maximal_Principle_implies_Zorn's_Lemma | [
"Hausdorff's Maximal Principle",
"Zorn's Lemma"
] | [
"Hausdorff's Maximal Principle",
"Zorn's Lemma"
] | [
"Definition:Partially Ordered Set",
"Hausdorff's Maximal Principle",
"Definition:Maximal Chain",
"Definition:Maximal Chain"
] |
proofwiki-14410 | Zorn's Lemma Implies Zermelo's Well-Ordering Theorem | Zorn's Lemma implies Zermelo's Well-Ordering Theorem. | Let $X$ be a set.
If $X = \O$ the theorem holds vacuously.
Assume $X$ is not empty.
Let $\WW$ be the collection of pairs $\tuple {W, \preceq}$ such that:
:$W \subseteq X$
:$\preceq$ well-orders $W$
Next, define the partial ordering $\preccurlyeq$ on $\WW$ by $\tuple {W, \preceq} \preccurlyeq \tuple {W', \preceq'}$ {{if... | [[Zorn's Lemma]] implies [[Zermelo's Well-Ordering Theorem]]. | Let $X$ be a [[Definition:Set|set]].
If $X = \O$ the theorem holds [[Definition:Vacuous Truth|vacuously]].
Assume $X$ is not [[Definition:Empty Set|empty]].
Let $\WW$ be the [[Definition:Collection|collection]] of [[Definition:Ordered Pair|pairs]] $\tuple {W, \preceq}$ such that:
:$W \subseteq X$
:$\preceq$ [[Defini... | Zorn's Lemma Implies Zermelo's Well-Ordering Theorem | https://proofwiki.org/wiki/Zorn's_Lemma_Implies_Zermelo's_Well-Ordering_Theorem | https://proofwiki.org/wiki/Zorn's_Lemma_Implies_Zermelo's_Well-Ordering_Theorem | [
"Axiom of Choice",
"Well-Orderings",
"Zermelo's Well-Ordering Theorem"
] | [
"Zorn's Lemma",
"Zermelo's Well-Ordering Theorem"
] | [
"Definition:Set",
"Definition:Vacuous Truth",
"Definition:Empty Set",
"Definition:Collection",
"Definition:Ordered Pair",
"Definition:Well-Ordering",
"Definition:Partial Ordering",
"Definition:Restriction of Ordering",
"Zorn's Lemma",
"Definition:Chain (Order Theory)",
"Definition:Upper Bound",
... |
proofwiki-14411 | Uncountable Sum as Series | Let $X$ be an uncountable set.
Let $f: X \to \closedint 0 {+\infty}$ be an extended real-valued function.
The uncountable sum:
:$\ds \sum_{x \mathop \in X} \map f x = \sup \set {\sum_{x \mathop \in F} \map f x : F \subseteq X, F \text{ finite} }$
is:
:if $f$ has uncountably infinite support, then $+\infty$
:Otherwise,... | Define:
:$A_n = \set {x \in X: \map f x >\dfrac 1 n, n \in \N_{\ge 1} }$
Then $\sequence {A_n}_{n \mathop \in \N} \uparrow A$ is an exhausting sequence of sets, where:
:$A = \set {x \in X: \map f x > 0}$
Suppose $A$ is uncountable.
From Countable Union of Countable Sets is Countable, necessarily there is some $A_{n_0}$... | Let $X$ be an [[Definition:Uncountable Set|uncountable set]].
Let $f: X \to \closedint 0 {+\infty}$ be an [[Definition:Extended Real-Valued Function|extended real-valued function]].
The [[Definition:Uncountable Sum|uncountable sum]]:
:$\ds \sum_{x \mathop \in X} \map f x = \sup \set {\sum_{x \mathop \in F} \map f x... | Define:
:$A_n = \set {x \in X: \map f x >\dfrac 1 n, n \in \N_{\ge 1} }$
Then $\sequence {A_n}_{n \mathop \in \N} \uparrow A$ is an [[Definition:Exhausting Sequence of Sets|exhausting sequence of sets]], where:
:$A = \set {x \in X: \map f x > 0}$
Suppose $A$ is [[Definition:Uncountable Set|uncountable]].
From [[C... | Uncountable Sum as Series | https://proofwiki.org/wiki/Uncountable_Sum_as_Series | https://proofwiki.org/wiki/Uncountable_Sum_as_Series | [
"Real Analysis",
"Uncountable Sum as Series"
] | [
"Definition:Uncountable/Set",
"Definition:Extended Real-Valued Function",
"Definition:Uncountable Sum",
"Definition:Uncountable/Set",
"Definition:Support of Mapping to Algebraic Structure/Real-Valued Function",
"Definition:Divergent Series",
"Definition:Series"
] | [
"Definition:Exhausting Sequence of Sets",
"Definition:Uncountable/Set",
"Countable Union of Countable Sets is Countable",
"Definition:Uncountable/Set",
"Definition:Countably Infinite/Set",
"Definition:Finite Subset",
"Definition:Finite Set",
"Definition:Supremum",
"Definition:Supremum",
"Definitio... |
proofwiki-14412 | Sum with Maximum is Maximum of Sum | Let $a, b, c \in \R$ be real numbers.
Then:
:$a + \max \set {b, c} = \max \set {a + b, a + c}$ | {{WLOG}}, there are two cases to consider:
:$(1): \quad b \ge c$
:$(2): \quad b < c$
First let $b \ge c$.
We have:
{{begin-eqn}}
{{eqn | l = b
| o = \ge
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = a + b
| o = \ge
| r = a + c
| c = Addition of Real Numbers is Compatible with Us... | Let $a, b, c \in \R$ be [[Definition:Real Number|real numbers]].
Then:
:$a + \max \set {b, c} = \max \set {a + b, a + c}$ | {{WLOG}}, there are two cases to consider:
:$(1): \quad b \ge c$
:$(2): \quad b < c$
First let $b \ge c$.
We have:
{{begin-eqn}}
{{eqn | l = b
| o = \ge
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = a + b
| o = \ge
| r = a + c
| c = [[Addition of Real Numbers is Compatible w... | Sum with Maximum is Maximum of Sum | https://proofwiki.org/wiki/Sum_with_Maximum_is_Maximum_of_Sum | https://proofwiki.org/wiki/Sum_with_Maximum_is_Maximum_of_Sum | [
"Max Operation"
] | [
"Definition:Real Number"
] | [
"Addition of Real Numbers is Compatible with Usual Ordering",
"Addition of Real Numbers is Compatible with Usual Ordering",
"Proof by Cases"
] |
proofwiki-14413 | Sum of Indexed Suprema | Let $\sequence {a_i}_{i \mathop \in I}$ be a family of elements of the real numbers $\R$ indexed by $I$.
Let $\sequence {b_j}_{j \mathop \in J}$ be a family of elements of the real numbers $\R$ indexed by $J$.
Let $\map R i$ and $\map S j$ be propositional functions of $i \in I$, $j \in J$.
Let $\ds \sup_{\map R i} a_i... | {{improve|The results upon which this depends need to be enhanced to encompass the full supremum, not just the max}}
{{begin-eqn}}
{{eqn | l = \paren {\sup_{\map R i} a_i} + \paren {\sup_{\map S j} b_j}
| r = \sup_{\map R i} \paren {a_i + \sup_{\map S j} b_j}
| c = Sum with Maximum is Maximum of Sum
}}
{{eq... | Let $\sequence {a_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family of elements]] of the [[Definition:Real Number|real numbers]] $\R$ [[Definition:Indexing Set|indexed]] by $I$.
Let $\sequence {b_j}_{j \mathop \in J}$ be a [[Definition:Indexed Family|family of elements]] of the [[Definition:Real Number|rea... | {{improve|The results upon which this depends need to be enhanced to encompass the full supremum, not just the max}}
{{begin-eqn}}
{{eqn | l = \paren {\sup_{\map R i} a_i} + \paren {\sup_{\map S j} b_j}
| r = \sup_{\map R i} \paren {a_i + \sup_{\map S j} b_j}
| c = [[Sum with Maximum is Maximum of Sum]]
}}... | Sum of Indexed Suprema | https://proofwiki.org/wiki/Sum_of_Indexed_Suprema | https://proofwiki.org/wiki/Sum_of_Indexed_Suprema | [
"Suprema"
] | [
"Definition:Indexing Set/Family",
"Definition:Real Number",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Real Number",
"Definition:Indexing Set",
"Definition:Propositional Function",
"Definition:Supremum of Set/Real Numbers/Propositional Function"
] | [
"Sum with Maximum is Maximum of Sum",
"Sum with Maximum is Maximum of Sum"
] |
proofwiki-14414 | Product of Indexed Suprema of Non-Negative Numbers | Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\family {b_j}_{j \mathop \in J}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $J$.
Let $\map R i$ and $\map S j$ be propositional functions of $i \in I$, $... | {{begin-eqn}}
{{eqn | l = \paren {\sup_{\map R i} a_i} \paren {\sup_{\map S j} b_j}
| r = \sup_{\map R i} \paren {a_i \times \sup_{\map S j} b_j}
| c = Product with Supremum is Supremum of Product of Non-Negative Numbers
}}
{{eqn | r = \sup_{\map R i} \paren {\sup_{\map S j} \paren {a_i b_j} }
| c = P... | Let $\family {a_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family of elements]] of the [[Definition:Positive Real Number|non-negative real numbers]] $\R_{\ge 0}$ [[Definition:Indexing Set|indexed]] by $I$.
Let $\family {b_j}_{j \mathop \in J}$ be a [[Definition:Indexed Family|family of elements]] of the [[... | {{begin-eqn}}
{{eqn | l = \paren {\sup_{\map R i} a_i} \paren {\sup_{\map S j} b_j}
| r = \sup_{\map R i} \paren {a_i \times \sup_{\map S j} b_j}
| c = [[Product with Supremum is Supremum of Product of Non-Negative Numbers]]
}}
{{eqn | r = \sup_{\map R i} \paren {\sup_{\map S j} \paren {a_i b_j} }
| c... | Product of Indexed Suprema of Non-Negative Numbers | https://proofwiki.org/wiki/Product_of_Indexed_Suprema_of_Non-Negative_Numbers | https://proofwiki.org/wiki/Product_of_Indexed_Suprema_of_Non-Negative_Numbers | [
"Suprema"
] | [
"Definition:Indexing Set/Family",
"Definition:Positive/Real Number",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Positive/Real Number",
"Definition:Indexing Set",
"Definition:Propositional Function",
"Definition:Supremum of Set/Real Numbers/Propositional Function"
] | [
"Product with Supremum is Supremum of Product of Non-Negative Numbers",
"Product with Supremum is Supremum of Product of Non-Negative Numbers"
] |
proofwiki-14415 | Change of Index Variable of Supremum | Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\map R i$ be a propositional functions of $i \in I$.
Let $\ds \sup_{\map R i} a_i$ be the indexed supremum on $\family {a_i}$.
Then:
:$\ds \sup_{\map R i} a_i = \sup_{\map R j} a_j$ | {{ProofWanted|If anybody is keen on proving obvious things}} | Let $\family {a_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family of elements]] of the [[Definition:Positive Real Number|non-negative real numbers]] $\R_{\ge 0}$ [[Definition:Indexing Set|indexed]] by $I$.
Let $\map R i$ be a [[Definition:Propositional Function|propositional functions]] of $i \in I$.
Let... | {{ProofWanted|If anybody is keen on proving obvious things}} | Change of Index Variable of Supremum | https://proofwiki.org/wiki/Change_of_Index_Variable_of_Supremum | https://proofwiki.org/wiki/Change_of_Index_Variable_of_Supremum | [
"Suprema"
] | [
"Definition:Indexing Set/Family",
"Definition:Positive/Real Number",
"Definition:Indexing Set",
"Definition:Propositional Function",
"Definition:Supremum of Set/Real Numbers/Propositional Function"
] | [] |
proofwiki-14416 | Permutation of Indices of Supremum | Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\map R i$ be a propositional functions of $i \in I$.
Let $\ds \sup_{\map R i} a_i$ be the indexed supremum on $\family {a_i}$.
Then:
:$\ds \sum_{\map R i} a_i = \sum_{\map R {\map \pi i} } a_... | {{begin-eqn}}
{{eqn | l = \sup_{\map R {\map \pi j} } a_{\map \pi j}
| r = \sup_{j \mathop \in I} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }
| c =
}}
{{eqn | r = \sup_{j \mathop \in I} \paren {\sup_{i \mathop \in I} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j} }
| c =
}}
{{eqn | r = \sup_{i \math... | Let $\family {a_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family of elements]] of the [[Definition:Positive Real Number|non-negative real numbers]] $\R_{\ge 0}$ [[Definition:Indexing Set|indexed]] by $I$.
Let $\map R i$ be a [[Definition:Propositional Function|propositional functions]] of $i \in I$.
Let... | {{begin-eqn}}
{{eqn | l = \sup_{\map R {\map \pi j} } a_{\map \pi j}
| r = \sup_{j \mathop \in I} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }
| c =
}}
{{eqn | r = \sup_{j \mathop \in I} \paren {\sup_{i \mathop \in I} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j} }
| c =
}}
{{eqn | r = \sup_{i \math... | Permutation of Indices of Supremum | https://proofwiki.org/wiki/Permutation_of_Indices_of_Supremum | https://proofwiki.org/wiki/Permutation_of_Indices_of_Supremum | [
"Suprema"
] | [
"Definition:Indexing Set/Family",
"Definition:Positive/Real Number",
"Definition:Indexing Set",
"Definition:Propositional Function",
"Definition:Supremum of Set/Real Numbers/Propositional Function",
"Definition:Permutation",
"Definition:Fiber of Truth"
] | [
"Change of Index Variable of Supremum"
] |
proofwiki-14417 | Exchange of Order of Supremum Operators | Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\family {b_j}_{j \mathop \in J}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $J$.
Let $\map R i$ and $\map S j$ be propositional functions of $i \in I$, $... | For any chosen $i$ and $j$, we have:
:$\ds a_{ij} \le \sup_{\map R i} a_{ij}$
Taking the supremum over $\map S j$ yields:
:$\ds \sup_{\map S j} a_{ij} \le \sup_{\map S j} \sup_{\map R i} a_{ij}$
for all $i$.
Hence the {{RHS}} remains an upper bound after taking the supremum over $\map R i$ on the {{LHS}}.
That is:
:$\d... | Let $\family {a_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family of elements]] of the [[Definition:Positive Real Number|non-negative real numbers]] $\R_{\ge 0}$ [[Definition:Indexing Set|indexed]] by $I$.
Let $\family {b_j}_{j \mathop \in J}$ be a [[Definition:Indexed Family|family of elements]] of the [[... | For any chosen $i$ and $j$, we have:
:$\ds a_{ij} \le \sup_{\map R i} a_{ij}$
Taking the [[Definition:Indexed Supremum by Propositional Function|supremum]] over $\map S j$ yields:
:$\ds \sup_{\map S j} a_{ij} \le \sup_{\map S j} \sup_{\map R i} a_{ij}$
for all $i$.
Hence the {{RHS}} remains an [[Definition:Upper Boun... | Exchange of Order of Supremum Operators | https://proofwiki.org/wiki/Exchange_of_Order_of_Supremum_Operators | https://proofwiki.org/wiki/Exchange_of_Order_of_Supremum_Operators | [
"Suprema"
] | [
"Definition:Indexing Set/Family",
"Definition:Positive/Real Number",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Positive/Real Number",
"Definition:Indexing Set",
"Definition:Propositional Function",
"Definition:Supremum of Set/Real Numbers/Propositional Function"
] | [
"Definition:Supremum of Set/Real Numbers/Propositional Function",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set/Real Numbers/Propositional Function"
] |
proofwiki-14418 | Contour Integration by Substitution | Let $f$ be a holomorphic function on a simply connected domain $V \subseteq \mathbb C$.
Let $\gamma$ be a contour in $V$ starting at $z_1$ and ending at $z_2$.
Let $U$ be a connected domain.
Let $\phi: U \to V$ be a holomorphic function with $\phi^{-1} \sqbrk {\set {z_1, z_2} } \ne \O$.
Let $\omega$ be a contour in $U$... | {{MissingLinks|relationship between holomorphic on s.c. domain and having a primitive}}
Because $V$ is simply connected, $f$ has a primitive.
Let one such primitive $F$ be established.
By the Fundamental Theorem of Calculus for Contour Integrals:
{{begin-eqn}}
{{eqn | l = I_1
| r = \int_\gamma \map f z \rd z
... | Let $f$ be a [[Definition:Holomorphic Complex Function|holomorphic function]] on a [[Definition:Simply Connected Domain|simply connected domain]] $V \subseteq \mathbb C$.
Let $\gamma$ be a [[Definition:Contour (Complex Plane)|contour]] in $V$ starting at $z_1$ and ending at $z_2$.
Let $U$ be a [[Definition:Connected ... | {{MissingLinks|relationship between holomorphic on s.c. domain and having a primitive}}
Because $V$ is [[Definition:Simply Connected Domain|simply connected]], $f$ has a [[Definition:Complex Primitive|primitive]].
Let one such [[Definition:Complex Primitive|primitive]] $F$ be established.
By the [[Fundamental Theor... | Contour Integration by Substitution | https://proofwiki.org/wiki/Contour_Integration_by_Substitution | https://proofwiki.org/wiki/Contour_Integration_by_Substitution | [
"Complex Analysis"
] | [
"Definition:Holomorphic Function/Complex Plane",
"Definition:Connected Domain (Complex Analysis)/Simply Connected Domain",
"Definition:Contour/Complex Plane",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Holomorphic Function/Complex Plane",
"Definition:Contour/Complex Plane",
"Definitio... | [
"Definition:Connected Domain (Complex Analysis)/Simply Connected Domain",
"Definition:Primitive (Calculus)/Complex",
"Definition:Primitive (Calculus)/Complex",
"Fundamental Theorem of Calculus for Contour Integrals",
"Definition:Primitive (Calculus)/Complex",
"Fundamental Theorem of Calculus for Contour I... |
proofwiki-14419 | Infimum of Set of Integers is Integer | Let $S \subset \Z$ be a non-empty subset of the set of integers.
Let $S$ be bounded below in the set of real numbers.
Then its infimum $\inf S$ is an integer. | By Infimum of Set of Integers equals Smallest Element, $S$ has a smallest element $n \in \Z$, that is equals to the infimum of $S$.
{{qed}} | Let $S \subset \Z$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Integer|set of integers]].
Let $S$ be [[Definition:Bounded Below Subset of Real Numbers|bounded below]] in the [[Definition:Real Number|set of real numbers]].
Then its [[Definition:Infimum of Subset of Rea... | By [[Infimum of Set of Integers equals Smallest Element]], $S$ has a [[Definition:Smallest Element|smallest element]] $n \in \Z$, that is equals to the [[Definition:Infimum of Subset of Real Numbers|infimum]] of $S$.
{{qed}} | Infimum of Set of Integers is Integer | https://proofwiki.org/wiki/Infimum_of_Set_of_Integers_is_Integer | https://proofwiki.org/wiki/Infimum_of_Set_of_Integers_is_Integer | [
"Integers"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Integer",
"Definition:Bounded Below Set/Real Numbers",
"Definition:Real Number",
"Definition:Infimum of Set/Real Numbers",
"Definition:Integer"
] | [
"Infimum of Set of Integers equals Smallest Element",
"Definition:Smallest Element",
"Definition:Infimum of Set/Real Numbers"
] |
proofwiki-14420 | Infimum of Set of Integers equals Smallest Element | Let $S \subset \Z$ be a non-empty subset of the set of integers.
Let $S$ be bounded below in the set of real numbers $\R$.
Then $S$ has a smallest element, and it is equal to the infimum $\sup S$. | By Set of Integers Bounded Below by Real Number has Smallest Element, $S$ has a smallest element, say $n \in S$.
By Smallest Element is Infimum, $n$ is the infimum of $S$.
{{qed}} | Let $S \subset \Z$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Integer|set of integers]].
Let $S$ be [[Definition:Bounded Below Subset of Real Numbers|bounded below]] in the [[Definition:Real Number|set of real numbers]] $\R$.
Then $S$ has a [[Definition:Smallest Elem... | By [[Set of Integers Bounded Below by Real Number has Smallest Element]], $S$ has a [[Definition:Smallest Element|smallest element]], say $n \in S$.
By [[Smallest Element is Infimum]], $n$ is the [[Definition:Infimum of Subset of Real Numbers|infimum]] of $S$.
{{qed}} | Infimum of Set of Integers equals Smallest Element | https://proofwiki.org/wiki/Infimum_of_Set_of_Integers_equals_Smallest_Element | https://proofwiki.org/wiki/Infimum_of_Set_of_Integers_equals_Smallest_Element | [
"Integers"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Integer",
"Definition:Bounded Below Set/Real Numbers",
"Definition:Real Number",
"Definition:Smallest Element",
"Definition:Infimum of Set/Real Numbers"
] | [
"Set of Integers Bounded Below by Real Number has Smallest Element",
"Definition:Smallest Element",
"Smallest Element is Infimum",
"Definition:Infimum of Set/Real Numbers"
] |
proofwiki-14421 | Greatest Element is Supremum | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a greatest element $M$.
Then $M$ is the supremum of $T$ in $S$. | Let $M$ be the greatest element of $T$.
Then by definition:
:$\forall x \in T: x \preceq M$
By definition of supremum, it is necessary to show that:
:$(1): \quad M$ is an upper bound of $T$ in $S$
:$(2): \quad M \preceq U$ for all upper bounds $U$ of $T$ in $S$.
By Greatest Element is Upper Bound, $M$ is an upper bound... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $T$ have a [[Definition:Greatest Element|greatest element]] $M$.
Then $M$ is the [[Definition:Supremum of Set|supremum]] of $T$ in $S$. | Let $M$ be the [[Definition:Greatest Element|greatest element]] of $T$.
Then by definition:
:$\forall x \in T: x \preceq M$
By definition of [[Definition:Supremum of Set|supremum]], it is necessary to show that:
:$(1): \quad M$ is an [[Definition:Upper Bound of Set|upper bound]] of $T$ in $S$
:$(2): \quad M \preceq ... | Greatest Element is Supremum | https://proofwiki.org/wiki/Greatest_Element_is_Supremum | https://proofwiki.org/wiki/Greatest_Element_is_Supremum | [
"Suprema",
"Greatest Elements"
] | [
"Definition:Ordered Set",
"Definition:Greatest Element",
"Definition:Supremum of Set"
] | [
"Definition:Greatest Element",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Greatest Element is Upper Bound",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set"
] |
proofwiki-14422 | Smallest Element is Infimum | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a smallest element $m$.
Then $m$ is the infimum of $T$ in $S$. | Let $M$ be the smallest element of $T$.
Then by definition:
:$\forall x \in T: m \preceq x$
By definition of infimum, it is necessary to show that:
:$(1): \quad m$ is a lower bound of $T$ in $S$
:$(2): \quad L \preceq m$ for all lower bounds $L$ of $T$ in $S$.
By Smallest Element is Lower Bound, $m$ is a lower bound of... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $T$ have a [[Definition:Smallest Element|smallest element]] $m$.
Then $m$ is the [[Definition:Infimum of Set|infimum]] of $T$ in $S$. | Let $M$ be the [[Definition:Smallest Element|smallest element]] of $T$.
Then by definition:
:$\forall x \in T: m \preceq x$
By definition of [[Definition:Infimum of Set|infimum]], it is necessary to show that:
:$(1): \quad m$ is a [[Definition:Lower Bound of Set|lower bound]] of $T$ in $S$
:$(2): \quad L \preceq m$ ... | Smallest Element is Infimum | https://proofwiki.org/wiki/Smallest_Element_is_Infimum | https://proofwiki.org/wiki/Smallest_Element_is_Infimum | [
"Infima",
"Smallest Elements"
] | [
"Definition:Ordered Set",
"Definition:Smallest Element",
"Definition:Infimum of Set"
] | [
"Definition:Smallest Element",
"Definition:Infimum of Set",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Smallest Element is Lower Bound",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set"
] |
proofwiki-14423 | Floor Function/Examples/Floor of 1.1 | :$\floor {1 \cdotp 1} = 1$ | We have that:
:$1 \le 1 \cdotp 1 < 2$
Hence $1$ is the floor of $1 \cdotp 1$ by definition.
{{qed}} | :$\floor {1 \cdotp 1} = 1$ | We have that:
:$1 \le 1 \cdotp 1 < 2$
Hence $1$ is the [[Definition:Floor Function|floor]] of $1 \cdotp 1$ by definition.
{{qed}} | Floor Function/Examples/Floor of 1.1 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_1.1 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_1.1 | [
"Examples of Floor Function"
] | [] | [
"Definition:Floor Function"
] |
proofwiki-14424 | Floor Function/Examples/Floor of -1.1 | :$\floor {-1 \cdotp 1} = -2$ | We have that:
:$-2 \le -1 \cdotp 1 < -1$
Hence $-2$ is the floor of $-1 \cdotp 1$ by definition.
{{qed}} | :$\floor {-1 \cdotp 1} = -2$ | We have that:
:$-2 \le -1 \cdotp 1 < -1$
Hence $-2$ is the [[Definition:Floor Function|floor]] of $-1 \cdotp 1$ by definition.
{{qed}} | Floor Function/Examples/Floor of -1.1 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_-1.1 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_-1.1 | [
"Examples of Floor Function"
] | [] | [
"Definition:Floor Function"
] |
proofwiki-14425 | Ceiling Function/Examples/Ceiling of -1.1 | :$\ceiling {-1 \cdotp 1} = -1$ | We have that:
:$-2 < -1 \cdotp 1 \le -1$
Hence $-1$ is the ceiling of $-1 \cdotp 1$ by definition.
{{qed}} | :$\ceiling {-1 \cdotp 1} = -1$ | We have that:
:$-2 < -1 \cdotp 1 \le -1$
Hence $-1$ is the [[Definition:Ceiling Function|ceiling]] of $-1 \cdotp 1$ by definition.
{{qed}} | Ceiling Function/Examples/Ceiling of -1.1 | https://proofwiki.org/wiki/Ceiling_Function/Examples/Ceiling_of_-1.1 | https://proofwiki.org/wiki/Ceiling_Function/Examples/Ceiling_of_-1.1 | [
"Examples of Ceiling Function"
] | [] | [
"Definition:Ceiling Function"
] |
proofwiki-14426 | Floor Function/Examples/Floor of 0.99999 | :$\floor {0 \cdotp 99999} = 0$ | We have that:
:$0 \le 0 \cdotp 99999 < 1$
Hence $0$ is the floor of $0 \cdotp 99999$ by definition.
{{qed}} | :$\floor {0 \cdotp 99999} = 0$ | We have that:
:$0 \le 0 \cdotp 99999 < 1$
Hence $0$ is the [[Definition:Floor Function|floor]] of $0 \cdotp 99999$ by definition.
{{qed}} | Floor Function/Examples/Floor of 0.99999 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_0.99999 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_0.99999 | [
"Examples of Floor Function"
] | [] | [
"Definition:Floor Function"
] |
proofwiki-14427 | Floor Function/Examples/Floor of Binary Logarithm of 35 | :$\floor {\lg 35} = 5$
where:
:$\lg x$ denotes the '''binary logarithm ($\log_2$)''' of $x$ | {{begin-eqn}}
{{eqn | l = \lg 32
| o = \le
| m = \lg 35
| mo= <
| r = \lg 64
| c =
}}
{{eqn | ll= \leadsto
| l = 5
| o = \le
| m = \lg 35
| mo= <
| r = 6
| c =
}}
{{end-eqn}}
Hence $5$ is the floor of $\lg 35$ by definition.
{{qed}} | :$\floor {\lg 35} = 5$
where:
:$\lg x$ denotes the '''[[Definition:Binary Logarithm|binary logarithm ($\log_2$)]]''' of $x$ | {{begin-eqn}}
{{eqn | l = \lg 32
| o = \le
| m = \lg 35
| mo= <
| r = \lg 64
| c =
}}
{{eqn | ll= \leadsto
| l = 5
| o = \le
| m = \lg 35
| mo= <
| r = 6
| c =
}}
{{end-eqn}}
Hence $5$ is the [[Definition:Floor Function|floor]] of $\lg 35$ by definiti... | Floor Function/Examples/Floor of Binary Logarithm of 35 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_Binary_Logarithm_of_35 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_Binary_Logarithm_of_35 | [
"Examples of Floor Function"
] | [
"Definition:General Logarithm/Binary"
] | [
"Definition:Floor Function"
] |
proofwiki-14428 | Modulo Operation/Examples/100 mod 3 | :$100 \bmod 3 = 1$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {100} 3 = 33 + \dfrac 1 3$
and so:
:$\floor {\dfrac {100} 3} = 33$
Thus:
{{begin-eqn}}
{{eqn | l = 100 \bmod 3
| r = 100 - 3 \times \floor {\dfrac {100} 3}
| c =
}}
{{eqn | r = 100 - 3 \times 33
... | :$100 \bmod 3 = 1$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {100} 3 = 33 + \dfrac 1 3$
and so:
:$\floor {\dfrac {100} 3} = 33$
Thus:
{{begin-eqn}}
{{eqn | l = 100 \bmod 3
| r = 100 - 3 \times \floor {\dfrac {100} 3}
| c =... | Modulo Operation/Examples/100 mod 3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/100_mod_3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/100_mod_3 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14429 | Modulo Operation/Examples/100 mod 7 | :$100 \bmod 7 = 2$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {100} 7 = 14 + \dfrac 2 7$
and so:
:$\floor {\dfrac {100} 7} = 14$
Thus:
{{begin-eqn}}
{{eqn | l = 100 \bmod 7
| r = 100 - 7 \times \floor {\dfrac {100} 7}
| c =
}}
{{eqn | r = 100 - 7 \times 14
... | :$100 \bmod 7 = 2$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {100} 7 = 14 + \dfrac 2 7$
and so:
:$\floor {\dfrac {100} 7} = 14$
Thus:
{{begin-eqn}}
{{eqn | l = 100 \bmod 7
| r = 100 - 7 \times \floor {\dfrac {100} 7}
| c =... | Modulo Operation/Examples/100 mod 7 | https://proofwiki.org/wiki/Modulo_Operation/Examples/100_mod_7 | https://proofwiki.org/wiki/Modulo_Operation/Examples/100_mod_7 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14430 | Modulo Operation/Examples/-100 mod 7 | :$-100 \bmod 7 = 5$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-100} 7 = -15 + \dfrac 5 7$
and so:
:$\floor {\dfrac {-100} 7} = -15$
Thus:
{{begin-eqn}}
{{eqn | l = -100 \bmod 7
| r = -100 - 7 \times \floor {\dfrac {-100} 7}
| c =
}}
{{eqn | r = -100 + 7 \tim... | :$-100 \bmod 7 = 5$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-100} 7 = -15 + \dfrac 5 7$
and so:
:$\floor {\dfrac {-100} 7} = -15$
Thus:
{{begin-eqn}}
{{eqn | l = -100 \bmod 7
| r = -100 - 7 \times \floor {\dfrac {-100} 7}
... | Modulo Operation/Examples/-100 mod 7 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-100_mod_7 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-100_mod_7 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14431 | Modulo Operation/Examples/-100 mod 0 | :$-100 \bmod 0 = -100$ | By definition of modulo $0$:
:$\forall x \in \R: x \bmod 0 = x$
Hence:
:$-100 \bmod 0 = -100$
{{qed}} | :$-100 \bmod 0 = -100$ | By definition of [[Definition:Modulo 0|modulo $0$]]:
:$\forall x \in \R: x \bmod 0 = x$
Hence:
:$-100 \bmod 0 = -100$
{{qed}} | Modulo Operation/Examples/-100 mod 0 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-100_mod_0 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-100_mod_0 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation/Modulo Zero"
] |
proofwiki-14432 | Modulo Operation/Examples/5 mod -3 | :$5 \bmod -3 = -1$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac 5 {-3} = -2 + \dfrac 1 3$
and so:
:$\floor {\dfrac 5 {-3} } = -2$
Thus:
{{begin-eqn}}
{{eqn | l = 5 \bmod -3
| r = 5 - \paren {-3} \times \floor {\dfrac 5 {-3} }
| c =
}}
{{eqn | r = 5 - \paren {-3... | :$5 \bmod -3 = -1$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac 5 {-3} = -2 + \dfrac 1 3$
and so:
:$\floor {\dfrac 5 {-3} } = -2$
Thus:
{{begin-eqn}}
{{eqn | l = 5 \bmod -3
| r = 5 - \paren {-3} \times \floor {\dfrac 5 {-3} }
... | Modulo Operation/Examples/5 mod -3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/5_mod_-3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/5_mod_-3 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14433 | Modulo Operation/Examples/18 mod -3 | :$18 \bmod -3 = 0$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {18} {-3} = -6 + \dfrac 0 3$
and so:
:$\floor {\dfrac {18} {-3} } = -6$
Thus:
{{begin-eqn}}
{{eqn | l = 18 \bmod -3
| r = 18 - \paren {-3} \times \floor {\dfrac {18} {-3} }
| c =
}}
{{eqn | r = 18 ... | :$18 \bmod -3 = 0$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {18} {-3} = -6 + \dfrac 0 3$
and so:
:$\floor {\dfrac {18} {-3} } = -6$
Thus:
{{begin-eqn}}
{{eqn | l = 18 \bmod -3
| r = 18 - \paren {-3} \times \floor {\dfrac {18} {... | Modulo Operation/Examples/18 mod -3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/18_mod_-3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/18_mod_-3 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14434 | Modulo Operation/Examples/-2 mod -3 | :$-2 \bmod -3 = -2$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-2} {-3} = 0 + \dfrac 2 3$
and so:
:$\floor {\dfrac {-2} {-3} } = 0$
Thus:
{{begin-eqn}}
{{eqn | l = -2 \bmod -3
| r = -2 - \paren {-3} \times \floor {\dfrac {-2} {-3} }
| c =
}}
{{eqn | r = -2 - ... | :$-2 \bmod -3 = -2$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-2} {-3} = 0 + \dfrac 2 3$
and so:
:$\floor {\dfrac {-2} {-3} } = 0$
Thus:
{{begin-eqn}}
{{eqn | l = -2 \bmod -3
| r = -2 - \paren {-3} \times \floor {\dfrac {-2} {-3... | Modulo Operation/Examples/-2 mod -3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-2_mod_-3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-2_mod_-3 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14435 | Modulo Operation/Examples/1.1 mod 1 | :$1 \cdotp 1 \bmod 1 = 0 \cdotp 1$ | By definition of modulo $1$:
:$x \bmod 1 = x - \floor x$
Thus:
{{begin-eqn}}
{{eqn | l = 1 \cdotp 1 \bmod 1
| r = 1 \cdotp 1 - \floor {1 \cdotp 1}
| c =
}}
{{eqn | r = 1 \cdotp 1 - 1
| c =
}}
{{eqn | r = 0 \cdotp 1
| c =
}}
{{end-eqn}}
{{qed}} | :$1 \cdotp 1 \bmod 1 = 0 \cdotp 1$ | By definition of [[Definition:Modulo 1|modulo $1$]]:
:$x \bmod 1 = x - \floor x$
Thus:
{{begin-eqn}}
{{eqn | l = 1 \cdotp 1 \bmod 1
| r = 1 \cdotp 1 - \floor {1 \cdotp 1}
| c =
}}
{{eqn | r = 1 \cdotp 1 - 1
| c =
}}
{{eqn | r = 0 \cdotp 1
| c =
}}
{{end-eqn}}
{{qed}} | Modulo Operation/Examples/1.1 mod 1 | https://proofwiki.org/wiki/Modulo_Operation/Examples/1.1_mod_1 | https://proofwiki.org/wiki/Modulo_Operation/Examples/1.1_mod_1 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation/Modulo One"
] |
proofwiki-14436 | Modulo Operation/Examples/0.11 mod 0.1 | :$0 \cdotp 11 \bmod 0 \cdotp 1 = 0 \cdotp 01$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
{{begin-eqn}}
{{eqn | l = \dfrac {0 \cdotp 11} {0 \cdotp 1}
| r = \dfrac {1 \cdotp 1} 1
| c =
}}
{{eqn | r = 1 \cdotp 1
| c =
}}
{{end-eqn}}
and so:
:$\floor {\dfrac {0 \cdotp 11} {0 \cdotp 1} } = 1$... | :$0 \cdotp 11 \bmod 0 \cdotp 1 = 0 \cdotp 01$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
{{begin-eqn}}
{{eqn | l = \dfrac {0 \cdotp 11} {0 \cdotp 1}
| r = \dfrac {1 \cdotp 1} 1
| c =
}}
{{eqn | r = 1 \cdotp 1
| c =
}}
{{end-eqn}}
and so:
:$\floor {\dfr... | Modulo Operation/Examples/0.11 mod 0.1/Proof 1 | https://proofwiki.org/wiki/Modulo_Operation/Examples/0.11_mod_0.1 | https://proofwiki.org/wiki/Modulo_Operation/Examples/0.11_mod_0.1/Proof_1 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14437 | Modulo Operation/Examples/0.11 mod -0.1 | :$0 \cdotp 11 \bmod -0 \cdotp 1 = -0 \cdotp 09$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
{{begin-eqn}}
{{eqn | l = \dfrac {0 \cdotp 11} {-0 \cdotp 1}
| r = \dfrac {1 \cdotp 1} {-1}
| c =
}}
{{eqn | r = -1 \cdotp 1
| c =
}}
{{end-eqn}}
and so:
:$\floor {\dfrac {0 \cdotp 11} {-0 \cdotp 1} ... | :$0 \cdotp 11 \bmod -0 \cdotp 1 = -0 \cdotp 09$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
{{begin-eqn}}
{{eqn | l = \dfrac {0 \cdotp 11} {-0 \cdotp 1}
| r = \dfrac {1 \cdotp 1} {-1}
| c =
}}
{{eqn | r = -1 \cdotp 1
| c =
}}
{{end-eqn}}
and so:
:$\floor ... | Modulo Operation/Examples/0.11 mod -0.1 | https://proofwiki.org/wiki/Modulo_Operation/Examples/0.11_mod_-0.1 | https://proofwiki.org/wiki/Modulo_Operation/Examples/0.11_mod_-0.1 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-14438 | Modulo Operation as Integer Difference by Quotient | Let $x, y, z \in \R$ be real numbers.
Let $y > 0$.
Let $0 \le z < y$.
Let:
:$\dfrac {x - z} y = k$
for some integer $k$.
Then:
:$z \equiv x \bmod y$
where $\bmod$ denotes the modulo operation. | We have:
{{begin-eqn}}
{{eqn | l = \dfrac {x - z} y
| r = k
| c =
}}
{{eqn | n = 1
| ll= \leadsto
| l = x
| r = z - k y
| c =
}}
{{end-eqn}}
We also have:
:$0 \le z < y$
Hence:
:$0 \le \dfrac z y < 1$
and so by definition of floor function:
:$(2): \quad \floor {\dfrac z y} = 0$
Thu... | Let $x, y, z \in \R$ be [[Definition:Real Number|real numbers]].
Let $y > 0$.
Let $0 \le z < y$.
Let:
:$\dfrac {x - z} y = k$
for some [[Definition:Integer|integer]] $k$.
Then:
:$z \equiv x \bmod y$
where $\bmod$ denotes the [[Definition:Modulo Operation|modulo operation]]. | We have:
{{begin-eqn}}
{{eqn | l = \dfrac {x - z} y
| r = k
| c =
}}
{{eqn | n = 1
| ll= \leadsto
| l = x
| r = z - k y
| c =
}}
{{end-eqn}}
We also have:
:$0 \le z < y$
Hence:
:$0 \le \dfrac z y < 1$
and so by definition of [[Definition:Floor Function|floor function]]:
:$(2)... | Modulo Operation as Integer Difference by Quotient | https://proofwiki.org/wiki/Modulo_Operation_as_Integer_Difference_by_Quotient | https://proofwiki.org/wiki/Modulo_Operation_as_Integer_Difference_by_Quotient | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Integer",
"Definition:Modulo Operation"
] | [
"Definition:Floor Function",
"Definition:By Hypothesis",
"Floor of Number plus Integer"
] |
proofwiki-14439 | Common Factor Cancelling in Congruence/Corollary 1/Warning | Let $a$ ''not'' be coprime to $m$.
Then it is not necessarily the case that:
:$x \equiv y \pmod m$ | Proof by Counterexample:
Let $a = 6, b = 21, x = 7, y = 12, m = 15$.
We note that $\map \gcd {6, 15} = 3$ and so $6$ and $15$ are not coprime.
We have that:
{{begin-eqn}}
{{eqn | l = 6
| o = \equiv
| r = 6
| rr= \pmod {15}
| c =
}}
{{eqn | l = 21
| o = \equiv
| r = 6
| rr= \pm... | Let $a$ ''not'' be [[Definition:Coprime Integers|coprime]] to $m$.
Then it is not necessarily the case that:
:$x \equiv y \pmod m$ | [[Proof by Counterexample]]:
Let $a = 6, b = 21, x = 7, y = 12, m = 15$.
We note that $\map \gcd {6, 15} = 3$ and so $6$ and $15$ are not [[Definition:Coprime Integers|coprime]].
We have that:
{{begin-eqn}}
{{eqn | l = 6
| o = \equiv
| r = 6
| rr= \pmod {15}
| c =
}}
{{eqn | l = 21
| ... | Common Factor Cancelling in Congruence/Corollary 1/Warning | https://proofwiki.org/wiki/Common_Factor_Cancelling_in_Congruence/Corollary_1/Warning | https://proofwiki.org/wiki/Common_Factor_Cancelling_in_Congruence/Corollary_1/Warning | [
"Common Factor Cancelling in Congruence"
] | [
"Definition:Coprime/Integers"
] | [
"Proof by Counterexample",
"Definition:Coprime/Integers"
] |
proofwiki-14440 | Congruence by Factors of Modulo/Warning | Let $r$ ''not'' be coprime to $s$.
Then it is not necessarily the case that:
:$a \equiv b \pmod {r s}$ {{iff}} $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$. | Proof by Counterexample:
Let $a = 30, b = 40, r = 2, s = 10$.
We have that:
{{begin-eqn}}
{{eqn | l = 30
| o = \equiv
| r = 40
| rr= \pmod 2
| c =
}}
{{eqn | l = 30
| o = \equiv
| r = 40
| rr= \pmod {10}
| c =
}}
{{eqn-intertext|But note that:}}
{{eqn | l = 30
| o... | Let $r$ ''not'' be [[Definition:Coprime Integers|coprime]] to $s$.
Then it is not necessarily the case that:
:$a \equiv b \pmod {r s}$ {{iff}} $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is [[Definition:Congruence (Number Theory)|congruent modulo $r$]] to $b$. | [[Proof by Counterexample]]:
Let $a = 30, b = 40, r = 2, s = 10$.
We have that:
{{begin-eqn}}
{{eqn | l = 30
| o = \equiv
| r = 40
| rr= \pmod 2
| c =
}}
{{eqn | l = 30
| o = \equiv
| r = 40
| rr= \pmod {10}
| c =
}}
{{eqn-intertext|But note that:}}
{{eqn | l = 30
... | Congruence by Factors of Modulo/Warning | https://proofwiki.org/wiki/Congruence_by_Factors_of_Modulo/Warning | https://proofwiki.org/wiki/Congruence_by_Factors_of_Modulo/Warning | [
"Congruence by Factors of Modulo"
] | [
"Definition:Coprime/Integers",
"Definition:Congruence (Number Theory)"
] | [
"Proof by Counterexample"
] |
proofwiki-14441 | Euler Phi Function of 2 | :$\map \phi 2 = 1$ | From Euler Phi Function of Prime:
:$\map \phi p = p - 1$
As $2$ is a prime number it follows that:
:$\map \phi 2 = 2 - 1 = 1$
{{qed}} | :$\map \phi 2 = 1$ | From [[Euler Phi Function of Prime]]:
:$\map \phi p = p - 1$
As $2$ is a [[Definition:Prime Number|prime number]] it follows that:
:$\map \phi 2 = 2 - 1 = 1$
{{qed}} | Euler Phi Function of 2 | https://proofwiki.org/wiki/Euler_Phi_Function_of_2 | https://proofwiki.org/wiki/Euler_Phi_Function_of_2 | [
"Examples of Euler Phi Function",
"2"
] | [] | [
"Euler Phi Function of Prime",
"Definition:Prime Number"
] |
proofwiki-14442 | Dirichlet Series Convergence Lemma/General | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$. | Let $s = \sigma + i t$
Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.
Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$
We may create a new Dirichlet series that converges at 0 by writing:
{{begin-eqn}}
{{eqn | l = \map g s
| r = \map f {s + s_0}
}}
{{eqn | r = \sum_{n \mathop = 1}... | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a [[Definition:General Dirichlet Series|general Dirichlet series]].
Let $\map f s$ [[Definition:Convergent Series|converge]] at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ [[Definition:Convergent Series|converge]] for all $s = \sigma + i t... | Let $s = \sigma + i t$
Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.
Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$
We may create a new Dirichlet series that converges at 0 by writing:
{{begin-eqn}}
{{eqn | l = \map g s
| r = \map f {s + s_0}
}}
{{eqn | r = \sum_{n \mathop ... | Dirichlet Series Convergence Lemma/General | https://proofwiki.org/wiki/Dirichlet_Series_Convergence_Lemma/General | https://proofwiki.org/wiki/Dirichlet_Series_Convergence_Lemma/General | [
"General Dirichlet Series"
] | [
"Definition:General Dirichlet Series",
"Definition:Convergent Series",
"Definition:Convergent Series"
] | [
"Cauchy's Convergence Criterion",
"Abel's Lemma/Formulation 2",
"Definition:Cauchy Sequence/Complex Numbers",
"Definition:Complex Modulus",
"Triangle Inequality for Integrals/Complex",
"Telescoping Series/Example 1",
"Category:General Dirichlet Series"
] |
proofwiki-14443 | Existence of Abscissa of Convergence | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.
Let the series $\ds \sum_{n \mathop = 1}^\infty \size {a_n n^{-s} }$ not converge for all $s \in \C$, or diverge for all $s \in \C$.
Then there exists a real number $\sigma_c$ such that $\map f s$ converges for all $s = \sigma + i t$ wit... | Let $S$ be the set of all complex numbers $s$ such that $\map f s$ converges.
By hypothesis, there is some $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ converges, so $S$ is not empty.
{{questionable|Which hypothesis?}}
Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Convergenc... | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
Let the series $\ds \sum_{n \mathop = 1}^\infty \size {a_n n^{-s} }$ not [[Definition:Convergent Series|converge]] for all $s \in \C$, or diverge for all $s \in \C$.
Then there exists a [[Definition:Rea... | Let $S$ be the set of all [[Definition:Complex Number|complex numbers]] $s$ such that $\map f s$ [[Definition:Convergent Series|converges]].
By hypothesis, there is some $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ [[Definition:Convergent Series|converges]], so $S$ is not [[Definition:Empty Set|empty]].
{... | Existence of Abscissa of Convergence | https://proofwiki.org/wiki/Existence_of_Abscissa_of_Convergence | https://proofwiki.org/wiki/Existence_of_Abscissa_of_Convergence | [
"Dirichlet Series"
] | [
"Definition:Dirichlet Series",
"Definition:Convergent Series",
"Definition:Real Number",
"Definition:Convergent Series",
"Definition:Convergent Series"
] | [
"Definition:Complex Number",
"Definition:Convergent Series",
"Definition:Convergent Series",
"Definition:Empty Set",
"Dirichlet Series Convergence Lemma",
"Definition:Convergent Series",
"Definition:Infimum of Set",
"Definition:Convergent Series",
"Dirichlet Series Convergence Lemma",
"Definition:... |
proofwiki-14444 | Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial | {{begin-eqn}}
{{eqn | l = n!
| r = \sum_{k \mathop \ge 0} \dfrac { {!k} \, n^{\underline k} } {k!}
| c =
}}
{{eqn | r = \dfrac { !0 \times n^{\underline 0} } {0!} + \dfrac { {!1} \times n^{\underline 1} } {1!} + \dfrac { {!2} \times n^{\underline 2} } {2!} + \dfrac { {!3} \times n^{\underline 3} } {3!} + \... | Let $n$ be a non-negative integer.
We assume a solution of the form:
:$(1): \quad n! = a_0 + a_1 n + a_2 n \paren {n - 1} + a_3 n \paren {n - 1} \paren {n - 2} + \cdots$
We can express $(1)$ using binomial coefficients:
:$(2): \quad n! = \ds \sum_k \dbinom n k k! a_k$
Then:
{{begin-eqn}}
{{eqn | l = \sum_n n! \binom m ... | {{begin-eqn}}
{{eqn | l = n!
| r = \sum_{k \mathop \ge 0} \dfrac { {!k} \, n^{\underline k} } {k!}
| c =
}}
{{eqn | r = \dfrac { !0 \times n^{\underline 0} } {0!} + \dfrac { {!1} \times n^{\underline 1} } {1!} + \dfrac { {!2} \times n^{\underline 2} } {2!} + \dfrac { {!3} \times n^{\underline 3} } {3!} + \... | Let $n$ be a [[Definition:Positive Integer|non-negative integer]].
We assume a solution of the form:
:$(1): \quad n! = a_0 + a_1 n + a_2 n \paren {n - 1} + a_3 n \paren {n - 1} \paren {n - 2} + \cdots$
We can express $(1)$ using [[Definition:Binomial Coefficient|binomial coefficients]]:
:$(2): \quad n! = \ds \sum_k... | Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial/Proof | https://proofwiki.org/wiki/Factorial_as_Sum_of_Series_of_Subfactorial_by_Falling_Factorial_over_Factorial | https://proofwiki.org/wiki/Factorial_as_Sum_of_Series_of_Subfactorial_by_Falling_Factorial_over_Factorial/Proof | [
"Factorials",
"Falling Factorials",
"Subfactorials",
"Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial"
] | [] | [
"Definition:Positive/Integer",
"Definition:Binomial Coefficient",
"Permutation of Indices of Summation"
] |
proofwiki-14445 | Factorial of Half | :$\left({\dfrac 1 2}\right)! = \dfrac {\sqrt \pi} 2$ | {{begin-eqn}}
{{eqn | l = \paren {\dfrac 1 2}!
| r = \map \Gamma {1 + \dfrac 1 2}
| c = Gamma Function Extends Factorial
}}
{{eqn | r = \dfrac 1 2 \map \Gamma {\dfrac 1 2}
| c = Gamma Difference Equation
}}
{{eqn | r = \dfrac 1 2 \sqrt {\pi}
| c = Gamma Function of One Half
}}
{{end-eqn}}
{{qed}... | :$\left({\dfrac 1 2}\right)! = \dfrac {\sqrt \pi} 2$ | {{begin-eqn}}
{{eqn | l = \paren {\dfrac 1 2}!
| r = \map \Gamma {1 + \dfrac 1 2}
| c = [[Gamma Function Extends Factorial]]
}}
{{eqn | r = \dfrac 1 2 \map \Gamma {\dfrac 1 2}
| c = [[Gamma Difference Equation]]
}}
{{eqn | r = \dfrac 1 2 \sqrt {\pi}
| c = [[Gamma Function of One Half]]
}}
{{end-... | Factorial of Half/Proof 1 | https://proofwiki.org/wiki/Factorial_of_Half | https://proofwiki.org/wiki/Factorial_of_Half/Proof_1 | [
"Factorials/Examples",
"Factorial of Half"
] | [] | [
"Gamma Function Extends Factorial",
"Gamma Difference Equation",
"Gamma Function of One Half"
] |
proofwiki-14446 | Factorial of Half | :$\left({\dfrac 1 2}\right)! = \dfrac {\sqrt \pi} 2$ | From Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta:
{{:Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta}}
Setting:
:$k = 2$
:$\alpha_1 = \alpha_2 = 0$
:$\beta_1 = -\dfrac 1 2, \beta_2 = \dfrac 1 2$
we see that:
:$\alpha_1 + \alpha_2 = \beta_1 + ... | :$\left({\dfrac 1 2}\right)! = \dfrac {\sqrt \pi} 2$ | From [[Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta]]:
{{:Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta}}
Setting:
:$k = 2$
:$\alpha_1 = \alpha_2 = 0$
:$\beta_1 = -\dfrac 1 2, \beta_2 = \dfrac 1 2$
we see that:
:$\alpha_1 + \alpha_2 = \bet... | Factorial of Half/Proof 2 | https://proofwiki.org/wiki/Factorial_of_Half | https://proofwiki.org/wiki/Factorial_of_Half/Proof_2 | [
"Factorials/Examples",
"Factorial of Half"
] | [] | [
"Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta",
"Gamma Difference Equation",
"Wallis's Product",
"Gamma Function Extends Factorial"
] |
proofwiki-14447 | Laplace Transform of Complex Power | Let $q$ be a constant complex number with $\map \Re q > -1$.
Let $t^q$ be the the principal branch of the $q$-th complex power whose domain contains the half-plane $\map \Re s > 0$.
Then $t^q$ has a Laplace transform given by:
:$\laptrans {t^q} = \dfrac {\map \Gamma {q + 1} } {s^{q + 1} }$
where $\Gamma$ denotes the ga... | By definition of Laplace transform for a function not continuous at zero:
:$\ds \laptrans {t^q} = \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \map I {\varepsilon, L}$
where:
:$\ds \map I {\varepsilon, L} = \int_\varepsilon^L t^q e^{-s t} \rd t$
Let $n \in \Z_{>0}$.
From Laplace Transform of Positive... | Let $q$ be a [[Definition:Constant|constant]] [[Definition:Complex Number|complex number]] with $\map \Re q > -1$.
Let $t^q$ be the [[Definition:Principal Branch of Complex Number|the principal branch]] of the $q$-th [[Definition:Power to Complex Number|complex power]] whose [[Definition:Domain of Mapping|domain]] con... | By definition of [[Definition:Laplace Transform/Discontinuity at Zero|Laplace transform for a function not continuous at zero]]:
:$\ds \laptrans {t^q} = \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \map I {\varepsilon, L}$
where:
:$\ds \map I {\varepsilon, L} = \int_\varepsilon^L t^q e^{-s t} \rd t$... | Laplace Transform of Complex Power | https://proofwiki.org/wiki/Laplace_Transform_of_Complex_Power | https://proofwiki.org/wiki/Laplace_Transform_of_Complex_Power | [
"Examples of Laplace Transforms",
"Examples of Ansatz"
] | [
"Definition:Constant",
"Definition:Complex Number",
"Definition:Power (Algebra)/Complex Number/Principal Branch",
"Definition:Power (Algebra)/Complex Number",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Complex Half-Plane",
"Definition:Laplace Transform",
"Definition:Gamma Function"
] | [
"Definition:Laplace Transform/Discontinuity at Zero",
"Laplace Transform of Positive Integer Power",
"Gamma Function Extends Factorial",
"Definition:Ansatz",
"Poles of Gamma Function",
"Definition:Complex Number/Wholly Real",
"Definition:Complex Number",
"Definition:Gamma Function/Integral Form",
"C... |
proofwiki-14448 | Integer to Power of Itself Less One Falling is Factorial | :$n^{\underline {n - 1} } = n!$ | {{begin-eqn}}
{{eqn | l = n^{\underline {n - 1} }
| r = \dfrac {n!} {\left({n - \left({n - 1}\right)}\right)!}
| c = Falling Factorial as Quotient of Factorials
}}
{{eqn | r = \dfrac {n!} {1!}
| c =
}}
{{eqn | r = n!
| c = Factorial of $1$
}}
{{end-eqn}}
{{qed}}
Category:Falling Factorials
bh4x... | :$n^{\underline {n - 1} } = n!$ | {{begin-eqn}}
{{eqn | l = n^{\underline {n - 1} }
| r = \dfrac {n!} {\left({n - \left({n - 1}\right)}\right)!}
| c = [[Falling Factorial as Quotient of Factorials]]
}}
{{eqn | r = \dfrac {n!} {1!}
| c =
}}
{{eqn | r = n!
| c = [[Factorial/Examples/1|Factorial of $1$]]
}}
{{end-eqn}}
{{qed}}
[[... | Integer to Power of Itself Less One Falling is Factorial | https://proofwiki.org/wiki/Integer_to_Power_of_Itself_Less_One_Falling_is_Factorial | https://proofwiki.org/wiki/Integer_to_Power_of_Itself_Less_One_Falling_is_Factorial | [
"Falling Factorials"
] | [] | [
"Falling Factorial as Quotient of Factorials",
"Factorial/Examples/1",
"Category:Falling Factorials"
] |
proofwiki-14449 | Stirling's Formula/Refinement/Proof 2 | A refinement of Stirling's Formula is:
:$n! = \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} + \map \OO {\dfrac 1 {n^2} } }$ | Let $z\in \R_{>0}$ and $n \in \N_{\ge 0}$
Let $\ds c_n = \ln \map \Gamma {z + n}$
We begin by observing:
{{begin-eqn}}
{{eqn | l = \map \Gamma {z + n}
| r = \map \Gamma {z + 1} \times \paren {z + 1} \times \paren {z + 2} \times \cdots \times \paren {z + n - 1}
| c = Gamma Difference Equation
}}
{{eqn | ll =... | A refinement of [[Stirling's Formula]] is:
:$n! = \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} + \map \OO {\dfrac 1 {n^2} } }$ | Let $z\in \R_{>0}$ and $n \in \N_{\ge 0}$
Let $\ds c_n = \ln \map \Gamma {z + n}$
We begin by observing:
{{begin-eqn}}
{{eqn | l = \map \Gamma {z + n}
| r = \map \Gamma {z + 1} \times \paren {z + 1} \times \paren {z + 2} \times \cdots \times \paren {z + n - 1}
| c = [[Gamma Difference Equation]]
}}
{{eqn ... | Stirling's Formula/Refinement/Proof 2 | https://proofwiki.org/wiki/Stirling's_Formula/Refinement/Proof_2 | https://proofwiki.org/wiki/Stirling's_Formula/Refinement/Proof_2 | [
"Stirling's Formula"
] | [
"Stirling's Formula"
] | [
"Gamma Difference Equation",
"Sum of Logarithms",
"Definition:Derivative",
"Definition:Finite Difference Operator",
"Primitive of Logarithm of x",
"Definition:Derivative",
"Definition:Finite Difference Operator",
"Primitive of Logarithm of x",
"Sum of Logarithms",
"Power Series Expansion for Logar... |
proofwiki-14450 | Recursive Form of Generalized Termial | The termial function as defined on the real numbers fulfils the identity:
:$x? = x + \paren {x - 1}?$ | By definition of the termial on the real numbers:
:$x? = \dfrac {x \paren {x + 1} } 2$
Thus:
{{begin-eqn}}
{{eqn | l = x? - x
| r = \dfrac {x \paren {x + 1} } 2 - x
| c = {{Defof|Termial/Real Numbers|Termial}}
}}
{{eqn | r = \dfrac {x \paren {x + 1} - 2 x} 2
| c =
}}
{{eqn | r = \dfrac {x^2 + x - 2 x... | The [[Definition:Termial|termial function]] as [[Definition:Termial/Real Numbers|defined]] on the [[Definition:Real Number|real numbers]] fulfils the identity:
:$x? = x + \paren {x - 1}?$ | By definition of the [[Definition:Termial/Real Numbers|termial]] on the [[Definition:Real Number|real numbers]]:
:$x? = \dfrac {x \paren {x + 1} } 2$
Thus:
{{begin-eqn}}
{{eqn | l = x? - x
| r = \dfrac {x \paren {x + 1} } 2 - x
| c = {{Defof|Termial/Real Numbers|Termial}}
}}
{{eqn | r = \dfrac {x \paren {... | Recursive Form of Generalized Termial | https://proofwiki.org/wiki/Recursive_Form_of_Generalized_Termial | https://proofwiki.org/wiki/Recursive_Form_of_Generalized_Termial | [
"Termial Function"
] | [
"Definition:Termial",
"Definition:Termial/Real Numbers",
"Definition:Real Number"
] | [
"Definition:Termial/Real Numbers",
"Definition:Real Number"
] |
proofwiki-14451 | Euler Form of Gamma Function at Positive Integers | The Euler form of the Gamma function:
:$\ds \map \Gamma z := \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \cdots \paren {z + m} }$
converges to the factorial function at positive integers:
:$\ds \lim_{m \mathop \to \infty} \frac {m^n m!} {\paren {n + 1} \paren {n + 2} \cdots \paren {n + m... | {{begin-eqn}}
{{eqn | l = \lim_{m \mathop \to \infty} \frac {m^n m!} {\paren {n + 1} \paren {n + 2} \cdots \paren {n + m} }
| r = \lim_{m \mathop \to \infty} \frac {m^n m!} {\frac {\paren {n + m}!} {n!} }
| c =
}}
{{eqn | r = n! \lim_{m \mathop \to \infty} \frac {m^n} {\paren {m + 1} \paren {m + 2} \cdots ... | The [[Definition:Euler Form of Gamma Function|Euler form of the Gamma function]]:
:$\ds \map \Gamma z := \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \cdots \paren {z + m} }$
converges to the [[Definition:Factorial|factorial function]] at [[Definition:Positive Integer|positive integers]]... | {{begin-eqn}}
{{eqn | l = \lim_{m \mathop \to \infty} \frac {m^n m!} {\paren {n + 1} \paren {n + 2} \cdots \paren {n + m} }
| r = \lim_{m \mathop \to \infty} \frac {m^n m!} {\frac {\paren {n + m}!} {n!} }
| c =
}}
{{eqn | r = n! \lim_{m \mathop \to \infty} \frac {m^n} {\paren {m + 1} \paren {m + 2} \cdots ... | Euler Form of Gamma Function at Positive Integers | https://proofwiki.org/wiki/Euler_Form_of_Gamma_Function_at_Positive_Integers | https://proofwiki.org/wiki/Euler_Form_of_Gamma_Function_at_Positive_Integers | [
"Gamma Function"
] | [
"Definition:Gamma Function/Euler Form",
"Definition:Factorial",
"Definition:Positive/Integer"
] | [
"Factorial of Integer plus Reciprocal of Integer"
] |
proofwiki-14452 | Gamma Function of Minus One Half | :$\map \Gamma {-\dfrac 1 2} = -2 \sqrt \pi$ | {{begin-eqn}}
{{eqn | l = \map \Gamma {-\dfrac 1 2}
| r = \frac {\map \Gamma {\frac 1 2} } {-1/2}
| c = Gamma Difference Equation
}}
{{eqn | r = -2 \, \map \Gamma {\frac 1 2}
| c =
}}
{{eqn | r = -2 \sqrt \pi
| c = Gamma Function of One Half
}}
{{end-eqn}}
{{qed}} | :$\map \Gamma {-\dfrac 1 2} = -2 \sqrt \pi$ | {{begin-eqn}}
{{eqn | l = \map \Gamma {-\dfrac 1 2}
| r = \frac {\map \Gamma {\frac 1 2} } {-1/2}
| c = [[Gamma Difference Equation]]
}}
{{eqn | r = -2 \, \map \Gamma {\frac 1 2}
| c =
}}
{{eqn | r = -2 \sqrt \pi
| c = [[Gamma Function of One Half]]
}}
{{end-eqn}}
{{qed}} | Gamma Function of Minus One Half | https://proofwiki.org/wiki/Gamma_Function_of_Minus_One_Half | https://proofwiki.org/wiki/Gamma_Function_of_Minus_One_Half | [
"Examples of Gamma Function Values"
] | [] | [
"Gamma Difference Equation",
"Gamma Function of One Half"
] |
proofwiki-14453 | Legendre's Theorem/Corollary | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $B$ be the binary representation of $n$.
Let $r$ be the number of unit digits in $B$.
Let $n!$ denote the factorial of $n$.
Then $2^{n - r}$ is a divisor of $n!$, but $2^{n - r + 1}$ is not. | $n$ can be represented as:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 1}^r 2^{e_j}
| c = where $e_1 > e_2 > \cdots > e_r > 0$
}}
{{eqn | r = 2^{e_1} + 2^{e_2} + \cdots + 2^{e_r}
| c =
}}
{{end-eqn}}
where all of $e_1, e_2, \ldots, e_r$ are integers.
Then $r$ is the sum of the digits in $n$ a... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $B$ be the [[Definition:Binary Notation|binary representation]] of $n$.
Let $r$ be the number of [[Definition:Unit (One)|unit digits]] in $B$.
Let $n!$ denote the [[Definition:Factorial|factorial of $n$]].
Then $2^{n... | $n$ can be represented as:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 1}^r 2^{e_j}
| c = where $e_1 > e_2 > \cdots > e_r > 0$
}}
{{eqn | r = 2^{e_1} + 2^{e_2} + \cdots + 2^{e_r}
| c =
}}
{{end-eqn}}
where all of $e_1, e_2, \ldots, e_r$ are [[Definition:Integer|integers]].
Then $r$ is the [... | Legendre's Theorem/Corollary | https://proofwiki.org/wiki/Legendre's_Theorem/Corollary | https://proofwiki.org/wiki/Legendre's_Theorem/Corollary | [
"Factorials"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Binary Notation",
"Definition:Unit (One)",
"Definition:Factorial",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Digit",
"Definition:Binary Notation",
"Legendre's Theorem"
] |
proofwiki-14454 | Diameter of N-Cube | Let $Q_n = \closedint {c - R} {c + R}^n$ be an $n$-cube in Euclidean $n$-Space equipped with the usual metric.
Then the diameter of $Q_n$ is given by:
:$\map \diam {Q_n} = 2 R \sqrt n$ | Write:
:$Q_n = \ds \prod_{i \mathop = 1}^n \closedint {c - R} {c + R}_i$
Let $x, y \in Q_n$
By the definition of the usual metric, the distance between any two points $x$ and $y$ is given by:
:$\ds \map d {y - x} = \paren {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2}^{1 / 2}$
By Positive Power Function on Non-negative... | Let $Q_n = \closedint {c - R} {c + R}^n$ be an [[Definition:N-Cube (Euclidean Space)|$n$-cube]] in [[Definition:Real Euclidean Space|Euclidean $n$-Space]] equipped with the [[Definition:Usual Metric|usual metric]].
Then the [[Definition:Diameter of Subset of Metric Space|diameter]] of $Q_n$ is given by:
:$\map \diam... | Write:
:$Q_n = \ds \prod_{i \mathop = 1}^n \closedint {c - R} {c + R}_i$
Let $x, y \in Q_n$
By the definition of the usual metric, the distance between any two points $x$ and $y$ is given by:
:$\ds \map d {y - x} = \paren {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2}^{1 / 2}$
By [[Positive Power Function on Non-n... | Diameter of N-Cube | https://proofwiki.org/wiki/Diameter_of_N-Cube | https://proofwiki.org/wiki/Diameter_of_N-Cube | [
"Real Analysis"
] | [
"Definition:N-Cube (Euclidean Space)",
"Definition:Euclidean Space/Real",
"Definition:Usual Metric",
"Definition:Diameter of Subset of Metric Space"
] | [
"Positive Power Function on Non-negative Reals is Strictly Increasing",
"Definition:Addition/Summand",
"Definition:Real Interval/Length",
"Sum of Identical Terms",
"Category:Real Analysis"
] |
proofwiki-14455 | Legendre's Theorem | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $p$ be a prime number.
Let $n$ be expressed in base $p$ representation.
Let $r$ be the digit sum of the representation of $n$ in base $p$.
Then $n!$ is divisible by $p^\mu$ but not by $p^{\mu + 1}$, where:
:$\mu = \dfrac {n - r} {p - 1}$ | $n$ can be represented as:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 0}^m a_j p^j
| c = where $0 \le a_j < p$
}}
{{eqn | r = a_0 + a_1 p + a_2 p^2 + \cdots + a_m p^m
| c = for some $m > 0$
}}
{{end-eqn}}
Using De Polignac's Formula, we may extract all the powers of $p$ from $n!$.
:$\mu = \ds... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n$ be expressed in [[Definition:Number Base|base $p$ representation]].
Let $r$ be the [[Definition:Digit Sum|digit sum]] of the representation of $n$ in [[Definit... | $n$ can be represented as:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 0}^m a_j p^j
| c = where $0 \le a_j < p$
}}
{{eqn | r = a_0 + a_1 p + a_2 p^2 + \cdots + a_m p^m
| c = for some $m > 0$
}}
{{end-eqn}}
Using [[De Polignac's Formula]], we may extract all the [[Definition:Integer Power|pow... | Legendre's Theorem | https://proofwiki.org/wiki/Legendre's_Theorem | https://proofwiki.org/wiki/Legendre's_Theorem | [
"Factorials"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Prime Number",
"Definition:Number Base",
"Definition:Digit Sum",
"Definition:Number Base",
"Definition:Divisor (Algebra)/Integer"
] | [
"De Polignac's Formula",
"Definition:Power (Algebra)/Integer",
"Definition:Prime Decomposition/Multiplicity",
"Sum of Geometric Sequence"
] |
proofwiki-14456 | Uniform Convergence of General Dirichlet Series | Let $\map \arg z$ denote the argument of the complex number $z \in \C$.
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converges uniformly for all $s$ such that:
:$\cmod {\map \arg {s - s_0}... | Let $s = \sigma + i t$
Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.
Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$
We may create a new Dirichlet series that converges at $0$ by writing:
{{begin-eqn}}
{{eqn | l = \map g s
| r = \map f {s + s_0}
}}
{{eqn | r = \sum_{n \mathop = ... | Let $\map \arg z$ denote the [[Definition:Argument of Complex Number|argument]] of the [[Definition:Complex Number|complex number]] $z \in \C$.
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a [[Definition:General Dirichlet Series|general Dirichlet series]].
Let $\map f s$ [[Definitio... | Let $s = \sigma + i t$
Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.
Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$
We may create a new Dirichlet series that converges at $0$ by writing:
{{begin-eqn}}
{{eqn | l = \map g s
| r = \map f {s + s_0}
}}
{{eqn | r = \sum_{n \matho... | Uniform Convergence of General Dirichlet Series | https://proofwiki.org/wiki/Uniform_Convergence_of_General_Dirichlet_Series | https://proofwiki.org/wiki/Uniform_Convergence_of_General_Dirichlet_Series | [
"General Dirichlet Series"
] | [
"Definition:Argument of Complex Number",
"Definition:Complex Number",
"Definition:General Dirichlet Series",
"Definition:Convergent Series",
"Definition:Uniform Convergence"
] | [
"Cauchy's Convergence Criterion",
"Abel's Lemma/Formulation 2",
"Triangle Inequality",
"Definition:Cauchy Sequence/Complex Numbers",
"Triangle Inequality for Integrals/Complex",
"Shape of Secant Function",
"Telescoping Series/Example 1"
] |
proofwiki-14457 | Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial/Condition for Convergence | Consider the series:
{{:Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial}}
This converges only when $n \in \Z_{\ge 0}$, that is, when $n$ is a non-negative integer. | First we show that this series converges when $n \in \Z_{\ge 0}$.
Consider the coefficients:
:$1, \paren {1 - \dfrac 1 {1!} }, \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} }, \ldots$
By Power Series Expansion for Exponential Function, they converge to $\dfrac 1 e$.
Starting from the $\paren {n + 1}$th term, there is a fac... | Consider the series:
{{:Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial}}
This [[Definition:Convergent Series|converges]] only when $n \in \Z_{\ge 0}$, that is, when $n$ is a [[Definition:Positive Integer|non-negative integer]]. | First we show that this series [[Definition:Convergent Series|converges]] when $n \in \Z_{\ge 0}$.
Consider the [[Definition:Coefficient|coefficients]]:
:$1, \paren {1 - \dfrac 1 {1!} }, \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} }, \ldots$
By [[Power Series Expansion for Exponential Function]], they [[Definition:Conv... | Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial/Condition for Convergence | https://proofwiki.org/wiki/Factorial_as_Sum_of_Series_of_Subfactorial_by_Falling_Factorial_over_Factorial/Condition_for_Convergence | https://proofwiki.org/wiki/Factorial_as_Sum_of_Series_of_Subfactorial_by_Falling_Factorial_over_Factorial/Condition_for_Convergence | [
"Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial"
] | [
"Definition:Convergent Series",
"Definition:Positive/Integer"
] | [
"Definition:Convergent Series",
"Definition:Coefficient",
"Power Series Expansion for Exponential Function",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Convergent Series",
"Definition:Divergent Series",
"Divergence Test",
"Power Series Expansion for Exponential Function",
"Definition... |
proofwiki-14458 | Bounded Subspace of Euclidean Space is Totally Bounded | Let $\struct {\R^n, \norm {\, \cdot \,} }$ be a Euclidean space, where $\norm {\, \cdot \,}$ denotes the usual metric.
Let $M$ be a metric subspace of $\struct {\R^n, \norm {\, \cdot \,} }$.
Let $M$ be bounded.
Then $M$ is totally bounded. | As $M$ is bounded, it has a finite diameter $R$.
Consider arbitary $x, y \in \R^n$, expressed in their usual coordinates.
{{begin-eqn}}
{{eqn | l = \norm {y - x}
| r = \sqrt {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2}
}}
{{eqn | o = \le
| r = \sqrt {n \max_i \set {\paren {y_i - x_i}^2} }
}}
{{eqn | r = \s... | Let $\struct {\R^n, \norm {\, \cdot \,} }$ be a [[Definition:Euclidean Space|Euclidean space]], where $\norm {\, \cdot \,}$ denotes the [[Definition:Usual Metric|usual metric]].
Let $M$ be a [[Definition:Metric Subspace|metric subspace]] of $\struct {\R^n, \norm {\, \cdot \,} }$.
Let $M$ be [[Definition:Bounded Metri... | As $M$ is [[Definition:Bounded Metric Space|bounded]], it has a finite [[Definition:Diameter of Subset of Metric Space|diameter]] $R$.
Consider arbitary $x, y \in \R^n$, expressed in their usual [[Definition:Coordinate|coordinates]].
{{begin-eqn}}
{{eqn | l = \norm {y - x}
| r = \sqrt {\sum_{i \mathop = 1}^n \p... | Bounded Subspace of Euclidean Space is Totally Bounded | https://proofwiki.org/wiki/Bounded_Subspace_of_Euclidean_Space_is_Totally_Bounded | https://proofwiki.org/wiki/Bounded_Subspace_of_Euclidean_Space_is_Totally_Bounded | [
"Bounded Metric Spaces"
] | [
"Definition:Euclidean Space",
"Definition:Usual Metric",
"Definition:Metric Subspace",
"Definition:Bounded Metric Space",
"Definition:Totally Bounded Metric Space"
] | [
"Definition:Bounded Metric Space",
"Definition:Diameter of Subset of Metric Space",
"Definition:Coordinate",
"Characterization of N-Cube",
"Diameter of N-Cube",
"Definition:Subdivision of Interval/Normal Subdivision",
"Definition:Real Interval/Length",
"Definition:Subinterval",
"Definition:Subinterv... |
proofwiki-14459 | Difference is Rational is Equivalence Relation | Define $\sim$ as the relation on real numbers given by:
:$x \sim y \iff x - y \in \Q$
That is, that the difference between $x$ and $y$ is rational.
Then $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | Define $\sim$ as the [[Definition:Relation|relation]] on [[Definition:Real Number|real numbers]] given by:
:$x \sim y \iff x - y \in \Q$
That is, that the [[Definition:Real Subtraction|difference]] between $x$ and $y$ is [[Definition:Rational Number|rational]].
Then $\sim$ is an [[Definition:Equivalence Relation|eq... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Difference is Rational is Equivalence Relation | https://proofwiki.org/wiki/Difference_is_Rational_is_Equivalence_Relation | https://proofwiki.org/wiki/Difference_is_Rational_is_Equivalence_Relation | [
"Examples of Equivalence Relations"
] | [
"Definition:Relation",
"Definition:Real Number",
"Definition:Subtraction/Real Numbers",
"Definition:Rational Number",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-14460 | Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta | :$\ds \prod_{n \mathop \ge 1} \dfrac {\paren {n + \alpha_1} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \cdots \paren {n + \beta_k} } = \dfrac {\map \Gamma {1 + \beta_1} \cdots \map \Gamma {1 + \beta_k} } {\map \Gamma {1 + \alpha_1} \cdots \map \Gamma {1 + \alpha_k} }$
where:
:$\alpha_1 + \cdots + \alpha_k = \... | First we note that if any of the $\beta$s is a negative integer, the {{LHS}} would have $0$ as its denominator, and so would be undefined.
We have from the Euler form of the Gamma function that:
:$\map \Gamma {1 + \beta_i} = \ds \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_i} m!} {\paren {1 + \beta_i} \paren {2 + \... | :$\ds \prod_{n \mathop \ge 1} \dfrac {\paren {n + \alpha_1} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \cdots \paren {n + \beta_k} } = \dfrac {\map \Gamma {1 + \beta_1} \cdots \map \Gamma {1 + \beta_k} } {\map \Gamma {1 + \alpha_1} \cdots \map \Gamma {1 + \alpha_k} }$
where:
:$\alpha_1 + \cdots + \alpha_k = ... | First we note that if any of the $\beta$s is a [[Definition:Negative Integer|negative integer]], the {{LHS}} would have $0$ as its [[Definition:Denominator|denominator]], and so would be undefined.
We have from the [[Definition:Euler Form of Gamma Function|Euler form of the Gamma function]] that:
:$\map \Gamma {1 + \b... | Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta | https://proofwiki.org/wiki/Infinite_Product_of_Product_of_Sequence_of_n_plus_alpha_over_Sequence_of_n_plus_beta | https://proofwiki.org/wiki/Infinite_Product_of_Product_of_Sequence_of_n_plus_alpha_over_Sequence_of_n_plus_beta | [
"Gamma Function"
] | [
"Definition:Negative/Integer"
] | [
"Definition:Negative/Integer",
"Definition:Fraction/Denominator",
"Definition:Gamma Function/Euler Form"
] |
proofwiki-14461 | Characterization of N-Cube | Let $\struct {\R^n, d}$ be a Euclidean $n$-Space equipped with the usual metric $d$.
Let $x, y \in \R^n$, where:
:$x = \tuple {x_1, x_2, \ldots, x_n}$
:$y = \tuple {y_1, y_2, \ldots, y_n}$
Let $R > 0$ be fixed.
Let:
:$\ds Q = \set {x, y \in \R^n: \sup_{x, y} \max_i \size {y_i - x_i} \le R}$
Then $Q$ is an $n$-cube. | For ease of presentation, denote:
:$y - x = r \in \R^n$
and:
:$y_j - x_j = r_j$
for $j = 1, 2, \ldots, n$.
{{begin-eqn}}
{{eqn | l = Q
| r = \set {r: \map {\sup_r} {\max_i \size {r_i} \le R} }
}}
{{eqn | r = \set {r: \map {\sup_r} {\map \max {\size {r_1}, \size {r_2}, \ldots, \size {r_n} } \le R} }
}}
{{eqn | r =... | Let $\struct {\R^n, d}$ be a [[Definition:Real Euclidean Space|Euclidean $n$-Space]] equipped with the [[Definition:Usual Metric|usual metric]] $d$.
Let $x, y \in \R^n$, where:
:$x = \tuple {x_1, x_2, \ldots, x_n}$
:$y = \tuple {y_1, y_2, \ldots, y_n}$
Let $R > 0$ be fixed.
Let:
:$\ds Q = \set {x, y \in \R^n: \sup_... | For ease of presentation, denote:
:$y - x = r \in \R^n$
and:
:$y_j - x_j = r_j$
for $j = 1, 2, \ldots, n$.
{{begin-eqn}}
{{eqn | l = Q
| r = \set {r: \map {\sup_r} {\max_i \size {r_i} \le R} }
}}
{{eqn | r = \set {r: \map {\sup_r} {\map \max {\size {r_1}, \size {r_2}, \ldots, \size {r_n} } \le R} }
}}
{{eqn ... | Characterization of N-Cube | https://proofwiki.org/wiki/Characterization_of_N-Cube | https://proofwiki.org/wiki/Characterization_of_N-Cube | [
"Metric Spaces",
"Real Analysis"
] | [
"Definition:Euclidean Space/Real",
"Definition:Usual Metric",
"Definition:N-Cube (Euclidean Space)"
] | [] |
proofwiki-14462 | Factorial of Integer plus Reciprocal of Integer | Let $x \in \Z$ be a positive integer.
Then:
:$\ds \lim_{n \mathop \to \infty} \dfrac {\paren {n + x}!} {n! n^x} = 1$ | We have that:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {n + x}!} {n! n^x}
| r = \dfrac {\paren {n + 1} \paren {n + 2} \cdots \paren {n + x} } {n^x}
| c =
}}
{{eqn | r = \paren {\dfrac {n + 1} n} \paren {\dfrac {n + 2} n} \cdots \paren {\dfrac {n + x} n}
| c =
}}
{{eqn | r = \paren {1 + \frac 1 n} \p... | Let $x \in \Z$ be a [[Definition:Positive Integer|positive integer]].
Then:
:$\ds \lim_{n \mathop \to \infty} \dfrac {\paren {n + x}!} {n! n^x} = 1$ | We have that:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {n + x}!} {n! n^x}
| r = \dfrac {\paren {n + 1} \paren {n + 2} \cdots \paren {n + x} } {n^x}
| c =
}}
{{eqn | r = \paren {\dfrac {n + 1} n} \paren {\dfrac {n + 2} n} \cdots \paren {\dfrac {n + x} n}
| c =
}}
{{eqn | r = \paren {1 + \frac 1 n} \p... | Factorial of Integer plus Reciprocal of Integer | https://proofwiki.org/wiki/Factorial_of_Integer_plus_Reciprocal_of_Integer | https://proofwiki.org/wiki/Factorial_of_Integer_plus_Reciprocal_of_Integer | [
"Factorials"
] | [
"Definition:Positive/Integer"
] | [] |
proofwiki-14463 | Lower and Upper Bound of Factorial | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\dfrac {n^n} {e^{n - 1} } \le n! \le \dfrac {n^{n + 1} } {e^{n - 1} }$ | We have:
{{begin-eqn}}
{{eqn | l = 1 + x
| o = \le
| r = e^x
| c = Exponential of $x$ not less than $1+x$
}}
{{eqn | ll= \leadsto
| l = 1 + \frac 1 k
| o = \le
| r = e^{1 / k}
| c =
}}
{{eqn | n = 1
| ll= \leadsto
| l = \frac {k + 1} k
| o = \le
| r = e... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Then:
:$\dfrac {n^n} {e^{n - 1} } \le n! \le \dfrac {n^{n + 1} } {e^{n - 1} }$ | We have:
{{begin-eqn}}
{{eqn | l = 1 + x
| o = \le
| r = e^x
| c = [[Exponential of x not less than 1+x|Exponential of $x$ not less than $1+x$]]
}}
{{eqn | ll= \leadsto
| l = 1 + \frac 1 k
| o = \le
| r = e^{1 / k}
| c =
}}
{{eqn | n = 1
| ll= \leadsto
| l = \frac... | Lower and Upper Bound of Factorial | https://proofwiki.org/wiki/Lower_and_Upper_Bound_of_Factorial | https://proofwiki.org/wiki/Lower_and_Upper_Bound_of_Factorial | [
"Factorials"
] | [
"Definition:Strictly Positive/Integer"
] | [
"Exponential of x not less than 1+x",
"Exponential of x not less than 1+x",
"Ordering of Reciprocals"
] |
proofwiki-14464 | Sum of Indices of Falling Factorial | :$x^{\underline {m + n} } = x^{\underline m} \paren {x - m}^{\underline n}$ | {{begin-eqn}}
{{eqn | l = x^{\underline {m + n} }
| r = \prod_{j \mathop = 0}^{m + n - 1} \paren {x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x - j} \prod_{j \mathop = m}^{m + n - 1} \paren {x - j}
| c =
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \p... | :$x^{\underline {m + n} } = x^{\underline m} \paren {x - m}^{\underline n}$ | {{begin-eqn}}
{{eqn | l = x^{\underline {m + n} }
| r = \prod_{j \mathop = 0}^{m + n - 1} \paren {x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x - j} \prod_{j \mathop = m}^{m + n - 1} \paren {x - j}
| c =
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \p... | Sum of Indices of Falling Factorial | https://proofwiki.org/wiki/Sum_of_Indices_of_Falling_Factorial | https://proofwiki.org/wiki/Sum_of_Indices_of_Falling_Factorial | [
"Falling Factorials"
] | [] | [
"Translation of Index Variable of Product"
] |
proofwiki-14465 | Sum of Indices of Rising Factorial | :$x^{\overline {m + n} } = x^{\overline m} \paren {x + m}^{\overline n}$ | {{begin-eqn}}
{{eqn | l = x^{\overline {m + n} }
| r = \prod_{j \mathop = 0}^{m + n - 1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = m}^{m + n - 1} \paren {x + j}
| c =
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \par... | :$x^{\overline {m + n} } = x^{\overline m} \paren {x + m}^{\overline n}$ | {{begin-eqn}}
{{eqn | l = x^{\overline {m + n} }
| r = \prod_{j \mathop = 0}^{m + n - 1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = m}^{m + n - 1} \paren {x + j}
| c =
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \par... | Sum of Indices of Rising Factorial | https://proofwiki.org/wiki/Sum_of_Indices_of_Rising_Factorial | https://proofwiki.org/wiki/Sum_of_Indices_of_Rising_Factorial | [
"Rising Factorials"
] | [] | [
"Translation of Index Variable of Product"
] |
proofwiki-14466 | Union of Indexed Family of Sets Equal to Union of Disjoint Sets | Let $\family {E_n}_{n \mathop \in \N}$ be a countable indexed family of sets where at least two $E_n$ are distinct.
Then there exists a countable indexed family of disjoint sets $\family {F_n}_{n \mathop \in \N}$ defined by:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
satisfying:
:$\... | Denote:
{{begin-eqn}}
{{eqn | l = E
| r = \bigcup_{k \mathop \in \N} E_k
}}
{{eqn | l = F
| r = \bigcup_{k \mathop \in \N} F_k
}}
{{end-eqn}}
where:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
We first show that $E = F$.
That $x \in E \implies x \in F$ follows from the co... | Let $\family {E_n}_{n \mathop \in \N}$ be a [[Definition:Countable Set|countable]] [[Definition:Indexed Family of Subsets|indexed family of sets]] where at least two $E_n$ are [[Definition:Distinct|distinct]].
Then there exists a [[Definition:Countable Set|countable]] [[Definition:Pairwise Disjoint Family|indexed fam... | Denote:
{{begin-eqn}}
{{eqn | l = E
| r = \bigcup_{k \mathop \in \N} E_k
}}
{{eqn | l = F
| r = \bigcup_{k \mathop \in \N} F_k
}}
{{end-eqn}}
where:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
We first show that $E = F$.
That $x \in E \implies x \in F$ follows from... | Union of Indexed Family of Sets Equal to Union of Disjoint Sets | https://proofwiki.org/wiki/Union_of_Indexed_Family_of_Sets_Equal_to_Union_of_Disjoint_Sets | https://proofwiki.org/wiki/Union_of_Indexed_Family_of_Sets_Equal_to_Union_of_Disjoint_Sets | [
"Indexed Families",
"Union of Indexed Family of Sets Equal to Union of Disjoint Sets"
] | [
"Definition:Countable Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Distinct",
"Definition:Countable Set",
"Definition:Pairwise Disjoint/Family",
"Definition:Disjoint Union (Set Theory)"
] | [
"Definition:Subset",
"Rule of Simplification",
"Definition:Set Equality",
"Definition:Pairwise Disjoint/Family",
"Well-Ordering Principle",
"Definition:Smallest Element",
"Definition:Distinct/Plural",
"Definition:Disjoint Sets",
"Definition:Disjoint Union (Set Theory)",
"Definition:Set",
"Defini... |
proofwiki-14467 | Borel Sigma-Algebra Generated by Closed Sets | Let $\map \BB {S, \tau}$ be a Borel $\sigma$-algebra generated by the set of open sets in $S$.
Then $\map \BB {S, \tau}$ is equivalently generated by the set of closed sets in $S$. | By definition, a closed set is the relative complement of an open set.
The result follows from Sigma-Algebra Generated by Complements of Generators.
{{qed}} | Let $\map \BB {S, \tau}$ be a [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] generated by the set of [[Definition:Open Set|open sets]] in $S$.
Then $\map \BB {S, \tau}$ is equivalently generated by the set of [[Definition:Closed Set|closed sets]] in $S$. | By definition, a [[Definition:Closed Set|closed set]] is the [[Definition:Relative Complement|relative complement]] of an [[Definition:Open Set|open set]].
The result follows from [[Sigma-Algebra Generated by Complements of Generators]].
{{qed}} | Borel Sigma-Algebra Generated by Closed Sets | https://proofwiki.org/wiki/Borel_Sigma-Algebra_Generated_by_Closed_Sets | https://proofwiki.org/wiki/Borel_Sigma-Algebra_Generated_by_Closed_Sets | [
"Sigma-Algebras",
"Borel Sigma-Algebras",
"Borel Sigma-Algebras"
] | [
"Definition:Borel Sigma-Algebra",
"Definition:Open Set",
"Definition:Closed Set"
] | [
"Definition:Closed Set",
"Definition:Relative Complement",
"Definition:Open Set",
"Sigma-Algebra Generated by Complements of Generators"
] |
proofwiki-14468 | Stirling Number of n with n-m is Polynomial in n of Degree 2m/Unsigned First Kind | Let $m \in \Z_{\ge 0}$.
The unsigned Stirling number of the first kind $\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$. | The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$. | Let $m \in \Z_{\ge 0}$.
The [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]] $\ds {n \brack n - m}$ is a [[Definition:Polynomial over Real Numbers|polynomial]] in $n$ of [[Definition:Degree of Polynomial|degree]] $2 m$. | The proof proceeds by [[Principle of Mathematical Induction|induction]] over $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds {n \brack n - m}$ is a [[Definition:Polynomial over Real Numbers|polynomial]] in $n$ of [[Definition:Degree of Polynomial|degree]] $2 m$. | Stirling Number of n with n-m is Polynomial in n of Degree 2m/Unsigned First Kind | https://proofwiki.org/wiki/Stirling_Number_of_n_with_n-m_is_Polynomial_in_n_of_Degree_2m/Unsigned_First_Kind | https://proofwiki.org/wiki/Stirling_Number_of_n_with_n-m_is_Polynomial_in_n_of_Degree_2m/Unsigned_First_Kind | [
"Stirling Numbers"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Poly... |
proofwiki-14469 | Unsigned Stirling Number of the First Kind of n with n-2 | :$\ds {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$ | The proof proceeds by induction. | :$\ds {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$ | The proof proceeds by [[Principle of Mathematical Induction|induction]]. | Unsigned Stirling Number of the First Kind of n with n-2 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n_with_n-2 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n_with_n-2 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14470 | Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum | Let $\MM$ be an infinite $\sigma$-algebra on a set $X$.
Then $\MM$ is has cardinality at least that of the cardinality of the continuum $\mathfrak c$:
:$\map \Card \MM \ge \mathfrak c$ | Let:
:$\mathbb M_\infty := \set {A \in \MM : \map \Card {\MM_A} = \infty}$
where:
:$\MM_A$ denotes the trace $\sigma$-algebra of $A$ in $\MM$
Let $A \in \mathbb M_\infty$.
Observe:
{{begin-eqn}}
{{eqn | l = \map \Card {\MM_B} + \map \Card {\MM_{A \mathop \setminus B} }
| o = \ge
| r = \map \Card {\MM_B \cup... | Let $\MM$ be an [[Definition:Infinite Set|infinite]] [[Definition:Sigma-Algebra|$\sigma$-algebra]] on a [[Definition:Set|set]] $X$.
Then $\MM$ is has [[Definition:Cardinality|cardinality]] at least that of the [[Definition:Cardinality of Continuum|cardinality of the continuum]] $\mathfrak c$:
:$\map \Card \MM \ge \m... | Let:
:$\mathbb M_\infty := \set {A \in \MM : \map \Card {\MM_A} = \infty}$
where:
:$\MM_A$ denotes the [[Definition:Trace Sigma-Algebra|trace $\sigma$-algebra]] of $A$ in $\MM$
Let $A \in \mathbb M_\infty$.
Observe:
{{begin-eqn}}
{{eqn | l = \map \Card {\MM_B} + \map \Card {\MM_{A \mathop \setminus B} }
| o = ... | Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum | https://proofwiki.org/wiki/Cardinality_of_Infinite_Sigma-Algebra_is_at_Least_Cardinality_of_Continuum | https://proofwiki.org/wiki/Cardinality_of_Infinite_Sigma-Algebra_is_at_Least_Cardinality_of_Continuum | [
"Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum",
"Cardinality of Continuum",
"Sigma-Algebras"
] | [
"Definition:Infinite Set",
"Definition:Sigma-Algebra",
"Definition:Set",
"Definition:Cardinality",
"Definition:Cardinality of Continuum"
] | [
"Definition:Trace Sigma-Algebra",
"Axiom:Axiom of Choice",
"Definition:Choice Function",
"Definition:Mapping",
"Definition:Projection (Mapping Theory)",
"Definition:Non-Empty Set",
"Definition:Pairwise Disjoint",
"Definition:Set",
"Definition:Injection",
"Power Set of Natural Numbers has Cardinali... |
proofwiki-14471 | Stirling Number of the Second Kind of n with n-2 | :$\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$ | The proof proceeds by induction. | :$\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$ | The proof proceeds by [[Principle of Mathematical Induction|induction]]. | Stirling Number of the Second Kind of n with n-2 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n_with_n-2 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n_with_n-2 | [
"Examples of Stirling Numbers of the Second Kind"
] | [] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14472 | Unsigned Stirling Number of the First Kind of n with n-3 | :$\ds {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$ | The proof proceeds by induction. | :$\ds {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$ | The proof proceeds by [[Principle of Mathematical Induction|induction]]. | Unsigned Stirling Number of the First Kind of n with n-3 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n_with_n-3 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n_with_n-3 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14473 | Stirling Number of the Second Kind of n with n-3 | :$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$ | The proof proceeds by induction. | :$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$ | The proof proceeds by [[Principle of Mathematical Induction|induction]]. | Stirling Number of the Second Kind of n with n-3 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n_with_n-3 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n_with_n-3 | [
"Examples of Stirling Numbers of the Second Kind"
] | [] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14474 | Duality Law for Stirling Numbers | For all integers $n, m \in \Z$:
:$\ds {n \brace m} = {-m \brack -n}$
where:
:$\ds {n \brace m}$ denotes a Stirling number of the second kind
:$\ds {n \brack m}$ denotes an unsigned Stirling number of the first kind. | {{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the negative integers. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}} | For all [[Definition:Integer|integers]] $n, m \in \Z$:
:$\ds {n \brace m} = {-m \brack -n}$
where:
:$\ds {n \brace m}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
:$\ds {n \brack m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirli... | {{ProofWanted|It is unclear exactly how the [[Definition:Stirling Numbers|Stirling numbers]] are [[Definition:Extension of Mapping|extended]] to the [[Definition:Negative Integer|negative integers]]. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}} | Duality Law for Stirling Numbers | https://proofwiki.org/wiki/Duality_Law_for_Stirling_Numbers | https://proofwiki.org/wiki/Duality_Law_for_Stirling_Numbers | [
"Stirling Numbers"
] | [
"Definition:Integer",
"Definition:Stirling Numbers of the Second Kind",
"Definition:Stirling Numbers of the First Kind/Unsigned"
] | [
"Definition:Stirling Numbers",
"Definition:Extension of Mapping",
"Definition:Negative/Integer"
] |
proofwiki-14475 | Power of Complex Number as Summation of Stirling Numbers of Second Kind | Let $z \in \C$ be a complex number whose real part is positive.
Then:
:$z^r = \ds \sum_{k \mathop \in \Z} {r \brace r - k} z^{\underline {r - k} }$
where:
:$\ds {r \brace r - k}$ denotes the extension of the Stirling numbers of the second kind to the complex plane
:$z^{\underline {r - k} }$ denotes $z$ to the $r - k$ f... | {{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the complex plane. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}} | Let $z \in \C$ be a [[Definition:Complex Number|complex number]] whose [[Definition:Real Part|real part]] is [[Definition:Positive Real Number|positive]].
Then:
:$z^r = \ds \sum_{k \mathop \in \Z} {r \brace r - k} z^{\underline {r - k} }$
where:
:$\ds {r \brace r - k}$ denotes the [[Definition:Stirling Numbers of th... | {{ProofWanted|It is unclear exactly how the [[Definition:Stirling Numbers|Stirling numbers]] are [[Definition:Extension of Mapping|extended]] to the [[Definition:Complex Plane|complex plane]]. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}} | Power of Complex Number as Summation of Stirling Numbers of Second Kind/Proof | https://proofwiki.org/wiki/Power_of_Complex_Number_as_Summation_of_Stirling_Numbers_of_Second_Kind | https://proofwiki.org/wiki/Power_of_Complex_Number_as_Summation_of_Stirling_Numbers_of_Second_Kind/Proof | [
"Stirling Numbers",
"Power of Complex Number as Summation of Stirling Numbers of Second Kind"
] | [
"Definition:Complex Number",
"Definition:Complex Number/Real Part",
"Definition:Positive/Real Number",
"Definition:Stirling Numbers of the Second Kind/Complex Numbers",
"Definition:Complex Number/Complex Plane",
"Definition:Falling Factorial"
] | [
"Definition:Stirling Numbers",
"Definition:Extension of Mapping",
"Definition:Complex Number/Complex Plane"
] |
proofwiki-14476 | Falling Factorial of Complex Number as Summation of Unsigned Stirling Numbers of First Kind | Let $z \in \C$ be a complex number whose real part is positive.
Then:
:$z^{\underline r} = \ds \sum_{k \mathop = 0}^m{r \brack r - k} \paren {-1}^k z^{r - k} + \map \OO {z^{r - m - 1} }$
where:
:$\ds {r \brack r - k}$ denotes the extension of the unsigned Stirling numbers of the first kind to the complex plane
:$z^{\un... | {{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the complex plane. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}} | Let $z \in \C$ be a [[Definition:Complex Number|complex number]] whose [[Definition:Real Part|real part]] is [[Definition:Positive Real Number|positive]].
Then:
:$z^{\underline r} = \ds \sum_{k \mathop = 0}^m{r \brack r - k} \paren {-1}^k z^{r - k} + \map \OO {z^{r - m - 1} }$
where:
:$\ds {r \brack r - k}$ denotes ... | {{ProofWanted|It is unclear exactly how the [[Definition:Stirling Numbers|Stirling numbers]] are [[Definition:Extension of Mapping|extended]] to the [[Definition:Complex Plane|complex plane]]. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}} | Falling Factorial of Complex Number as Summation of Unsigned Stirling Numbers of First Kind/Proof | https://proofwiki.org/wiki/Falling_Factorial_of_Complex_Number_as_Summation_of_Unsigned_Stirling_Numbers_of_First_Kind | https://proofwiki.org/wiki/Falling_Factorial_of_Complex_Number_as_Summation_of_Unsigned_Stirling_Numbers_of_First_Kind/Proof | [
"Falling Factorials",
"Stirling Numbers",
"Falling Factorial of Complex Number as Summation of Unsigned Stirling Numbers of First Kind"
] | [
"Definition:Complex Number",
"Definition:Complex Number/Real Part",
"Definition:Positive/Real Number",
"Definition:Stirling Numbers of the First Kind/Unsigned/Complex Numbers",
"Definition:Complex Number/Complex Plane",
"Definition:Falling Factorial",
"Definition:Big-O Notation"
] | [
"Definition:Stirling Numbers",
"Definition:Extension of Mapping",
"Definition:Complex Number/Complex Plane"
] |
proofwiki-14477 | Non-Divisbility of Binomial Coefficients of n by Prime | Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $p$ be a prime number.
Then:
:$\dbinom n k$ is not divisible by $p$ for any $k \in \Z_{\ge 0}$ where $0 \le k \le n$
{{iff}}:
:$n = a p^m - 1$ where $1 \le a < p$
for some $m \in \Z_{\ge 0}$. | The statement:
:$\dbinom n k$ is not divisible by $p$
is equivalent to:
:$\dbinom n k \not \equiv 0 \pmod p$
The corollary to Lucas' Theorem gives:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
where:
:$n, k \in \Z_{\ge 0}$ and $p$ is prime
:the representations of $n$ and $k$ to the base... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$\dbinom n k$ is not [[Definition:Divisor of Integer|divisible]] by $p$ for any $k \in \Z_{\ge 0}$ where $0 \le k \le n$
{{iff}}:
:$n = a p^m - 1$ where $1 \le a < p$
for some $... | The statement:
:$\dbinom n k$ is not [[Definition:Divisor of Integer|divisible]] by $p$
is equivalent to:
:$\dbinom n k \not \equiv 0 \pmod p$
The [[Lucas' Theorem/Corollary|corollary to Lucas' Theorem]] gives:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
where:
:$n, k \in \Z_{\ge 0}$... | Non-Divisbility of Binomial Coefficients of n by Prime | https://proofwiki.org/wiki/Non-Divisbility_of_Binomial_Coefficients_of_n_by_Prime | https://proofwiki.org/wiki/Non-Divisbility_of_Binomial_Coefficients_of_n_by_Prime | [
"Pascal's Triangle"
] | [
"Definition:Positive/Integer",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Divisor (Algebra)/Integer",
"Lucas' Theorem/Corollary",
"Definition:Prime Number",
"Definition:Number Base",
"Definition:Number Base",
"Sum of Geometric Sequence/Corollary 1",
"Definition:Digit",
"Definition:Digit",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/... |
proofwiki-14478 | Product of r Choose k with r Minus Half Choose k/Formulation 2 | Let $k \in \Z$, $r \in \R$.
:$\dbinom r k \dbinom {r - \frac 1 2} k = \dfrac {\dbinom {2 r} {2 k} \dbinom {2 k} k} {4^k}$
where $\dbinom r k$ denotes a binomial coefficient. | From Binomial Coefficient expressed using Beta Function:
:$(1): \quad \dbinom r k \dbinom {r - \frac 1 2} k = \dfrac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom r {k + 1} \dbinom {r - \frac 1 2} {k + 1}
| r... | Let $k \in \Z$, $r \in \R$.
:$\dbinom r k \dbinom {r - \frac 1 2} k = \dfrac {\dbinom {2 r} {2 k} \dbinom {2 k} k} {4^k}$
where $\dbinom r k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Binomial Coefficient expressed using Beta Function]]:
:$(1): \quad \dbinom r k \dbinom {r - \frac 1 2} k = \dfrac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom r {k + 1} \dbinom {r - \frac 1 2} {k + 1}
... | Product of r Choose k with r Minus Half Choose k/Formulation 2 | https://proofwiki.org/wiki/Product_of_r_Choose_k_with_r_Minus_Half_Choose_k/Formulation_2 | https://proofwiki.org/wiki/Product_of_r_Choose_k_with_r_Minus_Half_Choose_k/Formulation_2 | [
"Product of r Choose k with r Minus Half Choose k"
] | [
"Definition:Binomial Coefficient"
] | [
"Binomial Coefficient expressed using Beta Function",
"Beta Function of x with y+1 by x+y over y",
"Beta Function of x with y+1 by x+y over y",
"Beta Function of x with y+1 by x+y over y",
"Beta Function of x with y+1 by x+y over y",
"Binomial Coefficient expressed using Beta Function",
"Beta Function o... |
proofwiki-14479 | Summation Formula for Reciprocal of Binomial Coefficient | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x}
| r = \dfrac 1 {x \binom {n + x} n}
| c =
}}
{{eqn | r = \dfrac {n!} {x \paren {x + 1} \cdots \paren {x + n} }
| c =
}}
{{end-eqn}}
as long as the denominators are not zero. | First note that:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k k} {k + x} + \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k x} {k + x}
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \binom n k \paren {-1}^k \dfrac {k + x} {k + x}
| c =
}}
{{eqn | r ... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x}
| r = \dfrac 1 {x \binom {n + x} n}
| c =
}}
{{eqn | r = \dfrac {n!} {x \paren {x + 1} \cdots \paren {x + n} }
| c =
}}
{{end-eqn}}
as long as the [[Definition:Denominator|denominators]] are not [[Definition:... | First note that:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k k} {k + x} + \sum_{k \mathop \in \Z} \binom n k \dfrac {\paren {-1}^k x} {k + x}
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \binom n k \paren {-1}^k \dfrac {k + x} {k + x}
| c =
}}
{{eqn | r... | Summation Formula for Reciprocal of Binomial Coefficient/Proof 1 | https://proofwiki.org/wiki/Summation_Formula_for_Reciprocal_of_Binomial_Coefficient | https://proofwiki.org/wiki/Summation_Formula_for_Reciprocal_of_Binomial_Coefficient/Proof_1 | [
"Binomial Coefficients",
"Summation Formula for Reciprocal of Binomial Coefficient"
] | [
"Definition:Fraction/Denominator",
"Definition:Zero (Number)"
] | [
"Alternating Sum and Difference of Binomial Coefficients for Given n",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Zero Choose n",
"Binomial Coefficient with Zero",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Pascal's R... |
proofwiki-14480 | Summation Formula for Reciprocal of Binomial Coefficient | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x}
| r = \dfrac 1 {x \binom {n + x} n}
| c =
}}
{{eqn | r = \dfrac {n!} {x \paren {x + 1} \cdots \paren {x + n} }
| c =
}}
{{end-eqn}}
as long as the denominators are not zero. | Consider the value of $\map \Beta {x, n + 1}$, where $\Beta$ is the beta function.
We have:
{{begin-eqn}}
{{eqn | l = \map \Beta {x, n + 1}
| r = \int_0^1 t^{x - 1} \paren {1 - t}^n \rd t
| c = {{Defof|Beta Function|index = 1}}
}}
{{eqn | r = \int_0^1 t^{x - 1} \sum_{k \mathop \ge 0} \binom n k \paren {-t}^... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x}
| r = \dfrac 1 {x \binom {n + x} n}
| c =
}}
{{eqn | r = \dfrac {n!} {x \paren {x + 1} \cdots \paren {x + n} }
| c =
}}
{{end-eqn}}
as long as the [[Definition:Denominator|denominators]] are not [[Definition:... | Consider the value of $\map \Beta {x, n + 1}$, where $\Beta$ is the [[Definition:Beta Function|beta function]].
We have:
{{begin-eqn}}
{{eqn | l = \map \Beta {x, n + 1}
| r = \int_0^1 t^{x - 1} \paren {1 - t}^n \rd t
| c = {{Defof|Beta Function|index = 1}}
}}
{{eqn | r = \int_0^1 t^{x - 1} \sum_{k \mathop ... | Summation Formula for Reciprocal of Binomial Coefficient/Proof 2 | https://proofwiki.org/wiki/Summation_Formula_for_Reciprocal_of_Binomial_Coefficient | https://proofwiki.org/wiki/Summation_Formula_for_Reciprocal_of_Binomial_Coefficient/Proof_2 | [
"Binomial Coefficients",
"Summation Formula for Reciprocal of Binomial Coefficient"
] | [
"Definition:Fraction/Denominator",
"Definition:Zero (Number)"
] | [
"Definition:Beta Function",
"Binomial Theorem",
"Linear Combination of Integrals/Definite",
"Index Laws/Sum of Indices",
"Gamma Function Extends Factorial",
"Gamma Difference Equation",
"Gamma Difference Equation",
"Gamma Difference Equation"
] |
proofwiki-14481 | Sum over k of r Choose k by Minus r Choose m Minus 2k | Let $r \in \R$, $m \in \Z$.
:$\ds \sum_{k \mathop \in \Z} \binom r k \binom {-r} {m - 2 k} \paren {-1}^{m + k} = \binom r m$ | We have:
{{begin-eqn}}
{{eqn | l = \paren {1 - x^2}
| r = \paren {1 - x} \paren {1 + x}
| c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
| l = \paren {1 - x}^r
| r = \dfrac {\paren {1 - x^2}^r} {\paren {1 + x}^r}
| c =
}}
{{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}
| c... | Let $r \in \R$, $m \in \Z$.
:$\ds \sum_{k \mathop \in \Z} \binom r k \binom {-r} {m - 2 k} \paren {-1}^{m + k} = \binom r m$ | We have:
{{begin-eqn}}
{{eqn | l = \paren {1 - x^2}
| r = \paren {1 - x} \paren {1 + x}
| c = [[Difference of Two Squares]]
}}
{{eqn | ll= \leadsto
| l = \paren {1 - x}^r
| r = \dfrac {\paren {1 - x^2}^r} {\paren {1 + x}^r}
| c =
}}
{{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}
... | Sum over k of r Choose k by Minus r Choose m Minus 2k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_Minus_r_Choose_m_Minus_2k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_Minus_r_Choose_m_Minus_2k | [
"Binomial Coefficients"
] | [] | [
"Difference of Two Squares",
"Binomial Theorem"
] |
proofwiki-14482 | Sigma-Algebra Contains Generated Sigma-Algebra of Subset | Let $\sigma_\FF$ be a be a $\sigma$-algebra on a set $\FF$.
Let $\sigma_\FF$ contain a set of sets $\EE$.
Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.
Then $\map \sigma \EE \subseteq \sigma_\FF$ | {{MissingLinks}}
$\sigma_\FF$ is a $\sigma$-algebra containing $\EE$.
$\map \sigma \EE$ is a subset of ''all'' $\sigma$-algebras containing $\EE$, by definition of a generated $\sigma$-algebra.
Therefore it contains $\map \sigma \EE$.
{{qed}} | Let $\sigma_\FF$ be a be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on a [[Definition:Set|set]] $\FF$.
Let $\sigma_\FF$ [[Definition:Contain|contain]] a [[Definition:Set of Sets|set of sets]] $\EE$.
Let $\map \sigma \EE$ be the [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra gene... | {{MissingLinks}}
$\sigma_\FF$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]] containing $\EE$.
$\map \sigma \EE$ is a subset of ''all'' $\sigma$-algebras containing $\EE$, by definition of a generated $\sigma$-algebra.
Therefore it contains $\map \sigma \EE$.
{{qed}} | Sigma-Algebra Contains Generated Sigma-Algebra of Subset | https://proofwiki.org/wiki/Sigma-Algebra_Contains_Generated_Sigma-Algebra_of_Subset | https://proofwiki.org/wiki/Sigma-Algebra_Contains_Generated_Sigma-Algebra_of_Subset | [
"Sigma-Algebras"
] | [
"Definition:Sigma-Algebra",
"Definition:Set",
"Definition:Subset",
"Definition:Set of Sets",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Definition:Sigma-Algebra"
] |
proofwiki-14483 | Generated Sigma-Algebra Contains Generated Sigma-Algebra of Subset | Let $\map \sigma \FF$ be the $\sigma$-algebra generated by $\EE$.
Let $\map \sigma \FF$ contain a set of sets $\EE$.
Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.
Then $\map \sigma \EE \subseteq \map \sigma \FF$. | Apply Sigma-Algebra Contains Generated Sigma-Algebra of Subset to $\map \sigma \FF$.
{{qed}} | Let $\map \sigma \FF$ be the [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by $\EE$]].
Let $\map \sigma \FF$ [[Definition:Contain|contain]] a [[Definition:Set of Sets|set of sets]] $\EE$.
Let $\map \sigma \EE$ be the [[Definition:Sigma-Algebra Generated by Collection of Subs... | Apply [[Sigma-Algebra Contains Generated Sigma-Algebra of Subset]] to $\map \sigma \FF$.
{{qed}} | Generated Sigma-Algebra Contains Generated Sigma-Algebra of Subset | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Contains_Generated_Sigma-Algebra_of_Subset | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Contains_Generated_Sigma-Algebra_of_Subset | [
"Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Subsets",
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Subset",
"Definition:Set of Sets",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Sigma-Algebra Contains Generated Sigma-Algebra of Subset"
] |
proofwiki-14484 | Binomial Theorem/Abel's Generalisation/x+y = 0 | Consider Abel's Generalisation of Binomial Theorem:
{{:Abel's Generalisation of Binomial Theorem}}
This holds in the special case where $x + y = 0$. | As $x + y = 0$, we can substitute $y = -x$, and so:
:$\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k} = 0$
is to be proved.
So:
{{begin-eqn}}
{{eqn | l = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k}
| r = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-1}... | Consider [[Abel's Generalisation of Binomial Theorem]]:
{{:Abel's Generalisation of Binomial Theorem}}
This holds in the special case where $x + y = 0$. | As $x + y = 0$, we can substitute $y = -x$, and so:
:$\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k} = 0$
is to be proved.
So:
{{begin-eqn}}
{{eqn | l = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k}
| r = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren... | Binomial Theorem/Abel's Generalisation/x+y = 0 | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation/x+y_=_0 | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation/x+y_=_0 | [
"Binomial Theorem"
] | [
"Binomial Theorem/Abel's Generalisation"
] | [
"Sum over k of r Choose k by -1^r-k by Polynomial",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial"
] |
proofwiki-14485 | Existence of Minimal Uncountable Well-Ordered Set | There exists a minimal uncountable well-ordered set.
That is, there exists an uncountable well-ordered set $\Omega$ with the property that every initial segment in $\Omega$ is countable. | By the axiom of powers, there exists the power set $\powerset \N$, where $\N$ is the set of natural numbers.
By Power Set of Natural Numbers is not Countable, this set is uncountable.
By Zermelo's Well-Ordering Theorem, $\powerset \N$ can be endowed with a well-ordering.
Denote such an ordering with the symbol $\preccu... | There exists a [[Definition:Minimal Uncountable Well-Ordered Set|minimal uncountable well-ordered set]].
That is, there exists an [[Definition:Uncountable Set|uncountable]] [[Definition:Well-Ordered Set|well-ordered set]] $\Omega$ with the [[Definition:Property|property]] that every [[Definition:Initial Segment|initia... | By the [[Axiom:Axiom of Powers (Class Theory)|axiom of powers]], there exists the [[Definition:Power Set|power set]] $\powerset \N$, where $\N$ is the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]].
By [[Power Set of Natural Numbers is not Countable]], this [[Definition:Set|set]] is [[Definit... | Existence of Minimal Uncountable Well-Ordered Set/Proof Using Choice | https://proofwiki.org/wiki/Existence_of_Minimal_Uncountable_Well-Ordered_Set | https://proofwiki.org/wiki/Existence_of_Minimal_Uncountable_Well-Ordered_Set/Proof_Using_Choice | [
"Order Theory",
"Existence of Minimal Uncountable Well-Ordered Set"
] | [
"Definition:Minimal Uncountable Well-Ordered Set",
"Definition:Uncountable/Set",
"Definition:Well-Ordered Set",
"Definition:Property",
"Definition:Initial Segment",
"Definition:Countable Set"
] | [
"Axiom:Axiom of Powers/Class Theory",
"Definition:Power Set",
"Definition:Set",
"Definition:Natural Numbers",
"Power Set of Natural Numbers is Uncountable",
"Definition:Set",
"Definition:Uncountable/Set",
"Zermelo's Well-Ordering Theorem",
"Definition:Well-Ordering",
"Definition:Well-Ordering",
... |
proofwiki-14486 | Existence of Minimal Uncountable Well-Ordered Set | There exists a minimal uncountable well-ordered set.
That is, there exists an uncountable well-ordered set $\Omega$ with the property that every initial segment in $\Omega$ is countable. | By the axiom of powers, there exists the power set $\powerset \N$, where $\N$ is the set of natural numbers.
By Power Set of Natural Numbers is not Countable, this set is uncountable.
Consider the set of ordered pairs:
:$\AA = \set {\struct {A, \prec}: A \in \powerset \N}$
where:
:the first coordinate is a (possibly em... | There exists a [[Definition:Minimal Uncountable Well-Ordered Set|minimal uncountable well-ordered set]].
That is, there exists an [[Definition:Uncountable Set|uncountable]] [[Definition:Well-Ordered Set|well-ordered set]] $\Omega$ with the [[Definition:Property|property]] that every [[Definition:Initial Segment|initia... | By the [[Axiom:Axiom of Powers (Set Theory)|axiom of powers]], there exists the [[Definition:Power Set|power set]] $\powerset \N$, where $\N$ is the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]].
By [[Power Set of Natural Numbers is not Countable]], this set is [[Definition:Uncountable Set|u... | Existence of Minimal Uncountable Well-Ordered Set/Proof Without Using Choice | https://proofwiki.org/wiki/Existence_of_Minimal_Uncountable_Well-Ordered_Set | https://proofwiki.org/wiki/Existence_of_Minimal_Uncountable_Well-Ordered_Set/Proof_Without_Using_Choice | [
"Order Theory",
"Existence of Minimal Uncountable Well-Ordered Set"
] | [
"Definition:Minimal Uncountable Well-Ordered Set",
"Definition:Uncountable/Set",
"Definition:Well-Ordered Set",
"Definition:Property",
"Definition:Initial Segment",
"Definition:Countable Set"
] | [
"Axiom:Axiom of Powers/Set Theory",
"Definition:Power Set",
"Definition:Set",
"Definition:Natural Numbers",
"Power Set of Natural Numbers is Uncountable",
"Definition:Uncountable/Set",
"Definition:Ordered Pair",
"Definition:Coordinate System/Coordinate/Element of Ordered Pair",
"Definition:Empty Set... |
proofwiki-14487 | Binomial Theorem/Hurwitz's Generalisation | :$\ds \paren {x + y}^n = \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
where the summation ranges over all $2^n$ choices of $\epsilon_1, \ldots, \epsilon_n = 0$ or $1$ inde... | Follows from this formula:
:$(1): \quad \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y + \paren {1 - \epsilon_1} z_1 - \cdots + \paren {1 - \epsilon_n} z_n}^{n - \epsilon_1 - \cdots - \epsilon_n} = \paren {x + y} \paren {x + y + z_1 + \cdots + z_n}^{n -... | :$\ds \paren {x + y}^n = \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
where the [[Definition:Summation|summation]] ranges over all $2^n$ choices of $\epsilon_1, \ldots, \e... | Follows from [[Link required to result in Knuth: exercise 2.3.4.4-30|this formula]]:
:$(1): \quad \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y + \paren {1 - \epsilon_1} z_1 - \cdots + \paren {1 - \epsilon_n} z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}... | Binomial Theorem/Hurwitz's Generalisation | https://proofwiki.org/wiki/Binomial_Theorem/Hurwitz's_Generalisation | https://proofwiki.org/wiki/Binomial_Theorem/Hurwitz's_Generalisation | [
"Binomial Theorem"
] | [
"Definition:Summation"
] | [
"Link required to result in Knuth: exercise 2.3.4.4-30"
] |
proofwiki-14488 | Set of Relations can be Ordered by Subset Relation | Let $S \times T$ be the product of two sets.
Let $\RR$ be a set of relations on $S \times T$.
Then $\RR$ can be ordered by the subset relation. | Let $R$ be a relation on $S \times T$.
By the definition of relation, $R$ is associated with a subset $R \subseteq S \times T$.
Thus $\RR$ is a subset of the power set $\powerset {S \times T}$.
The result follows from Subset Relation is Ordering.
{{qed}} | Let $S \times T$ be the [[Definition:Cartesian Product|product]] of two [[Definition:Set|sets]].
Let $\RR$ be a [[Definition:Set|set]] of [[Definition:Relation|relations]] on $S \times T$.
Then $\RR$ can be [[Definition:Set Ordered by Subset Relation|ordered by the subset relation]]. | Let $R$ be a [[Definition:Relation|relation]] on $S \times T$.
By the definition of [[Definition:Relation|relation]], $R$ is associated with a [[Definition:Subset|subset]] $R \subseteq S \times T$.
Thus $\RR$ is a [[Definition:Subset|subset]] of the [[Definition:Power Set|power set]] $\powerset {S \times T}$.
The re... | Set of Relations can be Ordered by Subset Relation | https://proofwiki.org/wiki/Set_of_Relations_can_be_Ordered_by_Subset_Relation | https://proofwiki.org/wiki/Set_of_Relations_can_be_Ordered_by_Subset_Relation | [
"Order Theory",
"Subset Relation"
] | [
"Definition:Cartesian Product",
"Definition:Set",
"Definition:Set",
"Definition:Relation",
"Definition:Set Ordered by Subset Relation"
] | [
"Definition:Relation",
"Definition:Relation",
"Definition:Subset",
"Definition:Subset",
"Definition:Power Set",
"Subset Relation is Ordering"
] |
proofwiki-14489 | Set of Mappings can be Ordered by Subset Relation | Let $S \times T$ be the product of two sets.
Let $\FF$ be a set of mappings on $S \times T$.
Then $\FF$ can be ordered by the subset relation. | By the definition of mapping, a mapping is a specific type of relation.
The result then follows from Set of Relations can be Ordered by Subset Relation.
{{qed}}
Category:Order Theory
Category:Subset Relation
j0y0swkto559bco1v1p98qjrul6xn9j | Let $S \times T$ be the [[Definition:Cartesian Product|product]] of two [[Definition:Set|sets]].
Let $\FF$ be a [[Definition:Set|set]] of [[Definition:Mapping|mappings]] on $S \times T$.
Then $\FF$ can be [[Definition:Set Ordered by Subset Relation|ordered by the subset relation]]. | By the definition of [[Definition:Mapping|mapping]], a mapping is a specific type of [[Definition:Relation|relation]].
The result then follows from [[Set of Relations can be Ordered by Subset Relation]].
{{qed}}
[[Category:Order Theory]]
[[Category:Subset Relation]]
j0y0swkto559bco1v1p98qjrul6xn9j | Set of Mappings can be Ordered by Subset Relation | https://proofwiki.org/wiki/Set_of_Mappings_can_be_Ordered_by_Subset_Relation | https://proofwiki.org/wiki/Set_of_Mappings_can_be_Ordered_by_Subset_Relation | [
"Order Theory",
"Subset Relation"
] | [
"Definition:Cartesian Product",
"Definition:Set",
"Definition:Set",
"Definition:Mapping",
"Definition:Set Ordered by Subset Relation"
] | [
"Definition:Mapping",
"Definition:Relation",
"Set of Relations can be Ordered by Subset Relation",
"Category:Order Theory",
"Category:Subset Relation"
] |
proofwiki-14490 | Binomial Theorem/Abel's Generalisation/Negative n | Abel's Generalisation of Binomial Theorem:
{{:Abel's Generalisation of Binomial Theorem}}
does not hold for $n \in \Z_{< 0}$. | Putting $n = x = -1$ and $y = z = 1$ into the {{LHS}}
:$\paren {-1 + 1}^{-1} = \dfrac 1 0$
which is undefined.
Putting the same values into the {{RHS}} gives:
{{begin-eqn}}
{{eqn | o =
| r = \sum_k \dbinom {-1} k \paren {-1} \paren {-1 - k}^{k - 1} \paren {1 + k}^{-1 - k}
| c =
}}
{{eqn | r = \sum_k \pare... | [[Abel's Generalisation of Binomial Theorem]]:
{{:Abel's Generalisation of Binomial Theorem}}
does not hold for $n \in \Z_{< 0}$. | Putting $n = x = -1$ and $y = z = 1$ into the {{LHS}}
:$\paren {-1 + 1}^{-1} = \dfrac 1 0$
which is undefined.
Putting the same values into the {{RHS}} gives:
{{begin-eqn}}
{{eqn | o =
| r = \sum_k \dbinom {-1} k \paren {-1} \paren {-1 - k}^{k - 1} \paren {1 + k}^{-1 - k}
| c =
}}
{{eqn | r = \sum_k \... | Binomial Theorem/Abel's Generalisation/Negative n | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation/Negative_n | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation/Negative_n | [
"Binomial Theorem"
] | [
"Binomial Theorem/Abel's Generalisation"
] | [
"Translation of Index Variable of Summation/Corollary",
"Basel Problem"
] |
proofwiki-14491 | Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k | :$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$ | The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$ | :$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$ | The proof proceeds by [[Principle of Mathematical Induction|induction]] over $m$.
For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m +... | Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k | https://proofwiki.org/wiki/Summation_from_k_to_m_of_r_Choose_k_by_s_Choose_n-k_by_nr-(r+s)k | https://proofwiki.org/wiki/Summation_from_k_to_m_of_r_Choose_k_by_s_Choose_n-k_by_nr-(r+s)k | [
"Binomial Coefficients"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14492 | Factors of Binomial Coefficient/Corollary 1 | For all $r \in \R, k \in \Z$:
:$\paren {r - k} \dbinom r k = r \dbinom {r - 1} k$
from which:
:$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ (if $r \ne k$) | {{begin-eqn}}
{{eqn | l = r \binom {r - 1} k
| r = r \frac {\paren {r - 1} \paren {\paren {r - 1} - 1} \cdots \paren {\paren {r - 1} - k + 2} \paren {\paren {r - 1} - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}
| c =
}}
{{eqn | r = \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - k + 1} \p... | For all $r \in \R, k \in \Z$:
:$\paren {r - k} \dbinom r k = r \dbinom {r - 1} k$
from which:
:$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ (if $r \ne k$) | {{begin-eqn}}
{{eqn | l = r \binom {r - 1} k
| r = r \frac {\paren {r - 1} \paren {\paren {r - 1} - 1} \cdots \paren {\paren {r - 1} - k + 2} \paren {\paren {r - 1} - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}
| c =
}}
{{eqn | r = \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - k + 1} \p... | Factors of Binomial Coefficient/Corollary 1 | https://proofwiki.org/wiki/Factors_of_Binomial_Coefficient/Corollary_1 | https://proofwiki.org/wiki/Factors_of_Binomial_Coefficient/Corollary_1 | [
"Binomial Coefficients"
] | [] | [] |
proofwiki-14493 | Factors of Binomial Coefficient/Corollary 2 | For all $r \in \R, k \in \Z$:
:$\paren {r - k + 1} \dbinom r {k - 1} = k \dbinom r k$ | {{begin-eqn}}
{{eqn | l = \paren {r - k + 1} \dbinom r {k - 1}
| r = \paren {r - k + 1} \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - \paren {k - 1} + 1} } {\paren {k - 1} \paren {k - 2} \cdots 1}
| c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
}}
{{eqn | r = \frac {r \paren {r - 1} \... | For all $r \in \R, k \in \Z$:
:$\paren {r - k + 1} \dbinom r {k - 1} = k \dbinom r k$ | {{begin-eqn}}
{{eqn | l = \paren {r - k + 1} \dbinom r {k - 1}
| r = \paren {r - k + 1} \frac {r \paren {r - 1} \paren {r - 2} \cdots \paren {r - \paren {k - 1} + 1} } {\paren {k - 1} \paren {k - 2} \cdots 1}
| c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
}}
{{eqn | r = \frac {r \paren {r - 1} \... | Factors of Binomial Coefficient/Corollary 2 | https://proofwiki.org/wiki/Factors_of_Binomial_Coefficient/Corollary_2 | https://proofwiki.org/wiki/Factors_of_Binomial_Coefficient/Corollary_2 | [
"Binomial Coefficients"
] | [] | [
"Category:Binomial Coefficients"
] |
proofwiki-14494 | Summation from k to m of 2k-1 Choose k by 2n-2k Choose n-k by -1 over 2k-1 | :$\ds \sum_{k \mathop = 0}^m \binom {2 k - 1} k \binom {2 n - 2 k} {n - k} \dfrac {-1} {2 k - 1} = \dfrac {n - m} {2 n} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m} + \dfrac 1 2 \dbinom {2 n} n$ | From Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Set $r - \dfrac 1 2$ and $s = -\dfrac 1 2$.
:$\ds \sum_{k \mathop = 0}^m \dbinom {1/2} k ... | :$\ds \sum_{k \mathop = 0}^m \binom {2 k - 1} k \binom {2 n - 2 k} {n - k} \dfrac {-1} {2 k - 1} = \dfrac {n - m} {2 n} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m} + \dfrac 1 2 \dbinom {2 n} n$ | From [[Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k]]:
:$\ds \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$
Set $r - \dfrac 1 2$ and $s = -\dfrac 1 2$.
:$\ds \sum_{k \mathop = 0}^m \dbinom ... | Summation from k to m of 2k-1 Choose k by 2n-2k Choose n-k by -1 over 2k-1 | https://proofwiki.org/wiki/Summation_from_k_to_m_of_2k-1_Choose_k_by_2n-2k_Choose_n-k_by_-1_over_2k-1 | https://proofwiki.org/wiki/Summation_from_k_to_m_of_2k-1_Choose_k_by_2n-2k_Choose_n-k_by_-1_over_2k-1 | [
"Binomial Coefficients"
] | [] | [
"Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k",
"Binomial Coefficient of Minus Half",
"Binomial Coefficient of Half/Corollary",
"Binomial Coefficient of Minus Half",
"Binomial Coefficient of Half",
"Factors of Binomial Coefficient",
"Factors of Binomial Coefficient/Corollary 1"
] |
proofwiki-14495 | Inverse of Pascal's Triangle expressed as Matrix | Consider Pascal's triangle expressed as a (square) matrix $\mathbf M$, with the top left element holding $\dbinom 0 0$.
:$\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 ... | We have from Sum over $k$ of $\dbinom r k$ by $\dbinom {s + k} n$ by $\left({-1}\right)^{r - k}$: Corollary:
{{:Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary}}
By definition of matrix multiplication, this is the element $a_{r n}$ of the matrix formed by multiplying the two matrices above.
As can be seen,... | Consider [[Definition:Pascal's Triangle|Pascal's triangle]] expressed as a [[Definition:Square Matrix|(square) matrix]] $\mathbf M$, with the top left [[Definition:Element of Matrix|element]] holding $\dbinom 0 0$.
:$\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 1 & 0 & 0 & 0 & ... | We have from [[Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary|Sum over $k$ of $\dbinom r k$ by $\dbinom {s + k} n$ by $\left({-1}\right)^{r - k}$: Corollary]]:
{{:Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary}}
By definition of [[Definition:Matrix Product (Conventional)|matrix multiplicati... | Inverse of Pascal's Triangle expressed as Matrix | https://proofwiki.org/wiki/Inverse_of_Pascal's_Triangle_expressed_as_Matrix | https://proofwiki.org/wiki/Inverse_of_Pascal's_Triangle_expressed_as_Matrix | [
"Pascal's Triangle"
] | [
"Definition:Pascal's Triangle",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Element",
"Definition:Inverse Matrix",
"Definition:Matrix/Element",
"Definition:Matrix/Element",
"Definition:Matrix/Diagonal/Main"
] | [
"Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary",
"Definition:Matrix Product (Conventional)",
"Definition:Matrix/Element",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Unit Matrix"
] |
proofwiki-14496 | Binomial Coefficient of Half | Let $k \in \Z$.
:$\dbinom {\frac 1 2} k = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k} k$
where $\dbinom {\frac 1 2} k$ denotes a binomial coefficient. | {{begin-eqn}}
{{eqn | l = \dbinom {\frac 1 2} k
| r = \dfrac {1/2} {1/2 - k} \dbinom {1/2 - 1} k
| c = Factors of Binomial Coefficient: Corollary 1
}}
{{eqn | r = \dfrac 1 {1 - 2 k} \dbinom {-1/2} k
| c =
}}
{{eqn | r = \dfrac 1 {1 - 2 k} \dfrac {\left({-1}\right)^k} {4^k} \dbinom {2 k} k
| c =... | Let $k \in \Z$.
:$\dbinom {\frac 1 2} k = \dfrac {\left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k} k$
where $\dbinom {\frac 1 2} k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | {{begin-eqn}}
{{eqn | l = \dbinom {\frac 1 2} k
| r = \dfrac {1/2} {1/2 - k} \dbinom {1/2 - 1} k
| c = [[Factors of Binomial Coefficient/Corollary 1|Factors of Binomial Coefficient: Corollary 1]]
}}
{{eqn | r = \dfrac 1 {1 - 2 k} \dbinom {-1/2} k
| c =
}}
{{eqn | r = \dfrac 1 {1 - 2 k} \dfrac {\left(... | Binomial Coefficient of Half | https://proofwiki.org/wiki/Binomial_Coefficient_of_Half | https://proofwiki.org/wiki/Binomial_Coefficient_of_Half | [
"Examples of Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [
"Factors of Binomial Coefficient/Corollary 1",
"Binomial Coefficient of Minus Half"
] |
proofwiki-14497 | Binomial Coefficient of Half/Corollary | Let $k \in \Z_{\ge 0}$.
:$\dbinom {\frac 1 2} k = \dfrac {\paren {-1}^{k - 1} } {2^{2 k - 1} \paren {2 k - 1} } \dbinom {2 k - 1} k - \delta_{k 0}$
where:
:$\dbinom {\frac 1 2} k$ denotes a binomial coefficient
:$\delta_{k 0}$ denotes the Kronecker delta. | When $k > 0$ we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dbinom {2 k} k
| r = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dfrac {2 k} {2 k - k} \dbinom {2 k - 1} k
| c = {{Corollary|Factors of Binomial Coefficient|1}}
}}
{{eqn | r = \dfrac {\paren {-1... | Let $k \in \Z_{\ge 0}$.
:$\dbinom {\frac 1 2} k = \dfrac {\paren {-1}^{k - 1} } {2^{2 k - 1} \paren {2 k - 1} } \dbinom {2 k - 1} k - \delta_{k 0}$
where:
:$\dbinom {\frac 1 2} k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$\delta_{k 0}$ denotes the [[Definition:Kronecker Delta|Kronecker delt... | When $k > 0$ we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dbinom {2 k} k
| r = \dfrac {\paren {-1}^{k - 1} } {4^k \paren {2 k - 1} } \dfrac {2 k} {2 k - k} \dbinom {2 k - 1} k
| c = {{Corollary|Factors of Binomial Coefficient|1}}
}}
{{eqn | r = \dfrac {\paren {-... | Binomial Coefficient of Half/Corollary | https://proofwiki.org/wiki/Binomial_Coefficient_of_Half/Corollary | https://proofwiki.org/wiki/Binomial_Coefficient_of_Half/Corollary | [
"Examples of Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Kronecker Delta"
] | [
"Negated Upper Index of Binomial Coefficient"
] |
proofwiki-14498 | Inverse of Stirling's Triangle expressed as Matrix | Consider Stirling's triangle of the first kind (signed) expressed as a (square) matrix $\mathbf A$, with the top left element holding $\map s {0, 0}$.
:$\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 1 & 0 & 0 & 0 & 0 & \cdots \\
0 & -1 & 1 & 0 & ... | First note that from Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
From First Inversion Formula for Stirling Numbers:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
From Second Inversion Formula for Stirling N... | Consider [[Definition:Stirling's Triangle of the First Kind (Signed)|Stirling's triangle of the first kind (signed)]] expressed as a [[Definition:Square Matrix|(square) matrix]] $\mathbf A$, with the top left [[Definition:Element of Matrix|element]] holding $\map s {0, 0}$.
:$\begin{pmatrix}
1 & 0 & 0 & ... | First note that from [[Relation between Signed and Unsigned Stirling Numbers of the First Kind]]:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
From [[First Inversion Formula for Stirling Numbers]]:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
From [[Second Inversion Formula f... | Inverse of Stirling's Triangle expressed as Matrix | https://proofwiki.org/wiki/Inverse_of_Stirling's_Triangle_expressed_as_Matrix | https://proofwiki.org/wiki/Inverse_of_Stirling's_Triangle_expressed_as_Matrix | [
"Stirling Numbers"
] | [
"Definition:Stirling's Triangle of the First Kind (Signed)",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Element",
"Definition:Stirling's Triangle of the Second Kind",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Element"
] | [
"Relation between Signed and Unsigned Stirling Numbers of the First Kind",
"First Inversion Formula for Stirling Numbers",
"Second Inversion Formula for Stirling Numbers",
"Definition:Matrix Product (Conventional)",
"Definition:Matrix/Element",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Uni... |
proofwiki-14499 | Wosets are Isomorphic to Each Other or Initial Segments | Let $\struct {S, \preceq_S}$ and $\struct {T, \preceq_T}$ be well-ordered sets.
Then precisely one of the following hold:
:$\struct {S, \preceq_S}$ is order isomorphic to $\struct {T, \preceq_T}$
or:
:$\struct {S, \preceq_S}$ is order isomorphic to an initial segment in $\struct {T, \preceq_T}$
or:
:$\struct {T, \prece... | We assume $S \ne \O \ne T$; otherwise the theorem holds vacuously.
Define:
:$S' = S \cup \text{ initial segments in } S$
:$T' = T \cup \text{ initial segments in } T$
:$\FF = \set {f: S' \to T' \mid f \text{ is an order isomorphism} }$
We note that $\FF$ is non-empty, because it at least contains a trivial order isomo... | Let $\struct {S, \preceq_S}$ and $\struct {T, \preceq_T}$ be [[Definition:Well-Ordered Set|well-ordered sets]].
Then precisely one of the following hold:
:$\struct {S, \preceq_S}$ is [[Definition:Order Isomorphic Well-Orderings|order isomorphic]] to $\struct {T, \preceq_T}$
or:
:$\struct {S, \preceq_S}$ is [[Definit... | We assume $S \ne \O \ne T$; otherwise the theorem holds [[Definition:Vacuous Truth|vacuously]].
Define:
:$S' = S \cup \text{ initial segments in } S$
:$T' = T \cup \text{ initial segments in } T$
:$\FF = \set {f: S' \to T' \mid f \text{ is an order isomorphism} }$
We note that $\FF$ is [[Definition:Non-Empty Set|... | Wosets are Isomorphic to Each Other or Initial Segments/Proof Using Choice | https://proofwiki.org/wiki/Wosets_are_Isomorphic_to_Each_Other_or_Initial_Segments | https://proofwiki.org/wiki/Wosets_are_Isomorphic_to_Each_Other_or_Initial_Segments/Proof_Using_Choice | [
"Wosets are Isomorphic to Each Other or Initial Segments",
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Initial Segment",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Initial Segment"
] | [
"Definition:Vacuous Truth",
"Definition:Non-Empty Set",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Singleton",
"Definition:Initial Segment",
"Definition:Chain (Order Theory)",
"Antilexicographic Product of Totally Ordered Sets is Totally Ordered",
"Zorn's Lemma",
"Definition:Domain (... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.